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2013-I-11
| 2,013
| 11
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions: (a) If $16$ , $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and (b) There are three integers $0 < x < y < z < 14$ such that when $x$ , $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over. Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
| 148
|
I
|
[
"$N$ must be some multiple of $\\text{lcm}(14, 15, 16)= 2^{4}\\cdot 3\\cdot 5\\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$. $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $10$, and $12$ all divide $k$, so $x, y, z = 9, 11, 13$ We have the following three modulo equations: $nk\\equiv 3\\pmod{9}$ $nk\\equiv 3\\pmod{11}$ $nk\\equiv 3\\pmod{13}$ To solve the equations, you can notice the answer must be of the form $9\\cdot 11\\cdot 13\\cdot m + 3$ where $m$ is an integer. This must be divisible by $lcm$ $(14, 15, 16)$, which is $560\\cdot 3$. Therefore, $\\frac{9\\cdot 11\\cdot 13\\cdot m + 3}{560\\cdot 3} = q$, which is an integer. Factor out $3$ and divide to get $\\frac{429m+1}{560} = q$. Therefore, $429m+1=560q$. We can use Bezout's Identity or a Euclidean algorithm bash to solve for the least of $m$ and $q$. We find that the least $m$ is $171$ and the least $q$ is $131$. Since we want to factor $1680\\cdot q$, don't multiply: we already know that the prime factors of $1680$ are $2$, $3$, $5$, and $7$, and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = 148.",
"Note that the number of play blocks is a multiple of the LCM of $16$, $15$, and $14$. The value of this can be found to be $(16)(15)(7) = 1680$. This number is also divisible by $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $10$, and $12$, thus, the three numbers $x, y, z$ are $9, 11, 13$. Thus, $1680k\\equiv 3$ when taken mod $9$, $11$, $13$. Since $1680$ is congruent to $6$ mod $9$ and $3$ mod $13$, and congruent to $8$ mod $11$, the number $k$ must be a number that is congruent to $1$ mod $13$, $2$ mod $3$ (because $6$ is a multiple of $3$, which is a factor of $9$ that can be divided out) and cause $8$ to become $3$ when multiplied under modulo $11$. Looking at the last condition shows that $k\\equiv 10$ mod $11$ (after a bit of bashing) and is congruent to $1$ mod $13$ and $2$ mod $3$ as previously noted. Listing out the numbers congruent to $10$ mod $11$ and $1$ mod $13$ yield the following lists: $10$ mod $11$: $21$, $32$, $43$, $54$, $65$, $76$, $87$, $98$, $109$, $120$, $131$... $1$ mod $13$: $14$, $27$, $40$, $53$, $66$, $79$, $92$, $105$, $118$, $131$, $144$, $157$, $170$... Both lists contain $x$ elements where $x$ is the modulo being taken, thus, there must be a solution in these lists as adding $11(13)$ to this solution yields the next smallest solution. In this case, $131$ is the solution for $k$ and thus the answer is $1680(131)$. Since $131$ is prime, the sum of the prime factors is $2 + 3 + 5 + 7 + 131 = 148.",
"It is obvious that $N=a\\cdot 2^4 \\cdot 3\\cdot 5\\cdot 7$ and so the only mod $3$ number of students are $9, 11, 13$. Therefore, $N=1287\\cdot k+3$. Try some approaches and you will see that this one is one of the few successful ones: Start by setting the two $N$ equations together, then we get $1680a=1287k+3$. Divide by $3$. Note that since the RHS is $1\\pmod{3}$, and since $560$ is $2\\pmod{3}$, then $a=3b+2$, where $b$ is some nonnegative integer, because $a$ must be $2\\pmod{3}$. This reduces to $560 \\cdot 3b + 1119 = 429k$. Now, take out the $11!$ With the same procedure, $b=11c-1$, where $c$ is some nonnegative integer. You also get $c=13d+4$, at which point $k=171+560d$. $d$ cannot be equal to $0$. Therefore, $c=4, b=43, a=131$, and we know the prime factors of $N$ are $2, 3, 5, 7, 131$ so the answer is $148.",
"We start by noticing that $N = a\\textbf{lcm}(14, 15, 16) = 1680a$ for some integer $a$ in order to satisfy the first condition. Next, we satisfy the second condition. Since $x<y<z < 14$ must leave a remainder when dividing $1680a$, they are not divisors of $1680x$. Thus, we can eliminate all $y \\le 14$ s.t. $\\gcd(y, 1680x) \\ne 1$ which leaves $(x, y, z) = (9, 11, 13)$. Thus, $N = 1680a \\equiv 3 \\pmod 9 \\equiv 3 \\pmod {11} \\equiv 3 \\pmod {13}$. Now, we seek to find the least $a$ which satisfies this set of congruences. By Chinese Remainder Theorem on the first two congruences, we find that $a \\equiv 32 \\pmod {33}$ (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on $a \\equiv 32 \\pmod {33}$ and $1680a \\equiv 3 \\pmod {13}$ we find that $a \\equiv 131 \\pmod {429}$. Thus, the minimal value of $N$ is possible at $a = 131$. The prime factorization of this minimum value is $2^4 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 131$ and so the answer is $2 + 3 + 5 + 7 + 131 = 148.",
"As the problem stated, the number of boxes is definitely a multiple of $lcm(14,15,16)=1680$, so we assume total number of boxes is $1680k$ Then, according to $(b)$ statement, we get $1680k \\equiv 3 \\pmod x \\equiv 3 \\pmod {y} \\equiv 3 \\pmod {z}$. So we have $lcm(x,y,z)+3+m\\cdot lcm(x,y,z)=1680k$, we just write it to be $(1+n)lcm(x,y,z)=1680k-3$ Which tells that $x,y,z$ must be all odd number. Moreover, we can see $(1+m)lcm(x,y,z)$ can't be a multiple of $3,5,7$(as $1680$ is a multiple of $5,7$) which means that $lcm(x,y,z)=lcm(9,11,13)=1287$ We let $n=1+m$ Now, we write $1287n+3=1680k, 429n+1=560k$ It is true that $n\\equiv 1 \\pmod{10}$, let $n=10p+1$, it has $429p+43=56k$ Then, $p$ must be odd, let $p=2q+1$, it indicates $429q+236=28k$ Now, $q$ must be even, $q=2s$ tells $429s+118=14k$ Eventually, $s$ must be even, $s=2y$, $858y+59=7k$, $y=1$ is the smallest. This time, $k=131$ So the number of balls is $1680\\cdot 131=2^4\\cdot 3\\cdot 5\\cdot 7 \\cdot 131$, the desired value is $2+3+5+7+131=148 ~bluesoul",
"From part (a), we know that $2^4\\cdot3\\cdot5\\cdot7 | N$. From part (b), we know that $N \\equiv 3 \\pmod {1287}$. We can expand on part (a) by saying that $N = 1680k$ for some $k$. Rather than taking the three modulos together, we take them individually. \\[1680k \\equiv 6k \\equiv 3 \\pmod 9\\] \\[k \\equiv 2^{-1} \\pmod 9\\] The inverse of 2 mod 9 is easily seen to be $5$. \\[k \\equiv 5 \\pmod 9\\] Now moving to the second modulo which we leave as follows, \\[1680k \\equiv 8k \\equiv 3 \\pmod {11}\\] Now the last modulo, \\[1680k \\equiv 3k \\equiv 3 \\pmod {13}\\] \\[k \\equiv 1 \\pmod{13}\\] CRT on the first and the third one results in $k \\equiv 4 \\pmod {117}$. Now doing the second one and the one we just made, $k \\equiv 131 \\pmod{1287}$. Thus, the smallest value that works for $k = 131$. Thus $N = 2^4\\cdot3\\cdot5\\cdot7\\cdot131$ $2+3+5+7+131 = 148 ~YBSuburbanTea",
"From part (a), we know that $2^4\\cdot3\\cdot5\\cdot7 | N$. From part (b), we know that $N \\equiv 3 \\pmod {1287}$. We can expand on part (a) by saying that $N = 1680k$ for some $k$. Rather than taking the three modulos together, we take them individually. \\[1680k \\equiv 6k \\equiv 3 \\pmod 9\\] \\[k \\equiv 2^{-1} \\pmod 9\\] The inverse of 2 mod 9 is easily seen to be $5$. \\[k \\equiv 5 \\pmod 9\\] Now moving to the second modulo which we leave as follows, \\[1680k \\equiv 8k \\equiv 3 \\pmod {11}\\] Now the last modulo, \\[1680k \\equiv 3k \\equiv 3 \\pmod {13}\\] \\[k \\equiv 1 \\pmod{13}\\] CRT on the first and the third one results in $k \\equiv 4 \\pmod {117}$. Now doing the second one and the one we just made, $k \\equiv 131 \\pmod{1287}$. Thus, the smallest value that works for $k = 131$. Thus $N = 2^4\\cdot3\\cdot5\\cdot7\\cdot131$ $2+3+5+7+131 = 148 ~YBSuburbanTea"
] |
2013-I-12
| 2,013
| 12
|
Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^o$ and $\angle Q = 60^o$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\overline{RP}$ . There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ , where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$ .
| 21
|
I
|
[
"First, find that $\\angle R = 45^\\circ$. Draw $ABCDEF$. Now draw $\\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$. The height of $ABCDEF$ is $\\sqrt{3}$, so the length of base $QR$ is $2+\\sqrt{3}$. Let the equation of $\\overline{RP}$ be $y = x$. Then, the equation of $\\overline{PQ}$ is $y = -\\sqrt{3} (x - (2+\\sqrt{3})) \\to y = -x\\sqrt{3} + 2\\sqrt{3} + 3$. Solving the two equations gives $y = x = \\frac{\\sqrt{3} + 3}{2}$. The area of $\\bigtriangleup PQR$ is $\\frac{1}{2} * (2 + \\sqrt{3}) * \\frac{\\sqrt{3} + 3}{2} = \\frac{5\\sqrt{3} + 9}{4}$. $a + b + c + d = 9 + 5 + 3 + 4 = 021 -hwan",
"Use coordinates. Call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$. After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the coordinate of $R$. Draw a perpendicular of $F$, call it $H$, and note that $QP = 1 + \\sqrt{3}$ after using the trig functions for $75$ degrees. Now, get the lines for $QR$ and $RP$: $y=\\sqrt{3}x$ and $y=-(2+\\sqrt{3})x + (5+\\sqrt{3})$, whereupon we get the ordinate of $R$ to be $\\frac{3+2\\sqrt{3}}{2}$, and the area is $\\frac{5\\sqrt{3} + 9}{4}$, so our answer is $021.",
"Use coordinates. Call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$. After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the coordinate of $R$. Draw a perpendicular of $F$, call it $H$, and note that $QP = 1 + \\sqrt{3}$ after using the trig functions for $75$ degrees. Now, get the lines for $QR$ and $RP$: $y=\\sqrt{3}x$ and $y=-(2+\\sqrt{3})x + (5+\\sqrt{3})$, whereupon we get the ordinate of $R$ to be $\\frac{3+2\\sqrt{3}}{2}$, and the area is $\\frac{5\\sqrt{3} + 9}{4}$, so our answer is $021.",
"Angle chasing yields that both triangles $PAF$ and $PQR$ are $75$-$60$-$45$ triangles. First look at triangle $PAF$. Using Law of Sines, we find: $\\frac{\\frac{\\sqrt{6} + \\sqrt{2}}{4}}{1} = \\frac{\\frac{\\sqrt{2}}{2}}{PA}$ Simplifying, we find $PA = \\sqrt{3} - 1$. Since $\\angle{Q} = 60^\\circ$, WLOG assume triangle $BQC$ is equilateral, so $BQ = 1$. So $PQ = \\sqrt{3} + 1$. Apply Law of Sines again, $\\frac{\\frac{\\sqrt{2}}{2}}{\\sqrt{3} + 1} = \\frac{\\frac{\\sqrt{3}}{2}}{PR}$ Simplifying, we find $PR = \\frac{\\sqrt{6}}{2} \\cdot (1 + \\sqrt{3})$. $[PQR] = \\frac{1}{2} \\cdot PQ \\cdot PR \\cdot \\sin 75^\\circ$. Evaluating and reducing, we get $\\frac{9 + 5\\sqrt{3}}{4},$thus the answer is $021",
"Angle chasing yields that both triangles $PAF$ and $PQR$ are $75$-$60$-$45$ triangles. First look at triangle $PAF$. Using Law of Sines, we find: $\\frac{\\frac{\\sqrt{6} + \\sqrt{2}}{4}}{1} = \\frac{\\frac{\\sqrt{2}}{2}}{PA}$ Simplifying, we find $PA = \\sqrt{3} - 1$. Since $\\angle{Q} = 60^\\circ$, WLOG assume triangle $BQC$ is equilateral, so $BQ = 1$. So $PQ = \\sqrt{3} + 1$. Apply Law of Sines again, $\\frac{\\frac{\\sqrt{2}}{2}}{\\sqrt{3} + 1} = \\frac{\\frac{\\sqrt{3}}{2}}{PR}$ Simplifying, we find $PR = \\frac{\\sqrt{6}}{2} \\cdot (1 + \\sqrt{3})$. $[PQR] = \\frac{1}{2} \\cdot PQ \\cdot PR \\cdot \\sin 75^\\circ$. Evaluating and reducing, we get $\\frac{9 + 5\\sqrt{3}}{4},$thus the answer is $021",
"As we can see, the $75^\\circ$ angle of $\\angle P$ can be split into a $45^\\circ$ angle and a $30^\\circ$ angle. This allows us to drop an altitude from point $P$ for $\\triangle RPQ$ which intersects $\\overline{AF}$ at point $a$ and $\\overline{RQ}$ at point $b$. The main idea of our solution is to obtain enough sides of $\\triangle RPQ$ that we are able to directly figure out its area (specifically by figuring out side $\\overline{RQ}$ and $\\overline{Pb}$). We first begin by figuring out the length of $\\overline{PQ}$. This can be easily done, since $\\overline{AB}$ is simply $1$ (given in the problem) and $\\overline{BQ}=1$ because $\\triangle BCQ$ is an equilateral after some simple angle calculations. Now we need to find $\\overline{PA}$. This is when we bring in some simple algebra. PREPARATION: $\\overline{aF}=\\overline{Pa}$ (45-45-90 Right Triangle) $\\overline{Pa}=\\sqrt{3}\\overline{Aa}$ (30-60-90 Right Triangle) $\\overline{PA}=2\\overline{Aa}$ $\\overline{Aa}+\\overline{aF}=1$ SOLVING: $\\overline{Aa}+\\sqrt{3}\\overline{aF}=1$ so $\\overline{Aa}(\\sqrt{3}+1)=1$ $\\overline{Aa}=\\frac{1}{\\sqrt{3}+1}=\\frac{\\sqrt{3}-1}{2}$ Finally, $\\overline{PA}=2\\cdot\\frac{\\sqrt{3}-1}{2}=\\sqrt{3}-1$ Now, we can finally get the length of $\\overline{PQ}$ by adding up $\\overline{PA}+\\overline{AB}+\\overline{BQ}$, which is simply $(\\sqrt{3}-1)+(1)+(1)=\\sqrt{3}+1$ To get $\\overline{RQ}$ and $\\overline{Pb}$, we first work bit by bit. $\\overline{Qb}=\\frac{\\overline{PQ}}{2}=\\frac{\\sqrt{3}+1}{2}$ (30-60-90 Right Triangle) $\\overline{Pb}=\\sqrt{3}\\overline{Qb}=\\frac{\\sqrt{3}+3}{2}$ (same 30-60-90 Right Triangle) Since $\\overline{Pb}=\\overline{Rb}$ because of 45-45-90 right triangles, $\\overline{Rb}=\\frac{\\sqrt{3}+3}{2}$ too. Now, we can finally calculate $\\overline{RQ}$, and it is $\\overline{Rb}+\\overline{Qb}=\\frac{\\sqrt{3}+3}{2}+\\frac{\\sqrt{3}+1}{2}=\\sqrt{3}+2$. Finally, the area of $\\triangle PRQ$ can be calculated by $\\frac{1}{2}\\cdot\\overline{RQ}\\cdot\\overline{Pb}$, which is equal to $[\\triangle PRQ]=\\frac{1}{2} \\cdot (\\sqrt{3}+2) \\cdot \\frac{\\sqrt{3}+3}{2} =\\frac{9+5\\sqrt{3}}{4}$. So the final answer is $9+5+3+4=021. -by What do Humanitarians Eat?",
"As we can see, the $75^\\circ$ angle of $\\angle P$ can be split into a $45^\\circ$ angle and a $30^\\circ$ angle. This allows us to drop an altitude from point $P$ for $\\triangle RPQ$ which intersects $\\overline{AF}$ at point $a$ and $\\overline{RQ}$ at point $b$. The main idea of our solution is to obtain enough sides of $\\triangle RPQ$ that we are able to directly figure out its area (specifically by figuring out side $\\overline{RQ}$ and $\\overline{Pb}$). We first begin by figuring out the length of $\\overline{PQ}$. This can be easily done, since $\\overline{AB}$ is simply $1$ (given in the problem) and $\\overline{BQ}=1$ because $\\triangle BCQ$ is an equilateral after some simple angle calculations. Now we need to find $\\overline{PA}$. This is when we bring in some simple algebra. PREPARATION: $\\overline{aF}=\\overline{Pa}$ (45-45-90 Right Triangle) $\\overline{Pa}=\\sqrt{3}\\overline{Aa}$ (30-60-90 Right Triangle) $\\overline{PA}=2\\overline{Aa}$ $\\overline{Aa}+\\overline{aF}=1$ SOLVING: $\\overline{Aa}+\\sqrt{3}\\overline{aF}=1$ so $\\overline{Aa}(\\sqrt{3}+1)=1$ $\\overline{Aa}=\\frac{1}{\\sqrt{3}+1}=\\frac{\\sqrt{3}-1}{2}$ Finally, $\\overline{PA}=2\\cdot\\frac{\\sqrt{3}-1}{2}=\\sqrt{3}-1$ Now, we can finally get the length of $\\overline{PQ}$ by adding up $\\overline{PA}+\\overline{AB}+\\overline{BQ}$, which is simply $(\\sqrt{3}-1)+(1)+(1)=\\sqrt{3}+1$ To get $\\overline{RQ}$ and $\\overline{Pb}$, we first work bit by bit. $\\overline{Qb}=\\frac{\\overline{PQ}}{2}=\\frac{\\sqrt{3}+1}{2}$ (30-60-90 Right Triangle) $\\overline{Pb}=\\sqrt{3}\\overline{Qb}=\\frac{\\sqrt{3}+3}{2}$ (same 30-60-90 Right Triangle) Since $\\overline{Pb}=\\overline{Rb}$ because of 45-45-90 right triangles, $\\overline{Rb}=\\frac{\\sqrt{3}+3}{2}$ too. Now, we can finally calculate $\\overline{RQ}$, and it is $\\overline{Rb}+\\overline{Qb}=\\frac{\\sqrt{3}+3}{2}+\\frac{\\sqrt{3}+1}{2}=\\sqrt{3}+2$. Finally, the area of $\\triangle PRQ$ can be calculated by $\\frac{1}{2}\\cdot\\overline{RQ}\\cdot\\overline{Pb}$, which is equal to $[\\triangle PRQ]=\\frac{1}{2} \\cdot (\\sqrt{3}+2) \\cdot \\frac{\\sqrt{3}+3}{2} =\\frac{9+5\\sqrt{3}}{4}$. So the final answer is $9+5+3+4=021. -by What do Humanitarians Eat?",
"With some simple angle chasing we can show that $\\triangle OJL$ and $\\triangle MPL$ are congruent. This means we have a large equilateral triangle with side length $3$ and quadrilateral $OJQN$. We know that $[OJQN] = [\\triangle NQL] - [\\triangle OJL]$. Using Law of Sines and the fact that $\\angle N = 45^{\\circ}$ we know that $\\overline{NL} = \\sqrt{6}$ and the height to that side is $\\frac{\\sqrt{3} -1}{\\sqrt{2}}$ so $[\\triangle NQL] = \\frac{3-\\sqrt{3}}{2}$. Using an extremely similar process we can show that $\\overline{OJ} = 2-\\sqrt{3}$ which means the height to $\\overline{LJ}$ is $\\frac{2\\sqrt{3}-3}{2}$. So the area of $\\triangle OJL = \\frac{2\\sqrt{3}-3}{4}$. This means the area of quadrilateral $OJQN = \\frac{3-\\sqrt{3}}{2} - \\frac{2\\sqrt{3}-3}{4} = \\frac{9-4\\sqrt{3}}{4}$. So the area of our larger triangle is $\\frac{9-4\\sqrt{3}}{4} + \\frac{9\\sqrt{3}}{4} = \\frac{9+5\\sqrt{3}}{4}$. Therefore $9+5+3+4=021.",
"With some simple angle chasing we can show that $\\triangle OJL$ and $\\triangle MPL$ are congruent. This means we have a large equilateral triangle with side length $3$ and quadrilateral $OJQN$. We know that $[OJQN] = [\\triangle NQL] - [\\triangle OJL]$. Using Law of Sines and the fact that $\\angle N = 45^{\\circ}$ we know that $\\overline{NL} = \\sqrt{6}$ and the height to that side is $\\frac{\\sqrt{3} -1}{\\sqrt{2}}$ so $[\\triangle NQL] = \\frac{3-\\sqrt{3}}{2}$. Using an extremely similar process we can show that $\\overline{OJ} = 2-\\sqrt{3}$ which means the height to $\\overline{LJ}$ is $\\frac{2\\sqrt{3}-3}{2}$. So the area of $\\triangle OJL = \\frac{2\\sqrt{3}-3}{4}$. This means the area of quadrilateral $OJQN = \\frac{3-\\sqrt{3}}{2} - \\frac{2\\sqrt{3}-3}{4} = \\frac{9-4\\sqrt{3}}{4}$. So the area of our larger triangle is $\\frac{9-4\\sqrt{3}}{4} + \\frac{9\\sqrt{3}}{4} = \\frac{9+5\\sqrt{3}}{4}$. Therefore $9+5+3+4=021.",
"We can find that $AF || CD || QR$. This means that the perpendicular from $P$ to $QR$ is perpendicular to $AF$ as well, so let that perpendicular intersect $AF$ at $G$, and the perpendicular intersect $QR$ at $H$. Set $AP=x$. Note that $\\angle {PAG} = 60^\\circ$, so $AG=\\frac{x}{2}$ and $PG = GF = \\frac{x\\sqrt3}{2}$. Also, $1=AF=AG+GF=\\frac{x}{2} + \\frac{x\\sqrt{3}}{2}$, so $x=\\sqrt{3} - 1$. It's easy to calculate the area now, because the perpendicular from $P$ to $QR$ splits $\\triangle{PQR}$ into a $30-60-90$ (PHQ) and a $45-45-90$ (PHR). From these triangles' ratios, it should follow that $QH=\\frac{\\sqrt{3} + 1}{2}, PH=HR=\\frac{\\sqrt{3}+3}{2}$, so the area is $\\frac{1}{2} * PH * QR = \\frac{1}{2} * PH * (QH + HR) = \\frac{1}{2} * \\frac{\\sqrt{3} + 3}{2} * \\frac{2\\sqrt{3}+4}{2} = \\frac{9+5\\sqrt{3}}{4}. By Mathscienceclass",
"We can find that $AF || CD || QR$. This means that the perpendicular from $P$ to $QR$ is perpendicular to $AF$ as well, so let that perpendicular intersect $AF$ at $G$, and the perpendicular intersect $QR$ at $H$. Set $AP=x$. Note that $\\angle {PAG} = 60^\\circ$, so $AG=\\frac{x}{2}$ and $PG = GF = \\frac{x\\sqrt3}{2}$. Also, $1=AF=AG+GF=\\frac{x}{2} + \\frac{x\\sqrt{3}}{2}$, so $x=\\sqrt{3} - 1$. It's easy to calculate the area now, because the perpendicular from $P$ to $QR$ splits $\\triangle{PQR}$ into a $30-60-90$ (PHQ) and a $45-45-90$ (PHR). From these triangles' ratios, it should follow that $QH=\\frac{\\sqrt{3} + 1}{2}, PH=HR=\\frac{\\sqrt{3}+3}{2}$, so the area is $\\frac{1}{2} * PH * QR = \\frac{1}{2} * PH * (QH + HR) = \\frac{1}{2} * \\frac{\\sqrt{3} + 3}{2} * \\frac{2\\sqrt{3}+4}{2} = \\frac{9+5\\sqrt{3}}{4}. By Mathscienceclass",
"We can observe that $RD=DF$ (because $\\angle R$ & $\\angle RFD$ are both $45^\\circ$). Thus we know that $RD$ is equivalent to the height of the hexagon, which is $\\sqrt3$. Now we look at triangle $\\triangle AFP$ and apply the Law of Sines to it. $\\frac{1}{\\sin{75}}=\\frac{AP}{\\sin{45}}$. From here we can solve for $AP$ and get that $AP=\\sqrt{3}-1$. Now we use the Sine formula for the area of a triangle with sides $RQ$, $PQ$, and $\\angle {RQP}$ to get the answer. Setting $PQ=\\sqrt{3}+1$ and $QR=\\sqrt{3}+2$ we get the expression $\\frac{(\\sqrt{3}+1)(\\sqrt{3}+2)(\\frac{\\sqrt{3}}{2})}{2}$ which is $\\frac{9 + 5\\sqrt{3}}{4}$. Thus our final answer is $9+5+3+4=021. By AwesomeLife_Math",
"We can observe that $RD=DF$ (because $\\angle R$ & $\\angle RFD$ are both $45^\\circ$). Thus we know that $RD$ is equivalent to the height of the hexagon, which is $\\sqrt3$. Now we look at triangle $\\triangle AFP$ and apply the Law of Sines to it. $\\frac{1}{\\sin{75}}=\\frac{AP}{\\sin{45}}$. From here we can solve for $AP$ and get that $AP=\\sqrt{3}-1$. Now we use the Sine formula for the area of a triangle with sides $RQ$, $PQ$, and $\\angle {RQP}$ to get the answer. Setting $PQ=\\sqrt{3}+1$ and $QR=\\sqrt{3}+2$ we get the expression $\\frac{(\\sqrt{3}+1)(\\sqrt{3}+2)(\\frac{\\sqrt{3}}{2})}{2}$ which is $\\frac{9 + 5\\sqrt{3}}{4}$. Thus our final answer is $9+5+3+4=021. By AwesomeLife_Math",
"Note that $\\triangle PAF \\sim \\triangle PQR$. Let $X$ be the foot of an altitude dropped from $F$ to $PQ$. Using trigonometry, we find that $PA = \\sqrt3 - 1$ and $FX = \\frac{\\sqrt3}{2}$, thus the area of $\\triangle PAF$ is $\\frac{3 - \\sqrt3}{4}$. Since $QA$ is clearly $2$, the scale factor between $\\triangle PAF$ and $\\triangle PQR$ is $\\frac{\\sqrt3 + 1}{\\sqrt3 - 1}$, and thus $[PQR] = \\frac{3 - \\sqrt3}{4} \\cdot \\left( \\frac{\\sqrt3 + 1}{\\sqrt3 - 1} \\right)^2 = \\frac{9 + 5\\sqrt3}{4}$, and we extract $021. ~ostriches88",
"Note that $\\triangle PAF \\sim \\triangle PQR$. Let $X$ be the foot of an altitude dropped from $F$ to $PQ$. Using trigonometry, we find that $PA = \\sqrt3 - 1$ and $FX = \\frac{\\sqrt3}{2}$, thus the area of $\\triangle PAF$ is $\\frac{3 - \\sqrt3}{4}$. Since $QA$ is clearly $2$, the scale factor between $\\triangle PAF$ and $\\triangle PQR$ is $\\frac{\\sqrt3 + 1}{\\sqrt3 - 1}$, and thus $[PQR] = \\frac{3 - \\sqrt3}{4} \\cdot \\left( \\frac{\\sqrt3 + 1}{\\sqrt3 - 1} \\right)^2 = \\frac{9 + 5\\sqrt3}{4}$, and we extract $021. ~ostriches88"
] |
2013-I-13
| 2,013
| 13
|
Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ , $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$ .
| 961
|
I
|
[
"Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$. Also, the area we seek is simply the ratio $k=\\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$, because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90. For ease, all ratios I will use to solve this problem are with respect to the area of $[AB_0C_0]$. For example, if I say some area has ratio $\\frac{1}{2}$, that means its area is 45. Now note that $k=$ 1 minus ratio of $[B_1C_1A]$ minus ratio $[B_0C_0C_1]$. We see by similar triangles given that ratio $[B_0C_0C_1]$ is $\\frac{17^2}{25^2}$. Ratio $[B_1C_1A]$ is $(\\frac{336}{625})^2$, after seeing that $C_1C_0 = \\frac{289}{625}$, . Now it suffices to find 90 times ratio $[B_0B_1C_1]$, which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find $k$ and clearing out the $5^8$, we see that the answer is $90\\cdot \\frac{5^8-336^2-17^2\\cdot 5^4}{5^8-336^2}$, which gives $q= 961.",
"Using Heron's Formula we can get the area of the triangle $\\Delta AB_0C_0 = 90$. Since $\\Delta AB_0C_0 \\sim \\Delta B_0C_1C_0$ then the scale factor for the dimensions of $\\Delta B_0C_1C_0$ to $\\Delta AB_0C_0$ is $\\dfrac{17}{25}.$ Therefore, the area of $\\Delta B_0C_1C_0$ is $(\\dfrac{17}{25})^2(90)$. Also, the dimensions of the other sides of the $\\Delta B_0C_1C_0$ can be easily computed: $\\overline{B_0C_1}= \\dfrac{17}{25}(12)$ and $\\overline{C_1C_0} = \\dfrac{17^2}{25}$. This allows us to compute one side of the triangle $\\Delta AB_0C_0$, $\\overline{AC_1} = 25 - \\dfrac{17^2}{25} = \\dfrac{25^2 - 17^2}{25}$. Therefore, the scale factor $\\Delta AB_1C_1$ to $\\Delta AB_0C_0$ is $\\dfrac{25^2 - 17^2}{25^2}$ , which yields the length of $\\overline{B_1C_1}$ as $\\dfrac{25^2 - 17^2}{25^2}(17)$. Therefore, the scale factor for $\\Delta B_1C_2C_1$ to $\\Delta B_0C_1C_0$ is $\\dfrac{25^2 - 17^2}{25^2}$. Some more algebraic manipulation will show that $\\Delta B_nC_{n+1}C_n$ to $\\Delta B_{n-1}C_nC_{n-1}$ is still $\\dfrac{25^2 - 17^2}{25^2}$. Also, since the triangles are disjoint, the area of the union is the sum of the areas. Therefore, the area is the geometric series $\\dfrac{90 \\cdot 17^2}{25^2} \\sum_{n=0}^{\\infty} (\\dfrac{25^2-17^2}{25^2})^2$ At this point, it may be wise to \"simplify\" $25^2 - 17^2 = (25-17)(25+17) = (8)(42) = 336$. So the geometric series converges to $\\dfrac{90 \\cdot 17^2}{25^2} \\dfrac{1}{1 - \\dfrac{336^2}{625^2}} = \\dfrac{90 \\cdot 17^2}{25^2} \\dfrac{625^2}{625^2 - 336^2}$. Using the difference of squares, we get $\\dfrac{90 \\cdot 17^2}{25^2}\\dfrac{625^2}{(625 - 336)(625 + 336)}$, which simplifies to $\\dfrac{90 \\cdot 17^2}{25^2} \\dfrac{625^2}{(289)(961)}$. Cancelling all common factors, we get the reduced fraction $= \\dfrac{90 \\cdot 25^2}{31^2}$. So $\\frac{p}{q}=1-\\frac{90 \\cdot 25^2}{31^2}=\\frac{90 \\cdot 336}{961}$, yielding the answer $961.",
"For this problem, the key is to find the $\\frac{[\\triangle{B_nB_{n-1}C_n}]}{[\\triangle{AB_{n-1}C_{n-1}}]}$. The area of the biggest triangle is $90$ according to the Heron's formula easily Firstly, we discuss the ratio of $\\frac{[\\triangle{B_0C_1C_0}]}{[\\triangle{AB_0C_0}]}$ Since the problem said that two triangles are similar, so $\\frac{C_1C_0}{B_0C_0}=\\frac{17}{25}$, Getting that $C_1C_0=\\frac{289}{25}$, which is not hard to find that $AC_1=\\frac{336}{25}$, Since $\\frac{AB_1}{AB_0}=\\frac{AC_1}{AC_0}=\\frac{336}{625}$, we can find the ratio of $\\frac{[\\triangle{B_0B_1C_1}]}{[\\triangle{AB_0C_0}]}=\\frac{336}{625}\\cdot\\frac{289}{625}$, the common ratio between two similar triangles is $(\\frac{336}{625})^2$, the similar triangles means two consecutive $(\\triangle{AB_nC_n});(\\triangle{AB_{n+1}C_{n+1}})$ Now the whole summation of $S=1+(\\frac{336}{625})^2+(\\frac{336}{625})^3+....+(\\frac{336}{625})^n=\\frac{625^2}{961\\cdot289}$ The desired answer is $90\\cdot\\frac{336\\cdot289\\cdot625^2}{625^2\\cdot961\\cdot289}=\\frac{30240}{961}$ Which our answer is $961 ~bluesoul ~Marshall_Huang (some minor latex stuff}",
"Let $k$ be the coefficient of the similarity of triangles \\[\\triangle B_0 C_1 C_0 \\sim \\triangle AB_0 C_0 \\implies k = \\frac {B_0 C_0}{AC_0} = \\frac {17}{25}.\\] Then area $\\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \\implies \\frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$ The height of triangles $\\triangle B_0C_1A$ and $\\triangle AB_0C_0$ from $B_0$ is the same $\\implies \\frac {AC_1}{AC_0} = 1 – k^2.$ The coefficient of the similarity of triangles $\\triangle AB_1C_1 \\sim \\triangle AB_0C_0$ is $\\frac {AC_1}{AC_0} = 1 – k^2 \\implies \\frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$ Analogically the coefficient of the similarity of triangles $\\triangle AB_2C_2 \\sim \\triangle AB_0C_0$ is $(1 – k^2)^2 \\implies \\frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on. The yellow area $[Y]$ is $\\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \\frac {k^2}{1 – (1 – k^2)^2} = \\frac{1}{2 – k^2}.$ The required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \\cdot (1 – \\frac{1}{2 – k^2}) = [AB_0C_0 ] \\cdot \\frac {1 – k^2}{2 – k^2} = [AB_0C_0 ] \\cdot \\frac {25^2 – 17^2} {2 \\cdot 25^2 – 17^2} = [AB_0C_0 ] \\cdot \\frac {336}{961}.$ The number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $961. [email protected], vvsss"
] |
2013-I-14
| 2,013
| 14
|
For $\pi \le \theta < 2\pi$ , let \[P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots\] and \[Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots\] so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 36
|
I
|
[
"Noticing the $\\sin$ and $\\cos$ in both $P$ and $Q,$ we think of the angle addition identities: \\[\\sin(a + b) = \\sin a \\cos b + \\cos a \\sin b, \\cos(a + b) = \\cos a \\cos b - \\sin a \\sin b\\] With this in mind, we multiply $P$ by $\\sin \\theta$ and $Q$ by $\\cos \\theta$ to try and use some angle addition identities. Indeed, we get \\begin{align*} P \\sin \\theta + Q \\cos \\theta &= \\cos \\theta + \\dfrac{1}{2}(\\cos \\theta \\sin \\theta - \\sin \\theta \\cos \\theta) - \\dfrac{1}{4}(\\sin{2 \\theta} \\sin \\theta + \\cos{2 \\theta} \\cos{\\theta}) - \\cdots \\\\ &= \\cos \\theta - \\dfrac{1}{4} \\cos \\theta + \\dfrac{1}{8} \\sin{2 \\theta} + \\dfrac{1}{16} \\cos{3 \\theta} + \\cdots \\\\ &= \\cos \\theta - \\dfrac{1}{2}P \\end{align*} after adding term-by-term. Similar term-by-term adding yields \\[P \\cos \\theta + Q \\sin \\theta = -2(Q - 1).\\] This is a system of equations; rearrange and rewrite to get \\[P(1 + 2 \\sin \\theta) + 2Q \\cos \\theta = 2 \\cos \\theta\\] and \\[P \\cos^2 \\theta + Q \\cos \\theta(2 + \\sin \\theta) = 2 \\cos \\theta.\\] Subtract the two and rearrange to get \\[\\dfrac{P}{Q} = \\dfrac{\\cos \\theta}{2 + \\sin \\theta} = \\dfrac{2 \\sqrt{2}}{7}.\\] Then, square both sides and use Pythagorean Identity to get a quadratic in $\\sin \\theta.$ Factor that quadratic and solve for $\\sin \\theta = -17/19, 1/3.$ Since we're given $\\pi\\leq\\theta<2\\pi,$ $\\sin\\theta$ is nonpositive. We therefore use the negative solution, and our desired answer is $17 + 19 = 036",
"Use sum to product formulas to rewrite $P$ and $Q$ \\[P \\sin\\theta\\ + Q \\cos\\theta\\ = \\cos \\theta\\ - \\frac{1}{4}\\cos \\theta + \\frac{1}{8}\\sin 2\\theta + \\frac{1}{16}\\cos 3\\theta - \\frac{1}{32}\\sin 4\\theta + ...\\] Therefore, \\[P \\sin \\theta - Q \\cos \\theta = -2P\\] Using \\[\\frac{P}{Q} = \\frac{2\\sqrt2}{7}, Q = \\frac{7}{2\\sqrt2} P\\] Plug in to the previous equation and cancel out the \"P\" terms to get: \\[\\sin\\theta - \\frac{7}{2\\sqrt2} \\cos\\theta = -2\\] Then use the pythagorean identity to solve for $\\sin\\theta$, \\[\\sin\\theta = -\\frac{17}{19} \\implies 036\\]",
"Note that \\[e^{i\\theta}=\\cos(\\theta)+i\\sin(\\theta)\\] Thus, the following identities follow immediately: \\[ie^{i\\theta}=i(\\cos(\\theta)+i\\sin(\\theta))=-\\sin(\\theta)+i\\cos(\\theta)\\] \\[i^2 e^{i\\theta}=-e^{i\\theta}=-\\cos(\\theta)-i\\sin(\\theta)\\] \\[i^3 e^{i\\theta}=\\sin(\\theta)-i\\cos(\\theta)\\] Consider, now, the sum $Q+iP$. It follows fairly immediately that: \\[Q+iP=1+\\left(\\frac{i}{2}\\right)^1e^{i\\theta}+\\left(\\frac{i}{2}\\right)^2e^{2i\\theta}+\\ldots=\\frac{1}{1-\\frac{i}{2}e^{i\\theta}}=\\frac{2}{2-ie^{i\\theta}}\\] \\[Q+iP=\\frac{2}{2-ie^{i\\theta}}=\\frac{2}{2-(-\\sin(\\theta)+i\\cos(\\theta))}=\\frac{2}{(2+\\sin(\\theta))-i\\cos(\\theta)}\\] This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find: \\[Q+iP=\\frac{2}{(2+\\sin(\\theta))-i\\cos(\\theta)}\\left(\\frac{(2+\\sin(\\theta))+i\\cos(\\theta)}{(2+\\sin(\\theta))+i\\cos(\\theta)}\\right)\\] \\[Q+iP=\\frac{2((2+\\sin(\\theta))+i\\cos(\\theta))}{(2+\\sin(\\theta))^2+\\cos^2(\\theta)}\\] Comparing real and imaginary parts, we find: \\[\\frac{P}{Q}=\\frac{\\cos(\\theta)}{2+\\sin(\\theta)}=\\frac{2\\sqrt{2}}{7}\\] Squaring this equation and letting $\\sin^2(\\theta)=x$: $\\frac{P^2}{Q^2}=\\frac{\\cos^2(\\theta)}{4+4\\sin(\\theta)+\\sin^2(\\theta)}=\\frac{1-x^2}{4+4x+x^2}=\\frac{8}{49}$ Clearing denominators and solving for $x$ gives sine as $x=-\\frac{17}{19}$. $017+019=036",
"A bit similar to Solution 3. We use $\\phi = \\theta+90^\\circ$ because the progression cycles in $P\\in (\\sin 0\\theta,\\cos 1\\theta,-\\sin 2\\theta,-\\cos 3\\theta\\dots)$. So we could rewrite that as $P\\in(\\sin 0\\phi,\\sin 1\\phi,\\sin 2\\phi,\\sin 3\\phi\\dots)$. Similarly, $Q\\in (\\cos 0\\theta,-\\sin 1\\theta,-\\cos 2\\theta,\\sin 3\\theta\\dots)\\implies Q\\in(\\cos 0\\phi,\\cos 1\\phi, \\cos 2\\phi, \\cos 3\\phi\\dots)$. Setting complex $z=q_1+p_1i$, we get $z=\\frac{1}{2}\\left(\\cos\\phi+\\sin\\phi i\\right)$ $(Q,P)=\\sum_{n=0}^\\infty z^n=\\frac{1}{1-z}=\\frac{1}{1-\\frac{1}{2}\\cos\\phi-\\frac{i}{2}\\sin\\phi}=\\frac{1-0.5\\cos\\phi+0.5i\\sin\\phi}{\\dots}$. The important part is the ratio of the imaginary part $i$ to the real part. To cancel out the imaginary part from the denominator, we must add $0.5i\\sin\\phi$ to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find $\\frac{P}{Q}=\\tan\\text{arg}(\\Sigma)$ a PROPORTION of values. So denominators would cancel out. $\\frac{P}{Q}=\\frac{\\text{Re}\\Sigma}{\\text{Im}\\Sigma}=\\frac{0.5\\sin\\phi}{1-0.5\\cos\\phi}=\\frac{\\sin\\phi}{2-\\cos\\phi}=\\frac{2\\sqrt{2}}{7}$. Setting $\\sin\\theta=y$, we obtain \\[\\frac{\\sqrt{1-y^2}}{2+y}=\\frac{2\\sqrt{2}}{7}\\] \\[7\\sqrt{1-y^2}=2\\sqrt{2}(2+y)\\] \\[49-49y^2=8y^2+32y+32\\] \\[57y^2+32y-17=0\\rightarrow y=\\frac{-32\\pm\\sqrt{1024+4\\cdot 969}}{114}\\] \\[y=\\frac{-32\\pm\\sqrt{4900}}{114}=\\frac{-16\\pm 35}{57}\\]. Since $y<0$ because $\\pi<\\theta<2\\pi$, $y=\\sin\\theta=-\\frac{51}{57}=-\\frac{17}{19}$. Adding up, $17+19=036.",
"We notice $\\sin\\theta=-\\frac{i}{2}(e^{i\\theta}-e^{-i\\theta})$ and $\\cos\\theta=\\frac{1}{2}(e^{i\\theta}+e^{-i\\theta})$ We observe that both $P$ and $Q$ can be split into $2$ parts, namely the terms which contain the $\\cos$ and the terms which contain the $\\sin .$ The $\\cos$ part of $P$ can be expressed as: \\begin{align*}\\frac12\\cos\\theta-\\frac18\\cos3\\theta+\\cdots&=\\frac14\\left(e^{i\\theta}\\left(1-\\frac{e^{i2\\theta}}{4}+\\cdots\\right)+e^{-i\\theta}\\left(1-\\frac{e^{-i2\\theta}}{4}+\\cdots\\right)\\right) \\\\ &= \\frac{1}{4}\\left(\\frac{e^{i\\theta}}{1+\\frac{1}{4}e^{i2\\theta}}+\\frac{e^{-i\\theta}}{1+\\frac{1}{4}e^{-i2\\theta}}\\right)\\\\ &= \\frac{5(e^{i\\theta}+e^{-i\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}}.\\end{align*} Repeating the above process, we find that the $\\sin$ part of $P$ is \\[\\frac{2i(e^{i2\\theta}-e^{-i2\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}},\\] the $\\cos$ part of $Q$ is \\[\\frac{16+2(e^{i2\\theta}+e^{-i2\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}},\\] and finally, the $\\sin$ part of $Q$ is \\[\\frac{3i(e^{i\\theta}-e^{-i\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}}.\\] Converting back to trigonometric form, we have \\begin{align*}\\frac{2\\sqrt{2}}{7}&=\\frac{10\\cos{\\theta}-4\\sin{2\\theta}}{16+4\\cos{2\\theta}-6\\sin{\\theta}}\\\\ &=\\frac{5\\cos{\\theta}-2\\sin{2\\theta}}{8+2\\cos{2\\theta}-3\\sin{\\theta}}.\\end{align*} Using the $\\sin$ double identity and simplifying, we have \\[\\frac{2\\sqrt2}{7}=\\frac{\\cos{\\theta}(5-4\\sin{\\theta})}{10-4\\sin^2{\\theta}-3\\sin{\\theta}}.\\] Factoring the denominator, we have \\[10-4\\sin^2{\\theta}-3\\sin{\\theta}=(5-4\\sin\\theta)(2+\\sin\\theta).\\] Simplifying \\begin{align*}\\frac{2\\sqrt2}{7}&= \\frac{\\cos{\\theta}(5-4\\sin{\\theta})}{(5-4\\sin\\theta)(2+\\sin\\theta)}\\\\ &=\\frac{\\cos\\theta}{2+\\sin\\theta}.\\end{align*} We set $\\sin\\theta$ as $x$, and by the Pythagorean Identity, we have $57x^2+32x-17=0$. This factors into $(19x+17)(3x-1)=0$, which yields the 2 solutions $x=-\\frac{17}{19}, x=\\frac{1}{3}$. As $\\pi\\leq\\theta<2\\pi$, the latter root is erroneous, and we are left with $\\sin\\theta=-\\frac{17}{19}$. Thus, our final answer is $17+19=036. ~ASAB",
"We notice $\\sin\\theta=-\\frac{i}{2}(e^{i\\theta}-e^{-i\\theta})$ and $\\cos\\theta=\\frac{1}{2}(e^{i\\theta}+e^{-i\\theta})$ We observe that both $P$ and $Q$ can be split into $2$ parts, namely the terms which contain the $\\cos$ and the terms which contain the $\\sin .$ The $\\cos$ part of $P$ can be expressed as: \\begin{align*}\\frac12\\cos\\theta-\\frac18\\cos3\\theta+\\cdots&=\\frac14\\left(e^{i\\theta}\\left(1-\\frac{e^{i2\\theta}}{4}+\\cdots\\right)+e^{-i\\theta}\\left(1-\\frac{e^{-i2\\theta}}{4}+\\cdots\\right)\\right) \\\\ &= \\frac{1}{4}\\left(\\frac{e^{i\\theta}}{1+\\frac{1}{4}e^{i2\\theta}}+\\frac{e^{-i\\theta}}{1+\\frac{1}{4}e^{-i2\\theta}}\\right)\\\\ &= \\frac{5(e^{i\\theta}+e^{-i\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}}.\\end{align*} Repeating the above process, we find that the $\\sin$ part of $P$ is \\[\\frac{2i(e^{i2\\theta}-e^{-i2\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}},\\] the $\\cos$ part of $Q$ is \\[\\frac{16+2(e^{i2\\theta}+e^{-i2\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}},\\] and finally, the $\\sin$ part of $Q$ is \\[\\frac{3i(e^{i\\theta}-e^{-i\\theta})}{17+4e^{i2\\theta}+4e^{-i2\\theta}}.\\] Converting back to trigonometric form, we have \\begin{align*}\\frac{2\\sqrt{2}}{7}&=\\frac{10\\cos{\\theta}-4\\sin{2\\theta}}{16+4\\cos{2\\theta}-6\\sin{\\theta}}\\\\ &=\\frac{5\\cos{\\theta}-2\\sin{2\\theta}}{8+2\\cos{2\\theta}-3\\sin{\\theta}}.\\end{align*} Using the $\\sin$ double identity and simplifying, we have \\[\\frac{2\\sqrt2}{7}=\\frac{\\cos{\\theta}(5-4\\sin{\\theta})}{10-4\\sin^2{\\theta}-3\\sin{\\theta}}.\\] Factoring the denominator, we have \\[10-4\\sin^2{\\theta}-3\\sin{\\theta}=(5-4\\sin\\theta)(2+\\sin\\theta).\\] Simplifying \\begin{align*}\\frac{2\\sqrt2}{7}&= \\frac{\\cos{\\theta}(5-4\\sin{\\theta})}{(5-4\\sin\\theta)(2+\\sin\\theta)}\\\\ &=\\frac{\\cos\\theta}{2+\\sin\\theta}.\\end{align*} We set $\\sin\\theta$ as $x$, and by the Pythagorean Identity, we have $57x^2+32x-17=0$. This factors into $(19x+17)(3x-1)=0$, which yields the 2 solutions $x=-\\frac{17}{19}, x=\\frac{1}{3}$. As $\\pi\\leq\\theta<2\\pi$, the latter root is erroneous, and we are left with $\\sin\\theta=-\\frac{17}{19}$. Thus, our final answer is $17+19=036. ~ASAB",
"Follow solution 3, up to the point of using the geometric series formula \\[Q+iP=\\frac{1}{1+\\frac{\\sin(\\theta)}{2}-\\frac{Qi\\cos(\\theta)}{2}}\\] Moving everything to the other side, and considering only the imaginary part, we get \\[Pi+\\frac{Pi}{2}\\sin\\theta-\\frac{Qi}{2}\\cos\\theta = 0\\] We can then write $P = 2 \\sqrt{2} k$, and $Q = 7k$, ($k \\neq 0$). Thus, we can substitute and divide out by k. \\[2\\sqrt{2}+\\sqrt{2}\\sin\\theta-\\frac{7}{2}\\cos\\theta\\ =\\ 0\\] \\[2\\sqrt{2}+\\sqrt{2}\\sin\\theta-\\frac{7}{2}\\sqrt{1-\\sin^{2}\\theta}=\\ 0\\] \\[2\\sqrt{2}+\\sqrt{2}\\sin\\theta\\ =\\frac{7}{2}\\left(\\sqrt{1-\\sin^{2}\\theta}\\right)\\] \\[8+8\\sin\\theta+2\\sin^{2}\\theta=\\frac{49}{4}-\\frac{49}{7}\\sin^{2}\\theta\\] \\[\\frac{57}{4}\\sin^{2}\\theta+8\\sin\\theta-\\frac{17}{4} = 0\\] \\[57\\sin^{2}\\theta+32\\sin\\theta-17 = 0\\] \\[\\left(3\\sin\\theta-1\\right)\\left(19\\sin\\theta+17\\right) = 0\\] Since $\\pi \\le \\theta < 2\\pi$, we get $\\sin \\theta < 0$, and thus, $\\sin\\theta = \\frac{-17}{19} \\implies 036 -Alexlikemath"
] |
2013-I-15
| 2,013
| 15
|
Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions: (a) $0\le A<B<C\le99$ , (b) there exist integers $a$ , $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ , (c) $p$ divides $A-a$ , $B-b$ , and $C-c$ , and (d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$ .
| 272
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I
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[
"From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$. Condition $\\text{(c)}$ states that $p\\mid B-D-a$, $p | B-a+d$, and $p\\mid B+D-a-d$. We subtract the first two to get $p\\mid-d-D$, and we do the same for the last two to get $p\\mid 2d-D$. We subtract these two to get $p\\mid 3d$. So $p\\mid 3$ or $p\\mid d$. The second case is clearly impossible, because that would make $c=a+d>p$, violating condition $\\text{(b)}$. So we have $p\\mid 3$, meaning $p=3$. Condition $\\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\\in (1,0,2)\\rightarrow (-2,0,2)\\text{ }(D\\equiv 2\\text{ mod 3})$. Now we return to condition $\\text{(c)}$, which now implies that $(A,B,C)\\equiv(-2,0,2)\\pmod{3}$. Now, we set $B=3k$ for increasing positive integer values of $k$. $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$, giving us $1$ solution. If $B=6$, we get $2$ solutions, $(4,6,8)$ and $(1,6,11)$. Proceeding in the manner, we see that if $B=48$, we get 16 solutions. However, $B=51$ still gives $16$ solutions because $C_\\text{max}=2B-1=101>100$. Likewise, $B=54$ gives $15$ solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\\cdots+16+16+15+\\cdots+1=2\\cdot\\frac{16(17)}{2}=16\\cdot 17=272.",
"Let $(A, B, C)$ = $(B-x, B, B+x)$ and $(b, a, c) = (a-y, a, a+y)$. Now the 3 differences would be \\begin{align} \\label{1} &A-a = B-x-a \\\\ \\label{2} &B - b = B-a+y \\\\ \\label{3} &C - c = B+x-a-y \\end{align} Adding equations $(1)$ and $(3)$ would give $2B - 2a - y$. Then doubling equation $(2)$ would give $2B - 2a + 2y$. The difference between them would be $3y$. Since $p|\\{(1), (2), (3)\\}$, then $p|3y$. Since $p$ is prime, $p|3$ or $p|y$. However, since $p > y$, we must have $p|3$, which means $p=3$. If $p=3$, the only possible values of $(b, a, c)$ are $(0, 1, 2)$. Plugging this into our differences, we get \\begin{align*} &A-a = B-x-1 \\hspace{4cm}(4)\\\\ &B - b = B \\hspace{5.35cm}(5)\\\\ &C - c = B+x-2 \\hspace{4cm}(6) \\end{align*} The difference between $(4)$ and $(5)$ is $x+1$, which should be divisible by 3. So $x \\equiv 2 \\mod 3$. Also note that since $3|(5)$, $3|B$. Now we can try different values of $x$ and $B$: When $x=2$, $B=3, 6, ..., 96 \\Rightarrow 32$ triples. When $x=5$, $B=6, 9, ..., 93\\Rightarrow 30$ triples.. ... and so on until When $x=47$, $B=48, 51\\Rightarrow 2$ triple. So the answer is $32 + 30 + \\cdots + 2 = 272 ~SoilMilk"
] |
2013-II-1
| 2,013
| 1
|
Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ , $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$ .
| 275
|
II
|
[
"There are $24 \\cdot 60=1440$ normal minutes in a day , and $10 \\cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\\frac{1440}{1000}$, which simplifies to $\\frac{36}{25}$. This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\\text{6:36}$ AM, $6 \\cdot 60+36=396$ normal minutes pass. This can be viewed as $\\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \\cdot 11=275$ to $\\text{0:00}$ gives $\\text{2:75}$, so the answer is $275.",
"First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\\frac{1}{24}$ of a day, and 36 minutes is $\\frac{36}{60}=\\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \\cdot \\frac{1}{24} + \\frac{1}{24} \\cdot \\frac{3}{5}=\\frac{1}{4}+\\frac{1}{40}=\\frac{11}{40}$ of a day has passed. Now the metric timing equivalent of $\\frac{11}{40}$ of a day is $\\frac{11}{40}\\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $275 - mathleticguyyy"
] |
2013-II-2
| 2,013
| 2
|
Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ .
| 881
|
II
|
[
"To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\\log_{2^a}(\\log_{2^b}(2^{1000}))=1$ (because $2^0=1$). Doing this again, we get $\\log_{2^b}(2^{1000})=2^a$. Doing the process one more time, we finally eliminate all of the logs, getting ${(2^{b})}^{(2^a)}=2^{1000}$. Using the property that ${(a^x)^{y}}=a^{xy}$, we simplify to $2^{b\\cdot2^{a}}=2^{1000}$. Eliminating equal bases leaves $b\\cdot2^a=1000$. The largest $a$ such that $2^a$ divides $1000$ is $3$, so we only need to check $1$,$2$, and $3$. When $a=1$, $b=500$; when $a=2$, $b=250$; when $a=3$, $b=125$. Summing all the $a$'s and $b$'s gives the answer of $881 are positive).",
"We proceed as in Solution 1, raising $2$ to both sides to achieve $\\log_{2^a}(\\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\\log_{2^b}(2^{1000})=2^a$, then simplify to get $\\dfrac{1000}b=2^a$. At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$. To help us see the next step, we factorize $1000$: $\\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$; when $a=1$, $b=500$; when $a=2$, $b=250$; and finally, when $a=3$, $b=125.$ We sum all the pairs to get $881 ~Technodoggo"
] |
2013-II-3
| 2,013
| 3
|
A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$ -th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$ . Find $10h$ .
| 350
|
II
|
[
"We find that $T=10(1+2+\\cdots +119)$. From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$. The value of $\\frac{T}{2}$ is therefore $35700$. We find that $35700$ is $10(3570)=10\\cdot \\frac{k(k+1)}{2}$, so $3570=\\frac{k(k+1)}{2}$. As a result, $7140=k(k+1)$, which leads to $0=k^2+k-7140$. We notice that $k=84$, so the answer is $10(119-84)=350."
] |
2013-II-4
| 2,013
| 4
|
In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ .
| 40
|
II
|
[
"The distance from point $A$ to point $B$ is $\\sqrt{13}$. The vector that starts at point A and ends at point B is given by $B - A = (1, 2\\sqrt{3})$. Since the center of an equilateral triangle, $P$, is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to $\\overline{AB}$. The line perpendicular to $\\overline{AB}$ through the midpoint, $M = \\left(\\dfrac{3}{2},\\sqrt{3}\\right)$, $\\overline{AB}$ can be parameterized by $\\left(\\dfrac{2\\sqrt{3}}{\\sqrt{13}}, \\dfrac{-1}{\\sqrt{13}}\\right) t + \\left(\\dfrac{3}{2},\\sqrt{3}\\right)$. At this point, it is useful to note that $\\Delta BMP$ is a 30-60-90 triangle with $\\overline{MB}$ measuring $\\dfrac{\\sqrt{13}}{2}$. This yields the length of $\\overline{MP}$ to be $\\dfrac{\\sqrt{13}}{2\\sqrt{3}}$. Therefore, $P =\\left(\\dfrac{2\\sqrt{3}}{\\sqrt{13}},\\dfrac{-1}{\\sqrt{13}}\\right)\\left(\\dfrac{\\sqrt{13}}{2\\sqrt{3}}\\right) + \\left(\\dfrac{3}{2},\\sqrt{3}\\right) = \\left(\\dfrac{5}{2}, \\dfrac{5}{2\\sqrt{3}}\\right)$. Therefore $xy = \\dfrac{25\\sqrt{3}}{12}$ yielding an answer of $p + q + r = 25 + 3 + 12 = 040.",
"Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is $2 + 2\\sqrt{3}i$. Recall that a rotation of $\\theta$ radians counterclockwise is equivalent to multiplying a complex number by $e^{i\\theta}$, but here we require a clockwise rotation, so we multiply by $e^{-\\frac{i\\pi}{3}}$ to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. $\\left(\\frac{5}{2}, \\frac{5\\sqrt{3}}{6}\\right)$. Therefore $xy$ is $\\frac{25\\sqrt{3}}{12}$ and the answer is $25 + 12 + 3 = 040.",
"We can also consider the slopes of the lines. Midpoint $M$ of $AB$ has coordinates $\\left(\\frac{3}{2},\\ \\sqrt{3}\\right)$. Because line $AB$ has slope $2\\sqrt{3}$, the slope of line $MP$ is $-\\frac{1}{2\\sqrt{3}}$ (Because of perpendicular slopes). Since $\\Delta ABC$ is equilateral, and since point $P$ is the centroid, we can quickly calculate that $MP = \\frac{\\sqrt{39}}{6}$. Then, define $\\Delta x$ and $\\Delta y$ to be the differences between points $M$ and $P$. Because of the slope, it is clear that $\\Delta x = 2\\sqrt{3} \\Delta y$. We can then use the Pythagorean Theorem on line segment $MP$: $MP^2 = \\Delta x^2 + \\Delta y^2$ yields $\\Delta y = -\\frac{1}{2\\sqrt{3}}$ and $\\Delta x = 1$, after substituting $\\Delta x$. The coordinates of P are thus $\\left(\\frac{5}{2},\\ \\frac{5\\sqrt{3}}{6}\\right)$. Multiplying these together gives us $\\frac{25\\sqrt{3}}{12}$, giving us $040 as our answer.",
"Since $AC$ will be segment $AB$ rotated clockwise $60^{\\circ}$, we can use a rotation matrix to find $C$. We first translate the triangle $1$ unit to the left, so $A'$ lies on the origin, and $B' = (1, 2\\sqrt{3})$. Rotating clockwise $60^{\\circ}$ is the same as rotating $300^{\\circ}$ counter-clockwise, so our rotation matrix is $\\begin{bmatrix} \\cos{300^{\\circ}} & -\\sin{300^{\\circ}}\\\\ \\sin{300^{\\circ}} & \\cos{300^{\\circ}}\\\\ \\end{bmatrix} = \\begin{bmatrix} \\frac{1}{2} & \\frac{\\sqrt{3}}{2}\\\\ -\\frac{\\sqrt{3}}{2} & \\frac{1}{2}\\\\ \\end{bmatrix}$. Then $C' = \\begin{bmatrix} \\frac{1}{2} & \\frac{\\sqrt{3}}{2}\\\\ -\\frac{\\sqrt{3}}{2} & \\frac{1}{2}\\\\ \\end{bmatrix} \\cdot \\begin{bmatrix} 1\\\\ 2\\sqrt{3}\\\\ \\end{bmatrix} = \\begin{bmatrix} \\frac{7}{2}\\\\ \\frac{\\sqrt{3}}{2}\\\\ \\end{bmatrix}$. Thus, $C = (\\frac{9}{2}, \\frac{\\sqrt{3}}{2})$. Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then $P = (\\frac{1 + 2 + \\frac{9}{2}}{3}, \\frac{2\\sqrt{3} + \\frac{\\sqrt{3}}{2}}{3}) = (\\frac{5}{2}, \\frac{5\\sqrt{3}}{6})$. Our answer is $\\frac{5}{2} \\cdot \\frac{5\\sqrt{3}}{6} = \\frac{25\\sqrt{3}}{12}$. $25 + 3 + 12 = 40",
"We construct point $C$ by drawing two circles with radius $r = AB = \\sqrt{13}$. One circle will be centered at $A$, while the other is centered at $B$. The equations of the circles are: $(x - 1)^2 + y^2 = 13$ $(x - 2)^2 + (y - 2\\sqrt{3})^2 = 13$ Setting the LHS of each of these equations equal to each other and solving for $x$ yields after simplification: $x = \\frac{15}{2} - 2\\sqrt{3}y$ Plugging that into the first equation gives the following quadratic in $y$ after simplification: $y^2 - 2\\sqrt{3}y + \\frac{9}{4} = 0$ The quadratic formula gives $y = \\frac{\\sqrt{3}}{2}, \\frac{3\\sqrt{3}}{2}$. Since $x > 0$ and $x = \\frac{15}{2} - 2\\sqrt{3}y$, we pick $y = \\frac{\\sqrt{3}}{2}$ in the hopes that it will give $x > 0$. Plugging $y$ into the equation for $x$ yields $x = \\frac{9}{2}$. Thus, $C(\\frac{9}{2}, \\frac{\\sqrt{3}}{2})$. Averaging the coordinates of the vertices of equilateral triangle $ABC$ will give the center of mass of the triangle. Thus, $P(\\frac{5}{2}, \\frac{5\\sqrt{3}}{6})$, and the product of the coordinates is $\\frac{25\\sqrt{3}}{12}$, so the desired quantity is $040.",
"Labeling our points and sketching a graph we get that $C$ is to the right of $AB$. Of course, we need to find $C$. Note that the transformation from $A$ to $B$ is $[1,2\\sqrt{3}]$, and if we imagine a height dropped to $AB$ we see that a transformation from the midpoint $(\\frac{3}{2},\\sqrt {3})$ to $C$ is basically the first transformation, with $\\frac{\\sqrt{3}}{2}$ the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of $[3,\\frac{-\\sqrt{3}}{2}]$ we get that $C=(\\frac{9}{2},\\frac{\\sqrt{3}}{2})$ which means that $P=(\\frac{5}{2},\\frac{5\\sqrt{3}}{6})$. Then our answer is $40.",
"Transform this into the complex plane and let $a=1, b=2+2\\sqrt3 i$. We know that 3 complex numbers $a,b,c$ form an equilateral triangle if $a^2+b^2+c^2=ab+bc+ac$, so plugging in our values of $a,b$, we get $8\\sqrt3 i - 7 +c^2 = 2+2\\sqrt3 i + (3+2\\sqrt 3i)c.$ Solving for $c$ using Wolfram Alpha, we find that the solutions are $c=\\frac 92 + \\frac{i\\sqrt3}{2}, -\\frac 32 + \\frac{3i\\sqrt3}{2}$. The first one is in the first quadrant, so $C\\left( \\frac 92, \\frac{\\sqrt3}{2} \\right)$. The center is the average of the coordinates and we find that it is $\\left(\\frac{5}2, \\frac{5\\sqrt3}{6} \\right)$. Then $xy = \\frac{25\\sqrt3}{12} \\implies 40$. -bobthegod78, krwang, and Simplest14"
] |
2013-II-5
| 2,013
| 5
|
In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ .
| 20
|
II
|
[
"[asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$D$\", D, S); label(\"$M$\", M, S); label(\"$E$\", E, S); draw(A--D); draw(A--M); draw(A--E); [/asy] Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier. Let $M$ be the midpoint of $\\overline{DE}$. Then $\\Delta MCA$ is a 30-60-90 triangle with $MC = \\dfrac{3}{2}$, $AC = 3$ and $AM = \\dfrac{3\\sqrt{3}}{2}$. Since the triangle $\\Delta AME$ is right, then we can find the length of $\\overline{AE}$ by pythagorean theorem, $AE = \\sqrt{7}$. Therefore, since $\\Delta AME$ is a right triangle, we can easily find $\\sin(\\angle EAM) = \\dfrac{1}{2\\sqrt{7}}$ and $\\cos(\\angle EAM) = \\sqrt{1-\\sin(\\angle EAM)^2}=\\dfrac{3\\sqrt{3}}{2\\sqrt{7}}$. So we can use the double angle formula for sine, $\\sin(\\angle EAD) = 2\\sin(\\angle EAM)\\cos(\\angle EAM) = \\dfrac{3\\sqrt{3}}{14}$. Therefore, $a + b + c = 020.",
"We find that, as before, $AE = \\sqrt{7}$, and also the area of $\\Delta DAE$ is $\\dfrac{1}{3}$ the area of $\\Delta ABC$. Thus, using the area formula, $\\dfrac{1}{2} \\cdot 7 \\cdot \\sin(\\angle EAD) = \\dfrac{3\\sqrt{3}}{4}$, and $\\sin(\\angle EAD) = \\dfrac{3\\sqrt{3}}{14}$. Therefore, $a + b + c = 020",
"Let A be the origin of the complex plane, B be $1+i\\sqrt{3}$, and C be $2$. Also, WLOG, let D have a greater imaginary part than E. Then, D is $\\frac{4}{3}+\\frac{2i\\sqrt{3}}{3}$ and E is $\\frac{5}{3}+\\frac{i\\sqrt{3}}{3}$. Then, $\\sin(\\angle DAE) = Im\\left(\\dfrac{\\frac{4}{3}+\\frac{2i\\sqrt{3}}{3}}{ \\frac{5}{3}+\\frac{i\\sqrt{3}}{3}}\\right) = Im\\left(\\frac{26+6i\\sqrt{3}}{28}\\right) = \\frac{3\\sqrt{3}}{14}$. Therefore, $a + b + c = 020",
"Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, $AE^2=1+3^2-2(1)(3)\\cos(\\angle DAE)$ or $AE=\\sqrt7$ The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, $\\cos(\\angle DAE)=(1-7-7)/-2(\\sqrt7)(\\sqrt7)=13/14$ Since $\\sin^2(\\angle DAE)=1-cos^2(\\angle DAE)$ Then $\\sin^2(\\angle DAE)= 1-\\frac{169}{196}=\\frac{27}{196}$ So $\\sin(\\angle DAE)=\\frac{3\\sqrt{3}}{14}$. Therefore, $a + b + c = 020",
"Setting up a convinient coordinate system, we let $A$ be at point $(0, 0)$, $B$ be at point $(3, 3\\sqrt3)$, and $C$ be at point $(6, 0)$. Then $D$ and $E$ will be at points $(4, 2\\sqrt3)$ and $(5, \\sqrt3)$. Then $\\cos(\\angle DAE) = \\frac{\\vec{AD}\\cdot\\vec{AE}}{\\|\\vec{AD}\\| \\|\\vec{AE}\\|} = \\frac{4\\cdot5 + 2\\sqrt{3}\\cdot\\sqrt{3}}{28}=\\frac{13}{14}$. From here, we see that $\\sin(\\angle DAE) = \\sqrt{1-\\cos^2(\\angle DAE)} = \\frac{3\\sqrt3}{14}\\Longrightarrow020",
"Setting up a convinient coordinate system, we let $A$ be at point $(0, 0)$, $B$ be at point $(3, 3\\sqrt3)$, and $C$ be at point $(6, 0)$. Then $D$ and $E$ will be at points $(4, 2\\sqrt3)$ and $(5, \\sqrt3)$. Then $\\cos(\\angle DAE) = \\frac{\\vec{AD}\\cdot\\vec{AE}}{\\|\\vec{AD}\\| \\|\\vec{AE}\\|} = \\frac{4\\cdot5 + 2\\sqrt{3}\\cdot\\sqrt{3}}{28}=\\frac{13}{14}$. From here, we see that $\\sin(\\angle DAE) = \\sqrt{1-\\cos^2(\\angle DAE)} = \\frac{3\\sqrt3}{14}\\Longrightarrow020"
] |
2013-II-6
| 2,013
| 6
|
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.
| 282
|
II
|
[
"Solution 1 The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\\geq1000$, so $x\\geq\\frac{999}{2}\\implies x\\geq500$. $x=500$ does not work, so $x>500$. Let $n=x-500$. By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference of the two squares. Examine when $n^2=1000$. Then, $n=10\\sqrt{10}$. One example way to estimate $\\sqrt{10}$ follows. $3^2=9$, so $10=(x+3)^2=x^2+6x+9$. $x^2$ is small, so $10=6x+9$. $x=1/6\\implies \\sqrt{10}\\approx 19/6$. This is 3.16. Then, $n\\approx 31.6$. $n^2<1000$, so $n$ could be $31$. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$. Checking, $531^2=281961$ and $532^2=283024$. $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\\implies282. ~yuxiaomatt",
"The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\\geq1000$, so $x\\geq\\frac{999}{2}\\implies x\\geq500$. $x=500$ does not work, so $x>500$. Let $n=x-500$. By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference of the two squares. Examine when $n^2=1000$. Then, $n=10\\sqrt{10}$. One example way to estimate $\\sqrt{10}$ follows. $3^2=9$, so $10=(x+3)^2=x^2+6x+9$. $x^2$ is small, so $10=6x+9$. $x=1/6\\implies \\sqrt{10}\\approx 19/6$. This is 3.16. Then, $n\\approx 31.6$. $n^2<1000$, so $n$ could be $31$. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$. Checking, $531^2=281961$ and $532^2=283024$. $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\\implies282. ~yuxiaomatt",
"Let us first observe the difference between $x^2$ and $(x+1)^2$, for any arbitrary $x\\ge 0$. $(x+1)^2-x^2=2x+1$. So that means for every $x\\ge 0$, the difference between that square and the next square have a difference of $2x+1$. Now, we need to find an $x$ such that $2x+1\\ge 1000$. Solving gives $x\\ge \\frac{999}{2}$, so $x\\ge 500$. Now we need to find what range of numbers has to be square-free: $\\overline{N000}\\rightarrow \\overline{N999}$ have to all be square-free. Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$, $x^2=250000$, and $(x+1)^2=251001$. Notice that this does not fit the criteria, because $250000$ is a square, whereas $\\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$, but not there, such as $961$ or $974$. Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$, so all we need to do is addition. After making a list, we find that $531^2=281961$, while $532^2=283024$. It skipped $282000$, so our answer is $282. ~yuxiaomatt",
"Let $x$ be the number being squared. Based on the reasoning above, we know that $N$ must be at least $250$, so $x$ has to be at least $500$. Let $k$ be $x-500$. We can write $x^2$ as $(500+k)^2$, or $250000+1000k+k^2$. We can disregard $250000$ and $1000k$, since they won't affect the last three digits, which determines if there are any squares between $\\overline{N000}\\rightarrow \\overline{N999}$. So we must find a square, $k^2$, such that it is under $1000$, but the next square is over $1000$. We find that $k=31$ gives $k^2=961$, and so $(k+1)^2=32^2=1024$. We can be sure that this skips a thousand because the $1000k$ increments it up $1000$ each time. Now we can solve for $x$: $(500+31)^2=281961$, while $(500+32)^2=283024$. We skipped $282000$, so the answer is $282. ~yuxiaomatt",
"The goal is to find the least $N \\in \\mathbb{N}$ such that $\\exists m \\in \\mathbb{N}$ where $m^2 + 1 \\leq 1000N, 1000N + 1000 \\leq (m+1)^2$. Combining the two inequalities leads to $(m+1)^2 \\geq m^2 + 1001, m \\geq 500$. Let $m = k + 500$, where $k \\in \\mathbb{W}$, then the inequalities become, $N \\geq \\frac{(k+500)^2 + 1}{1000} = \\frac{k^2 + 1}{1000} + k + 250$, and $N \\leq \\frac{(k+501)^2}{1000} - 1 = \\frac{(k+1)^2}{1000} + k + 250.$ For $k=31$, one can verify that $N = 282$ is the unique integer satisfying the inequalities. For $k \\leq 30$, $k + 250 < \\frac{k^2 + 1}{1000} + k + 250 \\leq N$ $\\leq \\frac{(k+1)^2}{1000} + k + 250 \\leq \\frac{(30+1)^2}{1000} + k + 250 < k + 251$, i.e., $k + 250 < N < k + 251$, a contradiction. Note $k \\geq 32$ leads to larger $N$(s). Hence, the answer is $282. ~yuxiaomatt"
] |
2013-II-7
| 2,013
| 7
|
A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.
| 945
|
II
|
[
"There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\\cdot\\frac{10}{60} \\cdot (x-3t)=1775$, and simplify that we get $19x-21t=355$. Now the problem is to find a reasonable integer solution. Now we know $x= \\frac{355+21t}{19}$, so $19$ divides $355+21t$, AND as long as $t$ is a integer, $19$ must divide $2t+355$. Now, we suppose that $19m=2t+355$, similarly we get $t=\\frac{19m-355}{2}$, and so in order to get a minimum integer solution for $t$, it is obvious that $m=19$ works. So we get $t=3$ and $x=22$. One and a half hour's work should be $30x+15(x-t)$, so the answer is $945.",
"We start with the same approach as solution 1 to get $19x-21t=355$. Then notice that $21t + 355 \\equiv 0 \\pmod{19}$, or $2t-6 \\equiv 0 \\pmod{19}$, giving the smallest solution at $t=3$. We find that $x=22$. Then the number of files they sorted will be $30x+15(x-t)=660+285=945"
] |
2013-II-8
| 2,013
| 8
|
A hexagon that is inscribed in a circle has side lengths $22$ , $22$ , $20$ , $22$ , $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ .
| 272
|
II
|
[
"[asy] import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label(\"$A$\",A,W);label(\"$B$\",B,NW);label(\"$C$\",C,NE);label(\"$D$\",D,dir(0));label(\"$E$\",E,SE);label(\"$F$\",F,SW); [/asy] Solution 1 Let us call the hexagon $ABCDEF$, where $AB=CD=DE=AF=22$, and $BC=EF=20$. We can just consider one half of the hexagon, $ABCD$, to make matters simpler. Draw a line from the center of the circle, $O$, to the midpoint of $BC$, $X$. Now, draw a line from $O$ to the midpoint of $AB$, $Y$. Clearly, $\\angle BXO=90^{\\circ}$, because $BO=CO$, and $\\angle BYO=90^{\\circ}$, for similar reasons. Also notice that $\\angle AOX=90^{\\circ}$. Let us call $\\angle BOY=\\theta$. Therefore, $\\angle AOB=2\\theta$, and so $\\angle BOX=90-2\\theta$. Let us label the radius of the circle $r$. This means \\[\\sin{\\theta}=\\frac{BY}{r}=\\frac{11}{r}\\] \\[\\sin{(90-2\\theta)}=\\frac{BX}{r}=\\frac{10}{r}\\] Now we can use simple trigonometry to solve for $r$. Recall that $\\sin{(90-\\alpha)}=\\cos(\\alpha)$: That means $\\sin{(90-2\\theta)}=\\cos{2\\theta}=\\frac{10}{r}$. Recall that $\\cos{2\\alpha}=1-2\\sin^2{\\alpha}$: That means $\\cos{2\\theta}=1-2\\sin^2{\\theta}=\\frac{10}{r}$. Let $\\sin{\\theta}=x$. Substitute to get $x=\\frac{11}{r}$ and $1-2x^2=\\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\\left(\\frac{11}{r}\\right)^2=\\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \\[r^2-10r-242=0\\] Using the quadratic equation to solve, we get that $r=5+\\sqrt{267}$ (because $5-\\sqrt{267}$ gives a negative value), so the answer is $5+267=272. ~Puck_0",
"Let us call the hexagon $ABCDEF$, where $AB=CD=DE=AF=22$, and $BC=EF=20$. We can just consider one half of the hexagon, $ABCD$, to make matters simpler. Draw a line from the center of the circle, $O$, to the midpoint of $BC$, $X$. Now, draw a line from $O$ to the midpoint of $AB$, $Y$. Clearly, $\\angle BXO=90^{\\circ}$, because $BO=CO$, and $\\angle BYO=90^{\\circ}$, for similar reasons. Also notice that $\\angle AOX=90^{\\circ}$. Let us call $\\angle BOY=\\theta$. Therefore, $\\angle AOB=2\\theta$, and so $\\angle BOX=90-2\\theta$. Let us label the radius of the circle $r$. This means \\[\\sin{\\theta}=\\frac{BY}{r}=\\frac{11}{r}\\] \\[\\sin{(90-2\\theta)}=\\frac{BX}{r}=\\frac{10}{r}\\] Now we can use simple trigonometry to solve for $r$. Recall that $\\sin{(90-\\alpha)}=\\cos(\\alpha)$: That means $\\sin{(90-2\\theta)}=\\cos{2\\theta}=\\frac{10}{r}$. Recall that $\\cos{2\\alpha}=1-2\\sin^2{\\alpha}$: That means $\\cos{2\\theta}=1-2\\sin^2{\\theta}=\\frac{10}{r}$. Let $\\sin{\\theta}=x$. Substitute to get $x=\\frac{11}{r}$ and $1-2x^2=\\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\\left(\\frac{11}{r}\\right)^2=\\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \\[r^2-10r-242=0\\] Using the quadratic equation to solve, we get that $r=5+\\sqrt{267}$ (because $5-\\sqrt{267}$ gives a negative value), so the answer is $5+267=272. ~Puck_0",
"Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $J$. Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$. Expanding and combining like terms gives us the quadratic \\[r^2-10r-242=0\\] and solving for $r$ gives $r=5+\\sqrt{267}$. So the solution is $5+267=272. ~Puck_0",
"Join the diameter of the circle $AD$ and let the length be $d$. By Ptolemy's Theorem on trapezoid $ADEF$, $(AD)(EF) + (AF)(DE) = (AE)(DF)$. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then \\[20d + 22^2 = x^2\\] Since $\\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\\triangle AED$: \\[(AE)^2 + (ED)^2 = (AD)^2\\] \\[x^2 + 22^2 = d^2\\] From the above equations, we have: \\[x^2 = d^2 - 22^2 = 20d + 22^2\\] \\[d^2 - 20d = 2\\times22^2\\] \\[d^2 - 20d + 100 = 968+100 = 1068\\] \\[(d-10) = \\sqrt{1068}\\] \\[d = \\sqrt{1068} + 10 = 2\\times(\\sqrt{267}+5)\\] Since the radius is half the diameter, it is $\\sqrt{267}+5$, so the answer is $5+267 \\Rightarrow 272. ~Puck_0",
"As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \\[22*2x+440=\\sqrt{4x^2-400}\\sqrt{4x^2-484}\\], after simple factorization, we can get \\[x^4-342x^2-2420x=0\\], it is easy to see that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \\[x(x+10)(x^2-10x-242)\\]so we only need to find the solution for \\[x^2-10x-242=0\\] and we can get $x=(\\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \\Rightarrow 272. ~Puck_0",
"Let $\\angle{AOB} = \\theta$. So, we have $\\sin \\dfrac{\\theta}{2} = \\dfrac{11}{r}$ and $\\cos \\dfrac{\\theta}{2} = \\dfrac{\\sqrt{r^{2} - 121}}{r}$. So, $\\sin \\theta = 2 \\sin \\dfrac{\\theta}{2} \\cos \\dfrac{\\theta}{2} = \\dfrac{22 \\sqrt{r^{2} - 121}}{r^{2}}$. Let $H$ be the foot of the perpendicular from $B$ to $\\overline{AD}$. We have $BF = 2 BH = 2 r \\sin \\theta = \\dfrac{44 \\sqrt{r^{2} - 121}}{r}$. Using Pythagorean theorem on $\\triangle BCF$, to get $(\\dfrac{44 \\sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$, or $\\dfrac{44^{2}r^{2} - 44^{2} \\cdot 121}{r^{2}} + 20^{2} = 4r^{4}$. Multiplying by $r^{2}$, we get $44^{2} r^{2} - 44^{2} \\cdot 121 + 20^{2} r^{2} = 4r^{4}$. Rearranging and simplifying, we get a quadratic in $r^{2}$: \\[r^{4} - 584r^{2} + 242^{2} = 0 \\text{,}\\] which gives us $r^{2} = 292 \\pm 10\\sqrt{267}$. Because $r$ is in the form $p + \\sqrt{q}$, we know to choose the larger option, meaning $r^2 = 292 + 10\\sqrt{267}$, so $p\\sqrt{q} = 5\\sqrt{267}$ and $p^2 + q = 292$. By inspection, we get $(p, q) = (5, 267)$, so our answer is $5 + 267 = 272. ~Puck_0",
"Let $\\angle{AOB} = \\theta$. So, we have $\\sin \\dfrac{\\theta}{2} = \\dfrac{11}{r}$ and $\\cos \\dfrac{\\theta}{2} = \\dfrac{\\sqrt{r^{2} - 121}}{r}$. So, $\\sin \\theta = 2 \\sin \\dfrac{\\theta}{2} \\cos \\dfrac{\\theta}{2} = \\dfrac{22 \\sqrt{r^{2} - 121}}{r^{2}}$. Let $H$ be the foot of the perpendicular from $B$ to $\\overline{AD}$. We have $BF = 2 BH = 2 r \\sin \\theta = \\dfrac{44 \\sqrt{r^{2} - 121}}{r}$. Using Pythagorean theorem on $\\triangle BCF$, to get $(\\dfrac{44 \\sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$, or $\\dfrac{44^{2}r^{2} - 44^{2} \\cdot 121}{r^{2}} + 20^{2} = 4r^{4}$. Multiplying by $r^{2}$, we get $44^{2} r^{2} - 44^{2} \\cdot 121 + 20^{2} r^{2} = 4r^{4}$. Rearranging and simplifying, we get a quadratic in $r^{2}$: \\[r^{4} - 584r^{2} + 242^{2} = 0 \\text{,}\\] which gives us $r^{2} = 292 \\pm 10\\sqrt{267}$. Because $r$ is in the form $p + \\sqrt{q}$, we know to choose the larger option, meaning $r^2 = 292 + 10\\sqrt{267}$, so $p\\sqrt{q} = 5\\sqrt{267}$ and $p^2 + q = 292$. By inspection, we get $(p, q) = (5, 267)$, so our answer is $5 + 267 = 272. ~Puck_0",
"[asy] import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); draw(B--D); draw(A--C); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label(\"$A$\",A,W);label(\"$B$\",B,NW);label(\"$C$\",C,NE);label(\"$D$\",D,dir(0));label(\"$E$\",E,SE);label(\"$F$\",F,SW); [/asy] We know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$, and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \\sqrt{x^2-400},$ with hypotenuse $x$. By Ptolemy's we have \\[22(x+20)=\\sqrt{x^2-400}\\sqrt{x^2-484}\\]. We square both sides to get \\[484(x+20)^2=(x^2-400)(x^2-484) \\implies 484(x+20)=(x-20)(x^2-484) \\implies 484x=x^3-20x^2-484x \\implies x(x^2-20x-968)=0\\] We solve for $x$ via the Quadratic Formula and receive $x=10+2\\sqrt{267}$, but we must divide by $2$ since we want the radius, and hence $267+5=272 ~SirAppel"
] |
2013-II-9
| 2,013
| 9
|
A $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ .
| 106
|
II
|
[
"Firstly, we consider how many different ways possible to divide the $7\\times 1$ board. We ignore the cases of 1 or 2 pieces since we need at least one tile of each color. Three pieces: $5+1+1$, $4+2+1$, $4+1+2$, etc, $\\dbinom{6}{2}=15$ ways in total (just apply stars and bars here) Four pieces: $\\dbinom{6}{3}=20$ Five pieces: $\\dbinom{6}{4}=15$ Six pieces: $\\dbinom{6}{5}=6$ Seven pieces: $\\dbinom{6}{6}=1$ Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them: Three pieces: $3^3-3\\times 2^3+3=6$ Four pieces: $3^4-3\\times 2^4+3=36$ Five pieces: $3^5-3\\times 2^5+3=150$ Six pieces: $3^6-3\\times 2^6+3=540$ Seven pieces: $3^7-3\\times 2^7+3=1806$ Finally, we combine them together: $15\\times 6+20\\times 36+15\\times 150+6\\times 540+1\\times 1806= 8106$. So the answer is $106.",
"This solution is basically solution 1 with more things done at once. The game plan: $\\sum_{i=0}^{7} ($the amount of ways to divide the board into $i$ pieces$) \\cdot ($the amount of ways to color the respective divisions) The amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving $3^i-3\\cdot2^i+3$. Plus, we don't have to worry about the cases where $i=1$ or $i=2$ since they both give no solutions. So our equation becomes: $\\sum_{i=3}^{7} \\left(\\dbinom{6}{i}\\right)\\cdot\\left(3^i-3\\cdot2^i+3\\right)$ Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $106. ~Rowechen Zhong",
"This solution is basically solution 1 with more things done at once. The game plan: $\\sum_{i=0}^{7} ($the amount of ways to divide the board into $i$ pieces$) \\cdot ($the amount of ways to color the respective divisions) The amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving $3^i-3\\cdot2^i+3$. Plus, we don't have to worry about the cases where $i=1$ or $i=2$ since they both give no solutions. So our equation becomes: $\\sum_{i=3}^{7} \\left(\\dbinom{6}{i}\\right)\\cdot\\left(3^i-3\\cdot2^i+3\\right)$ Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $106. ~Rowechen Zhong",
"3 colors is a lot. How many ways can we tile an $n \\times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$ Now let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \\times 1$ based off a $n \\times 1$. There are 2 cases we should consider: 1) The $n \\times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$ 2) The $n \\times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \\times 1$: We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \\times 1$. This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$ Time to tackle the 3-color tiling. Again, we split into 2 cases: 1) The $n \\times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling. 2) The $n \\times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \\times 1$. This gives $4f_3(n)$ So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$. Finally, we just sorta bash through the computation to get $f_3(7)=8106",
"3 colors is a lot. How many ways can we tile an $n \\times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$ Now let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \\times 1$ based off a $n \\times 1$. There are 2 cases we should consider: 1) The $n \\times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$ 2) The $n \\times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \\times 1$: We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \\times 1$. This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$ Time to tackle the 3-color tiling. Again, we split into 2 cases: 1) The $n \\times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling. 2) The $n \\times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \\times 1$. This gives $4f_3(n)$ So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$. Finally, we just sorta bash through the computation to get $f_3(7)=8106",
"Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \\times 1$ board with no color restrictions (some colors may be unused) and then use PIE. By stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \\choose k-1}$. There are $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is \\begin{align*} \\chi(n, x) &:= \\sum_{k=1}^n {n-1 \\choose k-1} x^k \\\\ &= x\\sum_{k=0}^{n-1} {n-1 \\choose k} x^k \\\\ &= x(x+1)^{n-1}. \\end{align*} In the given problem, we have $n=7$. By PIE, we have \\begin{align*} &\\quad {3 \\choose 3} \\chi(7, 3) - {3 \\choose 2} \\, \\chi(7, 2) + {3 \\choose 1} \\, \\chi(7, 1) \\\\ &= 3 \\cdot 4^6 - 3(2 \\cdot 3^6) + 3(1 \\cdot 2^6) \\\\ &= 8106 \\rightarrow 106 overall ways to divide and color the board (without the color use restriction)."
] |
2013-II-10
| 2,013
| 10
|
Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ , $b$ , $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ .
| 146
|
II
|
[
"[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label(\"$O$\", (0,0), S);label(\"$B$\", B, SW);label(\"$A$\", A, S); dot((0,0)); [/asy] Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\\sqrt{13},0)$, and $A(4+\\sqrt{13},0)$ The equation for Circle O is $x^2+y^2=13$, and let the slope of the line $AKL$ be $k$, then the equation for line $AKL$ is $y=k(x-4-\\sqrt{13})$. Then we get $(k^2+1)x^2-2k^2(4+\\sqrt{13})x+k^2\\cdot (4+\\sqrt{13})^2-13=0$. According to Vieta's Formulas, we get $x_1+x_2=\\frac{2k^2(4+\\sqrt{13})}{k^2+1}$, and $x_1x_2=\\frac{(4+\\sqrt{13})^2\\cdot k^2-13}{k^2+1}$ So, $LK=\\sqrt{1+k^2}\\cdot \\sqrt{(x_1+x_2)^2-4x_1x_2}$ Also, the distance between $B$ and $LK$ is $\\frac{k\\times \\sqrt{13}-(4+\\sqrt{13})\\cdot k}{\\sqrt{1+k^2}}=\\frac{-4k}{\\sqrt{1+k^2}}$ So the area $S=0.5ah=\\frac{-4k\\sqrt{(16+8\\sqrt{13})k^2-13}}{k^2+1}$ Then the maximum value of $S$ is $\\frac{104-26\\sqrt{13}}{3}$ So the answer is $104+26+13+3=146.",
"[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label(\"$O$\", (0,0), S);label(\"$B$\", B, SW);label(\"$A$\", A, S); dot((0,0)); [/asy] Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\\sqrt{13},0)$, and $A(4+\\sqrt{13},0)$ The equation for Circle O is $x^2+y^2=13$, and let the slope of the line $AKL$ be $k$, then the equation for line $AKL$ is $y=k(x-4-\\sqrt{13})$. Then we get $(k^2+1)x^2-2k^2(4+\\sqrt{13})x+k^2\\cdot (4+\\sqrt{13})^2-13=0$. According to Vieta's Formulas, we get $x_1+x_2=\\frac{2k^2(4+\\sqrt{13})}{k^2+1}$, and $x_1x_2=\\frac{(4+\\sqrt{13})^2\\cdot k^2-13}{k^2+1}$ So, $LK=\\sqrt{1+k^2}\\cdot \\sqrt{(x_1+x_2)^2-4x_1x_2}$ Also, the distance between $B$ and $LK$ is $\\frac{k\\times \\sqrt{13}-(4+\\sqrt{13})\\cdot k}{\\sqrt{1+k^2}}=\\frac{-4k}{\\sqrt{1+k^2}}$ So the area $S=0.5ah=\\frac{-4k\\sqrt{(16+8\\sqrt{13})k^2-13}}{k^2+1}$ Then the maximum value of $S$ is $\\frac{104-26\\sqrt{13}}{3}$ So the answer is $104+26+13+3=146.",
"[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label(\"$O$\", (0,0), S);label(\"$B$\", B, SW);label(\"$A$\", A, S); dot((0,0)); [/asy] Draw $OC$ perpendicular to $KL$ at $C$. Draw $BD$ perpendicular to $KL$ at $D$. \\[\\frac{\\triangle OKL}{\\triangle BKL} = \\frac{OC}{BD} = \\frac{AO}{AB} = \\frac{4+\\sqrt{13}}{4}\\] Therefore, to maximize area of $\\triangle BKL$, we need to maximize area of $\\triangle OKL$. \\[\\triangle OKL = \\frac12 r^2 \\sin{\\angle KOL}\\] So when area of $\\triangle OKL$ is maximized, $\\angle KOL = \\frac{\\pi}{2}$. Eventually, we get \\[\\triangle BKL= \\frac12 \\cdot (\\sqrt{13})^2\\cdot(\\frac{4}{4+\\sqrt{13}})=\\frac{104-26\\sqrt{13}}{3}\\] So the answer is $104+26+13+3=146.",
"A rather easier solution is presented in the Girls' Angle WordPress: http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/",
"A rather easier solution is presented in the Girls' Angle WordPress: http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/",
"Let $N,M$ les on $AL$ such that $BM\\bot AL, ON\\bot AL$, call $BM=h, ON=k,LN=KN=d$ We call $\\angle{LON}=\\alpha$ By similar triangle, we have $\\frac{h}{k}=\\frac{4}{4+\\sqrt{13}}, h=\\frac{4k}{4+\\sqrt{13}}$. Then, we realize the area is just $dh=d\\cdot \\frac{4K}{4+\\sqrt{13}}$ As $\\sin \\alpha=\\frac{d}{\\sqrt{13}}, \\cos \\alpha=\\frac{k}{\\sqrt{13}}$. Now, we have to maximize $\\frac{52\\sin \\alpha \\cos \\alpha}{4+\\sqrt{13}}=\\frac{26\\sin 2\\alpha}{4+\\sqrt{13}}$, which is obviously reached when $\\alpha=45^{\\circ}$, the answer is $\\frac{104-26\\sqrt{13}}{3}$ leads to $146 ~bluesoul",
"Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H. \\[\\triangle ABD \\sim \\triangle AOC \\implies \\frac {h}{H} = \\frac {4}{4 + \\sqrt{13}}.\\] $KL$ is the base of triangles $\\triangle OKL$ and $\\triangle BKL \\implies \\frac {[BKL]}{[OKL]} = \\frac{h}{H} =$ const $\\implies$ The maximum possible area for $\\triangle BKL$ and $\\triangle OKL$ are at the same position of point $K$. $\\triangle OKL$ has sides $OK = OL = \\sqrt{13}\\implies \\max[\\triangle OKL] = \\frac {OK^2}{2} = \\frac {13}{2}$ in the case $\\angle KOL = 90^\\circ.$ It is possible – if we rotate such triangle, we can find position when $A$ lies on $KL.$ \\[\\max[\\triangle BKL] = \\max[\\triangle OKL] \\cdot \\frac {4}{4+\\sqrt{13}} = \\frac {26}{4+\\sqrt{13}} \\implies \\textbf{146}\\] [email protected], vvsss"
] |
2013-II-11
| 2,013
| 11
|
Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ .
| 399
|
II
|
[
"Any such function can be constructed by distributing the elements of $A$ on three tiers. The bottom tier contains the constant value, $c=f(f(x))$ for any $x$. (Obviously $f(c)=c$.) The middle tier contains $k$ elements $x\\ne c$ such that $f(x)=c$, where $1\\le k\\le 6$. The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier. There are $7$ choices for $c$. Then for a given $k$, there are $\\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier. Thus $N=7\\cdot\\sum_{k=1}^6\\tbinom6k\\cdot k^{6-k}=7399$, giving the answer $399.",
"Define the three layers as domain $x$, codomain $f(x)$, and codomain $f(f(x))$. Each one of them is contained in the set $A$. We know that $f(f(x))$ is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value $c$ in $7$ ways. So now, we choose the values that can be $f(x)$ for all those values should satisfy $f(f(x))=c$. Let $S$ be that set of values. First things first, we must have $c$ to be part of $S$, for the $S$ is part of the domain of $x$. Since the values in $i\\in S$ all satisfy $f(i) = c$, we have $c$ to be a value that $f(x)$ can be. Now, for the elements other than $5$: If we have $k$ elements other than $5$ that can be part of $S$, we will have $\\binom{6}{k}$ ways to choose those values. There will also be $k$ ways for each of the elements in $A$ other than $c$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $c$). There are $6-k$ elements in $A$ other than $c$ and those in set $S$. Thus, there should be $k^{6-k}$ ways to match the domain $x$ to the values of $f(x)$. Summing up all possible values of $k$ ($[1,6]$), we have \\[\\sum_{k=1}^6 \\binom{6}{k} k^{6-k} = 6\\cdot 1 + 15\\cdot 16 + 20\\cdot 27 + 15\\cdot 16 + 6\\cdot 5 + 1 = 1057\\] Multiplying that by the original $7$ for the choice of $c$, we have $7 \\cdot 1057 = 7399",
"It is clear that we must have one fixed point (that is, $f(x)=x$). WLOG, let $x=1$ be a fixed point, so $f(1)=1$. Now, let's do casework on how many of the inputs $2, 3, 4, 5, 6 ,7$ leads to $1$. Generally, if some values in that aforementioned list leads to $1$, then running it in the function again will yield $1$. All other values must be the the values that leads to $1$. For example: $\\textbf{Case 1:}$ All $6$ of $2, 3, 4, 5, 6, 7$ lead to $1$. In this case, there is only $1$ way. $\\textbf{Case 2:}$ $5$ of $6$ of $2, 3, 4, 5, 6, 7$ lead to $1$. In this case, we choose $5$ of the $6$ to lead to $1$: $6\\choose5$. Then, the other value that does not lead to $1$ should be one of the values that do: $5$ ways. $\\binom{6}{5}\\cdot5$ $\\textbf{Case 3:}$ $4$ of $6$ lead to $1$. Choose which lead to $1$: $6\\choose4$ then the other values: $4^2$ ways $\\binom{6}{4}\\cdot4^2$ $\\textbf{Case 4:}$ $3$ of $6$ lead to $1$. $\\binom{6}{3}\\cdot3^3$ $\\textbf{Case 5:}$ $2$ of $6$ lead to $1$. $\\binom{6}{2}\\cdot2^4$ $\\textbf{Case 6:}$ $1$ of $6$ lead to $1$. $\\binom{6}{1}\\cdot1^5$ Adding up all the cases, we have $1057$ cases. But don't forget to account for the WLOG and multiply by $7$, yielding us the final answer of $7399 ~xHypotenuse"
] |
2013-II-12
| 2,013
| 12
|
Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$ , where $a$ , $b$ , and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$ .
| 540
|
II
|
[
"Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from Vieta's formulas. Case 1: $f(z)=(z-r)(z-\\omega)(z-\\omega^*)$, where $r\\in \\mathbb{R}$, $\\omega$ is nonreal, and $\\omega^*$ is the complex conjugate of omega (note that we may assume that $\\Im(\\omega)>0$). The real root $r$ must be one of $-20$, $20$, $-13$, or $13$. By Viète's formulas, $a=-(r+\\omega+\\omega^*)$, $b=|\\omega|^2+r(\\omega+\\omega^*)$, and $c=-r|\\omega|^2$. But $\\omega+\\omega^*=2\\Re{(\\omega)}$ (i.e., adding the conjugates cancels the imaginary part). Therefore, to make $a$ an integer, $2\\Re{(\\omega)}$ must be an integer. Conversely, if $\\omega+\\omega^*=2\\Re{(\\omega)}$ is an integer, then $a,b,$ and $c$ are clearly integers. Therefore $2\\Re{(\\omega)}\\in \\mathbb{Z}$ is equivalent to the desired property. Let $\\omega=\\alpha+i\\beta$. Subcase 1.1: $|\\omega|=20$. In this case, $\\omega$ lies on a circle of radius $20$ in the complex plane. As $\\omega$ is nonreal, we see that $\\beta\\ne 0$. Hence $-20<\\Re{(\\omega)}< 20$, or rather $-40<2\\Re{(\\omega)}< 40$. We count $79$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $20$ with positive imaginary part. Subcase 1.2: $|\\omega|=13$. In this case, $\\omega$ lies on a circle of radius $13$ in the complex plane. As $\\omega$ is nonreal, we see that $\\beta\\ne 0$. Hence $-13<\\Re{(\\omega)}< 13$, or rather $-26<2\\Re{(\\omega)}< 26$. We count $51$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $13$ with positive imaginary part. Therefore, there are $79+51=130$ choices for $\\omega$. We also have $4$ choices for $r$, hence there are $4\\cdot 130=520$ total polynomials in this case. Case 2: $f(z)=(z-r_1)(z-r_2)(z-r_3)$, where $r_1,r_2,r_3$ are all real. In this case, there are four possible real roots, namely $\\pm 13, \\pm20$. Let $p$ be the number of times that $13$ appears among $r_1,r_2,r_3$, and define $q,r,s$ similarly for $-13,20$, and $-20$, respectively. Then $p+q+r+s=3$ because there are three roots. We wish to find the number of ways to choose nonnegative integers $p,q,r,s$ that satisfy that equation. By balls and urns, these can be chosen in $\\binom{6}{3}=20$ ways. Therefore, there are a total of $520+20=540 polynomials with the desired property.",
"This combinatorics problem involves counting, and casework is most appropriate. There are two cases: either all three roots are real, or one is real and there are two imaginary roots. Case 1: Three roots are of the set ${13, -13, 20, -20}$. By stars and bars, there is $\\binom{6}{3}=20$ ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves). Case 2: One real root: one of $13, -13, 20, -20$. Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of $20$ or $13$. Call the root $a+bi$, where $a$ is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of $(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)$ tells us that we just need $2a$ to be integral, because $a^2+b^2$ IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.) Therefore, when the norm is $20$, the $a$ term can range from $-19.5, -19, ...., 0, 0.5, ..., 19.5$ or $79$ solutions. When the norm is $13$, the $a$ term has $51$ possibilities from $-12.5, -12, ..., 12.5$. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, $4$, and you get $520$ for this case. And $520+20=540$ and we are done.",
"This combinatorics problem involves counting, and casework is most appropriate. There are two cases: either all three roots are real, or one is real and there are two imaginary roots. Case 1: Three roots are of the set ${13, -13, 20, -20}$. By stars and bars, there is $\\binom{6}{3}=20$ ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves). Case 2: One real root: one of $13, -13, 20, -20$. Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of $20$ or $13$. Call the root $a+bi$, where $a$ is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of $(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)$ tells us that we just need $2a$ to be integral, because $a^2+b^2$ IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.) Therefore, when the norm is $20$, the $a$ term can range from $-19.5, -19, ...., 0, 0.5, ..., 19.5$ or $79$ solutions. When the norm is $13$, the $a$ term has $51$ possibilities from $-12.5, -12, ..., 12.5$. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, $4$, and you get $520$ for this case. And $520+20=540$ and we are done.",
"If the polynomial has one real root and two complex roots, then it can be factored as $(z-r)(z^2+pz+q),$ where $r$ is real with $|r|=13,20$ and $p,q$ are integers with $p^2 <4q.$ The roots $z_1$ and $z_2$ are conjugates. We have $|z_1|^2=|z_2|^2=z_1z_2=q.$ So $q$ is either $20^2$ or $13^2$. The only requirement for $p$ is $p<\\sqrt{4q^2}=2q.$ All such quadratic equations are listed as follows: $z^2+pz+20^2,$ where $p=0,\\pm1,\\pm2,\\cdots,\\pm 39,$ $z^2+pz+13^2,$ where $p=0,\\pm1,\\pm2,\\cdots,\\pm 25$. Total of 130 equations, multiplied by 4 (the number of cases for real $r$, we have 520 equations, as indicated in the solution. -JZ",
"If the polynomial has one real root and two complex roots, then it can be factored as $(z-r)(z^2+pz+q),$ where $r$ is real with $|r|=13,20$ and $p,q$ are integers with $p^2 <4q.$ The roots $z_1$ and $z_2$ are conjugates. We have $|z_1|^2=|z_2|^2=z_1z_2=q.$ So $q$ is either $20^2$ or $13^2$. The only requirement for $p$ is $p<\\sqrt{4q^2}=2q.$ All such quadratic equations are listed as follows: $z^2+pz+20^2,$ where $p=0,\\pm1,\\pm2,\\cdots,\\pm 39,$ $z^2+pz+13^2,$ where $p=0,\\pm1,\\pm2,\\cdots,\\pm 25$. Total of 130 equations, multiplied by 4 (the number of cases for real $r$, we have 520 equations, as indicated in the solution. -JZ",
"There are two cases: either all the roots are real, or one is real and two are imaginary. Case 1: All roots are real. Then each of the roots is a member of the set $\\{-20, 20, -13, 13\\}$. It splits into three sub-cases: either no two are the same, exactly two are the same, or all three are the same. Sub-case 1.1: No two are the same. This is obviously $\\dbinom{4}{3}=4$. Sub-case 1.2: Exactly two are the same. There are four ways to choose the root that will repeat twice, and three ways to choose the remaining root. For this sub-case, $4\\cdot 3=12$. Sub-case 1.3: All three are the same. This is obviously $4$. Thus for case one, we have $4+12+4=20$ polynomials in $S$. We now have case two, which we state below. Case 2: Two roots are imaginary and one is real. Let these roots be $p-qi$, $p+qi$, and $r$. Then by Vieta's formulas $-(2p+r)=a$; $p^{2}+q^{2}+2pr=b$; $-\\left(p^{2}+q^{2}\\right)r=c$. Since $a$, $b$, $c$, and $r$ are integers, we have that $p=\\frac{1}{2}k$ for some integer $k$. Case two splits into two sub-cases now: Sub-case 2.1: $|p-qi|=|p+qi|=13$. Obviously, $|p|<13$. The $51$ cases in which $p$ is either $0,\\pm\\frac{1}{2},\\pm\\frac{2}{2},\\pm\\frac{3}{2},\\ldots,\\pm\\frac{25}{2}$ are acceptable. Each can pair with one value of $q$ and four values of $r$, adding $51\\cdot 4=204$ polynomials to $S$. Sub-case 2.2: $|p-qi|=|p+qi|=20$. Obviously, $|p|<20$. Here, the $79$ cases in which $p$ is either $0,\\pm\\frac{1}{2},\\pm\\frac{2}{2},\\pm\\frac{3}{2},\\ldots,\\pm\\frac{39}{2}$ are acceptable. Again, each can pair with a single value of $q$ as well as four values of $r$, adding $79\\cdot 4=316$ polynomials to $S$. Thus for case two, $204+316=520$ polynomials are part of $S$. All in all, $20+204+316=540 home."
] |
2013-II-13
| 2,013
| 13
|
In $\triangle ABC$ , $AC = BC$ , and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that $CE = \sqrt{7}$ and $BE = 3$ , the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ .
| 10
|
II
|
[
"We can set $AE=ED=m$. Set $BD=k$, therefore $CD=3k, AC=4k$. Thereafter, by Stewart's Theorem on $\\triangle ACD$ and cevian $CE$, we get $2m^2+14=25k^2$. Also apply Stewart's Theorem on $\\triangle CEB$ with cevian $DE$. After simplification, $2m^2=17-6k^2$. Therefore, $k=1, m=\\frac{\\sqrt{22}}{2}$. Finally, note that (using [] for area) $[CED]=[CAE]=3[EDB]=3[AEB]=\\frac{3}{8}[ABC]$, because of base-ratios. Using Heron's Formula on $\\triangle EDB$, as it is simplest, we see that $[ABC]=3\\sqrt{7}$, so your answer is $10$.",
"After drawing the figure, we suppose $BD=a$, so that $CD=3a$, $AC=4a$, and $AE=ED=b$. Using Law of Cosines for $\\triangle CED$ and $\\triangle AEC$,we get \\[b^2+7-2b\\sqrt{7}\\cdot \\cos(\\angle CED)=9a^2\\qquad (1)\\] \\[b^2+7+2b\\sqrt{7}\\cdot \\cos(\\angle CED)=16a^2\\qquad (2)\\] So, $(1)+(2)$, we get\\[2b^2+14=25a^2. \\qquad (3)\\] Using Law of Cosines in $\\triangle ACD$, we get \\[4b^2+9a^2-2\\cdot 2b\\cdot 3a\\cdot \\cos(\\angle ADC)=16a^2\\] So, \\[\\cos(\\angle ADC)=\\frac{4b^2-7a^2}{12ab}.\\qquad (4)\\] Using Law of Cosines in $\\triangle EDC$ and $\\triangle EDB$, we get \\[b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)\\] \\[b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)\\] $(5)+(6)$, and according to $(4)$, we can get \\[37a^2+2b^2=48. \\qquad (7)\\] Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, \\[4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2\\] then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $010.",
"Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. [asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,N);label(\"$D$\",D,NE);label(\"$L$\",L,NW);label(\"$M$\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"$X$\",X,S);label(\"$E$\",EE,NW);label(\"$Y$\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy] Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that \\[MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.\\] Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that \\[\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.\\] Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get \\[\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.\\] Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $010.",
"Let the coordinates of $A$, $B$ and $C$ be $(-a, 0)$, $(a, 0)$ and $(0, h)$ respectively. Then $D = \\Big(\\frac{3a}{4}, \\frac{h}{4}\\Big)$ and $E = \\Big(-\\frac{a}{8},\\frac{h}{8}\\Big).$ $EC^2 = 7$ implies $a^2 + 49h^2 = 448$; $EB^2 = 9$ implies $81a^2 + h^2 = 576.$ Solve this system of equations simultaneously, $a=\\sqrt{7}$ and $h=3$. Area of the triangle is $ah = 3\\sqrt{7}$, giving us an answer of $010.",
"[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,N);label(\"$D$\",D,NE);label(\"$L$\",L,NW);label(\"$M$\",M,S); pair EE=D/2; label(\"$\\sqrt{7}$\", C--EE, W); label(\"$x$\", D--B, E); label(\"$3x$\", C--D, E); label(\"$l$\", EE--D, N); label(\"$3$\", EE--B, N); label(\"$E$\",EE,NW); [/asy] Let $BD = x$. Then $CD = 3x$ and $AC = 4x$. Also, let $AE = ED = l$. Using Stewart's Theorem on $\\bigtriangleup CEB$ gives us the equation $(x)(3x)(4x) + (4x)(l^2) = 27x + 7x$ or, after simplifying, $4l^2 = 34 - 12x^2$. We use Stewart's again on $\\bigtriangleup CAD$: $(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)$, which becomes $2l^2 = 25x^2 - 14$. Substituting $2l^2 = 17 - 6x^2$, we see that $31x^2 = 31$, or $x = 1$. Then $l^2 = \\frac{11}{2}$. We now use Law of Cosines on $\\bigtriangleup CAD$. $(2l)^2 = (4x)^2 + (3x)^2 - 2(4x)(3x)\\cos C$. Plugging in for $x$ and $l$, $22 = 16 + 9 - 2(4)(3)\\cos C$, so $\\cos C = \\frac{1}{8}$. Using the Pythagorean trig identity $\\sin^2 + \\cos^2 = 1$, $\\sin^2 C = 1 - \\frac{1}{64}$, so $\\sin C = \\frac{3\\sqrt{7}}{8}$. $[ABC] = \\frac{1}{2} AC \\cdot BC \\sin C = (\\frac{1}{2})(4)(4)(\\frac{3\\sqrt{7}}{8}) = 3\\sqrt{7}$, and our answer is $3 + 7 = 010?",
"Let ABC be the reference triangle, with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. We can easily calculate $D=(0,\\frac{3}{4},\\frac{1}{4})$ and subsequently $E=(\\frac{1}{2},\\frac{3}{8},\\frac{1}{8})$. Using distance formula on $\\overline{EC}=(\\frac{1}{2},\\frac{3}{8},-\\frac{7}{8})$ and $\\overline{EB}=(\\frac{1}{2},-\\frac{5}{8},\\frac{1}{8})$ gives \\begin{align*} \\begin{cases} 7&=|EC|^2=-a^2 \\cdot \\frac{3}{8} \\cdot (-\\frac{7}{8})-b^2 \\cdot \\frac{1}{2} \\cdot (-\\frac{7}{8})-c^2 \\cdot \\frac{1}{2} \\cdot \\frac{3}{8} \\\\ 9&=|EB|^2=-a^2 \\cdot (-\\frac{5}{8}) \\cdot \\frac{1}{8}-b^2 \\cdot \\frac{1}{2} \\cdot \\frac{1}{8}-c^2 \\cdot \\frac{1}{2} \\cdot (-\\frac{5}{8}) \\\\ \\end{cases} \\end{align*} But we know that $a=b$, so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable: \\begin{align*} \\begin{cases} 7\\cdot 64&=3\\cdot 7\\cdot a^2+b^2\\cdot 4\\cdot 7-c^2\\cdot 4\\cdot 3\\\\ 9\\cdot 64&=5a^2-4b^2+4\\cdot 5\\cdot c^2 \\\\ \\end{cases} \\end{align*} \\begin{align*} \\begin{cases} 7\\cdot 64&=49a^2-12c^2 \\\\ 9\\cdot 64&=a^2+20c^2 \\\\ \\end{cases} \\end{align*} \\begin{align*} \\begin{cases} 5\\cdot 7\\cdot 64&=245a^2-60c^2 \\\\ 3\\cdot 9\\cdot 64&=3a^2+60c^2 \\\\ \\end{cases} \\end{align*} Then we add the equations to get \\begin{align*} 62\\cdot 64&=248a^2 \\\\ a^2 &=16 \\\\ a &=4 \\\\ \\end{align*} Then plugging gives $b=4$ and $c=2\\sqrt{7}$. Then the height from $C$ is $3$, and the area is $3\\sqrt{7}$ and our answer is $010.",
"Let ABC be the reference triangle, with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. We can easily calculate $D=(0,\\frac{3}{4},\\frac{1}{4})$ and subsequently $E=(\\frac{1}{2},\\frac{3}{8},\\frac{1}{8})$. Using distance formula on $\\overline{EC}=(\\frac{1}{2},\\frac{3}{8},-\\frac{7}{8})$ and $\\overline{EB}=(\\frac{1}{2},-\\frac{5}{8},\\frac{1}{8})$ gives \\begin{align*} \\begin{cases} 7&=|EC|^2=-a^2 \\cdot \\frac{3}{8} \\cdot (-\\frac{7}{8})-b^2 \\cdot \\frac{1}{2} \\cdot (-\\frac{7}{8})-c^2 \\cdot \\frac{1}{2} \\cdot \\frac{3}{8} \\\\ 9&=|EB|^2=-a^2 \\cdot (-\\frac{5}{8}) \\cdot \\frac{1}{8}-b^2 \\cdot \\frac{1}{2} \\cdot \\frac{1}{8}-c^2 \\cdot \\frac{1}{2} \\cdot (-\\frac{5}{8}) \\\\ \\end{cases} \\end{align*} But we know that $a=b$, so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable: \\begin{align*} \\begin{cases} 7\\cdot 64&=3\\cdot 7\\cdot a^2+b^2\\cdot 4\\cdot 7-c^2\\cdot 4\\cdot 3\\\\ 9\\cdot 64&=5a^2-4b^2+4\\cdot 5\\cdot c^2 \\\\ \\end{cases} \\end{align*} \\begin{align*} \\begin{cases} 7\\cdot 64&=49a^2-12c^2 \\\\ 9\\cdot 64&=a^2+20c^2 \\\\ \\end{cases} \\end{align*} \\begin{align*} \\begin{cases} 5\\cdot 7\\cdot 64&=245a^2-60c^2 \\\\ 3\\cdot 9\\cdot 64&=3a^2+60c^2 \\\\ \\end{cases} \\end{align*} Then we add the equations to get \\begin{align*} 62\\cdot 64&=248a^2 \\\\ a^2 &=16 \\\\ a &=4 \\\\ \\end{align*} Then plugging gives $b=4$ and $c=2\\sqrt{7}$. Then the height from $C$ is $3$, and the area is $3\\sqrt{7}$ and our answer is $010.",
"Let $C=(0,0), A=(x,y),$ and $B=(-x,y)$. It is trivial to show that $D=\\left(-\\frac{3}{4}x,\\frac{3}{4}y\\right)$ and $E=\\left(\\frac{1}{8}x,\\frac{7}{8}y\\right)$. Thus, since $BE=3$ and $CE=\\sqrt{7}$, we get that \\begin{align*} \\left(\\frac{1}{8}x\\right)^2+\\left(\\frac{7}{8}y\\right)^2&=7 \\\\ \\left(\\frac{9}{8}x\\right)^2+\\left(\\frac{1}{8}y\\right)^2&=9 \\\\ \\end{align*} Multiplying both equations by $64$, we get that \\begin{align*} x^2+49y^2&=448 \\\\ 81x^2+y^2&=576 \\\\ \\end{align*} Solving these equations, we get that $x=\\sqrt{7}$ and $y=3$. Thus, the area of $\\triangle ABC$ is $xy=3\\sqrt{7}$, so our answer is $010.",
"The main in solution is to prove that $\\angle BEC = 90^\\circ$. Let $M$ be midpoint $AB.$ Let $F$ be cross point of $AC$ and $BE.$ We use the formula for crossing segments in $\\triangle ABC$ and get: \\[\\frac {CF}{AF}= \\frac {DE}{AE} \\cdot (\\frac {CD}{BD} + 1) = 1 \\cdot (3 + 1) = 4.\\] \\[\\frac {FE }{BE}= \\frac {CD}{BD} : (\\frac {CF}{AF} + 1) = \\frac {3}{5} \\implies FE = \\frac {9}{5}.\\] \\[\\triangle BCF:\\hspace{5mm} BC = x, CF = \\frac {4}{5}x, EF = \\frac {9}{5}, BF = 3, CE = \\sqrt{7}.\\] By Stewart's Theorem on $\\triangle BCF$ and cevian $CE$, we get after simplification \\[x = 4 \\implies BC^2 = CE^2 + BE^2 \\implies \\angle BEC = 90^\\circ.\\] \\[AE = ED, AM = MB \\implies EM ||BC.\\] $\\angle BEC = \\angle CMB = 90^\\circ \\implies$ trapezium $BCEM$ is cyclic $\\implies$ \\[BM = CE, CM = BE \\implies [ABC] = CM \\cdot BM = 3 \\sqrt {7} \\implies 3+ 7 = \\textbf{010}.\\] [email protected], vvsss",
"Let $AB = 2x$ and let $y = BD.$ Then $CD = 3y$ and $AC = 4y.$ [asy] unitsize(1.5 cm); pair A, B, C, D, E; A = (-sqrt(7),0); B = (sqrt(7),0); C = (0,3); D = interp(B,C,1/4); E = (A + D)/2; draw(A--B--C--cycle); draw(A--D); draw(B--E--C); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, N); label(\"$D$\", D, NE); label(\"$E$\", E, NW); label(\"$2x$\", (A + B)/2, S); label(\"$y$\", (B + D)/2, NE); label(\"$3y$\", (C + D)/2, NE); label(\"$4y$\", (A + C)/2, NW); label(\"$3$\", (B + E)/2, N); label(\"$\\sqrt{7}$\", (C + E)/2, W); [/asy] By the Law of Cosines on triangle $ABC,$ \\[\\cos C = \\frac{16y^2 + 16y^2 - 4x^2}{2 \\cdot 4y \\cdot 4y} = \\frac{32y^2 - 4x^2}{32y^2} = \\frac{8y^2 - x^2}{8y^2}.\\]Then by the Law of Cosines on triangle $ACD,$ \\begin{align*} AD^2 &= 16y^2 + 9y^2 - 2 \\cdot 4y \\cdot 3y \\cdot \\cos C \\\\ &= 25y^2 - 24y^2 \\cdot \\frac{8y^2 - x^2}{8y^2} \\\\ &= 3x^2 + y^2. \\end{align*}Applying Stewart's Theorem to median $\\overline{BE}$ in triangle $ABD,$ we get \\[BE^2 + AE \\cdot DE = \\frac{AB^2 + BD^2}{2}.\\]Thus, \\[9 + \\frac{3x^2 + y^2}{4} = \\frac{4x^2 + y^2}{2}.\\]This simplifies to $5x^2 + y^2 = 36.$ Applying Stewart's Theorem to median $\\overline{CE}$ in triangle $ACD,$ we get \\[CE^2 + AE \\cdot DE = \\frac{AC^2 + CD^2}{2}.\\]Thus, \\[7 + \\frac{3x^2 + y^2}{4} = \\frac{16y^2 + 9y^2}{2}.\\]This simplifies to $3x^2 + 28 = 49y^2.$ Solving the system $5x^2 + y^2 = 36$ and $3x^2 + 28 = 49y^2,$ we find $x^2 = 7$ and $y^2 = 1,$ so $x = \\sqrt{7}$ and $y = 1.$ Plugging this back in for our equation for $\\cos C$ gives us $\\frac{1}{8}$, so $\\sin C = \\frac{3\\sqrt{7}}{8}.$ We can apply the alternative area of a triangle formula, where $AC \\cdot BC \\cdot \\sin C \\cdot \\frac{1}{2} = 3\\sqrt{7}.$ Therefore, our answer is $010."
] |
2013-II-14
| 2,013
| 14
|
For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$ .
| 512
|
II
|
[
"The Pattern We can find that $20\\equiv 6 \\pmod{7}$ $21\\equiv 5 \\pmod{8}$ $22\\equiv 6 \\pmod{8}$ $23\\equiv 7 \\pmod{8}$ $24\\equiv 6 \\pmod{9}$ $25\\equiv 7 \\pmod{9}$ $26\\equiv 8 \\pmod{9}$ Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get $99\\equiv 31 \\pmod{34}$ $100\\equiv 32 \\pmod{34}$ So the sum is $6+3\\times(6+...+31)+31+32=1512$,it is also 17+20+23+...+95, so the answer is $512 stated above. Kris17"
] |
2013-II-15
| 2,013
| 15
|
Let $A,B,C$ be angles of an acute triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ , $q$ , $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ .
| 222
|
II
|
[
"Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \\sin{A}$. By the Law of Sines, we must have $CA = \\sin{B}$ and $AB = \\sin{C}$. Now let us analyze the given: \\begin{align*} \\cos^2A + \\cos^2B + 2\\sin A\\sin B\\cos C &= 1-\\sin^2A + 1-\\sin^2B + 2\\sin A\\sin B\\cos C \\\\ &= 2-(\\sin^2A + \\sin^2B - 2\\sin A\\sin B\\cos C) \\end{align*} Now we can use the Law of Cosines to simplify this: \\[= 2-\\sin^2C\\] Therefore: \\[\\sin C = \\sqrt{\\dfrac{1}{8}},\\cos C = \\sqrt{\\dfrac{7}{8}}.\\] Similarly, \\[\\sin A = \\sqrt{\\dfrac{4}{9}},\\cos A = \\sqrt{\\dfrac{5}{9}}.\\] Note that the desired value is equivalent to $2-\\sin^2B$, which is $2-\\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\\dfrac{111-4\\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = 222.~bluesoul",
"Let us use the identity $\\cos^2A+\\cos^2B+\\cos^2C+2\\cos A \\cos B \\cos C=1$ . Add \\begin{align*} \\cos^2 C+2(\\cos A\\cos B-\\sin A \\sin B)\\cos C \\end{align*} to both sides of the first given equation. Thus, as \\begin{align*} \\cos A\\cos B-\\sin A\\sin B=\\cos (A+B)=-\\cos C ,\\end{align*} we have \\begin{align*} \\dfrac{15}{8}-2\\cos^2 C +\\cos^2 C=1, \\end{align*} so $\\cos C$ is $\\sqrt{\\dfrac{7}{8}}$ and therefore $\\sin C$ is $\\sqrt{\\dfrac{1}{8}}$. Similarily, we have $\\sin A =\\dfrac{2}{3}$ and $\\cos A=\\sqrt{\\dfrac{14}{9}-1}=\\sqrt{\\dfrac{5}{9}}$ and the rest of the solution proceeds as above. Solution 3 Let \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ ------ (1)}\\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\text{ ------ (2)}\\\\ \\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B &= x \\text{ ------ (3)}\\\\ \\end{align*} Adding (1) and (3) we get: \\[2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin A( \\sin B \\cos C + \\sin C \\cos B) = \\frac{15}{8} + x\\] or \\[2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin A \\sin (B+C) = \\frac{15}{8} + x\\] or \\[2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin ^2 A = \\frac{15}{8} + x\\] or \\[\\cos^2 B + \\cos^2 C = x - \\frac{1}{8} \\text{ ------ (4)}\\] Similarly adding (2) and (3) we get: \\[\\cos^2 A + \\cos^2 B = x - \\frac{4}{9} \\text{ ------ (5)}\\] Similarly adding (1) and (2) we get: \\[\\cos^2 A + \\cos^2 C = \\frac{14}{9} - \\frac{1}{8} \\text{ ------ (6)}\\] And (4) - (5) gives: \\[\\cos^2 C - \\cos^2 A = \\frac{4}{9} - \\frac{1}{8} \\text{ ------ (7)}\\] Now (6) - (7) gives: $\\cos^2 A = \\frac{5}{9}$ or $\\cos A = \\sqrt{\\dfrac{5}{9}}$ and $\\sin A = \\frac{2}{3}$ so $\\cos C$ is $\\sqrt{\\dfrac{7}{8}}$ and therefore $\\sin C$ is $\\sqrt{\\dfrac{1}{8}}$ Now $\\sin B = \\sin(A+C)$ can be computed first and then $\\cos^2 B$ is easily found. Thus $\\cos^2 B$ and $\\cos^2 C$ can be plugged into (4) above to give x = $\\dfrac{111-4\\sqrt{35}}{72}$. Hence the answer is = $111+4+35+72 = 222.~bluesoul",
"Let \\begin{align*} \\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C &= \\frac{15}{8} \\text{ ------ (1)}\\\\ \\cos^2 B + \\cos^2 C + 2 \\sin B \\sin C \\cos A &= \\frac{14}{9} \\text{ ------ (2)}\\\\ \\cos^2 C + \\cos^2 A + 2 \\sin C \\sin A \\cos B &= x \\text{ ------ (3)}\\\\ \\end{align*} Adding (1) and (3) we get: \\[2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin A( \\sin B \\cos C + \\sin C \\cos B) = \\frac{15}{8} + x\\] or \\[2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin A \\sin (B+C) = \\frac{15}{8} + x\\] or \\[2 \\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\sin ^2 A = \\frac{15}{8} + x\\] or \\[\\cos^2 B + \\cos^2 C = x - \\frac{1}{8} \\text{ ------ (4)}\\] Similarly adding (2) and (3) we get: \\[\\cos^2 A + \\cos^2 B = x - \\frac{4}{9} \\text{ ------ (5)}\\] Similarly adding (1) and (2) we get: \\[\\cos^2 A + \\cos^2 C = \\frac{14}{9} - \\frac{1}{8} \\text{ ------ (6)}\\] And (4) - (5) gives: \\[\\cos^2 C - \\cos^2 A = \\frac{4}{9} - \\frac{1}{8} \\text{ ------ (7)}\\] Now (6) - (7) gives: $\\cos^2 A = \\frac{5}{9}$ or $\\cos A = \\sqrt{\\dfrac{5}{9}}$ and $\\sin A = \\frac{2}{3}$ so $\\cos C$ is $\\sqrt{\\dfrac{7}{8}}$ and therefore $\\sin C$ is $\\sqrt{\\dfrac{1}{8}}$ Now $\\sin B = \\sin(A+C)$ can be computed first and then $\\cos^2 B$ is easily found. Thus $\\cos^2 B$ and $\\cos^2 C$ can be plugged into (4) above to give x = $\\dfrac{111-4\\sqrt{35}}{72}$. Hence the answer is = $111+4+35+72 = 222.~bluesoul",
"Let's take the first equation $\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos C = \\frac{15}{8}$. Substituting $180 - A - B$ for C, given A, B, and C form a triangle, and that $\\cos C = \\cos(A + B)$, gives us: $\\cos^2 A + \\cos^2 B + 2 \\sin A \\sin B \\cos (A+B) = \\frac{15}{8}$ Expanding out gives us $\\cos^2 A + \\cos^2 B + 2 \\sin^2 A \\sin^2 B - 2 \\sin A \\sin B \\cos A \\cos B = \\frac{15}{8}$. Using the double angle formula $\\cos^2 k = \\frac{\\cos (2k) + 1}{2}$, we can substitute for each of the squares $\\cos^2 A$ and $\\cos^2 B$. Next we can use the Pythagorean identity on the $\\sin^2 A$ and $\\sin^2 B$ terms. Lastly we can use the sine double angle to simplify. $\\cos^2 A + \\cos^2 B + 2(1 - \\cos^2 A)(1 - \\cos^2 B) - \\frac{1}{2} \\cdot \\sin 2A \\sin 2B = \\frac{15}{8}$. Expanding and canceling yields, and again using double angle substitution, $1 + 2 \\cdot \\frac{\\cos (2A) + 1}{2} \\cdot \\frac{\\cos (2B) + 1}{2} - \\frac{1}{2} \\cdot \\sin 2A \\sin 2B = \\frac{15}{8}$. Further simplifying yields: $\\frac{3}{2} + \\frac{\\cos 2A \\cos 2B - \\sin 2A \\sin 2B}{2} = \\frac{15}{8}$. Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation $2$ yields: $\\cos (2A + 2B) = \\frac{3}{4}$ and $\\cos (2B + 2C) = \\frac{1}{9}$. Substituting the identity $\\cos (2A + 2B) = \\cos(2C)$, we get: $\\cos (2C) = \\frac{3}{4}$ and $\\cos (2A) = \\frac{1}{9}$. Since the third expression simplifies to the expression $\\frac{3}{2} + \\frac{\\cos (2A + 2C)}{2}$, taking inverse cosine and using the angles in angle addition formula yields the answer, $\\frac{111 - 4\\sqrt{35}}{72}$, giving us the answer $222.~bluesoul",
"We will use the sum to product formula to simply these equations. Recall \\[2\\sin{\\frac{\\alpha+\\beta}{2}}\\sin{\\frac{\\alpha-\\beta}{2}} = \\cos{\\alpha}+\\cos{\\beta}.\\] Using this, let's rewrite the first equation: \\[\\cos^2(A) + \\cos^2(B) + 2 \\sin(A) \\sin(B) \\cos(C) = \\frac{15}{8}\\] \\[\\cos^2(A) + \\cos^2(B) + (\\cos(A+B)+\\cos(A-B))\\cos(C).\\] Now, note that $\\cos(C)=-\\cos(A+B)$. \\[\\cos^2(A) + \\cos^2(B) + (\\cos(A+B)+\\cos(A-B))(-\\cos(A+B))\\] \\[\\cos^2(A) + \\cos^2(B) - \\cos(A+B)\\cos(A-B)+cos^2(A+B)=\\frac{15}{8}.\\] We apply the sum to product formula again. \\[\\cos^2(A) + \\cos^2(B) - \\frac{\\cos(2A)+\\cos(2B)}{2}+cos^2(A+B)=\\frac{15}{8}.\\] Now, recall that $\\cos(2\\alpha)=2\\cos^2(\\alpha)-1$. We apply this and simplify our expression to get: \\[\\cos^2(A+B)=\\frac{7}{8}\\] \\[\\cos^2(C)=\\frac{7}{8}.\\] Analogously, \\[\\cos^2(A)=\\frac{5}{9}.\\] \\[\\cos^2(A+C)=\\frac{p-q\\sqrt{r}}{s}-1.\\] We can find this value easily by angle sum formula. After a few calculations, we get $\\frac{111 - 4\\sqrt{35}}{72}$, giving us the answer $222.~bluesoul",
"According to LOC $a^2+b^2-2ab\\cos{\\angle{c}}=c^2$, we can write it into $\\sin^2{\\angle{A}}+\\sin^2{\\angle{B}}-2\\sin{\\angle{A}}\\sin{\\angle{B}}\\cos{\\angle{C}}=\\sin^2{\\angle{C}}$. $\\sin^2{\\angle{A}}+\\sin^2{\\angle{B}}-2\\sin{\\angle{A}}\\sin{\\angle{B}}\\cos{\\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\\frac{15}{8}+sin^2C$ We can simplify to $2=sin^2C+\\frac{15}{8}$. Similarly, we can generalize $2=sin^2A+\\frac{14}{9}$. After solving, we can get that $sinA=\\frac{2}{3}; cosA=\\frac{\\sqrt{5}}{3}; sinC=\\frac{\\sqrt{2}}{4}; cosC=\\frac{\\sqrt{14}}{4}$ Assume the value we are looking for is $x$, we get $sin^2B+x=2$, while $sinB=sin(180^{\\circ}-A-C)=sin(A+C)$ which is $\\frac{2\\sqrt{14}+\\sqrt{10}}{12}$, so $x=\\frac{111 - 4\\sqrt{35}}{72}$, giving us the answer $222.~bluesoul"
] |
2014-I-1
| 2,014
| 1
|
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters. [asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy]
| 790
|
I
|
[
"The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\\dfrac{80}{3}$mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\\sqrt{50^2+\\left(\\frac{80}{3}\\right)^2}=\\frac{170}{3}$mm. Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side. Therefore, the minimum of the lace in millimeters is \\[6\\times \\dfrac{170}{3}+50+200\\times 2=790.\\]"
] |
2014-I-2
| 2,014
| 2
|
An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$ .
| 144
|
I
|
[
"First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$. The probability both are green is $\\frac{4}{10}\\cdot\\frac{16}{16+N}$, and the probability both are blue is $\\frac{6}{10}\\cdot\\frac{N}{16+N}$, so \\[\\frac{4}{10}\\cdot\\frac{16}{16+N}+\\frac{6}{10}\\cdot\\frac{N}{16+N}=\\frac{29}{50}\\] Solving this equation, \\[20\\left(\\frac{16}{16+N}\\right)+30\\left(\\frac{N}{16+N}\\right)=29\\] Multiplying both sides by $16+N$, we get \\begin{align*} 20\\cdot16+30\\cdot N&=29(16+N)\\\\ 320+30N&=464+29N\\\\ N&=144 \\end{align*}"
] |
2014-I-3
| 2,014
| 3
|
Find the number of rational numbers $r$ , $0<r<1,$ such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of $1000$ .
| 200
|
I
|
[
"Let the numerator and denominator $x,y$ with $\\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\\gcd(x,y) = 1$ then $\\gcd(x,y) =\\gcd(x,1000-x)= \\gcd(x,1000-x-(-1)x)=\\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\\frac{x}{y}<1$ and note by Euclidean algorithm $\\gcd(1000,1000-x)=1$, so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $200 desired fractions.",
"Let the numerator and denominator $x,y$ with $\\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\\gcd(x,y) = 1$ then $\\gcd(x,y) =\\gcd(x,1000-x)= \\gcd(x,1000-x-(-1)x)=\\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\\frac{x}{y}<1$ and note by Euclidean algorithm $\\gcd(1000,1000-x)=1$, so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $200 desired fractions.",
"If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of using $\\frac{n}{m}$ as the fraction to use the Euclidean Algorithm on, we can rewrite this as $\\frac{500-x}{500+x}$ or \\[\\gcd(500+x,500-x)=\\gcd((500+x)+(500-x),500-x)=\\gcd(1000,500-x).\\] Thus, we want $\\gcd(1000,500-x)=1$. You can either proceed as Solution $1$, or consider that no even numbers work, limiting us to $250$ choices of numbers and restricting $x$ to be odd. If $x$ is odd, $500-x$ is odd, so the only possible common factors $1000$ and $500-x$ can share are multiples of $5$. Thus, we want to avoid these. There are $50$ odd multiples of $5$ less than $500$, so the answer is $250-50=200.",
"Say $r=\\frac{d}{1000-d}$; then $1\\leq d\\leq499$. If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$. Since $1000=2^3\\cdot5^3$, that modulus can only be $2$ or $5$. This implies that if $d\\mid2$ or $d\\mid5$, the fraction is reducible. There are $249$ cases where $d\\mid2$, $99$ where $d\\mid5$, and $49$ where $d\\mid(2\\cdot5=10)$, so by PIE, the number of fails is $299$, so our answer is $200.",
"We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of $5$ because the denominator would also be a multiple of $5$. Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have $499$ fractions to start with, and $250$ with odd numerators. Subtract $50$ to account for the multiples of $5$, and we get $200.",
"We know that the set of these rational numbers is from $\\dfrac{1}{999}$ to $\\dfrac{499}{501}$ where each each element $\\dfrac{n}{m}$ has $n+m =1000$ and $\\dfrac{n}{m}$ is irreducible. We note that $\\dfrac{n}{m} =\\dfrac{1000-m}{m}=\\dfrac{1000}{m}-1$. Hence, $\\dfrac{n}{m}$ is irreducible if $\\dfrac{1000}{m}$ is irreducible, and $\\dfrac{1000}{m}$ is irreducible if $m$ is not divisible by $2$ or $5$. Thus, the answer to the question is the number of integers between $501$ and $999$ inclusive that are not divisible by $2$ or $5$. We note there are $499$ numbers between $501$ and $999$, and $249$ numbers are divisible by $2$ $99$ numbers are divisible by $5$ $49$ numbers are divisible by $10$ Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$, so our answer is $200 possible fractions satisfying the necessary conditions. Solution 1 is a more detailed solution utilizing Euler's totient.",
"We notice that there are a total of $400$ fractions that are in simplest form where the numerator and denominator add up to $1000$. Because the numerator and denominator have to be relatively prime, there are $\\varphi(1000)=400$ fractions. Half of these are greater than $1$, so the answer is $400\\div2=200 - bedwarsnoob ~MathFun1000 (Minor Edits)",
"Our fraction can be written in the form $\\frac{1000 - a}{a} = \\frac{1000}{a} - 1.$ Thus the fraction is reducible when $a$ divides $1000.$ We also want $500 < a < 1000.$ By PIE, the total values of $a$ that make the fraction reducible is, \\[249 + 99 - 49 = 299.\\] By complementary counting, the answer we want is $499 - 299 = 200",
"Suppose our fraction is $\\frac{a}{b}$. The given condition means $a+b=1000$. Now, if $a$ and $b$ share a common factor greater than $1$, then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even. There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$, which means the answer is $250-50=200 ~Geometry285",
"Suppose our fraction is $\\frac{a}{b}$. The given condition means $a+b=1000$. Now, if $a$ and $b$ share a common factor greater than $1$, then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even. There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$, which means the answer is $250-50=200 ~Geometry285"
] |
2014-I-4
| 2,014
| 4
|
Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 49
|
I
|
[
"For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\\dfrac{1}{3}$ mi/min. Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2. Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of $r_1 - \\dfrac{1}{3}$. Similarly, the second train has to cover a distance equal to its own length, at a rate of $r_2 + \\dfrac{1}{3}$. Since the times are equal and $d = rt$, we have that $\\dfrac{d}{r_1 - \\dfrac{1}{3}} = \\dfrac{d}{r_2 + \\dfrac{1}{3}}$. Solving for $r_1$ in terms of $r_2$, we get that $r_1 = r_2 + \\dfrac{2}{3}$. Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by $\\dfrac{1}{3}$, and decrease train 2's speed by $\\dfrac{1}{3}$. Thus, we have that $\\dfrac{d}{r_2 - \\dfrac{1}{3}} = 10\\dfrac{d}{r_1 + \\dfrac{1}{3}}$. Multiplying this out and simplifying, we get that $r_1 = 10r_2 - \\dfrac{11}{3}$. Since we now have 2 expressions for $r_1$ in terms of $r_2$, we can set them equal to each other: $r_2 + \\dfrac{2}{3} = 10r_2 - \\dfrac{11}{3}$. Solving for $r_2$, we get that $r_2 = \\dfrac{13}{27}$. Since we know that it took train 2 1 minute to pass Jon, we know that $1 = \\dfrac{d}{r_2 + \\dfrac{1}{3}}$. Plugging in $\\dfrac{13}{27}$ for $r_2$ and solving for $d$, we get that $d = \\dfrac{22}{27}$, and our answer is $27 + 22 = 049.",
"Using a similar approach to Solution 1, let the speed of the east bound train be $a$ and the speed of the west bound train be $b$. So $a-20=b+20$ and $a+20=10(b-20)$. From the first equation, $a=b+40$. Substituting into the second equation, \\[b+60=10b-200\\] \\[260=9b\\] \\[b=\\frac{260}{9}\\text{ mph}\\] This means that \\[a=\\frac{260}{9}+40=\\frac{620}{9}\\text{ mph}\\] Checking, we get that the common difference in Jon's speed and trains' speeds is $\\frac{440}{9}$ and the differences for Steve is $\\frac{800}{9}$ and $\\frac{80}{9}$. This question assumes the trains' lengths in MILES: \\[\\frac{440}{9}\\cdot \\frac{1}{60}=\\frac{440}{540}=\\frac{22}{27}\\text{ miles}\\] Adding up, we get $22+27=049.",
"Let the length of the trains be $L$, let the rate of the westward train be $W_{R}$, ;et the rate of the eastward train be $E_{R}$, and let the time it takes for the eastward train to pass Steve be $E_{T}$. We have that $L=(\\frac{1}{60})(W_{R}+20)$ $L=(\\frac{1}{60})(E_{R}-20)$. Adding both of the equations together, we get that $2L=\\frac{W_{R}}{60}+\\frac{E_{R}}{60}\\implies 120L=W_{R}+E_{R}$. Now, from the second part of the problem, we acquire that $L=(E_{T})(E_{R}+20)$ $L=(10E_{T})(W_{R}-20)$ Dividing the second equation by the first, we get that... $1=\\frac{10(W_{R}-20)}{E_{R}+20}\\implies E_{R}+20=10W_{R}-200\\implies E_{R}+220=10W_{R}\\implies E_{R}=10W_{R}-220$. Now, substituting into the $120L=W_{R}+E_{R}$. $120L=W_{R}+(10W_{R}-220)\\implies 120L= 11W_{R}-220\\implies W_{R}=\\frac{120L+220}{11}$. Finally, plugging this back into our very first equation.. $L=(\\frac{1}{60})((\\frac{120L+220}{11})+20)\\implies 660L=120L+440\\implies 540L=440\\implies L=\\frac{22}{27}$. Hence, the answer is $22+27=049."
] |
2014-I-5
| 2,014
| 5
|
Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called communal if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)
| 134
|
I
|
[
"By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers $2$ disjoint areas of the $12$-gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent to one another. We now count the number of ways to select a row of adjacent vertices. We notice that for any subset size between $1$ and $11$, there are $12$ possible subsets like this (this is true because we can pick any of the $12$ vertices as a \"starting\" vertex, include some number of vertices counterclockwise from that vertex, and generate all possible configurations). However, we also have to include the set of all $12$ vertices, as well as the empty set. Thus, the total number is $12\\cdot11 + 2 = 134."
] |
2014-I-6
| 2,014
| 6
|
The graphs $y = 3(x-h)^2 + j$ and $y = 2(x-h)^2 + k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ .
| 36
|
I
|
[
"Begin by setting $x$ to 0, then set both equations to $h^2=\\frac{2013-j}{3}$ and $h^2=\\frac{2014-k}{2}$, respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\\ge32$. We see that $h^2=\\frac{2014-k}{2}$, so we now need to find a positive integer $h$ which has positive integer x-intercepts for both equations. Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$. Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $036.",
"Let $x=0$ and $y=2013$ for the first equation, resulting in $j=2013-3h^2$. Substituting back in to the original equation, we get $y=3(x-h)^2+2013-3h^2$. Now we set $y$ equal to zero, since there are two distinct positive integer roots. Rearranging, we get $2013=3h^2-3(x-h)^2$, which simplifies to $671=h^2-(x-h)^2$. Applying difference of squares, we get $671=(2h-x)(x)$. Now, we know that $x$ and $h$ are both integers, so we can use the fact that $671=61\\times11$, and set $2h-x=11$ and $x=61$ (note that letting $x=11$ gets the same result). Therefore, $h=036 This answer however does not check out with the second equation which is why it is invalid.",
"Similar to the first two solutions, we deduce that $\\text{(-)}j$ and $\\text{(-)}k$ are of the form $3a^2$ and $2b^2$, respectively, because the roots are integers and so is the $y$-intercept of both equations. So the $x$-intercepts should be integers also. The first parabola gives \\[3h^2+j=3\\left(h^2-a^2\\right)=2013\\] \\[h^2-a^2=671\\] And the second parabola gives \\[2h^2+k=2\\left(h^2-b^2\\right)=2014\\] \\[h^2-b^2=1007\\] We know that $671=11\\cdot 61$ and that $1007=19\\cdot 53$. It is just a fitting coincidence that the average of $11$ and $61$ is the same as the average of $19$ and $53$. That is $036. \\[y=3(x-h)^2+j\\rightarrow y=3(x-11)(x-61)=3x^2-216x+2013\\] \\[y=2(x-h)^2+k\\rightarrow y=2(x-19)(x-53)=2x^2-144x+2014\\]",
"First, we expand both equations to get $y=3x^2-6hx+3h^2+j$ and $y=2x^2-4hx+2h^2+k$. The $y$-intercept for the first equation can be expressed as $3h^2+j$. From this, the x-intercepts for the first equation can be written as \\[x=h \\pm \\sqrt{(-6h)^2-4*3(3h^2+j)}=h \\pm \\sqrt{36h^2-12(2013)}=h \\pm \\sqrt{36h^2-24156}\\] Since the $x$-intercepts must be integers, $\\sqrt{36h^2-24156}$ must also be an integer. From solution 1, we know $h$ must be greater than or equal to 32. We can substitute increasing integer values for $h$ starting from 32; we find that $h=36$. We can test this result using the second equation, whose $x$-intercepts are \\[x=h \\pm \\sqrt{(-4h)^2-4*2(2h^2+k)}=h \\pm \\sqrt{16h^2-8(2014)}=h \\pm \\sqrt{16h^2-16112}\\] Substituting 36 in for $h$, we get $h=36 \\pm 68$, which satisfies the requirement that all x-intercepts must be (positive) integers. Thus, $h=036.",
"We have the equation $y=3(x-h)^2 + j.$ We know: $(x,y):(0,2013)$, so $h^2=2013/3 - j/3$ after plugging in the values and isolating $h^2$. Therefore, $h^2=671-j/3$. Lets call the x-intercepts $x_1$, $x_2$. Since both $x_1$ and $x_2$ are positive there is a relationship between $x_1$, $x_2$ and $h$. Namely, $x_1+x_2=2h$. The is because: $x_1-h=-(x_2-h)$, Similarly, we know: $(x,y):(x_1,0)$, so $j=-3(x_1-h)^2$. Combining the two equations gives us \\[h^2=671+(x_1-h)^2\\] \\[h^2=671+x_1^2-2x_1h+h^2\\] \\[h=(671+x_1^2)/2x_1.\\] Now since we have this relationship, $2h=x_1+x_2$, we can just multiply the last equation by 2(so that we get $2h$ on the left side) which gives us \\[2h=671/x_1+x_1^2/x^1\\] \\[2h=671/x_1+x_1\\] \\[x_1+x_2=671/x_1+x_1\\] \\[x_2=671/x_1\\] \\[x_1x_2=671.\\] Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$. Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$, so $h=36.",
"First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$. We can apply Vieta's. Let the roots of the first equation be $a, b$ and the roots of the second equation be $c, d$. Thus, we have that $a\\cdot b=1007$, $a+b=2h$ and $c\\cdot d=671$, $c+d=2h$. Simple evaluations finds that $h=36 ~Jske25",
"First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$. We can apply Vieta's. Let the roots of the first equation be $a, b$ and the roots of the second equation be $c, d$. Thus, we have that $a\\cdot b=1007$, $a+b=2h$ and $c\\cdot d=671$, $c+d=2h$. Simple evaluations finds that $h=36 ~Jske25"
] |
2014-I-7
| 2,014
| 7
|
Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.)
| 100
|
I
|
[
"Let $w = \\operatorname{cis}{(\\alpha)}$ and $z = 10\\operatorname{cis}{(\\beta)}$. Then, $\\dfrac{w - z}{z} = \\dfrac{\\operatorname{cis}{(\\alpha)} - 10\\operatorname{cis}{(\\beta)}}{10\\operatorname{cis}{\\beta}}$. Multiplying both the numerator and denominator of this fraction by $\\operatorname{cis}{(-\\beta)}$ gives us: $\\dfrac{w - z}{z} = \\dfrac{1}{10}\\operatorname{cis}{(\\alpha - \\beta)} - 1 = \\dfrac{1}{10}\\cos{(\\alpha - \\beta)} + \\dfrac{1}{10}i\\sin{(\\alpha - \\beta)} - 1$. We know that $\\tan{\\theta}$ is equal to the imaginary part of the above expression divided by the real part. Let $x = \\alpha - \\beta$. Then, we have that: $\\tan{\\theta} = \\dfrac{\\sin{x}}{\\cos{x} - 10}.$ We need to find a maximum of this expression, so we take the derivative: Note (not from author): To take the derivative, we need to use the Quotient Rule. In this case, \\[\\frac{d}{dx}\\left(\\frac{\\sin x}{\\cos x-10}\\right)=\\frac{\\cos x(\\cos x-10)-(-\\sin x)\\sin x}{(\\cos x-10)^2}=\\dfrac{1 - 10\\cos{x}}{(\\cos{x} - 10)^2}\\] Thus, we see that the maximum occurs when $\\cos{x} = \\dfrac{1}{10}$. Therefore, $\\sin{x} = \\pm\\dfrac{\\sqrt{99}}{10}$, and $\\tan{\\theta} = \\pm\\dfrac{\\sqrt{99}}{99}$. Thus, the maximum value of $\\tan^2{\\theta}$ is $\\dfrac{99}{99^2}$, or $\\dfrac{1}{99}$, and our answer is $1 + 99 = 100.",
"Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\\theta)$ is a measure of the angle if $z=10$ then $arg(\\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ [asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot(\"$A$\",A,dir(45)); dot(\"$B$\",B,dir(45)); dot(\"$O$\",O,dir(135)); dot(\"$ \\theta$\",O,(7,1.2)); label(\"$1$\", ( A--B )); label(\"$10$\",(O--A)); label(\"$t$\",(O--B)); [/asy] using the Law of Cosines we get: $1^2=10^2+t^2-t*10*2\\cos\\theta$ rearranging: \\[20t\\cos\\theta=t^2+99\\] solving for $\\cos\\theta$ we get: \\[\\frac{99}{20t}+\\frac{t}{20}=\\cos\\theta\\] if we want to maximize $\\theta$ we need to minimize $\\cos\\theta$ , using AM-GM inequality we get that the minimum value for $\\cos\\theta= 2\\left(\\sqrt{\\dfrac{99}{20t}\\dfrac{t}{20}}\\right)=2\\sqrt{\\dfrac{99}{400}}=\\dfrac{\\sqrt{99}}{10}$ hence using the identity $\\tan^2\\theta=\\sec^2\\theta-1$ we get $\\tan^2\\theta=\\frac{1}{99}$and our answer is $1 + 99 = 100",
"Note that $\\frac{w-z}{z}=\\frac{w}{z}-1$, and that $\\left|\\frac{w}{z}\\right|=\\frac{1}{10}$. Thus $\\frac{w}{z}-1$ is a complex number on the circle with radius $\\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\\omega$ denote this circle. Let $A$ and $C$ be the points that represent $\\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In order to maximize $\\tan^2(\\theta)$, we need to maximize $\\angle{AOC}$. This angle is maximized when $AO$ is tangent to $\\omega$. Using the Pythagorean Theorem, we get \\[AO^2=1^2-\\left(\\frac{1}{10}\\right)^2=\\frac{99}{100}\\] Thus \\[\\tan^2(\\theta)=\\frac{AC^2}{AO^2}=\\frac{1/100}{99/100}=\\frac{1}{99}\\] And the answer is $1+99=100."
] |
2014-I-8
| 2,014
| 8
|
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base 10, where digit $a$ is not zero. Find the three-digit number $abc$ .
| 937
|
I
|
[
"We have that $N^2 - N = N(N - 1)\\equiv 0\\mod{10000}$ Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$. Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$, which is impossible for a number that is divisible by either $2$ or $5$. Thus, one of them is divisible by $2^4 = 16$, and the other is divisible by $5^4 = 625$. Noting that $625 \\equiv 1\\mod{16}$, we see that $625$ would work for $N$, except the thousands digit is $0$. The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$. In order for this to happen, \\[N-1 \\equiv -1 \\pmod {16}.\\] Since $625 \\equiv 1 \\pmod{16}$, we know that $15 \\cdot 625 = 9375 \\equiv 15 \\equiv -1 \\mod{16}$. Thus, $N-1 = 9375$, so $N = 9376$, and our answer is $937.",
"We have that $N^2 - N = N(N - 1)\\equiv 0\\mod{10000}$ Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$. Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$, which is impossible for a number that is divisible by either $2$ or $5$. Thus, one of them is divisible by $2^4 = 16$, and the other is divisible by $5^4 = 625$. Noting that $625 \\equiv 1\\mod{16}$, we see that $625$ would work for $N$, except the thousands digit is $0$. The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$. In order for this to happen, \\[N-1 \\equiv -1 \\pmod {16}.\\] Since $625 \\equiv 1 \\pmod{16}$, we know that $15 \\cdot 625 = 9375 \\equiv 15 \\equiv -1 \\mod{16}$. Thus, $N-1 = 9375$, so $N = 9376$, and our answer is $937.",
"let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$. When we square $N$ we get that \\begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \\end{align*} However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \\[2000ad+2000bc+100c^2+200bd+20cd+d^2.\\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10. we first consider the ones digits: $d^2\\equiv d \\pmod{10}.$ This can happen for only 3 values : 1, 5 and 6. We can try to solve each case: Case 1 $(d=1)$ Considering the tenths place, we have that: $20cd=20c\\equiv 10c \\pmod {100}$ so $c= 0$. Considering the hundreds place we have that $200bd+100c^2= 200b \\equiv 100b \\pmod{1000}$ so again $b=0$ now considering the thousands place we have that $2000ad+2000bc = 2000a \\equiv 1000a \\pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$ Case 2 $(d=5)$ considering the tenths place we have that: $20cd+20=100c+20\\equiv 20 \\equiv 10c \\mod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$) so $c=2$ considering the hundreds place we have that $200bd+100c^2+100c= 1000b+600 \\equiv600\\equiv 100b \\pmod{1000}$ ( the extra $100c$ is carried from the tenths place) so $b=6$ now considering the thousands place we have that $2000ad+2000bc +1000b= 10000a+24000+ 6000\\equiv0\\equiv 1000a \\pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place) so a is equal 0 again Case 3$(d=6)$ considering the tenths place we have that: $20cd+30=120c+30\\equiv 30+20c \\equiv 10c \\pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$) if $c=7$ then we have $30+20 \\cdot 7 \\equiv 70\\equiv7 \\cdot 10 \\pmod{100}$ so $c=7$ considering the hundreds place we have that $200bd+100c^2+100c+100= 1200b+4900+800 \\equiv200b+700\\equiv 100b \\pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place) if $b=3$ then we have $700+200 \\cdot 3 \\equiv 300\\equiv3 \\cdot 100 \\pmod {1000}$ so $b=3$ now considering the thousands place we have that $2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\\equiv0\\equiv 2000a+1000\\equiv 1000a \\pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place) if $a=9$ then we have $2000 \\cdot 9+1000 \\equiv 9000\\equiv9 \\cdot 1000 \\pmod {1000}$ so $a=9$ so we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $937",
"let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$. When we square $N$ we get that \\begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \\end{align*} However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \\[2000ad+2000bc+100c^2+200bd+20cd+d^2.\\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10. we first consider the ones digits: $d^2\\equiv d \\pmod{10}.$ This can happen for only 3 values : 1, 5 and 6. We can try to solve each case: Case 1 $(d=1)$ Considering the tenths place, we have that: $20cd=20c\\equiv 10c \\pmod {100}$ so $c= 0$. Considering the hundreds place we have that $200bd+100c^2= 200b \\equiv 100b \\pmod{1000}$ so again $b=0$ now considering the thousands place we have that $2000ad+2000bc = 2000a \\equiv 1000a \\pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$ Case 2 $(d=5)$ considering the tenths place we have that: $20cd+20=100c+20\\equiv 20 \\equiv 10c \\mod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$) so $c=2$ considering the hundreds place we have that $200bd+100c^2+100c= 1000b+600 \\equiv600\\equiv 100b \\pmod{1000}$ ( the extra $100c$ is carried from the tenths place) so $b=6$ now considering the thousands place we have that $2000ad+2000bc +1000b= 10000a+24000+ 6000\\equiv0\\equiv 1000a \\pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place) so a is equal 0 again Case 3$(d=6)$ considering the tenths place we have that: $20cd+30=120c+30\\equiv 30+20c \\equiv 10c \\pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$) if $c=7$ then we have $30+20 \\cdot 7 \\equiv 70\\equiv7 \\cdot 10 \\pmod{100}$ so $c=7$ considering the hundreds place we have that $200bd+100c^2+100c+100= 1200b+4900+800 \\equiv200b+700\\equiv 100b \\pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place) if $b=3$ then we have $700+200 \\cdot 3 \\equiv 300\\equiv3 \\cdot 100 \\pmod {1000}$ so $b=3$ now considering the thousands place we have that $2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\\equiv0\\equiv 2000a+1000\\equiv 1000a \\pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place) if $a=9$ then we have $2000 \\cdot 9+1000 \\equiv 9000\\equiv9 \\cdot 1000 \\pmod {1000}$ so $a=9$ so we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $937",
"By the Chinese Remainder Theorem, the equation $N(N-1)\\equiv 0\\pmod{10000}$ is equivalent to the two equations: \\begin{align*} N(N-1)&\\equiv 0\\pmod{16},\\\\ N(N-1)&\\equiv 0\\pmod{625}. \\end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\\mod{16},N\\mod{625})\\in\\{(0,0),(0,1),(1,0),(1,1)\\}$. Let \\[\\varphi:\\mathbb Z/10000\\mathbb Z\\to\\mathbb Z/16\\mathbb Z\\times\\mathbb Z/625\\mathbb Z,\\] \\[x\\mapsto (x\\mod{16},x\\mod{625}).\\] The statement of the Chinese Remainder theorem is that $\\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\\varphi^{-1}(0,0)$, $\\varphi^{-1}(0,1)$, $\\varphi^{-1}(1,0)$, and $\\varphi^{-1}(1,1)$. Now we easily see that \\[\\varphi^{-1}(0,0)=0\\] and \\[\\varphi^{-1}(1,1)=1.\\] Noting that $625\\equiv 1\\pmod{16}$, it follows that \\[\\varphi^{-1}(1,0)=625.\\] To compute $\\varphi^{-1}(0,1)$, note that \\[(0,1)=15(1,0)+(1,1)\\] in \\[\\mathbb Z/16\\mathbb Z\\times\\mathbb Z/625\\mathbb Z,\\] so since $\\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \\[\\varphi^{-1}(0,1)=15\\varphi^{-1}(1,0)+\\varphi^{-1}(1,1)=15\\times 625+1=9376.\\] The four candidate digit strings $abcd$ are then $0000,0001,0625,9376$. Of those, only $9376$ has nonzero first digit, and therefore the answer is $937.",
"By the Chinese Remainder Theorem, the equation $N(N-1)\\equiv 0\\pmod{10000}$ is equivalent to the two equations: \\begin{align*} N(N-1)&\\equiv 0\\pmod{16},\\\\ N(N-1)&\\equiv 0\\pmod{625}. \\end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\\mod{16},N\\mod{625})\\in\\{(0,0),(0,1),(1,0),(1,1)\\}$. Let \\[\\varphi:\\mathbb Z/10000\\mathbb Z\\to\\mathbb Z/16\\mathbb Z\\times\\mathbb Z/625\\mathbb Z,\\] \\[x\\mapsto (x\\mod{16},x\\mod{625}).\\] The statement of the Chinese Remainder theorem is that $\\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\\varphi^{-1}(0,0)$, $\\varphi^{-1}(0,1)$, $\\varphi^{-1}(1,0)$, and $\\varphi^{-1}(1,1)$. Now we easily see that \\[\\varphi^{-1}(0,0)=0\\] and \\[\\varphi^{-1}(1,1)=1.\\] Noting that $625\\equiv 1\\pmod{16}$, it follows that \\[\\varphi^{-1}(1,0)=625.\\] To compute $\\varphi^{-1}(0,1)$, note that \\[(0,1)=15(1,0)+(1,1)\\] in \\[\\mathbb Z/16\\mathbb Z\\times\\mathbb Z/625\\mathbb Z,\\] so since $\\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \\[\\varphi^{-1}(0,1)=15\\varphi^{-1}(1,0)+\\varphi^{-1}(1,1)=15\\times 625+1=9376.\\] The four candidate digit strings $abcd$ are then $0000,0001,0625,9376$. Of those, only $9376$ has nonzero first digit, and therefore the answer is $937.",
"Note - $\\overline{abcd}$ means the number formed when the digits represented by $a$, $b$, $c$, and $d$ are substituted in. $\\overline{abcd}\\ne a\\times b\\times c\\times d$. WLOG, we can assume that $N$ is a 4-digit integer $\\overline{abcd}$. Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\\equiv d\\pmod{10}$, as the units digit of $\\overline{abcd}^2$ is affected only by $d$, regardless of $a$, $b$, or $c$. By checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$. Now, we seek to find $c$. Note that the only $\\overline{cd}$ that will satisfy $N$ will also satisfy $\\overline{cd}^2 \\equiv \\overline{cd}\\pmod{100}$, by the same reasoning as above - the last two digits of $\\overline{abcd}^2$ are only affected by $c$ and $d$. As we already have narrowed choices for $d$, we start reasoning out. First, we note that if $d=0$, then $c=0$, as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$, then recursively, $a$ and $b$ must be $0$, as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$. Now, let's assume that $d=1$. $\\overline{cd}$ is equal to $10c + d = 10c + 1$. Squaring this gives $100c^2 + 20c + 1$, and when modulo 100 is taken, it must equal $10c + 1$. As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \\equiv 20c + 1\\pmod{100}$, which must be equivalent to $10c + 1$. Note that this is really $\\overline{(2c)1}$ and $\\overline{c1}$, and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$.However, if we extend this process to find $b$ and $a$, we'd find that they are also 0. The only solution in this case is $1$, and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case. Let's assume that $d=5$. Note that $(10c + 5)^2 = 100c^2 + 100c + 25$, and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$, which works for now. Now, let $d=6$. Note that $(10c + 6)^2 = 100c^2 + 120c + 36$. Taking modulo 100, this reduces to $20c+36$, which must be equivalent to $10c+6$. Again, this is similar to $\\overline{(2c+3)6}$ and $\\overline{c6}$, so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$. Checking quickly $76^2 = 5776$, which works for now. Now, we find $b$, given two possibilities for $\\overline{cd}$. Start with $\\overline{cd} = 25$. $\\overline{bcd} = 100b + \\overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$, which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$. Therefore, the only $\\overline{bcd}$ that works here is $625$. $625^2 = 390625$. Now, assume that $\\overline{cd}=76$. We have $100b + 76$, and when squared, becomes $10000b^2 + 15200b + 5776$, which, modulo 1000, should be equivalent to $100b+76$. Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$. Using the same technique as before, we must equate the hundreds digit of $\\overline{(2b+7)76}$ to $\\overline{b76}$, or equate the units digit of $2b+7$ and $b$. Since $2b+7$ is necessarily odd, any possible $b$'s must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$. $376^2 = 141376$. Finally, we solve for $a$. Start with $\\overline{bcd}=625$. We have $1000a + 625$, which, squared, gives \\[1000000a^2 + 1250000a + 390625,\\] and reducing modulo 10000 gives simply 625. So $\\overline{abcd}=625$. However, that makes $a=0$. Therefore, no solutions exist in this case. We turn to our last case, $\\overline{bcd}=376$. We have \\[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\\] and reducing modulo $10000$ gives $2000a + 1376$, which must be equivalent to $1000a + 376$. So we must have $\\overline{(2a+1)376}$ being equivalent to $\\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$. Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$. Therefore, $\\overline{abcd}=9376$, so our answer is $937.",
"Note - $\\overline{abcd}$ means the number formed when the digits represented by $a$, $b$, $c$, and $d$ are substituted in. $\\overline{abcd}\\ne a\\times b\\times c\\times d$. WLOG, we can assume that $N$ is a 4-digit integer $\\overline{abcd}$. Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\\equiv d\\pmod{10}$, as the units digit of $\\overline{abcd}^2$ is affected only by $d$, regardless of $a$, $b$, or $c$. By checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$. Now, we seek to find $c$. Note that the only $\\overline{cd}$ that will satisfy $N$ will also satisfy $\\overline{cd}^2 \\equiv \\overline{cd}\\pmod{100}$, by the same reasoning as above - the last two digits of $\\overline{abcd}^2$ are only affected by $c$ and $d$. As we already have narrowed choices for $d$, we start reasoning out. First, we note that if $d=0$, then $c=0$, as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$, then recursively, $a$ and $b$ must be $0$, as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$. Now, let's assume that $d=1$. $\\overline{cd}$ is equal to $10c + d = 10c + 1$. Squaring this gives $100c^2 + 20c + 1$, and when modulo 100 is taken, it must equal $10c + 1$. As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \\equiv 20c + 1\\pmod{100}$, which must be equivalent to $10c + 1$. Note that this is really $\\overline{(2c)1}$ and $\\overline{c1}$, and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$.However, if we extend this process to find $b$ and $a$, we'd find that they are also 0. The only solution in this case is $1$, and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case. Let's assume that $d=5$. Note that $(10c + 5)^2 = 100c^2 + 100c + 25$, and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$, which works for now. Now, let $d=6$. Note that $(10c + 6)^2 = 100c^2 + 120c + 36$. Taking modulo 100, this reduces to $20c+36$, which must be equivalent to $10c+6$. Again, this is similar to $\\overline{(2c+3)6}$ and $\\overline{c6}$, so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$. Checking quickly $76^2 = 5776$, which works for now. Now, we find $b$, given two possibilities for $\\overline{cd}$. Start with $\\overline{cd} = 25$. $\\overline{bcd} = 100b + \\overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$, which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$. Therefore, the only $\\overline{bcd}$ that works here is $625$. $625^2 = 390625$. Now, assume that $\\overline{cd}=76$. We have $100b + 76$, and when squared, becomes $10000b^2 + 15200b + 5776$, which, modulo 1000, should be equivalent to $100b+76$. Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$. Using the same technique as before, we must equate the hundreds digit of $\\overline{(2b+7)76}$ to $\\overline{b76}$, or equate the units digit of $2b+7$ and $b$. Since $2b+7$ is necessarily odd, any possible $b$'s must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$. $376^2 = 141376$. Finally, we solve for $a$. Start with $\\overline{bcd}=625$. We have $1000a + 625$, which, squared, gives \\[1000000a^2 + 1250000a + 390625,\\] and reducing modulo 10000 gives simply 625. So $\\overline{abcd}=625$. However, that makes $a=0$. Therefore, no solutions exist in this case. We turn to our last case, $\\overline{bcd}=376$. We have \\[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\\] and reducing modulo $10000$ gives $2000a + 1376$, which must be equivalent to $1000a + 376$. So we must have $\\overline{(2a+1)376}$ being equivalent to $\\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$. Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$. Therefore, $\\overline{abcd}=9376$, so our answer is $937.",
"We are given that $abcd^2$ ends in $abcd$, which means that $d^2$ ends in $d$. The only possible values for which this works is $d=0,1,5,6$. Notice that if we set $d=0$, we will have $c=0, b=0, a=0$, which is not valid since $a\\neq0$. Furthermore, we can see that $d\\neq1$ by quickly going over $11^2, 12^2, ..., 19^2$ and seeing that none of those numbers work. From number sense, we know that $B25^2$ ends in $...625$ regardless of $B$. So, in this case, $b=6$. However, notice that $76^2$ ends in $...76$, and we can quickly find that $376^2=141376$. Going back to $A625^2$, we notice that no matter what $A$ is, $A625^2=...0625$. Since $A\\neq0$, this case is not valid (write out $A625^2$ and multiply to see that the thousands digit is $5a+3+2+5+5a=10a+10$ which always ends in $0$). We turn our focus to $A376^2$. Write out $376^2$ and multiply it out to see that the thousands digit is $6a+2+6+2+6a+1=12a+1$. There is an extra $1$ due to carry over. Quickly testing numbers, the only possible value for $A$ is $9$. So, $abcd=9376$ and our answer is $abc=937. ~hwan",
"We are given that $abcd^2$ ends in $abcd$, which means that $d^2$ ends in $d$. The only possible values for which this works is $d=0,1,5,6$. Notice that if we set $d=0$, we will have $c=0, b=0, a=0$, which is not valid since $a\\neq0$. Furthermore, we can see that $d\\neq1$ by quickly going over $11^2, 12^2, ..., 19^2$ and seeing that none of those numbers work. From number sense, we know that $B25^2$ ends in $...625$ regardless of $B$. So, in this case, $b=6$. However, notice that $76^2$ ends in $...76$, and we can quickly find that $376^2=141376$. Going back to $A625^2$, we notice that no matter what $A$ is, $A625^2=...0625$. Since $A\\neq0$, this case is not valid (write out $A625^2$ and multiply to see that the thousands digit is $5a+3+2+5+5a=10a+10$ which always ends in $0$). We turn our focus to $A376^2$. Write out $376^2$ and multiply it out to see that the thousands digit is $6a+2+6+2+6a+1=12a+1$. There is an extra $1$ due to carry over. Quickly testing numbers, the only possible value for $A$ is $9$. So, $abcd=9376$ and our answer is $abc=937. ~hwan"
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2014-I-9
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Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$ . Find $x_2(x_1+x_3)$ .
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[
"Substituting $n$ for $2014$, we get \\[\\sqrt{n}x^3 - (1+2n)x^2 + 2 = \\sqrt{n}x^3 - x^2 - 2nx^2 + 2\\] \\[= x^2(\\sqrt{n}x - 1) - 2(nx^2 - 1) = 0\\] Noting that $nx^2 - 1$ factors as a difference of squares to \\[(\\sqrt{n}x - 1)(\\sqrt{n}x+1)\\] we can factor the left side as \\[(\\sqrt{n}x - 1)(x^2 - 2(\\sqrt{n}x+1))\\] This means that $\\frac{1}{\\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\\sqrt{n}x - 2$. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\\sqrt{n}$, so the positive root must be greater than $2\\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \\frac{1}{\\sqrt{2014}} < 2\\sqrt{2014}$ is clearly true, $x_2 = \\frac{1}{\\sqrt{2014}}$ and $x_1 + x_3 = 2\\sqrt{2014}$. Multiplying these values together, we find that $x_2(x_1+x_3) = 002.",
"From Vieta's formulae, we know that \\[x_1x_2x_3 = \\dfrac{-2}{\\sqrt{2014}}\\] \\[x_1 + x_2 + x_3 = \\dfrac{4029}{\\sqrt{2014}}\\] and \\[x_1x_2 + x_2x_3 + x_1x_3 = 0\\] Thus, we know that \\[x_2(x_1 + x_3) = -x_1x_3\\] Now consider the polynomial with roots $x_1x_2, x_2x_3,$ and $x_1x_3$. Expanding the polynomial \\[(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)\\]we get the polynomial \\[x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2\\] Substituting the values obtained from Vieta's formulae, we find that this polynomial is \\[x^3 - \\dfrac{8058}{2014}x - \\dfrac{4}{2014}\\] We know $x_1x_3$ is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to \\[1007x^3 - 4029x - 2 = 0\\] Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the $x^3$ term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that $x = -2$ is a solution. Factoring it out, we get that \\[1007x^3 + 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)\\] Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, \\[x_1x_3 = -2\\] so \\[-x_1x_3 = 002\\]and we're done.",
"Observing the equation, we notice that the coefficient for the middle term $-4029$ is equal to \\[-2{\\sqrt{2014}}^2-1\\]. Also notice that the coefficient for the ${x^3}$ term is $\\sqrt{2014}$. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the $x$ term of the binomial would have a coefficient of $\\sqrt{2014}$. Similarly, the $x$ term of the trinomial would also have a coefficient of $\\sqrt{2014}$. The factored form of the expression would look something like the following: \\[({\\sqrt{2014}}x-a)(x^2-n{\\sqrt{2014}}x-b)\\] where ${a, b,c}$ are all positive integers (because the ${x^2}$ term of the original expression is negative, and the constant term is positive), and \\[{ab=2}\\] Multiplying this expression out gives \\[{{\\sqrt{2014}x^3-(2014n+a)x^2+(an{\\sqrt{2014}}-b{\\sqrt{2014}})x+ab}}\\] Equating this with the original expression gives \\[{2014n+a}=-4029\\] The only positive integer solutions of this expression is $(n, a)=(1, 2015)$ or $(2, 1)$. If $(n, a)=(1, 2015)$ then setting ${an{\\sqrt{2014}}-b{\\sqrt{2014}}}=0$ yields ${b=2015}$ and therefore ${ab=2015^2}$ which clearly isn't equal to $2$ as the constant term. Therefore, $(n, a)=(2, 1)$ and the factored form of the expression is: \\[({\\sqrt{2014}}x-1)(x^2-2{\\sqrt{2014}}x-2)\\] Therefore, one of the three roots of the original expression is \\[{x=\\dfrac{1}{\\sqrt{2014}}}\\] Using the quadratic formula yields the other two roots as \\[{x={\\sqrt{2014}}+{\\sqrt{2016}}}\\] and \\[{x={\\sqrt{2014}}-{\\sqrt{2016}}}\\] Arranging the roots in ascending order (in the order $x_1<x_2<x_3$), \\[{\\sqrt{2014}}-{\\sqrt{2016}}<\\dfrac{1}{\\sqrt{2014}}<{\\sqrt{2014}}+{\\sqrt{2016}}\\] Therefore, \\[x_2(x_1+x_3)=\\dfrac{1}{\\sqrt{2014}}{2\\sqrt{2014}}=002\\]",
"By Vieta's, we are seeking to find $x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\\frac{2}{\\sqrt{2014}x_2}$. Substitute $n=-x_1x_3$ and $x_2=\\frac{2}{\\sqrt{2014}n}$. Substituting this back into the original equation, we have $\\frac{4}{1007n^3}-\\frac{8058}{1007n^2}+2=0$, so $2n^3-\\frac{8058}{1007}n+\\frac{4}{1007}=2n^3-\\frac{8058n-4}{1007}=0$. Hence, $8058n-4\\equiv 2n-4 \\equiv 0 \\pmod{1007}$, and $n\\equiv 2\\pmod{1007}$. But since $n\\le 999$ because it is our desired answer, the only possible value for $n$ is $002 BEST PROOOFFFF Stormersyle & mathleticguyyy",
"Let $x =\\frac{y}{\\sqrt{2014}}.$ The original equation simplifies to $\\frac{y^3}{2014} -\\frac{4029y^2}{2014}+2 = 0 \\implies y^3 - 4029y^2 + 4028=0.$ Here we clearly see that $y=1$ is a root. Dividing $y-1$ from the sum we find that $(y-1)(y^2-4028y-4028)=0.$ From simple bounding we see that $y=1$ is the middle root. Therefore $x_{2}(x_{1}+x_{3}) =\\frac{1}{\\sqrt{2014}} \\cdot\\frac{4028}{\\sqrt{2014}} = 002",
"$\\sqrt{2014}$ occurs multiple times, so let k = $\\sqrt{2014}$. The equation becomes $0 = kx^3 - (2k^2 + 1)x^2 + 2$. Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula. We get the roots are: $y = \\frac{1}{x}$, and $y = \\frac{x}{2} - \\frac{1}{x}$. In the first case, $x = \\frac{1}{y} = \\frac{1}{\\sqrt{2014}}$. In the second case, $x^2 - 2\\sqrt{2014} - 2 = 0$. The solutions are $\\sqrt{2014} \\pm \\sqrt{2016}$. The sum of these 2 solutions is $2 \\sqrt{2014}$, and $\\frac{1}{\\sqrt{2014}}$ is the middle solution, and thus, $(x_1 + x_3) \\cdot x_2 = 2$",
"We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.) Let $p(x) = \\sqrt{2014}x^3-4029x^2+2=0$. We start by estimating $p(-1)$, $p(0)$, and $p(1)$ (A natural first step for function analysis.): $p(-1)\\approx -45-4029+2 \\approx -4000$ $p(0) = 2$ $p(1) \\approx 45-4029+2 \\approx -4000$ We conclude by Intermediate Value Theorem (or just common sense), that there is a root on $(-1, 0)$ and another root on $(0, 1)$. We know that $p(1) < 0$ and that $\\lim_{x\\to\\infty} p(x) = \\infty$. We conclude that the third root is on $(1, \\infty)$. Therefore, $x_1 \\in (-1, 0)$, $x_2 \\in (0, 1)$, and $x_3 \\in (1, \\infty)$. We will estimate $x_3$. For $x > 1$, the constant term of $p(x)$ is negligible. We simplify and get $p(x_3) \\approx \\sqrt{2014}x_{3}^3 - 4029x_{3}^2 = 0$. Solving for $x_3$ (We can divide by $x_{3}^2$ because we know $x_3 \\neq 0$), we get $x_3 \\approx \\frac{4029}{\\sqrt{2014}} \\approx 2\\sqrt{2014} \\approx 90$. We can intuitively bound $x_3$ between $88$ and $92$. We will now estimate $x_1$ and $x_2$. $x_1$ and $x_2$ are close to $0$. As a result, the $\\sqrt{2014}x^3$ term is negligible. We simplify and get $p(x) \\approx -4029x^2 + 2 = 0$. Solving for $x$, we get $x \\approx \\pm \\sqrt{\\frac{2}{4029}} \\approx \\pm \\sqrt{\\frac{1}{2014.5}} \\approx \\pm \\frac{1}{45}$. We can intuitively bound $x_1$ between $-\\frac{1}{43}$ and $-\\frac{1}{47}$. Similarly, we can intuitively bound $x_2$ between $\\frac{1}{47}$ and $\\frac{1}{43}$. We calculate that the minimum possible value of $x_2(x_1 + x_3)$ is $\\frac{88-\\frac{1}{43}}{47}$ and the maximum possible value is $\\frac{92-\\frac{1}{47}}{43}$. The only integer that falls is this range is $002. ~numerophile",
"We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.) Let $p(x) = \\sqrt{2014}x^3-4029x^2+2=0$. We start by estimating $p(-1)$, $p(0)$, and $p(1)$ (A natural first step for function analysis.): $p(-1)\\approx -45-4029+2 \\approx -4000$ $p(0) = 2$ $p(1) \\approx 45-4029+2 \\approx -4000$ We conclude by Intermediate Value Theorem (or just common sense), that there is a root on $(-1, 0)$ and another root on $(0, 1)$. We know that $p(1) < 0$ and that $\\lim_{x\\to\\infty} p(x) = \\infty$. We conclude that the third root is on $(1, \\infty)$. Therefore, $x_1 \\in (-1, 0)$, $x_2 \\in (0, 1)$, and $x_3 \\in (1, \\infty)$. We will estimate $x_3$. For $x > 1$, the constant term of $p(x)$ is negligible. We simplify and get $p(x_3) \\approx \\sqrt{2014}x_{3}^3 - 4029x_{3}^2 = 0$. Solving for $x_3$ (We can divide by $x_{3}^2$ because we know $x_3 \\neq 0$), we get $x_3 \\approx \\frac{4029}{\\sqrt{2014}} \\approx 2\\sqrt{2014} \\approx 90$. We can intuitively bound $x_3$ between $88$ and $92$. We will now estimate $x_1$ and $x_2$. $x_1$ and $x_2$ are close to $0$. As a result, the $\\sqrt{2014}x^3$ term is negligible. We simplify and get $p(x) \\approx -4029x^2 + 2 = 0$. Solving for $x$, we get $x \\approx \\pm \\sqrt{\\frac{2}{4029}} \\approx \\pm \\sqrt{\\frac{1}{2014.5}} \\approx \\pm \\frac{1}{45}$. We can intuitively bound $x_1$ between $-\\frac{1}{43}$ and $-\\frac{1}{47}$. Similarly, we can intuitively bound $x_2$ between $\\frac{1}{47}$ and $\\frac{1}{43}$. We calculate that the minimum possible value of $x_2(x_1 + x_3)$ is $\\frac{88-\\frac{1}{43}}{47}$ and the maximum possible value is $\\frac{92-\\frac{1}{47}}{43}$. The only integer that falls is this range is $002. ~numerophile",
"Let $a=x\\sqrt{2014}$. We have $\\sqrt{2014}x^3-4029x^2+2=ax^2-2a^2-x^2+2=x^2(a-1)-2(a^2-1)=x^2(a-1)-2(a-1)(a+1)=(a-1)(x^2-2a-2)=0$, so $a=x\\sqrt{2014}=1$ or $x^2-2a-2=x^2-2\\sqrt{2014}x-2=0$. As such, the solutions are at $x=\\frac{1}{\\sqrt{2014}}$ and $x=\\frac{2\\sqrt{2014}\\pm\\sqrt{4\\cdot2014+8}}{2}$. Note that $\\frac{2\\sqrt{2014}-\\sqrt{4\\cdot2014+8}}{2}<0<\\frac{1}{\\sqrt{2014}}<\\sqrt{2014}<\\frac{2\\sqrt{2014}+\\sqrt{4\\cdot2014+8}}{2}$, so $x_2(x_1+x_3)=\\frac{x_1+x_3}{\\sqrt{2014}}=\\frac{2\\sqrt{2014}}{\\sqrt{2014}}=2 by vietas.) -zhoujef000",
"Rewrite the polynomial as $nx^3 - (2n^2+1)x^2 + 2 = 0$. The answer probably holds for general $n$, so letting $n=0$, we have $x^2+2=0$. We'll assume that the missing root doesn't impact the final answer, so by Vietas, we have $002.",
"Rewrite the polynomial as $nx^3 - (2n^2+1)x^2 + 2 = 0$. The answer probably holds for general $n$, so letting $n=0$, we have $x^2+2=0$. We'll assume that the missing root doesn't impact the final answer, so by Vietas, we have $002."
] |
2014-I-10
| 2,014
| 10
|
A disk with radius $1$ is externally tangent to a disk with radius $5$ . Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$ . That is, if the center of the smaller disk has moved to the point $D$ , and the point on the smaller disk that began at $A$ has now moved to point $B$ , then $\overline{AC}$ is parallel to $\overline{BD}$ . Then $\sin^2(\angle BEA)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 58
|
I
|
[
"[asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label(\"$A$\",a,W,fontsize(9)); label(\"$B$\",b,NW,fontsize(9)); label(\"$C$\",c,E,fontsize(9)); label(\"$D$\",d,E,fontsize(9)); label(\"$E$\",e,SW,fontsize(9)); label(\"$F$\",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy] Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\\overset{\\frown}{AF}$ on the larger disk. Let $\\alpha = \\angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\\overset{\\frown}{BF}$ on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did,\\[\\overset{\\frown}{BF}=\\overset{\\frown}{AF}=5\\alpha\\]. Since $\\overline{AC} || \\overline{BD}$, \\[\\angle BDF \\cong \\angle FEA\\] so the angles of minor arc $\\overset{\\frown}{BF}$ and minor arc $\\overset{\\frown}{AF}$ are equal, so minor arc $\\overset{\\frown}{BF}$ has an angle of $\\alpha$. Since the smaller disk has a radius of $1$, the length of minor arc $\\overset{\\frown}{BF}$ is $\\alpha$. This means that $5\\alpha + \\alpha$ equals the circumference of the smaller disk, so $6\\alpha = 2\\pi$, or $\\alpha = \\frac{\\pi}{3}$. Now, to find $\\sin^2{\\angle BEA}$, we construct $\\triangle BDE$. Also, drop a perpendicular from $D$ to $\\overline{EA}$, and call this point $X$. Since $\\alpha = \\frac{\\pi}{3}$ and $\\angle DXE$ is right, \\[DE = 6\\] \\[EX = 3\\] and \\[DX = 3\\sqrt{3}\\] Now drop a perpendicular from $B$ to $\\overline{EA}$, and call this point $Y$. Since $\\overline{BD} || \\overline{EA}$, \\[XY = BD = 1\\] and \\[BY = DX = 3\\sqrt{3}\\] Thus, we know that \\[EY = EX - XY = 3 - 1 = 2\\] and by using the Pythagorean Theorem on $\\triangle BEY$, we get that \\[BE = \\sqrt{31}\\] Thus, \\[\\sin{\\angle BEA} = \\frac{\\sqrt{27}}{\\sqrt{31}}\\] so \\[\\sin^2{\\angle BEA} = \\frac{27}{31}\\] and our answer is $27 + 31 = 058.",
"First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate $5$ times (the circumference of the small circle is $2\\pi$ while the larger one has a circumference of $10\\pi$) plus the extra rotation the circle gets for rotating around the circle, for a total of $6$ times. Therefore, one rotation will bring point $D$ $60^\\circ$ from $C$. Now, draw $\\triangle DBE$, and call \\[\\angle BED = x^{\\circ}\\] We know that $\\overline{ED}$ is 6, and $\\overline{BD}$ is 1. Since $EC || BD$, \\[\\angle BDE = 60^\\circ\\] By the Law of Cosines, \\[\\overline{BE}^2=36+1-2\\times 6\\times 1\\times \\cos{60^\\circ} = 36+1-6=31\\] and since lengths are positive, \\[\\overline{BE}=\\sqrt{31}\\] By the Law of Sines, we know that \\[\\frac{1}{\\sin{x}}=\\frac{\\sqrt{31}}{\\sin{60^\\circ}}\\] so \\[\\sin{x} = \\frac{\\sin{60^\\circ}}{\\sqrt{31}} = \\frac{\\sqrt{93}}{62}\\] As $x$ is clearly between $0$ and $90^\\circ$, $\\cos{x}$ is positive. As $\\cos{x}=\\sqrt{1-\\sin^2{x}}$, \\[\\cos{x} = \\frac{11\\sqrt{31}}{62}\\] Now we use the angle sum formula to find the sine of $\\angle BEA$: \\[\\sin 60^\\circ\\cos x + \\cos 60^\\circ\\sin x = \\frac{\\sqrt{3}}{2}\\frac{11\\sqrt{31}}{62}+\\frac{1}{2}\\frac{\\sqrt{93}}{62}\\] \\[= \\frac{11\\sqrt{93}+\\sqrt{93}}{124} = \\frac{12\\sqrt{93}}{124} = \\frac{3\\sqrt{93}}{31} = \\frac{3\\sqrt{31}\\sqrt{3}}{31} = \\frac{3\\sqrt{3}}{\\sqrt{31}}\\] Finally, we square this to get \\[\\frac{9\\times 3}{31}=\\frac{27}{31}\\] so our answer is $27+31=058."
] |
2014-I-11
| 2,014
| 11
|
A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 391
|
I
|
[
"Perform the coordinate transformation $(x, y)\\rightarrow (x+y, x-y)$. Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\\langle 1, -1 \\rangle$, $\\langle 1, 1 \\rangle$, $\\langle -1, -1 \\rangle$, $\\langle -1, 1 \\rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$, which is just $\\binom63 = 20$. The probability of any of these sequences happening is $\\left(\\frac12\\right)^6$. Thus, the probability of ending on the x axis is $\\frac{20}{2^6}$. Similarly, the probability of ending on the y axis is the same. However, we overcount exactly one case: ending at $(0, 0)$. Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\\left(\\frac{20}{2^6}\\right)^2 = \\frac{25}{256}$. Using PIE, the total probability is $\\frac{20}{64} + \\frac{20}{64} - \\frac{25}{256} = \\frac{135}{256}$, giving an answer of $391. ~sampai7",
"Perform the coordinate transformation $(x, y)\\rightarrow (x+y, x-y)$. Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\\langle 1, -1 \\rangle$, $\\langle 1, 1 \\rangle$, $\\langle -1, -1 \\rangle$, $\\langle -1, 1 \\rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$, which is just $\\binom63 = 20$. The probability of any of these sequences happening is $\\left(\\frac12\\right)^6$. Thus, the probability of ending on the x axis is $\\frac{20}{2^6}$. Similarly, the probability of ending on the y axis is the same. However, we overcount exactly one case: ending at $(0, 0)$. Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\\left(\\frac{20}{2^6}\\right)^2 = \\frac{25}{256}$. Using PIE, the total probability is $\\frac{20}{64} + \\frac{20}{64} - \\frac{25}{256} = \\frac{135}{256}$, giving an answer of $391. ~sampai7",
"We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$, or $4096$. There are 4 possible cases that land along the line $y = x$: $x,y = \\pm 1; x,y = \\pm 2; x,y = \\pm 3;$ or $x = y = 0$. We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$, multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$. Case 1: The token ends at $(3, 3)$. In order for the token to end up here, it must have had 3 right moves, and 3 up moves. In other words, the total number of ways to get here is the ways to rearrange the letters in the sequence $RRRUUU$, which is ${6\\choose 3} = 20.$ Case 2: The token ends at $(2,2)$. In order for the token to end up here, it could have had 2 up moves, 3 right moves, and 1 left move; or 2 right moves, 3 up moves, and 1 down move. Thus, the total number of ways to get here is sum of the ways to rearrange the letters in the sequences $RRRLUU$ and $UUUDRR$, both of which are ${6\\choose 1}{5\\choose 2} = 60$, for a total of $120$ possibilities. Case 3: The token ends at $(1,1)$. In order for the token to end up here, it could have had: 1 right move, 3 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence $RUUUDD$ is ${6\\choose 1}{5\\choose 2} = 60.$ 1 up move, 3 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence $URRRLL$ is ${6\\choose 1}{5\\choose 2} = 60.$ 2 right moves, 1 left move, 2 up moves, and 1 down move. In this case, the number of ways to rearrange the letters in the sequence $UUDRRL$ is ${6\\choose 1}{5\\choose 1}{4\\choose 2} = 180.$ Thus, the total number of ways to end up at $(1,1)$ is $300$. Case 4: The token ends at $(0,0)$. In order for the token to end up here, it could have had: 3 right moves and 3 left moves. In this case, the number of ways to rearrange the letters in the sequence $RRRLLL$ is ${6\\choose 3} = 20.$ 3 up moves and 3 down moves. In this case, the number of ways to rearrange the letters in the sequence $UUUDDD$ is ${6\\choose 3} = 20.$ 1 right move, 1 left move, 2 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence $RLUUDD$ is ${6\\choose 1}{5\\choose 1}{4\\choose 2} = 180.$ 1 up move, 1 down move, 2 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence $RRLLUD$ is ${6\\choose 1}{5\\choose 1}{4\\choose 2} = 180.$ Thus, the total number of ways to end up at $(0,0)$ is $400$. Adding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$. Thus, our probability is $\\frac{2160}{4096}$. When this fraction is fully reduced, it is $\\frac{135}{256}$, so our answer is $135 + 256 = 391",
"We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$, or $4096$. There are 4 possible cases that land along the line $y = x$: $x,y = \\pm 1; x,y = \\pm 2; x,y = \\pm 3;$ or $x = y = 0$. We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$, multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$. Case 1: The token ends at $(3, 3)$. In order for the token to end up here, it must have had 3 right moves, and 3 up moves. In other words, the total number of ways to get here is the ways to rearrange the letters in the sequence $RRRUUU$, which is ${6\\choose 3} = 20.$ Case 2: The token ends at $(2,2)$. In order for the token to end up here, it could have had 2 up moves, 3 right moves, and 1 left move; or 2 right moves, 3 up moves, and 1 down move. Thus, the total number of ways to get here is sum of the ways to rearrange the letters in the sequences $RRRLUU$ and $UUUDRR$, both of which are ${6\\choose 1}{5\\choose 2} = 60$, for a total of $120$ possibilities. Case 3: The token ends at $(1,1)$. In order for the token to end up here, it could have had: 1 right move, 3 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence $RUUUDD$ is ${6\\choose 1}{5\\choose 2} = 60.$ 1 up move, 3 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence $URRRLL$ is ${6\\choose 1}{5\\choose 2} = 60.$ 2 right moves, 1 left move, 2 up moves, and 1 down move. In this case, the number of ways to rearrange the letters in the sequence $UUDRRL$ is ${6\\choose 1}{5\\choose 1}{4\\choose 2} = 180.$ Thus, the total number of ways to end up at $(1,1)$ is $300$. Case 4: The token ends at $(0,0)$. In order for the token to end up here, it could have had: 3 right moves and 3 left moves. In this case, the number of ways to rearrange the letters in the sequence $RRRLLL$ is ${6\\choose 3} = 20.$ 3 up moves and 3 down moves. In this case, the number of ways to rearrange the letters in the sequence $UUUDDD$ is ${6\\choose 3} = 20.$ 1 right move, 1 left move, 2 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence $RLUUDD$ is ${6\\choose 1}{5\\choose 1}{4\\choose 2} = 180.$ 1 up move, 1 down move, 2 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence $RRLLUD$ is ${6\\choose 1}{5\\choose 1}{4\\choose 2} = 180.$ Thus, the total number of ways to end up at $(0,0)$ is $400$. Adding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$. Thus, our probability is $\\frac{2160}{4096}$. When this fraction is fully reduced, it is $\\frac{135}{256}$, so our answer is $135 + 256 = 391",
"We split this into cases by making a chart. In each row, the entries $(\\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$. \\[\\begin{array}{ccccccccccccl} \\multicolumn{5}{c}{R (+)\\qquad L (-)}& |&\\multicolumn{5}{c}{U (+)\\qquad D (-)}\\\\ +1&& +1&& +1&| & +1&& +1&& +1\\\\ +1&& +1&& -1& | & +1&& +1&& -1\\\\ +1&& -1&& -1& | & +1&& -1&& -1\\\\ -1 && -1&& -1& | & -1&& -1&& -1\\\\ \\\\ +1&& +1&& +1&& -1 &|& +1&& +1\\\\ +1&& +1&& -1 && -1 &|& +1 && -1\\\\ +1&& -1&& -1 && -1 &|& -1 && -1 &&(\\times 2 \\text{ for symmetry by swapping }R-L\\text{ and }U-D)\\\\ \\\\ +1&& +1 &&+1 &&-1&& -1& |& +1\\\\ +1&& +1 &&-1&& -1&& -1 &|& -1&& (\\times 2\\text{ symmetry})\\\\ \\\\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\\times2 \\text{ symmetry})\\\\ \\end{array}\\] Note that to account for the cases when $x=-y$, we can simply multiply the $U-D$ steps by $-1$, so for example, the first row would become \\[+1 \\qquad+1\\qquad +1 \\ \\ \\ \\ |\\ \\ \\ -1\\qquad -1\\qquad -1.\\] Therefore, we must multiply the number of possibilities in each case by $2$, except for when $x=y=0$. Now, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$, $L$ represents $-1$ to the left of $|$, $U$ represents $+1$ to the right of $|$, and $D$ represents $-1$ to the right of $|$. Using multinomials, we compute the following numbers of possibilities for each case. \\[{6\\choose 3}\\cdot 2+ \\frac{6!}{2!2!}\\cdot 2 + \\frac{6!}{2!2!} \\cdot 2 + {6\\choose 3} \\cdot 2 = 2(20 + 180 + 180 + 20) = 800\\] \\[\\frac{6!}{3!2!}\\cdot 2 + \\frac{6!}{2!2!} + \\frac{6!}{3!2!}\\cdot 2 = 120 + 180 + 120 = 420\\ (\\times2\\text{ for symmetry})\\] \\[\\frac{6!}{3!2!} \\cdot 2 + \\frac{6!}{3!2!} \\cdot 2 = 120 + 120 = 240\\ (\\times2\\text{ for symmetry})\\] \\[{6\\choose 3} = 20\\ (\\times 2\\text{ for symmetry})\\] Thus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$. Because there are $4^6$ total possibilities, the probability is $\\frac{2160}{4^6} = \\frac{135}{256}$, so the answer is $391",
"We split this into cases by making a chart. In each row, the entries $(\\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$. \\[\\begin{array}{ccccccccccccl} \\multicolumn{5}{c}{R (+)\\qquad L (-)}& |&\\multicolumn{5}{c}{U (+)\\qquad D (-)}\\\\ +1&& +1&& +1&| & +1&& +1&& +1\\\\ +1&& +1&& -1& | & +1&& +1&& -1\\\\ +1&& -1&& -1& | & +1&& -1&& -1\\\\ -1 && -1&& -1& | & -1&& -1&& -1\\\\ \\\\ +1&& +1&& +1&& -1 &|& +1&& +1\\\\ +1&& +1&& -1 && -1 &|& +1 && -1\\\\ +1&& -1&& -1 && -1 &|& -1 && -1 &&(\\times 2 \\text{ for symmetry by swapping }R-L\\text{ and }U-D)\\\\ \\\\ +1&& +1 &&+1 &&-1&& -1& |& +1\\\\ +1&& +1 &&-1&& -1&& -1 &|& -1&& (\\times 2\\text{ symmetry})\\\\ \\\\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\\times2 \\text{ symmetry})\\\\ \\end{array}\\] Note that to account for the cases when $x=-y$, we can simply multiply the $U-D$ steps by $-1$, so for example, the first row would become \\[+1 \\qquad+1\\qquad +1 \\ \\ \\ \\ |\\ \\ \\ -1\\qquad -1\\qquad -1.\\] Therefore, we must multiply the number of possibilities in each case by $2$, except for when $x=y=0$. Now, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$, $L$ represents $-1$ to the left of $|$, $U$ represents $+1$ to the right of $|$, and $D$ represents $-1$ to the right of $|$. Using multinomials, we compute the following numbers of possibilities for each case. \\[{6\\choose 3}\\cdot 2+ \\frac{6!}{2!2!}\\cdot 2 + \\frac{6!}{2!2!} \\cdot 2 + {6\\choose 3} \\cdot 2 = 2(20 + 180 + 180 + 20) = 800\\] \\[\\frac{6!}{3!2!}\\cdot 2 + \\frac{6!}{2!2!} + \\frac{6!}{3!2!}\\cdot 2 = 120 + 180 + 120 = 420\\ (\\times2\\text{ for symmetry})\\] \\[\\frac{6!}{3!2!} \\cdot 2 + \\frac{6!}{3!2!} \\cdot 2 = 120 + 120 = 240\\ (\\times2\\text{ for symmetry})\\] \\[{6\\choose 3} = 20\\ (\\times 2\\text{ for symmetry})\\] Thus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$. Because there are $4^6$ total possibilities, the probability is $\\frac{2160}{4^6} = \\frac{135}{256}$, so the answer is $391",
"Denote $(x, y)_n$ the probability that starting from point $(x, y)$, the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$. We start by breaking this down: \\[(0, 0)_6 = \\frac14 \\cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$, so the equation simplifies to \\[(0, 0)_6 = (0, 1)_5\\] We now expand $(0, 1)_5$. \\[(0, 1)_5 = \\frac14 \\cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\\] First, we find $(0, 0)_4$. \\[(0, 0)_4 = (0, 1)_3\\] \\[(0, 1)_3 = \\frac14 \\cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\\] At this point, we can just count the possibilities to find $(0, 0)_2 = \\frac34$, $(0, 2)_2 = \\frac{7}{16}$, and $(1, 1)_2 = \\frac58$. Therefore, \\[(0, 1)_3 = \\frac14 \\cdot (\\frac34 + \\frac{7}{16} + 2 \\cdot \\frac58)\\] \\[(0, 1)_3 = \\frac{39}{64}\\] Next, we find $(0, 2)_4$. \\[(0, 2)_4 = \\frac14 \\cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\\] We already calculated $(0, 1)_3$, so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \\[(0, 3)_3 = \\frac14 \\cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\\] \\[(0, 3)_3 = \\frac14 \\cdot (\\frac{7}{16} + 0 + 2 \\cdot \\frac{1}{4})\\] \\[(0, 3)_3 = \\frac{15}{64}\\] \\[(1, 2)_3 = \\frac14 \\cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\\] \\[(1, 2)_3 = \\frac14 \\cdot (\\frac14 + \\frac58 + \\frac{7}{16} + \\frac12)\\] \\[(1, 2)_3 = \\frac{29}{64}\\] Therefore, \\[(0, 2)_4 = \\frac14 \\cdot (\\frac{39}{64} + \\frac{15}{64} + 2 \\cdot \\frac{29}{64})\\] \\[(0, 2)_4 = \\frac{7}{16}\\] Finally, we find $(1, 1)_4$. \\[(1, 1)_4 = \\frac12 \\cdot ((0, 1)_3 + (1, 2)_3)\\] \\[(1, 1)_4 = \\frac12 \\cdot (\\frac{39}{64} + \\frac{29}{64})\\] \\[(1, 1)_4 = \\frac{17}{32}\\] Putting it all together, \\[(0, 0)_6 = (0, 1)_5 =\\frac14 \\cdot (\\frac{39}{64} + \\frac{7}{16} + 2 \\cdot \\frac{17}{32})\\] \\[(0, 0)_6 = \\frac{135}{256}\\] Thus, the answer is $135 + 256 = 391.",
"Denote $(x, y)_n$ the probability that starting from point $(x, y)$, the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$. We start by breaking this down: \\[(0, 0)_6 = \\frac14 \\cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$, so the equation simplifies to \\[(0, 0)_6 = (0, 1)_5\\] We now expand $(0, 1)_5$. \\[(0, 1)_5 = \\frac14 \\cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\\] First, we find $(0, 0)_4$. \\[(0, 0)_4 = (0, 1)_3\\] \\[(0, 1)_3 = \\frac14 \\cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\\] At this point, we can just count the possibilities to find $(0, 0)_2 = \\frac34$, $(0, 2)_2 = \\frac{7}{16}$, and $(1, 1)_2 = \\frac58$. Therefore, \\[(0, 1)_3 = \\frac14 \\cdot (\\frac34 + \\frac{7}{16} + 2 \\cdot \\frac58)\\] \\[(0, 1)_3 = \\frac{39}{64}\\] Next, we find $(0, 2)_4$. \\[(0, 2)_4 = \\frac14 \\cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\\] We already calculated $(0, 1)_3$, so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \\[(0, 3)_3 = \\frac14 \\cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\\] \\[(0, 3)_3 = \\frac14 \\cdot (\\frac{7}{16} + 0 + 2 \\cdot \\frac{1}{4})\\] \\[(0, 3)_3 = \\frac{15}{64}\\] \\[(1, 2)_3 = \\frac14 \\cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\\] \\[(1, 2)_3 = \\frac14 \\cdot (\\frac14 + \\frac58 + \\frac{7}{16} + \\frac12)\\] \\[(1, 2)_3 = \\frac{29}{64}\\] Therefore, \\[(0, 2)_4 = \\frac14 \\cdot (\\frac{39}{64} + \\frac{15}{64} + 2 \\cdot \\frac{29}{64})\\] \\[(0, 2)_4 = \\frac{7}{16}\\] Finally, we find $(1, 1)_4$. \\[(1, 1)_4 = \\frac12 \\cdot ((0, 1)_3 + (1, 2)_3)\\] \\[(1, 1)_4 = \\frac12 \\cdot (\\frac{39}{64} + \\frac{29}{64})\\] \\[(1, 1)_4 = \\frac{17}{32}\\] Putting it all together, \\[(0, 0)_6 = (0, 1)_5 =\\frac14 \\cdot (\\frac{39}{64} + \\frac{7}{16} + 2 \\cdot \\frac{17}{32})\\] \\[(0, 0)_6 = \\frac{135}{256}\\] Thus, the answer is $135 + 256 = 391.",
"For any point $(x,y)$ on the coordinate plane, we can represent it as the term $a^xb^y$. Then, moving right is equivalent to multiplying by $a$, moving left is equivalent to multiplying by $1/a$, and similarly for up/down with $b$ and $1/b$. Because, on every move, the possibilities are these four directions, and each is equally likely, each move can be represented by multiplying by $a+\\frac{1}{a}+b+\\frac{1}{b}$. Thus, the possibilities after $6$ moves are the same as the coefficients of $\\left(a+\\frac{1}{a}+b+\\frac{1}{b}\\right)^6$. By symmetry, we just need to find the number of ways for the token to reach $(0,0),(1,1),(2,2),\\text{ and }(3,3)$. By the multinomial theorem, the coefficient of $a^3b^3$ is $\\binom{6}{3,3}=20$. Similarly, the coefficient of $a^2b^2$ is $2\\binom{6}{3,2,1}=120$. The coefficient of $ab$ is $\\binom{6}{2,2,1,1}+2\\binom{6}{3,2,1}=300$. Finally, the constant term is $2\\binom{6}{2,2,1,1}+2\\binom{6}{3,3}=400$. Because the total number of possibilities is $4^6$, the probability is \\[\\frac{4(20+120+300)+400}{4^6}=\\frac{5+30+75+25}{4^4}=\\frac{135}{256},\\] and the answer is $391$. --brainiacmaniac31",
"For any point $(x,y)$ on the coordinate plane, we can represent it as the term $a^xb^y$. Then, moving right is equivalent to multiplying by $a$, moving left is equivalent to multiplying by $1/a$, and similarly for up/down with $b$ and $1/b$. Because, on every move, the possibilities are these four directions, and each is equally likely, each move can be represented by multiplying by $a+\\frac{1}{a}+b+\\frac{1}{b}$. Thus, the possibilities after $6$ moves are the same as the coefficients of $\\left(a+\\frac{1}{a}+b+\\frac{1}{b}\\right)^6$. By symmetry, we just need to find the number of ways for the token to reach $(0,0),(1,1),(2,2),\\text{ and }(3,3)$. By the multinomial theorem, the coefficient of $a^3b^3$ is $\\binom{6}{3,3}=20$. Similarly, the coefficient of $a^2b^2$ is $2\\binom{6}{3,2,1}=120$. The coefficient of $ab$ is $\\binom{6}{2,2,1,1}+2\\binom{6}{3,2,1}=300$. Finally, the constant term is $2\\binom{6}{2,2,1,1}+2\\binom{6}{3,3}=400$. Because the total number of possibilities is $4^6$, the probability is \\[\\frac{4(20+120+300)+400}{4^6}=\\frac{5+30+75+25}{4^4}=\\frac{135}{256},\\] and the answer is $391$. --brainiacmaniac31"
] |
2014-I-12
| 2,014
| 12
|
Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ .
| 453
|
I
|
[
"The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$. Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$. Note also that the total number of cases is $4^4*4^4=4^8$. $f 1 g 1$: clearly, we choose one number as the range for $f$, one for $g$, yielding $12$ possibilities. $f 1 g 2$ with symmetry (WLOG $f$ has 1 element): start by selecting numbers for the ranges. This yields $4$ for the one number in $f$, and $3$ options for the two numbers for $g$. Afterwards, note that the function with 2 numbers in the range can have $4+6+4=14$ arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have $2*12*14$ possibilities, the 2 from symmetry. $f 2 g 2$: no symmetry, still easy! Just note that we have $6$ choices of which numbers go to $f$ and $g$, and within each, $14*14=196$ choices for the orientation of each of the two numbers. That's $6*196$ possibilities. $f 1 g 3$: again, symmetrical (WLOG $f$ has one element): $4$ ways to select the single element for $f$, and then find the number of ways to distribute the $3$ distinct numbers in the range for $g$. The only arrangement for the frequency of each number is ${1, 1, 2}$ in some order. Therefore, we have $3$ ways to choose which number is the one represented twice, and then note that there are $12$ ways to arrange these! The number of possibilities in this situation is $2 * 4 * 3 * 12$. Total, divided by $4^8$, gets $\\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}$, with numerator $453.",
"The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$. Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$. Note also that the total number of cases is $4^4*4^4=4^8$. $f 1 g 1$: clearly, we choose one number as the range for $f$, one for $g$, yielding $12$ possibilities. $f 1 g 2$ with symmetry (WLOG $f$ has 1 element): start by selecting numbers for the ranges. This yields $4$ for the one number in $f$, and $3$ options for the two numbers for $g$. Afterwards, note that the function with 2 numbers in the range can have $4+6+4=14$ arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have $2*12*14$ possibilities, the 2 from symmetry. $f 2 g 2$: no symmetry, still easy! Just note that we have $6$ choices of which numbers go to $f$ and $g$, and within each, $14*14=196$ choices for the orientation of each of the two numbers. That's $6*196$ possibilities. $f 1 g 3$: again, symmetrical (WLOG $f$ has one element): $4$ ways to select the single element for $f$, and then find the number of ways to distribute the $3$ distinct numbers in the range for $g$. The only arrangement for the frequency of each number is ${1, 1, 2}$ in some order. Therefore, we have $3$ ways to choose which number is the one represented twice, and then note that there are $12$ ways to arrange these! The number of possibilities in this situation is $2 * 4 * 3 * 12$. Total, divided by $4^8$, gets $\\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}$, with numerator $453.",
"We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value. We proceed with casework to determine the number of possible $f$ with range 1, 2, etc. Range 1: There are 4 possibilities: all elements output to 1, 2, 3, or 4. Range 2: We have ${{4}\\choose {2}} = 6$ ways to choose the two output elements for $f$. At this point we have two possibilities: either $f$ has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and ${{4}\\choose {1}} = 4$ ways to rearrange the 4 elements, for a total of $6 * 4 * 2 = 48$ ways for this case. For the second case, there are ${{4}\\choose {2}} = 6$ ways to rearrange the 4 elements, for a total of $6 * 6 = 36$ ways for this case. Adding these two, we get a total of $36 + 48 = 84$ total possibilities. Range 3: We have ${{4}\\choose {3}} = 4$ ways to choose the three output elements for $f$. We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are ${{4}\\choose{1}}*{{3}\\choose{1}} = 12$ ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is $4 * 3 * 12 = 144$ ways. Range 4: Since we know the elements present, we have $4!$ ways to arrange them, or 24 ways. (To check, $4 + 84 + 144 + 24 = 256$, which is the total number of possibilities). We now break $f$ down by cases, and count the number of $g$ whose ranges are disjoint from $f$'s. Case 1: $f$'s range contains 1 element We know that there are 3 possibilities for $g$ with 1 element. Since half the possibilities for $g$ with two elements will contain the element in $f$, there are $84/2 = 42$ possibilities for $g$ with 2 elements. Since $3/4$ the possibilities for $g$ with 3 elements will contain the element in $f$, there are $144/4 = 36$ possibilities for $g$ with 3 elements. Clearly, no 4-element range for $g$ is possible, so the total number of ways for this case to happen is $4(3 + 42 + 36) = 324$. Case 2: $f$'s range contains 2 elements We know that there are 2 possibilities for $g$ with 1 element. If $g$ has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for $g$ is $84/6 = 14$. No 3-element or 4-element ranges for $g$ are possible. Thus, the total number of ways for this to happen is $84(2 + 14) = 1344$. Case 3: $f$'s range contains 3 elements In this case, there is only 1 possibility for $g$ - all the output values are the element that does not appear in $f$'s range. Thus, the total number of ways for this to happen is $144$. Summing the cases We find that the probability of $f$ and $g$ having disjoint ranges is equal to: $\\dfrac{324 + 1344 + 144}{256^2}=\\dfrac{1812}{2^{16}}= \\dfrac{453}{2^{14}}$ Thus, our final answer is $453 having disjoint ranges is then \\[\\frac{4\\cdot 3^4 + 84\\cdot 2^4 + 144\\cdot 1^4}{(4^4)^2} = \\frac{453}{2^{14}}.\\]",
"We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value. We proceed with casework to determine the number of possible $f$ with range 1, 2, etc. Range 1: There are 4 possibilities: all elements output to 1, 2, 3, or 4. Range 2: We have ${{4}\\choose {2}} = 6$ ways to choose the two output elements for $f$. At this point we have two possibilities: either $f$ has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and ${{4}\\choose {1}} = 4$ ways to rearrange the 4 elements, for a total of $6 * 4 * 2 = 48$ ways for this case. For the second case, there are ${{4}\\choose {2}} = 6$ ways to rearrange the 4 elements, for a total of $6 * 6 = 36$ ways for this case. Adding these two, we get a total of $36 + 48 = 84$ total possibilities. Range 3: We have ${{4}\\choose {3}} = 4$ ways to choose the three output elements for $f$. We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are ${{4}\\choose{1}}*{{3}\\choose{1}} = 12$ ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is $4 * 3 * 12 = 144$ ways. Range 4: Since we know the elements present, we have $4!$ ways to arrange them, or 24 ways. (To check, $4 + 84 + 144 + 24 = 256$, which is the total number of possibilities). We now break $f$ down by cases, and count the number of $g$ whose ranges are disjoint from $f$'s. Case 1: $f$'s range contains 1 element We know that there are 3 possibilities for $g$ with 1 element. Since half the possibilities for $g$ with two elements will contain the element in $f$, there are $84/2 = 42$ possibilities for $g$ with 2 elements. Since $3/4$ the possibilities for $g$ with 3 elements will contain the element in $f$, there are $144/4 = 36$ possibilities for $g$ with 3 elements. Clearly, no 4-element range for $g$ is possible, so the total number of ways for this case to happen is $4(3 + 42 + 36) = 324$. Case 2: $f$'s range contains 2 elements We know that there are 2 possibilities for $g$ with 1 element. If $g$ has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for $g$ is $84/6 = 14$. No 3-element or 4-element ranges for $g$ are possible. Thus, the total number of ways for this to happen is $84(2 + 14) = 1344$. Case 3: $f$'s range contains 3 elements In this case, there is only 1 possibility for $g$ - all the output values are the element that does not appear in $f$'s range. Thus, the total number of ways for this to happen is $144$. Summing the cases We find that the probability of $f$ and $g$ having disjoint ranges is equal to: $\\dfrac{324 + 1344 + 144}{256^2}=\\dfrac{1812}{2^{16}}= \\dfrac{453}{2^{14}}$ Thus, our final answer is $453 having disjoint ranges is then \\[\\frac{4\\cdot 3^4 + 84\\cdot 2^4 + 144\\cdot 1^4}{(4^4)^2} = \\frac{453}{2^{14}}.\\]",
"As before, there are 4 functions with a range of size 1, 84 with a range of size 2, and 144 with a range of size 3. If the range of $f$ has size $k$, the codomain of $g$ is restricted to a set of size $4 - k$. Any function from $A$ into this codomain will do, so there are $(4 - k)^4$ possibilities for $g$ given a function $f$. The probability of $f$ and $g$ having disjoint ranges is then \\[\\frac{4\\cdot 3^4 + 84\\cdot 2^4 + 144\\cdot 1^4}{(4^4)^2} = \\frac{453}{2^{14}}.\\]",
"As before, there are 4 functions with a range of size 1, 84 with a range of size 2, and 144 with a range of size 3. If the range of $f$ has size $k$, the codomain of $g$ is restricted to a set of size $4 - k$. Any function from $A$ into this codomain will do, so there are $(4 - k)^4$ possibilities for $g$ given a function $f$. The probability of $f$ and $g$ having disjoint ranges is then \\[\\frac{4\\cdot 3^4 + 84\\cdot 2^4 + 144\\cdot 1^4}{(4^4)^2} = \\frac{453}{2^{14}}.\\]"
] |
2014-I-13
| 2,014
| 13
|
On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$ . [asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]
| 850
|
I
|
[
"Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\\overline{EG}$ and $\\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\\overline{AD}$. [asy] size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(S--R--T, gray+0.4); dotfactor = 3; dot(\"$A$\",A,dir(135)); dot(\"$B$\",B,dir(215)); dot(\"$C$\",C,dir(305)); dot(\"$D$\",D,dir(45)); dot(\"$H$\",H,dir(90)); dot(\"$F$\",F,dir(270)); dot(\"$E$\",E,dir(180)); dot(\"$G$\",G,dir(0)); dot(\"$Q$\",O,dir(-90)); dot(\"$R$\",R,dir(-180)); dot(\"$S$\",S,dir(0)); dot(\"$T$\",T,dir(90)); pair P = extension(F,H,E,G); dot(\"$P$\",P,dir(180+60)); [/asy] The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is \\[\\frac{275+405}{269+275+405+411}=\\frac 12,\\]so $Q$ is the center of $ABCD$. Thus $R$, $Q$, $S$ are collinear, and $RT=QS=\\tfrac 12 s$. Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\\tfrac 35$, so $RS=\\tfrac 35s$ and $RQ=\\tfrac{1}{10}s$. Because $\\triangle QSG \\cong \\triangle RTH$, the area of $DHPG$ is the sum \\[[DHPG]=[DTRS]+[RPQ].\\] Rectangle $DTRS$ has area $RS\\cdot RT = \\tfrac 35s\\cdot \\tfrac 12 s = \\tfrac{3}{10}s^2$. If $\\angle QRP = \\theta$ , then $\\triangle RPQ$ has area \\[[RPQ]= \\tfrac 12 \\cdot \\tfrac 1{10}s\\sin\\theta \\cdot \\tfrac 1{10}s\\cos\\theta = \\tfrac 1{400}s^2\\sin 2\\theta.\\]Therefore the area of $[DHPG]$ is $s^2(\\tfrac 3{10}+\\tfrac 1{400}\\sin 2\\theta)$. Because the area of trapezoid $CDHF$ is $\\tfrac 35 s^2$, the area of $CGPF$ is $s^2(\\tfrac 3{10}-\\tfrac 1{400}\\sin 2\\theta)$. Because these areas are in the ratio $411:405=(408+3):(408-3)$, it follows that \\[\\frac{\\frac 1{400}\\sin 2\\theta}{\\frac 3{10}}=\\frac 3{408},\\]from which we get $\\sin 2\\theta = \\tfrac {15}{17}$. Note that $\\theta =\\angle RHT > \\angle QAT = 45^\\circ$, so $\\cos 2\\theta = -\\sqrt{1-\\sin^2 2\\theta}= -\\tfrac 8{17}$ and $\\sin^2\\theta = \\tfrac{1}{2}(1-\\cos 2\\theta) = \\tfrac{25}{34}$. Then \\[[ABCD]=s^2 = EG^2\\sin^2\\theta = 34^2 \\cdot \\tfrac {25}{34} = 850.\\]",
"Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\\overline{EG}$ and $\\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\\overline{AD}$. [asy] size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(S--R--T, gray+0.4); dotfactor = 3; dot(\"$A$\",A,dir(135)); dot(\"$B$\",B,dir(215)); dot(\"$C$\",C,dir(305)); dot(\"$D$\",D,dir(45)); dot(\"$H$\",H,dir(90)); dot(\"$F$\",F,dir(270)); dot(\"$E$\",E,dir(180)); dot(\"$G$\",G,dir(0)); dot(\"$Q$\",O,dir(-90)); dot(\"$R$\",R,dir(-180)); dot(\"$S$\",S,dir(0)); dot(\"$T$\",T,dir(90)); pair P = extension(F,H,E,G); dot(\"$P$\",P,dir(180+60)); [/asy] The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is \\[\\frac{275+405}{269+275+405+411}=\\frac 12,\\]so $Q$ is the center of $ABCD$. Thus $R$, $Q$, $S$ are collinear, and $RT=QS=\\tfrac 12 s$. Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\\tfrac 35$, so $RS=\\tfrac 35s$ and $RQ=\\tfrac{1}{10}s$. Because $\\triangle QSG \\cong \\triangle RTH$, the area of $DHPG$ is the sum \\[[DHPG]=[DTRS]+[RPQ].\\] Rectangle $DTRS$ has area $RS\\cdot RT = \\tfrac 35s\\cdot \\tfrac 12 s = \\tfrac{3}{10}s^2$. If $\\angle QRP = \\theta$ , then $\\triangle RPQ$ has area \\[[RPQ]= \\tfrac 12 \\cdot \\tfrac 1{10}s\\sin\\theta \\cdot \\tfrac 1{10}s\\cos\\theta = \\tfrac 1{400}s^2\\sin 2\\theta.\\]Therefore the area of $[DHPG]$ is $s^2(\\tfrac 3{10}+\\tfrac 1{400}\\sin 2\\theta)$. Because the area of trapezoid $CDHF$ is $\\tfrac 35 s^2$, the area of $CGPF$ is $s^2(\\tfrac 3{10}-\\tfrac 1{400}\\sin 2\\theta)$. Because these areas are in the ratio $411:405=(408+3):(408-3)$, it follows that \\[\\frac{\\frac 1{400}\\sin 2\\theta}{\\frac 3{10}}=\\frac 3{408},\\]from which we get $\\sin 2\\theta = \\tfrac {15}{17}$. Note that $\\theta =\\angle RHT > \\angle QAT = 45^\\circ$, so $\\cos 2\\theta = -\\sqrt{1-\\sin^2 2\\theta}= -\\tfrac 8{17}$ and $\\sin^2\\theta = \\tfrac{1}{2}(1-\\cos 2\\theta) = \\tfrac{25}{34}$. Then \\[[ABCD]=s^2 = EG^2\\sin^2\\theta = 34^2 \\cdot \\tfrac {25}{34} = 850.\\]",
"Let $s$ be the side length of $ABCD$, let $[ABCD]=1360a$. Let $Q$ and $R$ be the midpoints of $\\overline{EG}$ and $\\overline{FH}$, respectively; because $269+411=275+405$, $Q$ is also the center of the square. Draw $\\overline{IJ} \\parallel \\overline{HF}$ through $Q$, with $I$ on $\\overline{AD}$, $J$ on $\\overline{BC}$. [asy] size(150); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S; Q=MP(\"Q\",origin,down); A=MP(\"A\",(-1,1),dir(135)); B=MP(\"B\",(-1,-1),dir(225)); C=MP(\"C\",(1,-1),dir(-45)); D=MP(\"D\",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP(\"E\",extension(A,B,Q,dir(theta)),left); J=MP(\"J\",extension(B,C,Q,dir(90+theta)),down); F=MP(\"F\",J+(shift*left),down); G=MP(\"G\",extension(C,D,Q,dir(theta)),right); I=MP(\"I\",extension(A,D,Q,dir(90+theta)),up); H=MP(\"H\",I+(shift*left),up); P=MP(\"P\",extension(E,G,F,H),2*dir(-110)); R=MP(\"R\",extension(F,H,Q,left),left); draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray); [/asy] Segments $\\overline{EG}$ and $\\overline{IJ}$ divide the square into four congruent quadrilaterals, each of area $\\tfrac 14 [ABCD]=340a$. Then \\[[HFJI]=[ABJI]-[ABFH]=136a.\\] The fraction of the total area occupied by parallelogram $HFJI$ is $\\tfrac 1{10}$, so $RQ=\\tfrac{s}{10}$. Because $[HFJI]= HF\\cdot PQ$, with $HF=34$, we get $PQ=4a$. Now \\[[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\\tfrac 12[IJFH] = 3a,\\] and because $[PQR]=\\tfrac 12 \\cdot PQ\\cdot PR$, with $PQ=4a$, we get $PR=\\tfrac 32$. By Pythagoras' Theorem on $\\triangle PQR$, we get \\[16a^2+\\frac 94 =\\tfrac{68}{5}a,\\quad \\text{i.e.}\\quad 320a^2-272a+45=0,\\] with roots $a=\\tfrac 9{40}$ or $a=\\tfrac58$. The former leads to a square with diagonal less than $34$, which can't be, since $EG=FH=34$; therefore $a=\\tfrac 58$ and $[ABCD]=850$.",
"$269+275+405+411=1360$, a multiple of $17$. In addition, $EG=FH=34$, which is $17\\cdot 2$. Therefore, we suspect the square of the \"hypotenuse\" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$. All of these triples are primitive: \\[17=1^2+4^2\\] \\[34=3^2+5^2\\] \\[51=\\emptyset\\] \\[68=\\emptyset\\text{ others}\\] \\[85=2^2+9^2=6^2+7^2\\] \\[102=\\emptyset\\] \\[119=\\emptyset \\dots\\] The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$: \\[\\sqrt{17}\\rightarrow 34\\implies 8\\sqrt{17}\\implies A=\\textcolor{red}{1088}\\] \\[\\sqrt{34}\\rightarrow 34\\implies 5\\sqrt{34}\\implies A=850\\] \\[\\sqrt{85}\\rightarrow 34\\implies \\{18\\sqrt{85}/5,14\\sqrt{85}/5\\}\\implies A=\\textcolor{red}{1101.6,666.4}\\] Thus, $850 is the only valid answer.",
"$269+275+405+411=1360$, a multiple of $17$. In addition, $EG=FH=34$, which is $17\\cdot 2$. Therefore, we suspect the square of the \"hypotenuse\" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$. All of these triples are primitive: \\[17=1^2+4^2\\] \\[34=3^2+5^2\\] \\[51=\\emptyset\\] \\[68=\\emptyset\\text{ others}\\] \\[85=2^2+9^2=6^2+7^2\\] \\[102=\\emptyset\\] \\[119=\\emptyset \\dots\\] The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$: \\[\\sqrt{17}\\rightarrow 34\\implies 8\\sqrt{17}\\implies A=\\textcolor{red}{1088}\\] \\[\\sqrt{34}\\rightarrow 34\\implies 5\\sqrt{34}\\implies A=850\\] \\[\\sqrt{85}\\rightarrow 34\\implies \\{18\\sqrt{85}/5,14\\sqrt{85}/5\\}\\implies A=\\textcolor{red}{1101.6,666.4}\\] Thus, $850 is the only valid answer.",
"Continue in the same way as solution 1 to get that $POK$ has area $3a$, and $OK = \\frac{d}{10}$. You can then find $PK$ has length $\\frac 32$. Then, if we drop a perpendicular from $H$ to $BC$ at $L$, We get $\\triangle HLF \\sim \\triangle OPK$. Thus, $LF = \\frac{15\\cdot 34}{d}$, and we know $HL = d$, and $HF = 34$. Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem. \\[\\frac{15^2 \\cdot 34^2}{d^2} + d^2 = 34^2\\] \\[d^4 - 34^2 d^2 + 15^2 \\cdot 34^2 = 0\\] \\[(d^2 - 34 \\cdot 25)(d^2 - 34 \\cdot 9) = 0\\] $d^2 = 34 \\cdot 9$ is extraneous, so $d^2 = 34 \\cdot 25$. Since the area is $d^2$, we have it is equal to $34 \\cdot 25 = 850 -Alexlikemath"
] |
2014-I-14
| 2,014
| 14
|
Let $m$ be the largest real solution to the equation \[\dfrac{3}{x-3} + \dfrac{5}{x-5} + \dfrac{17}{x-17} + \dfrac{19}{x-19} = x^2 - 11x - 4\] There are positive integers $a, b,$ and $c$ such that $m = a + \sqrt{b + \sqrt{c}}$ . Find $a+b+c$ .
| 263
|
I
|
[
"The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\\frac{3}{x-3}$, then the fraction becomes of the form $\\frac{x}{x - 3}$. A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is true given the answer format) we can cancel the common factor of $x$ from both sides of the equation. $\\frac{1}{x - 3} + \\frac{1}{x - 5} + \\frac{1}{x - 17} + \\frac{1}{x - 19} = x - 11$ Then, if we make the substitution $y = x - 11$, we can further simplify. $\\frac{1}{y + 8} + \\frac{1}{y + 6} + \\frac{1}{y - 6} + \\frac{1}{y - 8} =y$ If we group and combine the terms of the form $y - n$ and $y + n$, we get this equation: $\\frac{2y}{y^2 - 64} + \\frac{2y}{y^2 - 36} = y$ Then, we can cancel out a $y$ from both sides, knowing that $x = 11$ is not a possible solution given the answer format. After we do that, we can make the final substitution $z = y^2$. $\\frac{2}{z - 64} + \\frac{2}{z - 36} = 1$ $2z - 128 + 2z - 72 = (z - 64)(z - 36)$ $4z - 200 = z^2 - 100z + 64(36)$ $z^2 - 104z + 2504 = 0$ Using the quadratic formula, we get that the largest solution for $z$ is $z = 52 + 10\\sqrt{2}$. Then, repeatedly substituting backwards, we find that the largest value of $x$ is $11 + \\sqrt{52 + \\sqrt{200}}$. The answer is thus $11 + 52 + 200 = 263",
"Proceed as with Solution 1 until we get the following. $\\frac{2y}{y^2 - 64} + \\frac{2y}{y^2 - 36} = y \\implies$ $\\frac{1}{y^2 - 64} + \\frac{1}{y^2 - 36} = \\frac{1}{2}.$ Here, we may also use a slightly different substitution, $z = y^2 - 50.$ This gives: $\\frac{1}{z - 14} + \\frac{1}{z + 14} = \\frac{1}{2} \\implies$ $\\frac{2z}{z^2 - 196} = \\frac{1}{2} \\implies$ $z^2 - 4z - 196 = 0.$ We now have a simpler quadratic, eliminating tedious and potentially error-prone calculations. Following through, we get $z = 2 + \\sqrt{200} \\implies y = \\sqrt{52 + \\sqrt{200}}$ as desired."
] |
2014-I-15
| 2,014
| 15
|
In $\triangle ABC, AB = 3, BC = 4,$ and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B, \overline{BC}$ at $B$ and $D,$ and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4},$ length $DE=\frac{a\sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ .
| 41
|
I
|
[
"Since $\\angle DBE = 90^\\circ$, $DE$ is the diameter of $\\omega$. Then $\\angle DFE=\\angle DGE=90^\\circ$. But $DF=FE$, so $\\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$, we have that $EG=4x$, $DE=5x$, and $DF=EF=\\frac{5x}{\\sqrt{2}}$. Note that $\\triangle DGE \\sim \\triangle ABC$ by SAS similarity, so $\\angle BAC = \\angle GDE$ and $\\angle ACB = \\angle DEG$. Since $DEFG$ is a cyclic quadrilateral, $\\angle BAC = \\angle GDE=180^\\circ-\\angle EFG = \\angle AFE$ and $\\angle ACB = \\angle DEG = \\angle GFD$, implying that $\\triangle AFE$ and $\\triangle CDF$ are isosceles. As a result, $AE=CD=\\frac{5x}{\\sqrt{2}}$, so $BE=3-\\frac{5x}{\\sqrt{2}}$ and $BD =4-\\frac{5x}{\\sqrt{2}}$. Finally, using the Pythagorean Theorem on $\\triangle BDE$, \\[\\left(3-\\frac{5x}{\\sqrt{2}}\\right)^2 + \\left(4-\\frac{5x}{\\sqrt{2}}\\right)^2 = (5x)^2\\] Solving for $x$, we get that $x=\\frac{5\\sqrt{2}}{14}$, so $DE= 5x =\\frac{25 \\sqrt{2}}{14}$. Thus, the answer is $25+2+14=041.",
"[asy] pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot(\"$A$\",A,dir(135)); dot(\"$B$\",B,dir(215)); dot(\"$C$\",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot(\"$D$\",D,dir(270)); dot(\"$E$\",E,dir(180)); dot(\"$F$\",F,dir(90)); dot(\"$G$\",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy] First we note that $\\triangle DEF$ is an isosceles right triangle with hypotenuse $\\overline{DE}$ the same as the diameter of $\\omega$. We also note that $\\triangle DGE \\sim \\triangle ABC$ since $\\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$. From congruent arc intersections, we know that $\\angle GED \\cong \\angle GBC$, and that from similar triangles $\\angle GED$ is also congruent to $\\angle GCB$. Thus, $\\triangle BGC$ is an isosceles triangle with $BG = GC$, so $G$ is the midpoint of $\\overline{AC}$ and $AG = GC = 5/2$. Similarly, we can find from angle chasing that $\\angle ABF = \\angle EDF = \\frac{\\pi}4$. Therefore, $\\overline{BF}$ is the angle bisector of $\\angle B$. From the angle bisector theorem, we have $\\frac{AF}{AB} = \\frac{CF}{CB}$, so $AF = 15/7$ and $CF = 20/7$. Lastly, we apply power of a point from points $A$ and $C$ with respect to $\\omega$ and have $AE \\times AB=AF \\times AG$ and $CD \\times CB=CG \\times CF$, so we can compute that $EB = \\frac{17}{14}$ and $DB = \\frac{31}{14}$. From the Pythagorean Theorem, we result in $DE = \\frac{25 \\sqrt{2}}{14}$, so $a+b+c=25+2+14= 041. ~First",
"Call $DE=x$ and as a result $DF=EF=\\frac{x\\sqrt{2}}{2}, EG=\\frac{4x}{5}, GD=\\frac{3x}{5}$. Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$nd paragraph of Solution $2$) we get $AG=\\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\\frac{5}{14}$ so you can use Ptolemy to get $x=\\frac{25\\sqrt{2}}{14} \\implies 041. ~First",
"See inside the $\\triangle DEF$, we can find that $AG>AF$ since if $AG<AF$, we can see that Ptolemy Theorem inside cyclic quadrilateral $EFGD$ doesn't work. Now let's see when $AG>AF$, since $\\frac{DG}{EG} = \\frac{3}{4}$, we can assume that $EG=4x;GD=3x;ED=5x$, since we know $EF=FD$ so $\\triangle EFD$ is isosceles right triangle. We can denote $DF=EF=\\frac{5x\\sqrt{2}}{2}$.Applying Ptolemy Theorem inside the cyclic quadrilateral $EFGD$ we can get the length of $FG$ can be represented as $\\frac{x\\sqrt{2}}{2}$. After observing, we can see $\\angle AFE=\\angle EDG$, whereas $\\angle A=\\angle EDG$ so we can see $\\triangle AEF$ is isosceles triangle. Since $\\triangle ABC$ is a $3-4-5$ triangle so we can directly know that the length of AF can be written in the form of $3x\\sqrt{2}$. Denoting a point $J$ on side $AC$ with that $DJ$ is perpendicular to side $AC$. Now with the same reason, we can see that $\\triangle DJG$ is a isosceles right triangle, so we can get $GJ=\\frac{3x\\sqrt{2}}{2}$ while the segment $CJ$ is $2x\\sqrt{2}$ since its 3-4-5 again. Now adding all those segments together we can find that $AC=5=7x\\sqrt{2}$ and $x=\\frac{5\\sqrt{2}}{14}$ and the desired $ED=5x=\\frac{25\\sqrt{2}}{14}$ which our answer is $041 ~bluesoul",
"The main element of the solution is the proof that $BF$ is bisector of $\\angle B.$ Let $O$ be the midpoint of $DE.$ $\\angle EBF = 90^\\circ \\implies$ $O$ is the center of the circle $BDGFE.$ $\\angle EOF = 90^\\circ \\implies \\overset{\\Large\\frown} {EF} = 90^\\circ \\implies \\angle EBF = 45^\\circ \\implies$ BF is bisector of $\\angle ABC\\implies BF = \\frac {2AB \\cdot BC}{AB+BC} \\cos 45^\\circ =\\frac {12 \\cdot \\sqrt{2}}{7}.$ \\[\\angle EGD = 90^\\circ, \\frac {EG}{GD}=\\frac{4}{3} \\implies\\] \\[\\angle GED = \\angle GCD =\\gamma \\implies \\overset{\\Large\\frown} {DG} = 2\\gamma.\\] \\[2\\angle ACB = \\overset{\\Large\\frown} {BEF} - \\overset{\\Large\\frown} {DG} \\implies \\overset{\\Large\\frown} {BEF} = 4 \\gamma \\implies\\] \\[\\angle BOF = 4 \\gamma \\implies \\angle OBF = \\angle OFB = 90^\\circ – 2 \\gamma.\\] Let $BO = EO = DO = r \\implies BF = 2 r \\cos(90^\\circ – 2\\gamma) =$ \\[=2 r \\sin 2\\gamma = 4r \\sin \\gamma \\cdot \\cos \\gamma = 4 r\\cdot \\frac {3}{5} \\cdot \\frac {4}{5} = \\frac {48}{25} = \\frac {12 \\cdot \\sqrt{2}}{7}\\implies\\] \\[r = \\frac {25 \\cdot \\sqrt{2}}{28}\\implies ED = 2r = \\frac {25 \\cdot \\sqrt{2}}{14}\\implies \\textbf{041}.\\] [email protected], vvsss",
"The main element of the solution is the proof that $G$ is midpoint of $AC.$ As in Solution 5 we get $\\angle GED = \\angle DBG =\\gamma \\implies$ $\\triangle BCG$ is isosceles triangle with $BG=CG.$ Similarly $BG = AG \\implies AG = CG = BG = \\frac {AC}{2} =\\frac {5}{2}.$ \\[\\overset{\\Large\\frown} {FG} = 90^\\circ – \\overset{\\Large\\frown} {GD} = 90^\\circ – 2\\gamma \\implies\\] \\[\\overset{\\Large\\frown} {BFG} = 4\\gamma + 90^\\circ – 2\\gamma = 90^\\circ + 2\\gamma \\implies\\] \\[\\angle BOG = 90^\\circ + 2\\gamma \\implies \\angle BGO = \\angle GBO = 45^\\circ - \\gamma.\\] Let $\\hspace{10mm} BO = EO = DO = r \\implies$ \\[BG = 2 r \\cos(45^\\circ – \\gamma) = 2 r (\\sin \\gamma + \\cos \\gamma)\\frac {\\sqrt {2}}{2} =\\] \\[r \\biggl(\\frac {3}{5} + \\frac {4}{5}\\biggr) \\sqrt {2} = r \\frac {7 \\sqrt{2}}{5} = \\frac {5}{2}\\implies\\] \\[r = \\frac {25 \\cdot \\sqrt{2}}{28}\\implies ED = \\frac {25 \\cdot \\sqrt{2}}{14}\\implies \\textbf{041}.\\] [email protected], vvsss",
"Let $(BEFGD) = \\omega$. By Incenter-Excenter(Fact $5$), $F$ is the angle bisector of $\\angle B$. Then by Ratio Lemma we have \\[\\frac{AG}{CG} = \\frac{\\sin(ABG)}{\\sin(CBG)} \\cdot \\frac{AB}{BC} = \\frac{\\sin(GDE)}{\\sin(DEG)} \\cdot \\frac{3}{4} = 1\\] Thus, $G$ is the midpoint of $AC$. We can calculate $AF$ and $CF$ to be $\\frac{15}{7}$ and $\\frac{20}{7}$ respectively. And then by Power of a Point, we have $\\newline$ \\[\\operatorname{Pow}_{\\omega}(A) = AE \\cdot AB = AF \\cdot AG \\implies AE = \\frac{25}{14}\\] And then similarly, we have $CD = AE = \\frac{25}{14}$. $\\newline$ Then $EB = \\frac{17}{14}$ and $DB = \\frac{31}{14}$ and by Pythagorean we have $DE = \\frac{25\\sqrt{2}}{14}$, so our answer is $\\textbf{041} ~dolphinday"
] |
2014-II-1
| 2,014
| 1
|
Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.
| 334
|
II
|
[
"From the given information, we can see that Abe can paint $\\frac{1}{15}$ of the room in an hour, Bea can paint $\\frac{1}{15}\\times\\frac{3}{2} = \\frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\\frac{1}{15}\\times 2 = \\frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has painted $\\frac{1}{15}\\times\\frac{3}{2}=\\frac{1}{10}$ of the room. Working together, Abe and Bea can paint $\\frac{1}{15}+\\frac{1}{10}=\\frac{1}{6}$ of the room in an hour, so it takes then $\\frac{2}{5}\\div \\frac{1}{6}= \\frac{12}{5}$ hours to finish the first half of the room. All three working together can paint $\\frac{1}{6}+\\frac{2}{15}=\\frac{3}{10}$ of the room in an hour, and it takes them $\\frac{1}{2}\\div \\frac{3}{10}=\\frac{5}{3}$ hours to finish the room. The total amount of time they take is \\[90+\\frac{12}{5}\\times 60+\\frac{5}{3}\\times 60 = 90+ 144 + 100 = 334 \\text{\\ minutes.}\\]"
] |
2014-II-2
| 2,014
| 2
|
Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$ . The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 76
|
II
|
[
"We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram. [asy] pair A,B,C,D,E,F,G; A=(0,55); B=(60,55); C=(60,0); D=(0,0); draw(A--B--C--D--A); E=(30,35); F=(20,20); G=(40,20); draw(circle(E,15)); draw(circle(F,15)); draw(circle(G,15)); draw(\"$A$\",(30,52)); draw(\"$B$\",(7,7)); draw(\"$C$\",(53,7)); draw(\"100\",(5,60)); draw(\"10\",(30,40)); draw(\"10\",(15,15)); draw(\"10\",(45,15)); draw(\"14\",(30,16)); draw(\"14\",(38,29)); draw(\"14\",(22,29)); draw(\"$x$\",(30,25)); draw(\"$y$\",(10,45)); [/asy] Let $x$ be the number of men with all three risk factors. Since \"the probability that a randomly selected man has all three risk factors, given that he has A and B is $\\frac{1}{3}$,\" we can tell that $x = \\frac{1}{3}(x+14)$, since there are $x$ people with all three factors and 14 with only A and B. Thus $x=7$. Let $y$ be the number of men with no risk factors. It now follows that \\[y= 100 - 3 \\cdot 10 - 3 \\cdot 14 - 7 = 21.\\] The number of men with risk factor A is $10+2 \\cdot 14+7 = 45$ (10 with only A, 28 with A and one of the others, and 7 with all three). Thus the number of men without risk factor $A$ is 55, so the desired conditional probability is $21/55$. So the answer is $21+55=076."
] |
2014-II-3
| 2,014
| 3
|
A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$ . [asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]
| 720
|
II
|
[
"When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below. [asy] pair R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); draw(T--R,red); draw(X--Z,red); dot(\" \",R,NW); dot(\" \",S,NW); dot(\" \",T,NW); dot(\" \",X,NW); dot(\" \",Y,NW); dot(\" \",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy] By Heron's Formula, the area of each isosceles triangle is $\\sqrt{(30)(12)(12)(6)}=\\sqrt{180\\times 12^2}=72\\sqrt{5}$. So the area of both is $144\\sqrt{5}$. From the rectangle, our original area is $36a$. The area of the rectangle in the hexagon is $24a$. So we have \\[24a+144\\sqrt{5}=36a\\implies 12a=144\\sqrt{5}\\implies a=12\\sqrt{5}\\implies a^2=720.\\]",
"Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$. Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitude of the end triangles is $\\sqrt{3^2-2^2}=\\sqrt{5}$ given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area $\\sqrt{5}$ x $4$. Then, the area of the hexagon is $4p+4\\sqrt{5}$, and the area of the rectangle is $6p$. Equating them, $p=2\\sqrt{5}$. Multiply by scale factor of 6 and square it to get $36(20)= 720 \\implies a^2=720. ~BJHHar"
] |
2014-II-4
| 2,014
| 4
|
The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy $0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},$ where $a$ , $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ .
| 447
|
II
|
[
"Notice repeating decimals can be written as the following: $0.\\overline{ab}=\\frac{10a+b}{99}$ $0.\\overline{abc}=\\frac{100a+10b+c}{999}$ where a,b,c are the digits. Now we plug this back into the original fraction: $\\frac{10a+b}{99}+\\frac{100a+10b+c}{999}=\\frac{33}{37}$ Multiply both sides by $999*99.$ This helps simplify the right side as well because $999=111*9=37*3*9$: $9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$ Dividing both sides by $9$ and simplifying gives: $2210a+221b+11c=99^2=9801$ At this point, seeing the $221$ factor common to both a and b is crucial to simplify. This is because taking $mod 221$ to both sides results in: $2210a+221b+11c \\equiv 9801 \\mod 221 \\iff 11c \\equiv 77 \\mod 221$ Notice that we arrived to the result $9801 \\equiv 77 \\mod 221$ by simply dividing $9801$ by $221$ and seeing $9801=44*221+77.$ Okay, now it's pretty clear to divide both sides by $11$ in the modular equation but we have to worry about $221$ being multiple of $11.$ Well, $220$ is a multiple of $11$ so clearly, $221$ couldn't be. Also, $221=13*17.$ Now finally we simplify and get: $c \\equiv 7 \\mod 221$ But we know $c$ is between $0$ and $9$ because it is a digit, so $c$ must be $7.$ Now it is straightforward from here to find $a$ and $b$: $2210a+221b+11(7)=9801 \\iff 221(10a+b)=9724 \\iff 10a+b=44$ and since a and b are both between $0$ and $9$, we have $a=b=4$. Finally we have the $3$ digit integer $447",
"Note that $\\frac{33}{37}=\\frac{891}{999} = 0.\\overline{891}$. Also note that the period of $0.abab\\overline{ab}+0.abcabc\\overline{abc}$ is at most $6$. Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$. Adding the two, we get \\[\\begin{array}{ccccccc}&a&b&a&b&a&b\\\\ +&a&b&c&a&b&c\\\\ \\hline &8&9&1&8&9&1\\end{array}\\] From this, we can see that $a=4$, $b=4$, and $c=7$, so our desired answer is $447",
"Noting as above that $0.\\overline{ab} = \\frac{10a + b}{99}$ and $0.\\overline{abc} = \\frac{100a + 10b + c}{999}$, let $u = 10a + b$. Then \\[\\frac{u}{99} + \\frac{10u + c}{999} = \\frac{33}{37}\\] \\[\\frac{u}{11} + \\frac{10u + c}{111} = \\frac{9\\cdot 33}{37}\\] \\[\\frac{221u + 11c}{11\\cdot 111} = \\frac{9\\cdot 33}{37}\\] \\[221u + 11c = \\frac{9\\cdot 33\\cdot 11\\cdot 111}{37}\\] \\[221u + 11c = 9\\cdot 33^2.\\] Solving for $c$ gives \\[c = 3\\cdot 9\\cdot 33 - \\frac{221u}{11}\\] \\[c = 891 - \\frac{221u}{11}\\] Because $c$ must be integer, it follows that $u$ must be a multiple of $11$ (because $221$ clearly is not). Inspecting the equation, one finds that only $u = 44$ yields a digit $c, 7$. Thus $abc = 10u + c = 447",
"We note as above that $0.\\overline{ab} = \\frac{10a + b}{99}$ and $0.\\overline{abc} = \\frac{100a + 10b + c}{999},$ so \\[\\frac{10a + b}{99} + \\frac{100a + 10b + c}{999} = \\frac{33}{37} = \\frac{891}{999}.\\] As $\\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $11$ to cancel, so $11$ divides $10a + b.$ It follows that $a = b,$ so $\\frac{10a + b}{99} = \\frac{11a}{99} = \\frac{111a}{999},$ so \\[\\frac{111a}{999} + \\frac{110a+c}{999} = \\frac{891}{999}.\\] Then $111a + 110a + c = 891,$ or $221a + c = 891.$ Thus $a = b = 4$ and $c = 7,$ so the three-digit integer $abc$ is $447"
] |
2014-II-5
| 2,014
| 5
|
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ .
| 420
|
II
|
[
"Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$, and the remaining root of $q(x)$ is $-(r+s+1)$. The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$, and equating the two coefficients gives \\[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\\]from which $s = \\tfrac 12 (5r+13)$. Substitution should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $420.",
"Let $r$, $s$, and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$. Also, \\[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\\] Set up a similar equation for $s$: \\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.\\] Simplifying and adding the equations gives \\begin{align}\\tag{*} r^2 - s^2 + 4r + 3s + 49 &= 0 \\end{align} Now, let's deal with the $ax$ terms. Plugging the roots $r$, $s$, and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$, $s-3$, and $-1-r-s$ into $q(x)$ yields another long polynomial. Equating the coefficients of $x$ in both polynomials, we get: \\[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\\] which eventually simplifies to \\[s = \\frac{13 + 5r}{2}.\\] Substitution into (*) should give $r = -5$ and $r = 1$, corresponding to $s = -6$ and $s = 9$, and $|b| = 330, 90$, for an answer of $420.",
"The roots of $p(x)$ are $r$, $s$, and $-r-s$ since they sum to $0$ by Vieta's Formula (coefficient of $x^2$ term is $0$). Similarly, the roots of $q(x)$ are $r + 4$, $s - 3$, and $-r-s-1$, as they too sum to $0$. Then: $a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and $a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$. From these equations, we can write that \\[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a\\] and simplifying gives \\[2s-5r-13=0 \\Rightarrow s = \\frac{5r+13}{2}.\\] We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get \\[rs(r+s) = b\\] \\[(r+4)(s-3)(r+s+1)=b + 240.\\] Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$. Expanding, simplifying, substituting $s = \\frac{5r+13}{2}$, and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$, so $r = -5, 1$. Then $s = -6, 9$. Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$, or $b = (1)(9)(1 + 9) = 90$. The answer is $|-330|+|90| = 420.",
"By Vieta's, we know that the sum of roots of $p(x)$ is $0$. Therefore, the roots of $p$ are $r, s, -r-s$. By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$. Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$. Since $p(x)$ and $q(x)$ have the same coefficient for $x$, we can go ahead and match those up to get \\begin{align*} rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\\\ 0 &= -13 - 5r + 2s \\\\ s &= \\frac{5r + 13}{2} \\end{align*} At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$. Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$. Therefore, $q(s) = 240$. If we plug that into our expression, we get that \\begin{align*} q(s) &= 3(s - r - 4)(r + 2s + 1) \\\\ 240 &= 3(s - r - 4)(r + 2s + 1) \\\\ 240 &= 3\\left( \\frac{3r + 5}{2} \\right)(6r + 14) \\\\ 80 &= (3r + 5)(3r + 7) \\\\ 0 &= r^2 + 4r - 5 \\end{align*} This tells us that $(r, s) = (1, 9)$ or $(-5, -6)$. Since $-b$ is the product of the roots, we have that the two possibilities are $1 \\cdot 9 \\cdot (-10) = -90$ and $(-5)(-6)11 = 330$. Adding the absolute values of these gives us $420."
] |
2014-II-6
| 2,014
| 6
|
Charles has two six-sided dice. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 167
|
II
|
[
"The probability that he rolls a six twice when using the fair die is $\\frac{1}{6}\\times \\frac{1}{6}=\\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\\frac{2}{3}\\times \\frac{2}{3}=\\frac{4}{9}=\\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die. Therefore the probability that he is using the fair die is $\\frac{1}{17}$, and the probability that he is using the biased die is $\\frac{16}{17}$. The probability of rolling a third six is \\[\\frac{1}{17}\\times \\frac{1}{6} + \\frac{16}{17} \\times \\frac{2}{3} = \\frac{1}{102}+\\frac{32}{51}=\\frac{65}{102}\\] Therefore, our desired $p+q$ is $65+102= 167",
"This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)! The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus $\\frac{\\frac{1}{2}(\\frac{2}{3})^3+\\frac{1}{2}(\\frac{1}{6})^3}{\\frac{1}{2}(\\frac{2}{3})^2+\\frac{1}{2}(\\frac{1}{6})^2}= \\frac{65}{102}$. The answer is therefore $65+102= 167. ~th1nq3r Note: I have just found out that this is also the official solution. So I did not STEAL it, but as a coincidence, have come across the EXACT SAME SOLUTION. LIKE EXACTLY THE SAME. I AM SLIGHTLY FRIGHTENED. :/",
"This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)! The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus $\\frac{\\frac{1}{2}(\\frac{2}{3})^3+\\frac{1}{2}(\\frac{1}{6})^3}{\\frac{1}{2}(\\frac{2}{3})^2+\\frac{1}{2}(\\frac{1}{6})^2}= \\frac{65}{102}$. The answer is therefore $65+102= 167. ~th1nq3r Note: I have just found out that this is also the official solution. So I did not STEAL it, but as a coincidence, have come across the EXACT SAME SOLUTION. LIKE EXACTLY THE SAME. I AM SLIGHTLY FRIGHTENED. :/"
] |
2014-II-7
| 2,014
| 7
|
Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which $\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.$
| 21
|
II
|
[
"First, let's split it into two cases to get rid of the absolute value sign $\\left |\\sum_{k=1}^n\\log_{10}f(k)\\right|=1 \\iff \\sum_{k=1}^n\\log_{10}f(k)=1,-1$ Now we simplify using product-sum logarithmic identites: $\\log_{10}{f(1)}+\\log_{10}{f(2)}+...+\\log_{10}{f(n)}=\\log_{10}{f(1)\\cdot f(2) \\cdot ... \\cdot f(n)}=1,-1$ Note that the exponent $\\cos{\\pi(x)}$ is either $-1$ if $x$ is odd or $1$ if $x$ is even. Writing out the first terms we have $\\frac{1}{(2)(3)}(3)(4)\\frac{1}{(4)(5)} \\ldots$ This product clearly telescopes (i.e. most terms cancel) and equals either $10$ or $\\frac{1}{10}$. But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where $n$ is odd and another where $n$ is even. $\\textbf{Case 1: Odd n}$ For odd $n$, it telescopes to $\\frac{1}{2(n+2)}$ where $n$ is clearly $3$. $\\textbf{Case 2: Even n}$ For even $n$, it telescopes to $\\frac{n+2}{2}$ where $18$ is the only possible $n$ value. Thus the answer is $021",
"Note that $\\cos(\\pi x)$ is $-1$ when $x$ is odd and $1$ when $x$ is even. Also note that $x^2+3x+2=(x+1)(x+2)$ for all $x$. Therefore \\[\\log_{10}f(x)=\\log_{10}(x+1)+\\log_{10}(x+2)\\ \\ \\ \\text{if }x \\text{ is even}\\] \\[\\log_{10}f(x)=-\\log_{10}(x+1)-\\log_{10}(x+2)\\ \\ \\ \\text{if }x \\text{ is odd}\\] Because of this, $\\sum_{k=1}^n\\log_{10}f(k)$ is a telescoping series of logs, and we have \\[\\sum_{k=1}^n\\log_{10}f(k)= \\log_{10}(n+2)-\\log_{10}2=\\log_{10}\\frac{n+2}{2}\\ \\ \\ \\text{if }n \\text{ is even}\\] \\[\\sum_{k=1}^n\\log_{10}f(k)= -\\log_{10}(n+2)-\\log_{10}2=-\\log_{10}2(n+2)\\ \\ \\ \\text{if }n \\text{ is odd}\\] Setting each of the above quantities to $1$ and $-1$ and solving for $n$, we get possible values of $n=3$ and $n=18$ so our desired answer is $3+18=021"
] |
2014-II-8
| 2,014
| 8
|
Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle $D$ is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ .
| 254
|
II
|
[
"[asy] import graph; size(7.99cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000)); draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670)); draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416)); draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432)); draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666)); draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227)); draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002)); dot((7.780000000000009,-1.320000000000002),dotstyle); label(\"$C$\", (7.707377218471464,-1.576266740352400), NE * labelscalefactor); dot((7.271934046987930,-1.319179731427737),dotstyle); label(\"$D$\", (7.303064016111554,-1.276266740352400), NE * labelscalefactor); dot((9.198812158392690,-0.8249788848962227),dotstyle); label(\"$E$\", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor); dot((9.198009254448635,-1.322289365031666),dotstyle); label(\"$F$\", (9.225301294671791,-1.276266740352400), NE * labelscalefactor); dot((9.779997393419690,-1.323228980404432),dotstyle); label(\"$B$\", (9.810012253929656,-1.276266740352400), NE * labelscalefactor); dot((5.780002606580324,-1.316771019595571),dotstyle); label(\"$A$\", (5.812051070003994,-1.276266740352400), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $\\triangle CEF$ gives $CF=\\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so \\[DF=DC+CF=2-3r+\\sqrt{4-4r}.\\] Noting that $DE=4r$, we can now use the Pythagorean theorem in $\\triangle DEF$ to get \\[(2-3r+\\sqrt{4-4r})^2+r^2=16r^2.\\] Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\\sqrt{240}-14$ for a final answer of $254. Notice that C, E and the point of tangency to circle C for circle E will be collinear because C and E intersect the tangent line at a right angle, implying they must be on the same line.",
"Consider a reflection of circle $E$ over diameter $\\overline{AB}$. By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii $r$, $r$, and $3r$, and the big circle has radius $2$. Descartes' Circle Theorem gives $\\left(\\frac{1}{r}+\\frac{1}{r}+\\frac{1}{3r}-\\frac12\\right)^2 = 2\\left(\\left(\\frac{1}{r}\\right)^2+\\left(\\frac{1}{r}\\right)^2+\\left(\\frac{1}{3r}\\right)^2+\\left(-\\frac12\\right)^2\\right)$ Note that the big circle has curvature $-\\frac12$ because it is internally tangent. Solving gives $3r=\\sqrt{240}-14$ for a final answer of $254.",
"We use the notation of Solution 1 for triangle $\\triangle DEC$ \\[\\sin \\angle EDC = \\frac {EF}{DE} = \\frac {1}{4} \\implies \\cos \\angle EDC = \\frac {\\sqrt{15}}{4}.\\] We use Cosine Law for $\\triangle DEC$ and get: \\[(4r)^2 +(2 – 3r)^2 – 2 \\cdot 4r \\cdot (2 – 3r) \\cdot \\frac {\\sqrt{15}}{4} = (2 – r)^2\\]. \\[(24 + 6 \\sqrt{15} ) r^2 = (8 + 4 \\sqrt {15})r \\implies 3r = 4 \\sqrt{15} – 14 \\implies \\textbf{254}.\\] [email protected], vvsss",
"This problem can be very easily solved using Descartes' Circle Theorem. It states that if we have 4 circles that are all tangent with each other, $(k_1 + k_2 + k_3 + k_4)^{2} = 2(k_1^{2} + k_2^{2} + k_3^{2} + k_4^{2})$, where $k_i$ is the curvature of circle $i$, meaning $k_i = \\dfrac{1}{r}$. When three of the circles are internally tangent to the fourth one, the fourth circle has a negative curvature. Suppose we reflect Circle $E$ over $\\overline{AB}$. Now, we have our four circles to apply that theorem. First, lets scale our image down such that Circle $C$ has radius $1$, for ease of computation. Let the radius of Circle $D$ be $r$, so Circle $E$ has radius $\\dfrac{r}{3}$. Then, we have that $(-1 + \\dfrac{1}{r} + \\dfrac{3}{r} + \\dfrac{3}{r})^{2} = 2(1 + \\dfrac{1}{r^{2}} + \\dfrac{9}{r^{2}} + \\dfrac{9}{r^{2}})$. This simplifies to $\\dfrac{49}{r^{2}} - \\dfrac{14}{r} + 1 = \\dfrac{2r^{2} + 38}{r^{2}}$. Multiplying both sides by $r^{2}$, we get that $49 - 14r + r^{2} = 2r^{2} + 38$, or $r^2 + 14r - 11 = 0$. We get $r = -7 \\pm 2\\sqrt{15}$, but we want the positive solution, which is $r = 2\\sqrt{15} - 7$. We have to rescale back up, so we get $r = 4\\sqrt{15} - 14 = \\sqrt{240} - 14$, so we get that our answer is $240 + 14 = 254. ~Puck_0",
"This solution focuses on the area of $\\triangle DEC$. Because $EF$ is perpendicular to $AB$, it is an altitude of $\\triangle DEC$. Therefore, we can express the area of $\\triangle DEC$ as $\\frac{1}{2}\\cdot EF \\cdot DC$. We can also express the area of $\\triangle DEC$ using Heron's Formula. Let $r$ equal the radius of circle $E$. Then $DC$ = $CA - DA = 2-3r$. We also know that $CE = 2-r$ and $DE = 3r+r=4r$. The semi-perimeter of $\\triangle DEC$ is $(DC+CE+DE)/2 = 2$. Applying Heron's Formula, we get \\[\\sqrt{2(2-4r)(r)(3r)} = \\sqrt{6r^2(2-4r)}.\\] We set this equal to $\\frac{1}{2}r(2-3r)$. \\[\\sqrt{6r^2(2-4r)} = \\frac{1}{2}r(2-3r).\\] This simplifies to the quadratic equation $0 = 9r^2+84r-44$. Remember that we are solving for $3r$, which we will set equal to $x$. Then we now have the equation $0 = x^2 +28r - 44$. Applying the quadratic formula, we get $x=-14 \\pm 4\\sqrt{15}$. We want the positive solution, so we take $4\\sqrt{15}-14 = \\sqrt{240}-14$. Our answer is therefore $240 + 14 = 254. ~lprado",
"This solution focuses on the area of $\\triangle DEC$. Because $EF$ is perpendicular to $AB$, it is an altitude of $\\triangle DEC$. Therefore, we can express the area of $\\triangle DEC$ as $\\frac{1}{2}\\cdot EF \\cdot DC$. We can also express the area of $\\triangle DEC$ using Heron's Formula. Let $r$ equal the radius of circle $E$. Then $DC$ = $CA - DA = 2-3r$. We also know that $CE = 2-r$ and $DE = 3r+r=4r$. The semi-perimeter of $\\triangle DEC$ is $(DC+CE+DE)/2 = 2$. Applying Heron's Formula, we get \\[\\sqrt{2(2-4r)(r)(3r)} = \\sqrt{6r^2(2-4r)}.\\] We set this equal to $\\frac{1}{2}r(2-3r)$. \\[\\sqrt{6r^2(2-4r)} = \\frac{1}{2}r(2-3r).\\] This simplifies to the quadratic equation $0 = 9r^2+84r-44$. Remember that we are solving for $3r$, which we will set equal to $x$. Then we now have the equation $0 = x^2 +28r - 44$. Applying the quadratic formula, we get $x=-14 \\pm 4\\sqrt{15}$. We want the positive solution, so we take $4\\sqrt{15}-14 = \\sqrt{240}-14$. Our answer is therefore $240 + 14 = 254. ~lprado"
] |
2014-II-9
| 2,014
| 9
|
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
| 581
|
II
|
[
"We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \\cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \\cdot 4$ or $40$ have $4$ adjacent, and $10 \\cdot {5\\choose 2}$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \\cdot {4\\choose2}$ or $60$ have $4$ adjacent, $\\frac{10 \\cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \\cdot \\left({5\\choose2} - 3\\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\\choose7}$ or $120$, ${10\\choose8}$ or $45$, ${10\\choose9}$ or $10$, and ${10\\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = 581",
"We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \\cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \\cdot 4$ or $40$ have $4$ adjacent, and $10 \\cdot {5\\choose 2}$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \\cdot {4\\choose2}$ or $60$ have $4$ adjacent, $\\frac{10 \\cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \\cdot \\left({5\\choose2} - 3\\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\\choose7}$ or $120$, ${10\\choose8}$ or $45$, ${10\\choose9}$ or $10$, and ${10\\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = 581",
"Starting with small cases, we see that four chairs give $4 + 1 = 5$, five chairs give $5 + 5 + 1 = 11$, and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$ However, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$, and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = 581.",
"Starting with small cases, we see that four chairs give $4 + 1 = 5$, five chairs give $5 + 5 + 1 = 11$, and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$ However, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$, and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = 581.",
"It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$, where $T_n$ counts the ways to select a subset of chairs $\\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row. Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$, and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$. Now we know the complement is $274 + 81 + 88 = 443$, and subtracting from $2^{10} = 1024$ gives $1024 - 443 = 581.",
"It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$, where $T_n$ counts the ways to select a subset of chairs $\\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row. Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$, and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$. Now we know the complement is $274 + 81 + 88 = 443$, and subtracting from $2^{10} = 1024$ gives $1024 - 443 = 581.",
"Let's calculate the complement. As mentioned in solution $3$, the number of ways to have a subset $n$ chairs in a line with no 3 consectuive chairs satisfies $T_n = T_{n-1} + T_{n-2} + T_{n-3}$. Setting $T_{1} = 2, T_{2} = 4$, and $T_{3} = 7$, we get that $T_{10} = 504$. Since this is in a line and not a circle, we must subtract the cases that would include 3 consecutive chairs if the endpoints of the line were put together. If chairs $1$, $2$ and $10$ are in the subset, that would not work. The same goes for if chairs $1$, $9$ and $10$ were in the subset. If chairs $1$, $2$ and $10$ are in the set then chair $3$ must not be in the set. However, chair $9$ could be in or not in the set because we only want to count what cases where no $3$ chairs are consecutive in the line but there would be consecutive chairs in a circle. If chair $9$ is not included, there are $T_{5} = 24$ ways. If chair $9$ is in the set then there are $T_{4} = 13$ ways. So we must subtract $2 \\cdot (24 + 13) = 74$ However we are counting the case where chairs $1$, $2$, $9$ and $10$ are included twice. So we only have to subtract $74 - 13 = 61$. $504 - 61 = 443$. This entire time we were calculating the complement so $2^{10} - 443 = 581 ~sdfgfjh"
] |
2014-II-10
| 2,014
| 10
|
Let $z$ be a complex number with $|z|=2014$ . Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$ . Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$ , where $n$ is an integer. Find the remainder when $n$ is divided by $1000$ .
| 147
|
II
|
[
"Note that the given equality reduces to \\[\\frac{1}{w+z} = \\frac{w+z}{wz}\\] \\[wz = {(w+z)}^2\\] \\[w^2 + wz + z^2 = 0\\] \\[\\frac{w^3 - z^3}{w-z} = 0\\] \\[w^3 = z^3, w \\neq z\\] Now, let $w = r_w e^{i \\theta_w}$ and likewise for $z$. Consider circle $O$ with the origin as the center and radius 2014 on the complex plane. It is clear that $z$ must be one of the points on this circle, as $|z| = 2014$. By DeMoivre's Theorem, the complex modulus of $w$ is cubed when $w$ is cubed. Thus $w$ must lie on $O$, since its the cube of its modulus, and thus its modulus, must be equal to $z$'s modulus. Again, by DeMoivre's Theorem, $\\theta_w$ is tripled when $w$ is cubed and likewise for $z$. For $w$, $z$, and the origin to lie on the same line, $3 \\theta_w$ must be some multiple of 360 degrees apart from $3 \\theta_z$ , so $\\theta_w$ must differ from $\\theta_z$ by some multiple of 120 degrees. Now, without loss of generality, assume that $z$ is on the real axis. (The circle can be rotated to put $z$ in any other location.) Then there are precisely two possible distinct locations for $w$; one is obtained by going 120 degrees clockwise from $z$ about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points. Let the two possible locations for $w$ be $W_1$ and $W_2$ and the location of $z$ be point $Z$. Note that by symmetry, $W_1W_2Z$ is equilateral, say, with side length $x$. We know that the circumradius of this equilateral triangle is $2014$, so using the formula $\\frac{abc}{4R} = [ABC]$ and that the area of an equilateral triangle with side length $s$ is $\\frac{s^2\\sqrt{3}}{4}$, so we have \\[\\frac{x^3}{4R} = \\frac{x^2\\sqrt{3}}{4}\\] \\[x = R \\sqrt{3}\\] \\[\\frac{x^2\\sqrt{3}}{4} = \\frac{3R^2 \\sqrt{3}}{4}\\] Since we're concerned with the non-radical part of this expression and $R = 2014$, \\[\\frac{3R^2}{4} \\equiv 3 \\cdot 1007^2 \\equiv 3 \\cdot 7^2 \\equiv 147",
"Note that the given equality reduces to \\[\\frac{1}{w+z} = \\frac{w+z}{wz}\\] \\[wz = {(w+z)}^2\\] \\[w^2 + wz + z^2 = 0\\] \\[\\frac{w^3 - z^3}{w-z} = 0\\] \\[w^3 = z^3, w \\neq z\\] Now, let $w = r_w e^{i \\theta_w}$ and likewise for $z$. Consider circle $O$ with the origin as the center and radius 2014 on the complex plane. It is clear that $z$ must be one of the points on this circle, as $|z| = 2014$. By DeMoivre's Theorem, the complex modulus of $w$ is cubed when $w$ is cubed. Thus $w$ must lie on $O$, since its the cube of its modulus, and thus its modulus, must be equal to $z$'s modulus. Again, by DeMoivre's Theorem, $\\theta_w$ is tripled when $w$ is cubed and likewise for $z$. For $w$, $z$, and the origin to lie on the same line, $3 \\theta_w$ must be some multiple of 360 degrees apart from $3 \\theta_z$ , so $\\theta_w$ must differ from $\\theta_z$ by some multiple of 120 degrees. Now, without loss of generality, assume that $z$ is on the real axis. (The circle can be rotated to put $z$ in any other location.) Then there are precisely two possible distinct locations for $w$; one is obtained by going 120 degrees clockwise from $z$ about the circle and the other by moving the same amount counter-clockwise. Moving along the circle with any other multiple of 120 degrees in any direction will result in these three points. Let the two possible locations for $w$ be $W_1$ and $W_2$ and the location of $z$ be point $Z$. Note that by symmetry, $W_1W_2Z$ is equilateral, say, with side length $x$. We know that the circumradius of this equilateral triangle is $2014$, so using the formula $\\frac{abc}{4R} = [ABC]$ and that the area of an equilateral triangle with side length $s$ is $\\frac{s^2\\sqrt{3}}{4}$, so we have \\[\\frac{x^3}{4R} = \\frac{x^2\\sqrt{3}}{4}\\] \\[x = R \\sqrt{3}\\] \\[\\frac{x^2\\sqrt{3}}{4} = \\frac{3R^2 \\sqrt{3}}{4}\\] Since we're concerned with the non-radical part of this expression and $R = 2014$, \\[\\frac{3R^2}{4} \\equiv 3 \\cdot 1007^2 \\equiv 3 \\cdot 7^2 \\equiv 147",
"Assume $z = 2014$. Then \\[\\frac{1}{2014 + w} = \\frac{1}{2014} + \\frac{1}{w}\\] \\[2014w = w(2014 + w) + 2014(2014 + w)\\] \\[2014w = 2014w + w^2 + 2014^2 + 2014w\\] \\[0 = w^2 + 2014w + 2014^2\\] \\[w = \\frac{-2014 \\pm \\sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \\pm 1007\\sqrt{3}i\\] Thus $P$ is an isosceles triangle with area $\\frac{1}{2}(2014 - (-1007))(2\\cdot 1007\\sqrt{3}) = 3021\\cdot 1007\\sqrt{3}$ and $n \\equiv 7\\cdot 21\\equiv 147",
"Assume $z = 2014$. Then \\[\\frac{1}{2014 + w} = \\frac{1}{2014} + \\frac{1}{w}\\] \\[2014w = w(2014 + w) + 2014(2014 + w)\\] \\[2014w = 2014w + w^2 + 2014^2 + 2014w\\] \\[0 = w^2 + 2014w + 2014^2\\] \\[w = \\frac{-2014 \\pm \\sqrt{2014^2 - 4(2014^2)}}{2} = -1007 \\pm 1007\\sqrt{3}i\\] Thus $P$ is an isosceles triangle with area $\\frac{1}{2}(2014 - (-1007))(2\\cdot 1007\\sqrt{3}) = 3021\\cdot 1007\\sqrt{3}$ and $n \\equiv 7\\cdot 21\\equiv 147",
"Notice that \\[\\frac1{w+z} = \\frac{w+z}{wz} \\implies 0 = w^2 + wz + z^2 = \\frac{w^3-z^3}{w-z}.\\] Hence, $w=ze^{2\\pi i/3},ze^{4\\pi i/3}$, and $P$ is an equilateral triangle with circumradius $2014$. Then, \\[[P]=\\frac{3}{2}\\cdot 2014^2\\cdot\\sin\\frac{\\pi}3=3\\cdot 1007^2\\sqrt3,\\] and the answer is $3\\cdot 1007^2\\equiv 3\\cdot 7^2\\equiv147.",
"Notice that \\[\\frac1{w+z} = \\frac{w+z}{wz} \\implies 0 = w^2 + wz + z^2 = \\frac{w^3-z^3}{w-z}.\\] Hence, $w=ze^{2\\pi i/3},ze^{4\\pi i/3}$, and $P$ is an equilateral triangle with circumradius $2014$. Then, \\[[P]=\\frac{3}{2}\\cdot 2014^2\\cdot\\sin\\frac{\\pi}3=3\\cdot 1007^2\\sqrt3,\\] and the answer is $3\\cdot 1007^2\\equiv 3\\cdot 7^2\\equiv147.",
"I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: \\[\\frac1{z+w}=\\frac1z+\\frac1w\\] \\[\\frac w{z+w}=\\frac wz+1\\] \\[\\frac w{z+w}=\\frac{w+z}z\\] \\[\\frac{w-0}{w-(-z)}=\\frac{(-z)-w}{(-z)-0}\\] We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with $W$ ($w$), $V$ ($-z$), and $O$ ($0$), we can interpret the above equation to mean that the $\\angle OWV=\\angle OVW$, and hence triangle $OWV$ is isosceles, so length $OW$ = $OV$. Rearranging the equation \\[\\frac{w-0}{w-(-z)}=\\frac{(-z)-w}{(-z)-0},\\] we find that \\[(w-0)((-z)-0)=-(w-(-z))^2.\\] Taking the magnitude of both sides and using the fact that $OW=OV\\implies |w-0|=|(-z)-0|$, we find that \\[|w-0|^2=|w-(-z)|^2,\\] so length $OW=VW$ and triangle $OWV$ is equilateral. There are only two possible $W$'s for which $OWV$ is equilateral, lying on either side of $OV$. After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that $z$ and the two $w$'s form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is $2014\\sqrt3$ and hence its area is $\\frac{\\sqrt3(2014\\sqrt3)^2}4=147. ~SymbolicPermutation",
"I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: \\[\\frac1{z+w}=\\frac1z+\\frac1w\\] \\[\\frac w{z+w}=\\frac wz+1\\] \\[\\frac w{z+w}=\\frac{w+z}z\\] \\[\\frac{w-0}{w-(-z)}=\\frac{(-z)-w}{(-z)-0}\\] We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with $W$ ($w$), $V$ ($-z$), and $O$ ($0$), we can interpret the above equation to mean that the $\\angle OWV=\\angle OVW$, and hence triangle $OWV$ is isosceles, so length $OW$ = $OV$. Rearranging the equation \\[\\frac{w-0}{w-(-z)}=\\frac{(-z)-w}{(-z)-0},\\] we find that \\[(w-0)((-z)-0)=-(w-(-z))^2.\\] Taking the magnitude of both sides and using the fact that $OW=OV\\implies |w-0|=|(-z)-0|$, we find that \\[|w-0|^2=|w-(-z)|^2,\\] so length $OW=VW$ and triangle $OWV$ is equilateral. There are only two possible $W$'s for which $OWV$ is equilateral, lying on either side of $OV$. After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that $z$ and the two $w$'s form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is $2014\\sqrt3$ and hence its area is $\\frac{\\sqrt3(2014\\sqrt3)^2}4=147. ~SymbolicPermutation"
] |
2014-II-11
| 2,014
| 11
|
In $\triangle RED$ , $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$ . $RD=1$ . Let $M$ be the midpoint of segment $\overline{RD}$ . Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$ . Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$ . Then $AE=\frac{a-\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$ .
| 56
|
II
|
[
"Let $P$ be the foot of the perpendicular from $A$ to $\\overline{CR}$, so $\\overline{AP}\\parallel\\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\\overline{CR}$, and $\\overline{PM}\\parallel\\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \\frac{CD}{2}$. We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \\frac{1}{2}$ and $EO = RO = \\frac{\\sqrt{3}}{2}$, so $D\\left(\\frac{1}{2}, 0\\right)$, $E\\left(-\\frac{\\sqrt{3}}{2}, 0\\right)$, and $R\\left(0, \\frac{\\sqrt{3}}{2}\\right).$ $M =$ midpoint$(D, R) = \\left(\\frac{1}{4}, \\frac{\\sqrt{3}}{4}\\right)$ and the slope of $ME = \\frac{\\frac{\\sqrt{3}}{4}}{\\frac{1}{4} + \\frac{\\sqrt{3}}{2}} = \\frac{\\sqrt{3}}{1 + 2\\sqrt{3}}$, so the slope of $RC = -\\frac{1 + 2\\sqrt{3}}{\\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\\frac{x(1 + 2\\sqrt{3})}{\\sqrt{3}} = \\frac{\\sqrt{3}}{2}$ up. Thus, $x = \\frac{\\frac{3}{2}}{1 + 2\\sqrt{3}} = \\frac{3}{4\\sqrt{3} + 2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $056\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]",
"Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED - x$ Using Law of Sine in $\\triangle RED$, we find $RE$ = $\\frac{\\sqrt{6}}{2}$. We got the three sides of $\\triangle AER$. Now using the Law of Cosines on $\\angle AER$. There we can equate $x$ and solve for it. We got $AE=x=\\frac{\\sqrt{3}-1}{4\\sqrt{3}+2}$. Then rationalize the denominator, we get $AE = \\frac{7 - \\sqrt{27}}{22}$.",
"Let $P$ be the foot of the perpendicular from $A$ to $\\overline{CR}$, so $\\overline{AP}\\parallel\\overline{EM}$. Since $\\triangle ARC$ is isosceles, $P$ is the midpoint of $\\overline{CR}$, and by midpoint theorem $\\overline{PM}\\parallel\\overline{CD}$. Thus, $APME$ is a parallelogram and therefore $AE = PM = \\tfrac 12 CD$. [asy] unitsize(8cm); pair a, d, r, e, m, cm, c,p; d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m); draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed); pair[] PPAP = {a, d, r, e, m, c, p}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"$A$\", a, E); label(\"$E$\", e, S); label(\"$C$\", c, S); label(\"$D$\", d, SW); label(\"$M$\", m, NW); label(\"$R$\", r, N); label(\"$P$\", p, NW); MA(\"60^\\circ\",black,c,d,m,0.07, black); [/asy] We can now use coordinates with $D(0,0)$ as origin and $DE$ along the $x$-axis. Let $RD=4$ instead of $1$ (in the end we will scale down by $4$). Since $\\angle D = 60^\\circ$, we get $R(2,2\\sqrt{3})$, and therefore $M(1, \\sqrt{3})$. We use sine-law in $\\triangle RED$ to find the coordinates $E(2+2\\sqrt{3}, 0)$:\\[DE =4\\cdot \\frac{\\sin 75^\\circ}{\\sin 45^\\circ} = 4(\\sin 30^\\circ + \\cos 30^\\circ) = 2+2\\sqrt{3}.\\] Since slope$(ME)= -\\sqrt{3}/(1+2\\sqrt{3})$, and $RC\\perp ME$, it follows that slope$(RC)=(1+2\\sqrt{3})/\\sqrt{3}$. If $C(c,0)$ then we have\\[\\frac{2\\sqrt{3}}{2-c}=\\frac{1+2\\sqrt{3}}{\\sqrt{3}}\\qquad \\Longrightarrow\\qquad c=\\frac{4\\sqrt{3}-4}{1+2\\sqrt{3}}\\] Now $\\tfrac 12 CD = \\tfrac 12c =(2\\sqrt{3}-2)/(1+2\\sqrt{3})= \\tfrac 1{11}(14-6\\sqrt{3})$. Scaling down by $4$, we get $AE=\\tfrac 1{22}(7-3\\sqrt{3})$, so our answer is $7+27+22=056$.",
"Define the foot of the altitude from $R$ to $ED$ as $X.$ Suppose the median $EM$ is equal to $m,$ and the angle $\\angle{MED}$ as $\\theta.$ Using the Law of Cosines in triangle $EDM,$ we get $m^2=1+\\frac{\\sqrt{3}}{4}.$ By Law of Sines in $EMD,$ $\\frac{m}{\\sin{60^{\\circ}}} = \\frac{\\frac{1}{2}}{\\sin\\theta}.$ This gives $\\sin\\theta = \\frac{\\sqrt{3}}{4m},$ and solving for $\\cos\\theta$ gives $\\cos\\theta = \\sqrt{1-\\frac{3}{16m^2}} = \\frac{\\sqrt{16m^2-3}}{4m}.$ Using the established value of $m^2,$ we have $\\cos{\\theta} = \\frac{\\sqrt{13+4\\sqrt{3}}}{4m} = \\frac{1+2\\sqrt{3}}{4m}.$ Thus, we have $\\tan\\theta = \\frac{\\sqrt{3}}{1+2\\sqrt{3}}.$ Noting that $\\angle{XRC}=\\theta$ as well, this gives $CX = RX \\tan \\theta = \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{1+2\\sqrt{3}} = \\frac{3}{2+4\\sqrt{3}}.$ Suppose $AE=n.$ Now, using Pythagorean in triangle $AXR$ gives \\[\\left(n + \\frac{\\sqrt{3}}{2}\\right)^2 + \\frac{3}{4} = \\left(n + \\frac{\\sqrt{3}}{2} + \\frac{3}{2+4\\sqrt{3}}\\right)^2.\\] Solving, we have \\[\\frac{3}{4} = \\left(\\frac{3}{2+4\\sqrt{3}}\\right) \\left(2n + \\sqrt{3} + \\frac{3}{2+4\\sqrt{3}}\\right),\\] \\[2n + \\sqrt{3} + \\frac{3}{2+4\\sqrt{3}} = \\frac{1+2\\sqrt{3}}{2},\\] \\[2n + \\frac{3}{2+4\\sqrt{3}} = \\frac{1}{2}.\\] Isolating $2n$ and rationalizing the denominator returns the final answer of $n=\\frac{7-\\sqrt{27}}{22},$ thus giving an answer of $056 ~anduran",
"Define the foot of the altitude from $R$ to $ED$ as $X.$ Suppose the median $EM$ is equal to $m,$ and the angle $\\angle{MED}$ as $\\theta.$ Using the Law of Cosines in triangle $EDM,$ we get $m^2=1+\\frac{\\sqrt{3}}{4}.$ By Law of Sines in $EMD,$ $\\frac{m}{\\sin{60^{\\circ}}} = \\frac{\\frac{1}{2}}{\\sin\\theta}.$ This gives $\\sin\\theta = \\frac{\\sqrt{3}}{4m},$ and solving for $\\cos\\theta$ gives $\\cos\\theta = \\sqrt{1-\\frac{3}{16m^2}} = \\frac{\\sqrt{16m^2-3}}{4m}.$ Using the established value of $m^2,$ we have $\\cos{\\theta} = \\frac{\\sqrt{13+4\\sqrt{3}}}{4m} = \\frac{1+2\\sqrt{3}}{4m}.$ Thus, we have $\\tan\\theta = \\frac{\\sqrt{3}}{1+2\\sqrt{3}}.$ Noting that $\\angle{XRC}=\\theta$ as well, this gives $CX = RX \\tan \\theta = \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{1+2\\sqrt{3}} = \\frac{3}{2+4\\sqrt{3}}.$ Suppose $AE=n.$ Now, using Pythagorean in triangle $AXR$ gives \\[\\left(n + \\frac{\\sqrt{3}}{2}\\right)^2 + \\frac{3}{4} = \\left(n + \\frac{\\sqrt{3}}{2} + \\frac{3}{2+4\\sqrt{3}}\\right)^2.\\] Solving, we have \\[\\frac{3}{4} = \\left(\\frac{3}{2+4\\sqrt{3}}\\right) \\left(2n + \\sqrt{3} + \\frac{3}{2+4\\sqrt{3}}\\right),\\] \\[2n + \\sqrt{3} + \\frac{3}{2+4\\sqrt{3}} = \\frac{1+2\\sqrt{3}}{2},\\] \\[2n + \\frac{3}{2+4\\sqrt{3}} = \\frac{1}{2}.\\] Isolating $2n$ and rationalizing the denominator returns the final answer of $n=\\frac{7-\\sqrt{27}}{22},$ thus giving an answer of $056 ~anduran"
] |
2014-II-12
| 2,014
| 12
|
Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1$ . Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}$ . Find $m$ .
| 399
|
II
|
[
"Note that $\\cos{3C}=-\\cos{(3A+3B)}$. Thus, our expression is of the form $\\cos{3A}+\\cos{3B}-\\cos{(3A+3B)}=1$. Let $\\cos{3A}=x$ and $\\cos{3B}=y$. Using the fact that $\\cos(3A+3B)=\\cos 3A\\cos 3B-\\sin 3A\\sin 3B=xy-\\sqrt{1-x^2}\\sqrt{1-y^2}$, we get $x+y-xy+\\sqrt{1-x^2}\\sqrt{1-y^2}=1$, or $\\sqrt{1-x^2}\\sqrt{1-y^2}=xy-x-y+1=(x-1)(y-1)$. Squaring both sides, we get $(1-x^2)(1-y^2) = [(x-1)(y-1)]^2$. Cancelling factors, $(1+x)(1+y) = (1-x)(1-y)$. Notice here that we cancelled out one factor of (x-1) and (y-1), which implies that (x-1) and (y-1) were not 0. If indeed they were 0 though, we would have $cos(3A)-1=0, cos(3A)=1$ For this we could say that A must be 120 degrees for this to work. This is one case. The B case follows in the same way, where B must be equal to 120 degrees. This doesn't change the overall solution though, as then the other angles are irrelevant (this is the largest angle, implying that this will have the longest side and so we would want to have the 120 degreee angle opposite of the unknown side). Expanding, $1+x+y+xy=1-x-y+xy\\rightarrow x+y=-x-y$. Simplification leads to $x+y=0$. Therefore, $\\cos(3C)=1$. So $\\angle C$ could be $0^\\circ$ or $120^\\circ$. We eliminate $0^\\circ$ and use law of cosines to get our answer: \\[m=10^2+13^2-2\\cdot 10\\cdot 13\\cos\\angle C\\] \\[\\rightarrow m=269-260\\cos 120^\\circ=269-260\\left(\\text{-}\\frac{1}{2}\\right)\\] \\[\\rightarrow m=269+130=399\\] $399",
"WLOG, let C be the largest angle in the triangle. As above, we can see that $\\cos3A+\\cos3B-\\cos(3A+3B)=1$ Expanding, we get $\\cos3A+\\cos3B-\\cos3A\\cos3B+\\sin3A\\sin3B=1$ $\\cos3A\\cos3B-\\cos3A-\\cos3B+1=\\sin3A\\sin3B$ $(\\cos3A-1)(\\cos3B-1)=\\sin3A\\sin3B$ CASE 1: If $\\sin 3A = 0$ or $\\sin 3B = 0$, This implies one or both of A or B are 60 or 120. If one of A or B is 120, we have a contradiction, since C must be the largest angle. Otherwise, if one of A or B is 60, WLOG, assume A = 60, we would have $\\cos(3B) + \\cos(3C) = 2$, and thus, cos(3B) and cos(3C) both equal 1, implying $B = C = 120$, a contradiction to the fact that the sum of the angles of a triangle must be 180 degrees. CASE 2: If $\\sin 3A \\neq 0$ and $\\sin 3B \\neq 0$ $\\frac{\\cos3A-1}{\\sin3A}\\cdot\\frac{\\cos3B-1}{\\sin3B}=1$ $\\tan{\\frac{3A}{2}}\\tan{\\frac{3B}{2}}=1$ Note that $\\tan{x}=\\frac{1}{\\tan(90-x)}$, or $\\tan{x}\\tan(90-x)=1$ Thus $\\frac{3A}{2}+\\frac{3B}{2}=90$, or $A+B=60$. Now we know that $C=120$, so we can just use the Law of Cosines to get $399 -Alexlikemath",
"\\[\\cos3A+\\cos3B=1-\\cos(3C)=1+\\cos(3A+3B)\\] \\[2\\cos\\frac{3}{2}(A+B)\\cos\\frac{3}{2}(A-B)=2\\cos^2\\frac{3}{2}(A+B)\\] If $\\cos\\frac{3}{2}(A+B) = 0$, then $\\frac{3}{2}(A+B)=90$, $A+B=60$, so $C=120$; otherwise, \\[2\\cos\\frac{3}{2}(A-B)=2 \\cos\\frac{3}{2}(A+B)\\] \\[\\sin\\frac{3}{2}A\\sin\\frac{3}{2}B=0\\] so either $\\sin\\frac{3}{2}A=0$ or $\\sin\\frac{3}{2}B=0$, i.e., either $A=120$ or $B=120$. In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. $399 -Mathdummy",
"Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \\sqrt{269-260\\cos{A}}$. We notice that $\\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$, so $269-260\\cos{A}$ should be an integer. We can then assume that $A$ is $120$ degrees so $\\cos{A} = \\frac{-1}{2}$. We do this because $120$ degrees is a \"common\" value and it makes the value inside the square root an integer. Plugging this into $269-260\\cos{A}$ for $m$ we get that it is $399. -srisainandan6",
"Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \\sqrt{269-260\\cos{A}}$. We notice that $\\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$, so $269-260\\cos{A}$ should be an integer. We can then assume that $A$ is $120$ degrees so $\\cos{A} = \\frac{-1}{2}$. We do this because $120$ degrees is a \"common\" value and it makes the value inside the square root an integer. Plugging this into $269-260\\cos{A}$ for $m$ we get that it is $399. -srisainandan6"
] |
2014-II-13
| 2,014
| 13
|
Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$ , no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 28
|
II
|
[
"Label the left shoes be $L_1,\\dots, L_{10}$ and the right shoes $R_1,\\dots, R_{10}$. Notice that there are $10!$ possible pairings. Let a pairing be \"bad\" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad. Note that, in order to have a bad pairing, there must exist a collection of $k<5$ pairs that includes both the left and the right shoes of $k$ adults; in other words, it is bad if it is possible to pick $k$ pairs and properly redistribute all of its shoes to exactly $k$ people. Thus, if a left shoe is a part of a bad collection, its corresponding right shoe must also be in the bad collection (and vice versa). To search for bad collections, we can start at an arbitrary right shoe (say $R_1$), check the left shoe it is paired with (say $L_i$), and from the previous observation, we know that $R_i$ must also be in the bad collection. Then we may check the left shoe paired with $R_i$, find its counterpart, check its left pair, find its counterpart, etc. until we have found $L_1$. We can imagine each right shoe \"sending\" us to another right shoe (via its paired left shoe) until we reach the starting right shoe, at which point we know that we have found a bad collection if we have done this less than $5$ times. Effectively we have just traversed a cycle. (Note: This is the cycle notation of permutations.) The only condition for a bad pairing is that there is a cycle with length less than $5$; thus, we need to count pairings where every cycle has length at least $5$. This is only possible if there is a single cycle of length $10$ or two cycles of length $5$. The first case yields $9!$ working pairings. The second case yields $\\frac{{10\\choose 5}}{2}\\cdot{4!}^2=\\frac{10!}{2 \\cdot {5!}^2} \\cdot {4!}^2$ pairings. Therefore, taking these cases out of a total of $10!$, the probability is $\\frac{1}{10}+\\frac{1}{50} = \\frac{3}{25}$, for an answer of $028."
] |
2014-II-14
| 2,014
| 14
|
In $\triangle ABC$ , $AB=10$ , $\measuredangle A=30^{\circ}$ , and $\measuredangle C=45^{\circ}$ . Let $H$ , $D$ , and $M$ be points on line $\overline{BC}$ such that $AH\perp BC$ , $\measuredangle BAD=\measuredangle CAD$ , and $BM=CM$ . Point $N$ is the midpoint of segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp BC$ . Then $AP^2=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 77
|
II
|
[
"Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$. We will use this point later in the problem. As we can see, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$ $AHC$ is a $45-45-90$ triangle, so $\\angle{HAB}=15^\\circ$. $AHD$ is $30-60-90$ triangle. $AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ triangle also. Then if we use those informations we get $AD=2HD$ and $PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$. Now we know that $HM=AP$, we can find for $HM$ which is simpler to find. We can use point $B$ to split it up as $HM=HB+BM$, We can chase those lengths and we would get $AB=10$, so $OB=5$, so $BC=5\\sqrt{2}$, so $BM=\\dfrac{1}{2} \\cdot BC=\\dfrac{5\\sqrt{2}}{2}$ We can also use Law of Sines: \\[\\frac{BC}{AB}=\\frac{\\sin\\angle A}{\\sin\\angle C}\\] \\[\\frac{BC}{10}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{2}}{2}}\\implies BC=5\\sqrt{2}\\] Then using right triangle $AHB$, we have $HB=10 \\sin 15^\\circ$ So $HB=10 \\sin 15^\\circ=\\dfrac{5(\\sqrt{6}-\\sqrt{2})}{2}$. And we know that $AP = HM = HB + BM = \\frac{5(\\sqrt6-\\sqrt2)}{2} + \\frac{5\\sqrt2}{2} = \\frac{5\\sqrt6}{2}$. Finally if we calculate $(AP)^2$. $(AP)^2=\\dfrac{150}{4}=\\dfrac{75}{2}$. So our final answer is $75+2=77$. $m+n=077 -Gamjawon -edited by srisainandan6 to clarify and correct a small mistake",
"Here's a solution that doesn't need $\\sin 15^\\circ$. As above, get to $AP=HM$. As in the figure, let $O$ be the foot of the perpendicular from $B$ to $AC$. Then $BCO$ is a 45-45-90 triangle, and $ABO$ is a 30-60-90 triangle. So $BO=5$ and $AO=5\\sqrt{3}$; also, $CO=5$, $BC=5\\sqrt2$, and $MC=\\dfrac{BC}{2}=5\\dfrac{\\sqrt2}{2}$. But $MO$ and $AH$ are parallel, both being orthogonal to $BC$. Therefore $MH:AO=MC:CO$, or $MH=\\dfrac{5\\sqrt3}{\\sqrt2}$, and we're done.",
"Break our diagram into 2 special right triangle by dropping an altitude from $B$ to $AC$ we then get that \\[AC=5+5\\sqrt{3}, BC=5\\sqrt{2}.\\] Since $\\triangle{HCA}$ is a 45-45-90, \\[HC=\\frac{5\\sqrt2+5\\sqrt6}{2}\\] $MC=\\frac{BC}{2},$ \\[HM=\\frac{5\\sqrt6}{2}\\] \\[HN=\\frac{5\\sqrt6}{4}\\] We know that $\\triangle{AHD}\\simeq \\triangle{PND}$ and are 30-60-90. Thus, \\[AP=2 \\cdot HN=\\frac{5\\sqrt6}{2}.\\] $(AP)^2=\\dfrac{150}{4}=\\dfrac{75}{2}$. So our final answer is $75+2=077.",
"[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr); draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr); draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr); draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq); draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq); draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq); draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq); draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq); /* dots and labels */ dot((-1.4934334172297545,2.6953043701763835),dotstyle); label(\"$A$\", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor); dot((1.1286284157632023,-6.954814372303504),dotstyle); label(\"$B$\", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor); dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle); label(\"$C$\", (8.292420110475998,-6.741880204396438), NE * labelscalefactor); dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle); label(\"$H$\", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor); dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle); label(\"$E$\", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor); dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle); label(\"$L$\", (4.705181710331174,6.441792132441429), NE * labelscalefactor); dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle); label(\"$O$\", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor); dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle); label(\"$D$\", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor); dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle); label(\"$F$\", (4.750879014791618,-7.701523598065745), NE * labelscalefactor); dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle); label(\"$G$\", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor); dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle); label(\"$K$\", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor); dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle); label(\"$P$\", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] Draw the $45-45-90 \\triangle AHC$. Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\\triangle ABC$. It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$. Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$. Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \\parallel HG$. From here, it is clear that $AK = KG$; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$! After this point, note that $AP = PF$. It is easily derived that the circumradius of $\\triangle ABC$ is $\\frac{10}{\\sqrt{2}}$. Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $077. ~awang11's sol",
"Let $BO \\perp AC, O \\in AC.$ Let $ME \\perp BC, E \\in AD.$ $MB = MC, \\angle C = 45^\\circ \\implies$ points $M, E, O$ are collinear. $HN = NM, AH||NP||ME \\implies AP = PE.$ In $\\triangle ABO \\hspace{10mm} \\angle A = 30^\\circ \\implies AO = AB \\cos 60^\\circ = 5 \\sqrt{3}.$ In $\\triangle AEO \\hspace{10mm} \\angle A = 15^\\circ, \\angle O = 90^\\circ + 45^\\circ = 135^\\circ \\implies$ \\[\\angle AEO = 30^\\circ \\implies\\] \\[AE = AO \\frac {\\sin 135^\\circ}{\\sin 30^\\circ} = 5 \\sqrt{3} \\cdot \\sqrt{2} = 5 \\sqrt{6} \\implies\\] \\[AP = 5 \\sqrt {\\frac {3}{2}} \\implies AP^2 = \\frac {75}{2} \\implies \\textbf{077}.\\] [email protected], vvsss"
] |
2014-II-15
| 2,014
| 15
|
For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k$ . Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2$ . Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0$ . Find the smallest positive integer $t$ such that $x_t=2090$ .
| 149
|
II
|
[
"Note that for any $x_i$, for any prime $p$, $p^2 \\nmid x_i$. This provides motivation to translate $x_i$ into a binary sequence $y_i$. Let the prime factorization of $x_i$ be written as $p_{a_1} \\cdot p_{a_2} \\cdot p_{a_3} \\cdots$, where $p_i$ is the $i$th prime number. Then, for every $p_{a_k}$ in the prime factorization of $x_i$, place a $1$ in the $a_k$th digit of $y_i$. This will result in the conversion $x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \\cdots$. Multiplication for the sequence $x_i$ will translate to addition for the sequence $y_i$. Thus, we see that $x_{n+1}X(x_n) = x_np(x_n)$ translates into $y_{n+1} = y_n+1$. Since $x_0=1$, and $y_0=0$, $x_i$ corresponds to $y_i$, which is $i$ in binary. Since $x_{10010101_2} = 2 \\cdot 5 \\cdot 11 \\cdot 19 = 2090$, $t = 10010101_2$ = $149.",
"We go through the terms and look for a pattern. We find that $x_0 = 1$ $x_8 = 7$ $x_1 = 2$ $x_9 = 14$ $x_2 = 3$ $x_{10} = 21$ $x_3 = 6$ $x_{11} = 42$ $x_4 = 5$ $x_{12} = 35$ $x_5 = 10$ $x_{13} = 70$ $x_6 = 15$ $x_{14} = 105$ $x_7 = 30$ $x_{15} = 210$ Commit to the bash. Eventually, you will receive that $x_{149} = 2090$, so $149",
"We go through the terms and look for a pattern. We find that $x_0 = 1$ $x_8 = 7$ $x_1 = 2$ $x_9 = 14$ $x_2 = 3$ $x_{10} = 21$ $x_3 = 6$ $x_{11} = 42$ $x_4 = 5$ $x_{12} = 35$ $x_5 = 10$ $x_{13} = 70$ $x_6 = 15$ $x_{14} = 105$ $x_7 = 30$ $x_{15} = 210$ Commit to the bash. Eventually, you will receive that $x_{149} = 2090$, so $149",
"Let $P_k$ denote the $k$th prime. Lemma: $x_{n+2^{k-1}} = P_k \\cdot x_{n}$ for all $0 \\leq n \\leq 2^{k-1} - 1.$ $\\mathbf{\\mathrm{Proof:}}$ We can prove this using induction. Assume that $x_{2^{k-1}-1} = \\prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \\frac{x_np(x_n)}{X(x_n)}$, we would “start fresh” for $x_{2^{k-1}} = P_k.$ It is then easy to see that then $\\frac{x_n}{P_k}$ just cycles through the previous $x_{2^{k-1}}$ terms of $\\{ x_n \\},$ since the recursion process is the same and $P_k$ being a factor of $x_n$ is not affected until $n = 2 \\cdot {2^{k-1}} = 2^k,$ when given our assumption $x_{2^k - 1} = \\prod_{j=1}^{k} P_j$ and $n = 2^k$ is now the least $n$ such that \\[P_{k+1} = p(x_{2^{n-1}}),\\] in which $P_a = p(x_n)$ where $a > k$ is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, $x_{2^k} = P_{k+1},$ and the values of $x_n$ just starts cycling through the previous $x_{2^k}$ terms again until $x_{2^{k+1}}$ when a new prime shows up in the prime factorization of $x_n,$ when it starts cycling again, and so on. Using our base cases of $x_0$ and $x_1,$ our induction is complete. Now, it is easy to see that $2090 = 2 \\cdot 5 \\cdot 11 \\cdot 19 = P_1 \\cdot P_3 \\cdot P_5 \\cdot P_8,$ and by Lemma #1, the least positive integer $n$ such that $19 | x_n$ is $2^7.$ By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just $2^7 + 2^4 + 2^2 + 2^0,$ or $10010101_2 = 149 -fidgetboss_4000",
"Let $P_k$ denote the $k$th prime. Lemma: $x_{n+2^{k-1}} = P_k \\cdot x_{n}$ for all $0 \\leq n \\leq 2^{k-1} - 1.$ $\\mathbf{\\mathrm{Proof:}}$ We can prove this using induction. Assume that $x_{2^{k-1}-1} = \\prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \\frac{x_np(x_n)}{X(x_n)}$, we would “start fresh” for $x_{2^{k-1}} = P_k.$ It is then easy to see that then $\\frac{x_n}{P_k}$ just cycles through the previous $x_{2^{k-1}}$ terms of $\\{ x_n \\},$ since the recursion process is the same and $P_k$ being a factor of $x_n$ is not affected until $n = 2 \\cdot {2^{k-1}} = 2^k,$ when given our assumption $x_{2^k - 1} = \\prod_{j=1}^{k} P_j$ and $n = 2^k$ is now the least $n$ such that \\[P_{k+1} = p(x_{2^{n-1}}),\\] in which $P_a = p(x_n)$ where $a > k$ is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, $x_{2^k} = P_{k+1},$ and the values of $x_n$ just starts cycling through the previous $x_{2^k}$ terms again until $x_{2^{k+1}}$ when a new prime shows up in the prime factorization of $x_n,$ when it starts cycling again, and so on. Using our base cases of $x_0$ and $x_1,$ our induction is complete. Now, it is easy to see that $2090 = 2 \\cdot 5 \\cdot 11 \\cdot 19 = P_1 \\cdot P_3 \\cdot P_5 \\cdot P_8,$ and by Lemma #1, the least positive integer $n$ such that $19 | x_n$ is $2^7.$ By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just $2^7 + 2^4 + 2^2 + 2^0,$ or $10010101_2 = 149 -fidgetboss_4000"
] |
2015-I-1
| 2,015
| 1
|
The expressions $A$ = $1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$ and $B$ = $1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers $A$ and $B$ .
| 722
|
I
|
[
"We have \\[|A-B|=|1+3(4-2)+5(6-4)+ \\cdots + 37(38-36)-39(1-38)|\\]\\[\\implies |2(1+3+5+7+ \\cdots +37)-1-39(37)|\\]\\[\\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\\implies 722\\]",
"We see that $A=(1\\times 2)+(3\\times 4)+(5\\times 6)+\\cdots +(35\\times 36)+(37\\times 38)+39$ and $B=1+(2\\times 3)+(4\\times 5)+(6\\times 7)+\\cdots +(36\\times 37)+(38\\times 39)$. Therefore, $B-A=-38+(2\\times 2)+(2\\times 4)+(2\\times 6)+\\cdots +(2\\times 36)+(2\\times 38)$ $=-38+4\\times (1+2+3+\\cdots+19)$ $=-38+4\\times\\frac{20\\cdot 19}{2}=-38+760=722",
"For those that aren't shrewd enough to recognize the above, we may use Newton's Little Formula to semi-bash the equations. We write down the pairs of numbers after multiplication and solve each layer: \\[2, 12, 30, 56, 90...(39)\\] \\[10, 18, 26, 34...\\] \\[8, 8, 8...\\] and \\[(1) 6, 20, 42, 72...\\] \\[14, 22, 30...\\] \\[8, 8, 8...\\] Then we use Newton's Little Formula for the sum of $n$ terms in a sequence. Notice that there are $19$ terms in each sequence, plus the tails of $39$ and $1$ on the first and second equations, respectively. So, \\[2\\binom{19}{1}+10\\binom{19}{2}+8\\binom{19}{3}+1\\] \\[6\\binom{19}{1}+14\\binom{19}{2}+8\\binom{19}{3}+39\\] Subtracting $A$ from $B$ gives: \\[4\\binom{19}{1}+4\\binom{19}{2}-38\\] Which unsurprisingly gives us $722 -jackshi2006",
"For those that aren't shrewd enough to recognize the above, we may use Newton's Little Formula to semi-bash the equations. We write down the pairs of numbers after multiplication and solve each layer: \\[2, 12, 30, 56, 90...(39)\\] \\[10, 18, 26, 34...\\] \\[8, 8, 8...\\] and \\[(1) 6, 20, 42, 72...\\] \\[14, 22, 30...\\] \\[8, 8, 8...\\] Then we use Newton's Little Formula for the sum of $n$ terms in a sequence. Notice that there are $19$ terms in each sequence, plus the tails of $39$ and $1$ on the first and second equations, respectively. So, \\[2\\binom{19}{1}+10\\binom{19}{2}+8\\binom{19}{3}+1\\] \\[6\\binom{19}{1}+14\\binom{19}{2}+8\\binom{19}{3}+39\\] Subtracting $A$ from $B$ gives: \\[4\\binom{19}{1}+4\\binom{19}{2}-38\\] Which unsurprisingly gives us $722 -jackshi2006"
] |
2015-I-2
| 2,015
| 2
|
The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 139
|
I
|
[
"One of the best ways to solve this problem is to use PIE, or the Principle of Inclusion and Exclusion. To start off, we know that the denominator, or the total ways to pick $3$ officials, is $\\binom{9}{3} = 84$. Now, we find the number of ways that at least $2$ officials are from the same country and then subtract the number of ways all $3$ officials are from the same country. To start with at least $2$ officials, we know: There are $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ \"extra\" delegates to choose to be the third sleeper once both from Mexico are sleeping. There are $3\\times7=21$ ways to pick from Canada, as we choose $2$ of the $3$ Canadians ($\\binom{3}{2} = 3$) and then there are $7$ other options for the third sleeper. Lastly, there are $6\\times7=42$ ways to choose for the United States. We can choose two American officials with $\\binom{4}{2} = 6$. Then, there are $7$ options for the third sleeper. Now, we want to find the number of ways to have three sleepers from the same country. There are no ways for the $3$ sleepers to be from Mexico because there are only $2$ Mexican officials. Hence, we get $0$ ways. There is only $1$ way to pick all $3$ from Canada because there are exactly $3$ Canadian officials. We now consider the number of times we originally counted this, which after inspection, is $3$, so we have $1 * 3 = 3$. Lastly, there are $4$ ways to choose all $3$ officials from the United States ($\\binom{4}{3} = 4$). Once again, we counted this $3$ times, so we have $4*3 = 12$. Thus, the fraction is $\\frac{7+21+42-0-3-12}{84} = \\frac{55}{84}$, and our answer is $55+84=139 ways. Solution by: armang32324",
"One of the best ways to solve this problem is to use PIE, or the Principle of Inclusion and Exclusion. To start off, we know that the denominator, or the total ways to pick $3$ officials, is $\\binom{9}{3} = 84$. Now, we find the number of ways that at least $2$ officials are from the same country and then subtract the number of ways all $3$ officials are from the same country. To start with at least $2$ officials, we know: There are $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ \"extra\" delegates to choose to be the third sleeper once both from Mexico are sleeping. There are $3\\times7=21$ ways to pick from Canada, as we choose $2$ of the $3$ Canadians ($\\binom{3}{2} = 3$) and then there are $7$ other options for the third sleeper. Lastly, there are $6\\times7=42$ ways to choose for the United States. We can choose two American officials with $\\binom{4}{2} = 6$. Then, there are $7$ options for the third sleeper. Now, we want to find the number of ways to have three sleepers from the same country. There are no ways for the $3$ sleepers to be from Mexico because there are only $2$ Mexican officials. Hence, we get $0$ ways. There is only $1$ way to pick all $3$ from Canada because there are exactly $3$ Canadian officials. We now consider the number of times we originally counted this, which after inspection, is $3$, so we have $1 * 3 = 3$. Lastly, there are $4$ ways to choose all $3$ officials from the United States ($\\binom{4}{3} = 4$). Once again, we counted this $3$ times, so we have $4*3 = 12$. Thus, the fraction is $\\frac{7+21+42-0-3-12}{84} = \\frac{55}{84}$, and our answer is $55+84=139 ways. Solution by: armang32324",
"The total number of ways to pick $3$ officials from $9$ total is $\\binom{9}{3} = 84$, which will be our denominator. Now we can consider the number of ways for exactly two sleepers to be from the same country for each country individually and add them to find our numerator: There are $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ \"extra\" delegates to choose to be the third sleeper once both from Mexico are sleeping. There are $3\\times6=18$ ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are $9-3=6$ \"extra\" people to select one more sleeper from. Lastly, there are $6\\times5=30$ ways to choose for the United States. It is easy to count $6$ different ways to pick $2$ of the $4$ Americans, and each time you do there are $9-4=5$ officials left over to choose from. Thus, the fraction is $\\frac{7+18+30}{84} = \\frac{55}{84}$. Since this does not reduce, the answer is $55+84=139.",
"Like in the solution above, there are $84$ ways to pick $3$ delegates. We can use casework to find the probability that there aren't exactly $2$ sleepers from a county, then subtract from $1$. If no country has at least $2$ delegates sleeping, then every country must have $1$ delegate sleeping. There are $2*3*4=24$ ways for this to happen. If all $3$ sleeping delegates are from Canada, there are $\\binom{3}{3} = 1$ way. If all $3$ are from the US, there are $\\binom{4}{3} = 4$ ways. So, the probability that there are not exactly $2$ sleepers from one country is $\\frac{24+1+4}{84} = \\frac{29}{84}$, and the probability that exactly $2$ are from the same country is $1- \\frac{29}{84} = \\frac{55}{84}.$ Our answer is $55+84=139.",
"Let us take this step-by-step. The probability of having exactly 2 Mexican sleepers is $\\frac{2}{9} \\cdot \\frac{1}{8} \\cdot 3$. The probability of having exactly 2 Canadian sleepers is $\\frac{3}{9} \\cdot \\frac{2}{8} \\cdot \\frac{6}{7} \\cdot 3$. The probability of having exactly 2 American sleepers is $\\frac{4}{9} \\cdot \\frac{3}{8} \\cdot \\frac{5}{7} \\cdot 3$. Thus, adding these up our total probability is $\\frac{55}{84}$. Adding the numerator and denominator gives us our desired $55+84=139 ~SirAppel",
"Let us take this step-by-step. The probability of having exactly 2 Mexican sleepers is $\\frac{2}{9} \\cdot \\frac{1}{8} \\cdot 3$. The probability of having exactly 2 Canadian sleepers is $\\frac{3}{9} \\cdot \\frac{2}{8} \\cdot \\frac{6}{7} \\cdot 3$. The probability of having exactly 2 American sleepers is $\\frac{4}{9} \\cdot \\frac{3}{8} \\cdot \\frac{5}{7} \\cdot 3$. Thus, adding these up our total probability is $\\frac{55}{84}$. Adding the numerator and denominator gives us our desired $55+84=139 ~SirAppel"
] |
2015-I-3
| 2,015
| 3
|
There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$ .
| 307
|
I
|
[
"Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd. Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$ or synthetic divison once it is realized that $a = 1$ is a root): \\begin{align*} a^3-1 &= 16p\\\\ (a-1)(a^2+a+1) &= 16p\\\\ \\end{align*} Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$. However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$. Then our other factor, $a^2+a+1$, is the prime $p$: \\begin{align*} (a-1)(a^2+a+1) &= 16p\\\\ (17-1)(17^2+17+1) &=16p\\\\ p = 289+17+1 &= 307 \\end{align*}",
"Observe that this is the same as $16p+1=n^3$ for some integer $n$. So: \\begin{align*} 16p &= n^3-1\\\\ 16p &= n^3-1^3\\\\ 16p &= (n-1)(n^2+n+1)\\\\ \\end{align*} Observe that either $p=n-1$ or $p=n^2+n+1$ because $p$ and $16$ share no factors ($p$ can't be $2$). Let $p=n-1$. Then: \\begin{align*} p &= n-1\\\\ 16 &= n^2+n+1\\\\ n^2+n &= 15\\\\ n(n+1) &= 15\\\\ \\end{align*} Which is impossible for integer n. So $p=n^2+n+1$ and \\begin{align*} 16 &= n-1\\\\ n &= 17\\\\ p &= 17^2+17+1\\\\ p = 289+17+1 &= 307\\\\ \\end{align*} - firebolt360",
"Observe that this is the same as $16p+1=n^3$ for some integer $n$. So: \\begin{align*} 16p &= n^3-1\\\\ 16p &= n^3-1^3\\\\ 16p &= (n-1)(n^2+n+1)\\\\ \\end{align*} Observe that either $p=n-1$ or $p=n^2+n+1$ because $p$ and $16$ share no factors ($p$ can't be $2$). Let $p=n-1$. Then: \\begin{align*} p &= n-1\\\\ 16 &= n^2+n+1\\\\ n^2+n &= 15\\\\ n(n+1) &= 15\\\\ \\end{align*} Which is impossible for integer n. So $p=n^2+n+1$ and \\begin{align*} 16 &= n-1\\\\ n &= 17\\\\ p &= 17^2+17+1\\\\ p = 289+17+1 &= 307\\\\ \\end{align*} - firebolt360",
"Since $16p+1$ is odd, let $16p+1 = (2a+1)^3$. Therefore, $16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1$. From this, we get $8p=a(4a^2+6a+3)$. We know $p$ is a prime number and it is not an even number. Since $4a^2+6a+3$ is an odd number, we know that $a=8$. Therefore, $p=4a^2+6a+3=4*8^2+6*8+3=307.",
"Let $16p+1=a^3$. Realize that $a$ congruent to $1\\mod 4$, so let $a=4n+1$. Expansion, then division by 4, gets $16n^3+12n^2+3n=4p$. Clearly $n=4m$ for some $m$. Substitution and another division by 4 gets $256m^3+48m^4+3m=p$. Since $p$ is prime and there is a factor of $m$ in the LHS, $m=1$. Therefore, $p=307.",
"Notice that $16p+1$ must be in the form $(a+1)^3 = a^3 + 3a^2 + 3a + 1$. Thus $16p = a^3 + 3a^2 + 3a$, or $16p = a\\cdot (a^2 + 3a + 3)$. Since $p$ must be prime, we either have $p = a$ or $a = 16$. Upon further inspection and/or using the quadratic formula, we can deduce $p \\neq a$. Thus we have $a = 16$, and $p = 16^2 + 3\\cdot 16 + 3 = 307.",
"Notice that the cube 16p+1 is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here Hence, $p=307 is our only answer pi_is_3.141",
"If $16p+1 =k$, we have $k \\equiv 1 \\mod 16$, so $k^3 \\equiv 1 \\mod 16$. If $k=1$ we have $p=0$, which is not prime. If $k=17$ we have $16p+1=4913$, or $p=307",
"Notice that: \\begin{align*} 1^3 &= 0 +1\\\\ 2^3 &= 1*7+1 \\\\ 3^3 &= 2*13+1\\\\ 4^3 &= 3*21+1\\\\ 5^3 &= 4*31+1\\\\ 6^3 &= 5*43+1 \\end{align*} Here, we can see a clear pattern that $n^3=(n-1)p+1$, where $p$ is some positive (not necessarily prime) integer. Hence, the equation $16p+1=a^3$ can interpret as $17^3 = 16p+1$. Solving it, we got $p=307$. After checking all possible divisors, we will find that $307$ is prime. Hence, we got $p=307.",
"Notice that: \\begin{align*} 1^3 &= 0 +1\\\\ 2^3 &= 1*7+1 \\\\ 3^3 &= 2*13+1\\\\ 4^3 &= 3*21+1\\\\ 5^3 &= 4*31+1\\\\ 6^3 &= 5*43+1 \\end{align*} Here, we can see a clear pattern that $n^3=(n-1)p+1$, where $p$ is some positive (not necessarily prime) integer. Hence, the equation $16p+1=a^3$ can interpret as $17^3 = 16p+1$. Solving it, we got $p=307$. After checking all possible divisors, we will find that $307$ is prime. Hence, we got $p=307.",
"We see that $16p+1=n^3$ for a positive integer $n$. Subtracting $1$, we can turn this equation into a modular congruence, since $n^3-1$ must be a multiple of $16$. Since $n^3-1\\equiv0\\pmod{16}$, $n^3\\equiv1\\pmod{16}$. We observe that $n=1$ is a solution to this congruence, which doesn't work. The next, or most obvious number to try is $n=17$. Plugging this in to our original equation, we get $17^3=16p+1$, yielding $p=307, which is prime. -among us (countmath1)",
"We see that $16p+1=n^3$ for a positive integer $n$. Subtracting $1$, we can turn this equation into a modular congruence, since $n^3-1$ must be a multiple of $16$. Since $n^3-1\\equiv0\\pmod{16}$, $n^3\\equiv1\\pmod{16}$. We observe that $n=1$ is a solution to this congruence, which doesn't work. The next, or most obvious number to try is $n=17$. Plugging this in to our original equation, we get $17^3=16p+1$, yielding $p=307, which is prime. -among us (countmath1)",
"Recognizing that AIME answers are $0$ through $999$, the numbers whose cube could even be in contention to be equal to $16p + 1$ are $4-25$. The cubes of $1-3$ are all below $17$. We might consider $1$, but that would result in a $p$ of $0$, which is not prime and does not follow our given conditions. We can also see that $16p + 1$ must be an odd number. Hence, we brute force by looking at all cubes of odd numbers in $5-25$ until we get that the cube of $17$, $4913$, works when $p=307 Solution by: armang32324",
"Recognizing that AIME answers are $0$ through $999$, the numbers whose cube could even be in contention to be equal to $16p + 1$ are $4-25$. The cubes of $1-3$ are all below $17$. We might consider $1$, but that would result in a $p$ of $0$, which is not prime and does not follow our given conditions. We can also see that $16p + 1$ must be an odd number. Hence, we brute force by looking at all cubes of odd numbers in $5-25$ until we get that the cube of $17$, $4913$, works when $p=307 Solution by: armang32324"
] |
2015-I-4
| 2,015
| 4
|
Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$ .
| 507
|
I
|
[
"Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\\sqrt3)$ and $E=(18,2\\sqrt3)$ as well. Therefore, $M=(9,\\sqrt3)$ and $N=(14,4\\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives \\[x=\\dfrac 1 2 |16\\sqrt3+36\\sqrt3+0-(0+14\\sqrt3+64\\sqrt3)| =13\\sqrt3,\\] so $x^2=507",
"Note that $AB=DB=16$ and $BE=BC=4$. Also, $\\angle ABE = \\angle DBC = 120^{\\circ}$. Thus, $\\triangle ABE \\cong \\triangle DBC$ by SAS. From this, it is clear that a $60^{\\circ}$ rotation about $B$ will map $\\triangle ABE$ to $\\triangle DBC$. This rotation also maps $M$ to $N$. Thus, $BM=BN$ and $\\angle MBN=60^{\\circ}$. Thus, $\\triangle BMN$ is equilateral. Using the Law of Cosines on $\\triangle ABE$, \\[AE^2 = 16^2 + 4^2 - 2\\cdot 16\\cdot 4\\cdot\\left(-\\frac{1}{2}\\right)\\] \\[AE = 4\\sqrt{21}\\] Thus, $AM=ME=2\\sqrt{21}$. Using Stewart's Theorem on $\\triangle ABE$, \\[AM\\cdot ME\\cdot AE + AE\\cdot BM^2 = BE^2\\cdot AM + BA^2\\cdot ME\\] \\[BM = 2\\sqrt{13}\\] Calculating the area of $\\triangle BMN$, \\[[BMN] = \\frac{\\sqrt{3}}{4} BM^2\\] \\[[BMN] = 13\\sqrt{3}\\] Thus, $x=13\\sqrt{3}$, so $x^2 = 507$. Our final answer is $507 - SuperJJ",
"$AB = BD, BE = BC, \\angle ABE = \\angle CBD \\implies \\triangle ABE \\cong \\triangle DBC$ Medians are equal, so $MB = BN, \\angle ABM = \\angle DBN \\implies$ $\\angle MBN = \\angle ABD - \\angle ABM + \\angle DBN = 60^\\circ \\implies$ $\\triangle MNB$ is equilateral triangle. The height of $\\triangle BCE$ is $2 \\sqrt{3},$ distance from $A$ to midpoint $BC$ is $16 + 2 = 18 \\implies \\frac {AE^2}{4} =\\frac{ (16 + 2)^2 +2^2 \\cdot 3}{4} = 81 + 3 = 84.$ $BM$ is the median of $\\triangle ABE \\implies$ $BM^2 = \\frac {AB^2}{2} + \\frac {BE^2}{2} - \\frac {AE^2}{4}=16 \\cdot 8 + 4 \\cdot 2 - 84 = 52.$ The area of $\\triangle BMN$ \\[[BMN] = \\frac{\\sqrt{3}}{4} BM^2 =13 \\sqrt{3} \\implies \\textbf{507}.\\] [email protected], vvsss"
] |
2015-I-5
| 2,015
| 5
|
In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday, Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random, and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 341
|
I
|
[
"The order of the days doesn't matter, therefore we can solve backwards. Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\\dfrac{1}{9}$. Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\\dfrac{6}{7}.$ The only \"hard\" part is the third and fourth sock. But that is simple casework. If the third sock's color matches the color of one of the first two socks (which occurs with probability $\\dfrac{2}{6} = \\dfrac{1}{3}$), then the fourth sock can be arbitrary. Otherwise (with probability $\\dfrac{2}{3}$), the fourth sock can be chosen with probability $\\dfrac{4}{5}$ (5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus \\[\\frac{1}{9} \\cdot \\frac{6}{7} \\cdot \\left(\\dfrac{1}{3} + \\dfrac{2}{3} \\cdot \\dfrac{4}{5}\\right) = \\frac{26}{315}.\\] The sum is therefore $26+315=341",
"The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are $4$ pairs, the number of ways to do this is $\\dbinom{8}{2}-4$. Then, there are two pairs and two nonmatching socks for you to pick from on Monday, a total of $6$ socks. Since you don't want to pick a pair, the number of ways to do this is $\\dbinom{6}{2}-2$. Thus the answer is \\[\\dfrac{\\left(5\\right)\\left(\\dbinom{8}{2}-4\\right)\\left(\\dbinom{6}{2}-2\\right)}{\\dbinom{10}{2}\\dbinom{8}{2}\\dbinom{6}{2}}=\\dfrac{26}{315}.\\] $26 + 315 = 341.",
"For the first sock, note that to pick two different socks, can complementary count to get the total, $\\binom{10}{2}$ minus the number of pairs (5) to get \\[\\frac{\\binom{10}{2} - 5}{\\binom{10}{2}}\\] The next steps aren't quite as simple, though. WLOG suppose the socks are (a, a, b, b, c, c, d, d, e, e) and that we chose ab on the first day. This leaves abccddee as the remaining socks. We can think of our next choice as how many \"pairs\" we \"break\". Note that right now, in abccddee there are 3 pairs, namely {cc, dd, ee}. Our next choice could either leave all of these pairs (if we choose ab), leave 2 of these pairs, (choose say ac to keep dd and ee as pairs), or leave only 1 as a pair, say choosing de to get abccde. \\[\\bold{\\text{Case 1: Break zero pairs}}.\\] In this case, we want to keep 3 pairs remaining, so our only choice is to choose AB, which would make the remaining ccddee. The probability that we choose AB is $\\frac{1}{\\binom{8}{2}}$ and the probability that we get a pair after this is $\\frac{3}{\\binom{6}{2}}$ since we just choose CC, DD, or EE. This case has a probability $\\frac{1}{\\binom{8}{2}}\\frac{3}{\\binom{6}{2}}.$ \\[\\bold{\\text{Case 2: Break one pair}}.\\] In this case, we want to choose A and one of {C,C,D,D,E,E} or B and one of {C,C,D,D,E,E}. This yields $\\frac{6 + 6}{\\binom{8}{2}}$ probability. After this, we end up with 2 pairs, so the total probability of choosing a pair on Wednesday in this case is $\\frac{6 + 6}{\\binom{8}{2}}\\frac{2}{\\binom{6}{2}}.$ \\[\\bold{\\text{case 3: Break 2 pairs}}.\\] In this case we'd choose two distinct characters from {C,C,D,D,E,E} which would be $\\frac{\\binom{6}{2} - 3}{\\binom{6}{2}}.$ After this, only one pair remains so the total probability for this case is $\\frac{\\binom{6}{2} - 3}{\\binom{6}{2}} \\frac{6}{\\binom{8}{2}}\\frac{1}{\\binom{6}{2}}.$ Adding all the cases and multiplying by $\\frac{\\binom{10}{2} - 5}{\\binom{10}{2}}$ yields \\[\\frac{\\binom{10}{2} - 5}{\\binom{10}{2}}\\left(\\frac{1}{\\binom{8}{2}}\\frac{3}{\\binom{6}{2}}+\\frac{6 + 6}{\\binom{8}{2}}\\frac{2}{\\binom{6}{2}}+ \\frac{\\binom{6}{2} - 3}{\\binom{6}{2}}\\frac{6}{\\binom{8}{2}}\\frac{1}{\\binom{6}{2}}\\right) = \\frac{26}{315} \\rightarrow 26 + 315 = 341\\]"
] |
2015-I-6
| 2,015
| 6
|
Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$ . [asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315));[/asy]
| 58
|
I
|
[
"Let $O$ be the center of the circle with $ABCDE$ on it. Let $x$ be the degree measurement of $\\overarc{ED}=\\overarc{DC}=\\overarc{CB}=\\overarc{BA}$ in circle $O$ and $y$ be the degree measurement of $\\overarc{EF}=\\overarc{FG}=\\overarc{GH}=\\overarc{HI}=\\overarc{IA}$ in circle $C$. $\\angle ECA$ is, therefore, $5y$ by way of circle $C$ and \\[\\frac{360-4x}{2}=180-2x\\] by way of circle $O$. $\\angle ABD$ is $180 - \\frac{3x}{2}$ by way of circle $O$, and \\[\\angle AHG = 180 - \\frac{3y}{2}\\] by way of circle $C$. This means that: \\[180-\\frac{3x}{2}=180-\\frac{3y}{2}+12\\] which when simplified yields \\[\\frac{3x}{2}+12=\\frac{3y}{2}\\] or \\[x+8=y\\] Since: \\[5y=180-2x\\] and \\[5x+40=180-2x\\] So: \\[7x=140\\Longleftrightarrow x=20\\] \\[y=28\\] $\\angle BAG$ is equal to $\\angle BAE$ + $\\angle EAG$, which equates to $\\frac{3x}{2} + y$. Plugging in yields $30+28$, or $058.",
"Let $m$ be the degree measurement of $\\angle GCH$. Since $G,H$ lie on a circle with center $C$, $\\angle GHC=\\frac{180-m}{2}=90-\\frac{m}{2}$. Since $\\angle ACH=2 \\angle GCH=2m$, $\\angle AHC=\\frac{180-2m}{2}=90-m$. Adding $\\angle GHC$ and $\\angle AHC$ gives $\\angle AHG=180-\\frac{3m}{2}$, and $\\angle ABD=\\angle AHG+12=192-\\frac{3m}{2}$. Since $AE$ is parallel to $BD$, $\\angle DBA=180-\\angle ABD=\\frac{3m}{2}-12=$$\\overarc{BE}$. We are given that $A,B,C,D,E$ are evenly distributed on a circle. Hence, $\\overarc{ED}=\\overarc{DC}=\\overarc{CB}=\\overarc{BA}$$=\\frac{\\angle DBA}{3}=\\frac{m}{2}-4$ Here comes the key: Draw a line through $C$ parallel to $AE$, and select a point $X$ to the right of point $C$. $\\angle ACX$ = $\\overarc{AB}$ + $\\overarc{BC}$ = $m-8$. Let the midpoint of $\\overline{HG}$ be $Y$, then $\\angle YCX=\\angle ACX+\\angle ACY=(m-8)+\\frac{5m}{2}=90$. Solving gives $m=28$ The rest of the solution proceeds as in solution 1, which gives $058",
"Let $\\angle GAH = \\varphi \\implies \\overset{\\Large\\frown} {GH} = 2\\varphi \\implies$ \\[\\overset{\\Large\\frown} {EF} = \\overset{\\Large\\frown} {FG} = \\overset{\\Large\\frown} {HI} = \\overset{\\Large\\frown} {IA} = 2\\varphi \\implies\\] \\[\\angle AGH = 2\\varphi, \\angle ACE = 10 \\varphi.\\] \\[BD||GH \\implies \\angle AJB = \\angle AGH = 2 \\varphi.\\] \\[\\triangle AHG: \\hspace{10mm} \\angle AHG = \\beta = 180^\\circ – 3 \\varphi.\\] $\\hspace{10mm} \\triangle ABJ: \\hspace{10mm} \\angle BAG + \\angle ABD = \\alpha + \\gamma = 180^\\circ + 2 \\varphi.$ Let arc $\\overset{\\Large\\frown} {AB} = 2\\psi \\implies$ $\\angle ACE = \\frac {360^\\circ – 8 \\psi}{2}= 180^\\circ – 4 \\psi, \\angle ABD = \\gamma =\\frac {360^\\circ – 6 \\psi}{2} =180^\\circ – 3 \\psi.$ $\\gamma – \\beta = 3(\\varphi – \\psi) = 12^\\circ \\implies \\psi = \\varphi – 4^\\circ \\implies 10 \\varphi = 180^\\circ – 4(\\varphi – 4^\\circ) \\implies 14 \\varphi = 196^\\circ \\implies \\varphi = 14^\\circ.$ Therefore $\\gamma = 180^\\circ – 3 \\cdot (14^\\circ – 4^\\circ) = 150^\\circ \\implies \\alpha = 180^\\circ + 2 \\cdot 14^\\circ – 150^\\circ = \\textbf{058}"
] |
2015-I-7
| 2,015
| 7
|
In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$ . [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
| 539
|
I
|
[
"Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$. [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label(\"$A$\",A,NW); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$D$\",D,NE); label(\"$E$\",E,dir(90)); label(\"$F$\",F,NE); label(\"$G$\",G,NE); label(\"$H$\",H,W); label(\"$J$\",J,S); label(\"$K$\",K,SE); label(\"$L$\",L,SE); label(\"$M$\",M,dir(90)); label(\"$N$\",N,dir(180)); [/asy] Now, examine $BC$. We know $BC=BJ+JC$, and triangles $\\Delta BHJ$ and $\\Delta JFC$ are similar to $\\Delta EDC$ since they are $1-2-\\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$. \\[BJ=\\frac{1}{\\sqrt{5}}HJ=\\frac{x}{\\sqrt{5}}=\\frac{x\\sqrt{5}}{5}, JC=\\frac{\\sqrt{5}}{2}JF=\\frac{x\\sqrt{5}}{2}\\Rightarrow BC=\\frac{7x\\sqrt{5}}{10}\\] Now examine $AB$. We can express this length in terms of $x,y$ since $AB=AN+NH+HB$. By using similar triangles as in the first part, we have \\[AB=\\frac{1}{\\sqrt{5}}y+\\frac{\\sqrt{5}}{2}y+\\frac{2}{\\sqrt{5}}x\\] \\[AB=BC\\Rightarrow \\frac{7y\\sqrt{5}}{10}+\\frac{2x\\sqrt{5}}{5}=\\frac{7x\\sqrt{5}}{10}\\Rightarrow \\frac{7y\\sqrt{5}}{10}=\\frac{3x\\sqrt{5}}{10}\\Rightarrow 7y=3x\\] Now, it is trivial to see that $[FJGH]=\\left(\\frac{x}{y}\\right)^2[KLMN]=\\left(\\frac{7}{3}\\right)^2\\cdot 99=539",
"[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label(\"$A$\",A,NW); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$D$\",D,NE); label(\"$E$\",E,dir(90)); label(\"$F$\",F,NE); label(\"$G$\",G,NE); label(\"$H$\",H,W); label(\"$J$\",J,S); label(\"$K$\",K,SE); label(\"$L$\",L,SE); label(\"$M$\",M,dir(90)); label(\"$N$\",N,dir(180)); label(\"$P$\",P,dir(235)); [/asy] We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\\sqrt5$. Since angles $\\angle DCE$ and $\\angle FCJ$ are complementary, we have that $\\triangle CDE \\sim \\triangle JFC$ (and similarly the rest of the triangles are $1-2-\\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us: \\[JC = \\sqrt5 \\cdot FC = \\sqrt5 \\cdot FJ/2 = \\frac{b\\sqrt 5}{2}\\] and \\[BJ = \\frac{1}{\\sqrt5} \\cdot HJ = \\frac{b}{\\sqrt5}\\] Since $BC = CJ + BJ$, \\[2a = \\frac{b\\sqrt 5}{2} + \\frac{b}{\\sqrt5}\\] Solving for $b$ in terms of $a$ yields \\[b = \\frac{4a\\sqrt5}{7}\\] We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$ as in the diagram at the top This gives that \\[AM = 2 \\cdot AN = 2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5}\\] and \\[ME = \\sqrt5 \\cdot MP = \\sqrt5 \\cdot \\frac{EP}{2} = \\sqrt5 \\cdot \\frac{LG}{2} = \\sqrt5 \\cdot \\frac{HG - HK - KL}{2} = \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2}\\] Since $AE$ = $AM + ME$, we get \\[2 \\cdot \\frac{3\\sqrt{11}}{\\sqrt5} + \\sqrt{5} \\cdot \\frac{\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}}{2} = a\\] \\[\\Rightarrow 12\\sqrt{11} + 5(\\frac{4a\\sqrt5}{7} - \\frac{9\\sqrt{11}}{2}) = 2\\sqrt5a\\] \\[\\Rightarrow \\frac{-21}{2}\\sqrt{11} + \\frac{20a\\sqrt5}{7} = 2\\sqrt5a\\] \\[\\Rightarrow -21\\sqrt{11} = 2\\sqrt5a\\frac{14 - 20}{7}\\] \\[\\Rightarrow \\frac{49\\sqrt{11}}{4} = \\sqrt5a\\] \\[\\Rightarrow 7\\sqrt{11} = \\frac{4a\\sqrt{5}}{7}\\] So our final answer is $(7\\sqrt{11})^2 = 539.",
"This is a relatively quick solution but a fakesolve. We see that with a ruler, $KL = \\frac{3}{2}$ cm and $HG = \\frac{7}{2}$ cm. Thus if $KL$ corresponds with an area of $99$, then $HG$ ($FGHJ$'s area) would correspond with $99*(\\frac{7}{3})^2 = 539 - aops5234"
] |
2015-I-8
| 2,015
| 8
|
For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ .
| 695
|
I
|
[
"You know whatever $n$ is, it has to have 3 digits, because if it had only two, the maximum of $s(n)$ is 18 and because in AIME all answers are up to three digits. Now let $n=100a_2+10a_1+a_0$ So first we know, $a_2+a_1+a_0=20$. Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned \"put one number over the other\" method, not mental calculation), and observe what you do at each step. (1)$\\textcolor{red}{*}$ None of the digits get carried over to the next space: So this means $a_2<2, a_1<4$ and $a_0<6$. So $s(864+n)=(8+a_2)+(6+a_1)+(4+a_0)=38$ So it doesn't work. Now: (2) $a_2+8$ is the only one that carries over So this means $a_2>1, a_1<4$ and $a_0<6$. So $s(864+n)=1+(8+a_2-10)+(6+a_1)+(a_0+4)=29$ (3)$\\textcolor{red}{*}$ $a_0+4$ is the only one that carries over. So $s(864+n)=(8+a_2)+(6+a_1+1)+(4+a_0-10)=29$ (4)The first and second digit carry over (but not the third) $s(864+n)=1+(8+a_2-10+1)+(6+a_1-10)+(4+a_0)=20$ Aha! This case works but we still have to make sure it's possible for $a_2+a_1+a_0=20$ (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, $a_2>0, a_1>3$ and $a_0<6$. The smallest value we can get with this is 695. Let's see if we can find a smaller one: (5)The first and third digit carry over (but not the second) $s(864+n)=1+(8+a_2-10)+(7+a_1)+(4+a_0-10)=20$ The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929 (6) All the digits carry over $s(864+n)=1+(9+a_2-10)+(7+a_1-10)+(4+a_0-10)=\\text{Way less than 20}$ So the answer is $695), and therefore the first digit must ALWAYS carry. This is presumably the reason where the case where the second and third digits were carried but not the first was omitted, but for this reason cases (1) and (3) are also unnecessary.",
"First, it is easy to verify that $695$ works and that no other numbers beginning with the digit 6 work (i.e. $686, 677, 668, 659$ do not work). Suppose by contradiction that there is a smaller valid $n$, where the leading digit of the three-digit number $n$ is 5 or less. (Two-digit $n$ obviously do not work because 9 + 9 < 20.) Clearly $n > 200$ because the smallest three-digit number whose digits sum to 20 is $299$. Also, because the second digit is at most 9, the units digit is at least 6, which means that the addition $N = n + 864$ regroups in the ones place. Then the units digit of $N$ is clearly less than 4. But as $1000 < 200 + 864 < N < 600 + 864 = 1464$, the sum of the thousands digit and the hundredth digit is at most 5. Because the second digit is at most 9, the sum of the digits of $N$ is at most $5 + 9 + 4 < 20$, contradiction. Hence $695 is the answer.",
"First of all, notice that the smallest $n$ with $s(n) = 20$ is $299$. Also, if $s(n + 864) = 20$, $s(n - 136) = 19$ (because subtracting $1000$ from the number removes the $1$ in the thousands place). After checking $s(n - 136)$ for various $n$ with $s(n) = 20$, we see that we need to have a carry when subtracting $136$. To have this, we must either have a $2$ in the tens place or a $5$ in the units place. The minimum $n$ for the former is $929$, and for the latter it is $695$. We check and see that $s(695-136) = s(559) = 19$, so our answer is $695.",
"Observation (Lemma) : If r is the number of regroups in the addition of n+k, $S(n+k) = S(n)+S(k)-9r$. Proof : When you add two numbers, and you do a carry, you are taking away 10 from 1 column, and adding 1 to another column, giving a net loss of 9 to the total. Thus, we can see that we need to regroup exactly twice when we add 864. And, the lowest possible n is 299, so let's start from there. 299 gets three regroups, so we are going to need to take away from digits, and dump the excess in the hundred's place, since the hundreds are going to regroup anyways. So, if we take away from the tens digit, we need to take away until we get 2 in the tens digit(since the ones will regroup). So, we get the number 929, which works (929+864 = 1793), but is not the smallest. If we take away from the ones digit, we only have to take away 4, turning the unit's place to 5. 5+4 is 9, so it won't regroup. Dump the ones into the hundred's place, and we get the number $695 -AlexLikeMath",
"Although this solution doesn't directly solve the problem, it greatly hastens the bashing process. Call the three digits a, b, and c. When you add each of 8, 6, and 4 to a, b, and c the resultant will either get smaller or larger, depending on the original number. e.g. If c is 7, then adding 4 will reduce the 7 to a 1, whilst leaving a one for b. If c is 3, then adding 4 will simply add four to the total, and make the 3 a 7. Each of 8, 6, and 4 all can reduce the original number by a certain amount and can increase the original number by a certain amount. 8 can reduce by 2 for all numbers greater than 1, 6 can reduce numbers by 4, and 4 can reduce numbers by 6. Possibilities: -2, +8, (although this becomes obviously impossible later on) -4, +6, -6, +4 Also, realize that if the number is reduced, then a one will be carried to the following decimal place on the left, consequently reducing that amount they reduce. It's like a puzzle! Within no time you should find that if you add 4 to c, subtract 4 from b, subtract 1 from a, and leave a 1 in the thousands place, the total is equated to zero. This is optimal because most of the addition is kept to the left, where the effect to real value is less. (e.g. 299 is smaller than 992) Now you have +1, -1, -4, +4 in the decimals, and a VERY fast trial and error gives $695 -Jackshi2006 Postscript by Jackshi2006 When I revisited the problem I realized that you can actually list out every possible number for n. 695 stands out very clearly as the smallest, because the only other possibility is 595, which doesn’t add to 20 to begin with.",
"First, note that to compute $s(m+n)$ (for any positive integers $m$ and $n$), one can simply find the sum of $s(m)$ and $s(n)$ minus 9 times the number of times one regroups when adding $m$ to $n$. One can see why this is by noticing that if one were to \"forget\" to regroup, and leave, say, a 10 in the ones' place, the sum of the digits would be 9 higher than if one did regroup. Anyway, one can see that the smallest 3-digit number (on AIME, all the answers are integers from 0 to 999) whose digits sum to 20 is 299. If we add this number to 864, we have to regroup 3 times, so $s(299+864)$ will be $9=9\\cdot3-(8+6+4)$ smaller than $s(299)$. We want this difference to be 0, so we need to find a way to only regroup two times. We now notice that regrouping the hundreds is inevitable, so we must either prevent regrouping the ones or the tens. Preventing regrouping the tens would require moving many of the tens to the hundreds' place (the ones' place is already full), which is bad when we are trying to minimize the number, but preventing regrouping the ones requires moving fewer ones to the hundreds' place. We find that to preventing regrouping the ones, the ones' place of our number must be at most 5 (a larger number would sum to ten when added to 4). Because we want to minimize the number of ones we move to the hundreds' place, we leave exactly 5 by moving four ones to the hundreds' place: $299\\to695. ~ SymbolicPermutation",
"First, note that to compute $s(m+n)$ (for any positive integers $m$ and $n$), one can simply find the sum of $s(m)$ and $s(n)$ minus 9 times the number of times one regroups when adding $m$ to $n$. One can see why this is by noticing that if one were to \"forget\" to regroup, and leave, say, a 10 in the ones' place, the sum of the digits would be 9 higher than if one did regroup. Anyway, one can see that the smallest 3-digit number (on AIME, all the answers are integers from 0 to 999) whose digits sum to 20 is 299. If we add this number to 864, we have to regroup 3 times, so $s(299+864)$ will be $9=9\\cdot3-(8+6+4)$ smaller than $s(299)$. We want this difference to be 0, so we need to find a way to only regroup two times. We now notice that regrouping the hundreds is inevitable, so we must either prevent regrouping the ones or the tens. Preventing regrouping the tens would require moving many of the tens to the hundreds' place (the ones' place is already full), which is bad when we are trying to minimize the number, but preventing regrouping the ones requires moving fewer ones to the hundreds' place. We find that to preventing regrouping the ones, the ones' place of our number must be at most 5 (a larger number would sum to ten when added to 4). Because we want to minimize the number of ones we move to the hundreds' place, we leave exactly 5 by moving four ones to the hundreds' place: $299\\to695. ~ SymbolicPermutation",
"Bashing out modulo $9$ and getting lucky we get that if the $8$ and $6$ carry over when adding $n$ and $864$, that $100a+10b+c \\equiv 1+100(a-1)+10(b-4)+c+4 \\pmod{9}$ such that $n=100a+10b+c$ and after maximizing $b$ and $c$ such that $c<6$ to not make the $4$ carry over to minimize $a$ we get that $695. ~SirAppel",
"Bashing out modulo $9$ and getting lucky we get that if the $8$ and $6$ carry over when adding $n$ and $864$, that $100a+10b+c \\equiv 1+100(a-1)+10(b-4)+c+4 \\pmod{9}$ such that $n=100a+10b+c$ and after maximizing $b$ and $c$ such that $c<6$ to not make the $4$ carry over to minimize $a$ we get that $695. ~SirAppel",
"Alternatively, you can use python to solve the problem: def sum_of_digits(number): total = 0 while number > 0: digit = number % 10 total += digit number //= 10 return total for i in range(1000): digitsum = sum_of_digits(i) newdigitsum = sum_of_digits(i+864) if digitsum == newdigitsum: if digitsum == 20: print(i) ~yeetdayeet",
"Alternatively, you can use python to solve the problem: def sum_of_digits(number): total = 0 while number > 0: digit = number % 10 total += digit number //= 10 return total for i in range(1000): digitsum = sum_of_digits(i) newdigitsum = sum_of_digits(i+864) if digitsum == newdigitsum: if digitsum == 20: print(i) ~yeetdayeet"
] |
2015-I-9
| 2,015
| 9
|
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$ . Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$ . Find the number of such sequences for which $a_n=0$ for some $n$ .
| 494
|
I
|
[
"Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let $|y-x|>1$, and $|z-y|>1$. Then, $a_4 \\ge 2z$, $a_5 \\ge 4z$, and $a_6 \\ge 4z$. However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$, $|y-x|=2$. Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: $z>1$, $|y-x|>1$, and $|z-y|>1$; and $z=1$, $|y-x|>2$, and $|z-y|>1$. For the first one, $a_4 \\ge 2z$, $a_5 \\ge 4z$, $a_6 \\ge 8z$, and $a_7 \\ge 16z$, by which point we see that this function diverges. For the second one, $a_4 \\ge 3$, $a_5 \\ge 6$, $a_6 \\ge 18$, and $a_7 \\ge 54$, by which point we see that this function diverges. Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$, $|y-x|=2$. (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+16-2=494).",
"Note that the only way for a $0$ to be produced at $a_n$ is if either $a_{n-1} = 0$ or $a_{n-2} = a_{n-3}$. Since the first one will eventually get to the first three assuming that there is no $a_{n-2} = a_{n-3}$ for any $n$, that is not possible because $a_1 , a_2 , a_3 >= 1$. Therefore, we need $a_{n - 2} = a_{n - 3}$. If $2$ consecutive numbers out of $a_1 , a_2 , a_3$ are equal, then those cases work($a_1$ and $a_2$ or $a_2$ and $a_3$ $\\textbf{NOT}$ $a_1$ and $a_3$). This is simply $10 \\cdot 10 + 10 \\cdot 10 - 10 = 190$ by PIE. Now, note that if any of the first three numbers have difference of $1$, we have another working case. First, we calculate how many there are given exactly one of $a_1,a_2$ or $a_2,a_3$ have difference $1$. Given $3$ numbers such that the first $2$ have difference $1$, exactly $4$ permutations work(assuming the numbers are $x,y,$ and $z$ such that $|x-y| = 1$): $x,y,z$; $y,x,z$; $z,x,y$; and $z,y,x$. If the two consecutive numbers are $1$ and $2$, then the last number has $7$ possiblities: $4,5,6,\\cdots , 10$. This is symmetric for $9$ and $10$. If the consecutive numbers are $(2,3),\\cdots , (8,9)$, there are $6$ possibilities($10$ minus the numbers themselves and the numbers directly above and below). Note that we are not counting any cases already counted in the first case. Therefore, this case gives you $4(7 + 6 * 7 + 7) = 224$. Now we consider the case that there both adjacent $a$s have increments of $1$. \\[+1 , +1 \\rightarrow 8\\] \\[-1 , -1 \\rightarrow 8\\] \\[+1 , -1 \\rightarrow 9\\] \\[-1 , +1 \\rightarrow 9\\] Therefore this gives $224 + 8 + 8 + 9 + 9 = 258$. However, note that we have to add the case where you have $3$ consecutive numbers in arrangement such that only $2$ consecutive numbers have difference $1$. For example, $1,3,2$ is one such triple. There are $8$ triples of consecutive numbers and $4$ ways to arrange each one(e.x: $(1,3,2);(3,1,2);(2,1,3);(2,3,1)$). This adds on 32 working cases, so this case gives $258 + 32 = 290$. Note that there is an ultraspecial(Yes, I know that's not a word) case where we generate a pair of $a_i$ that have difference one. This can only happen if $a_3 = 1$ and $a_1$ and $a_2$ have difference $2$. This contributes $14$ cases($2 * 8$ and then subtract $2$ because of the cases $3,1,1$ and $4,2,1$). Therefore, our answer is $190 + 290 + 14 = 494. Solution by hyxue"
] |
2015-I-10
| 2,015
| 10
|
Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$ .
| 72
|
I
|
[
"Let $f(x)$ = $ax^3+bx^2+cx+d$. Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing $12$ and $-12$, it is easy to see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$; otherwise more bends would be required in the graph. Since only the absolute value of $f(0)$ is required, there is no loss of generalization by stating that $f(1)=12$, and $f(2)=-12$. This provides the following system of equations. \\[a + b + c + d = 12\\] \\[8a + 4b + 2c + d = -12\\] \\[27a + 9b + 3c + d = -12\\] \\[125a + 25b + 5c + d = 12\\] \\[216a + 36b + 6c + d = 12\\] \\[343a + 49b + 7c + d = -12\\] Using any four of these functions as a system of equations yields $d = |f(0)| = 072",
"By drawing the function, and similar to Solution 1, WLOG let $f(1)=f(5)=f(6)=12$. Then, $f(2)=f(3)=f(7)$. Set $g(x)+12=f(x)$. Then the roots of $g(x)$ are $1,5,6$. So, $g(x)=a(x-1)(x-5)(x-6)$. Plug in $x=2$ to find a. We know \\[-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a\\]. So, $a=-2$. Thus, $f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12$, and then $|f(0)|=60+12=072.",
"Without loss of generality, let $f(1) = 12$. (If $f(1) = -12$, then take $-f(x)$ as the polynomial, which leaves $|f(0)|$ unchanged.) Because $f$ is third-degree, write \\[f(x) - 12 = a(x - 1)(x - b)(x - c)\\] \\[f(x) + 12 = a(x - d)(x - e)(x - f)\\] where $\\{b, c, d, e, f \\}$ clearly must be a permutation of $\\{2, 3, 5, 6, 7\\}$ from the given condition. Thus $b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.$ However, subtracting the two equations gives $-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]$, so comparing $x^2$ coefficients gives $1 + b + c = d + e + f$ and thus both values equal to $\\dfrac{24}{2} = 12$. As a result, $\\{b, c \\} = \\{5, 6 \\}$. As a result, $-24 = a (12)$ and so $a = -2$. Now, we easily deduce that $f(0) = (-2) \\cdot (-1) \\cdot (-5) \\cdot (-6) + 12 = 72,$ and so removing the without loss of generality gives $|f(0)| = 072, which is our answer.",
"The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$. Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$. Hence $|f(0)|=\\sqrt{g(0)+144}=\\sqrt{1260k+144}$. Note that $g(x)=(f(x)+12)(f(x)-12)$. Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\\neq 0$ (else $f$ is not cubic) where $\\{q,r,s,t,u,v\\}$ is the same as the set $\\{1,2,3,5,6,7\\}$. Subtracting the second equation from the first, expanding, and collecting like terms, we have that \\[24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)\\] which must hold for all $x$. Since $a\\neq 0$ we have that (1) $t+u+v=q+r+s$, (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$. Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\\{2,3,7\\}$ and $\\{1,5,6\\}$. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that $|a|=\\frac{24}{|tuv-qrs|}=\\frac{24}{12}=2$. Since $a$ is the leading coefficient of $f(x)$, the leading coefficient of $(f(x))^2$ is $a^2=|a|^2=2^2=4$. Thus the leading coefficient of $g(x)$ is 4, i.e. $k=4$. Then from earlier, $|f(0)|=\\sqrt{g(0)+144}=\\sqrt{1260k+144}=\\sqrt{1260\\cdot4+144}=\\sqrt{5184}=72$ so that the answer is $072.",
"Express $f(x)$ in terms of powers of $(x-4)$: \\[f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d\\] By the same argument as in the first Solution, we see that $f(x)$ is an odd function about the line $x=4$, so its coefficients $b$ and $d$ are 0. From there it is relatively simple to solve $f(2)=f(3)=-12$ (as in the above solution, but with a smaller system of equations): \\[a(1)^3 + c(1) = -12\\] \\[a(2)^3 + c(2) = -12\\] $a=2$ and $c=-14$ \\[|f(0)| = |2(-4)^3 - 14(-4)| = 072\\]",
"Because a cubic must come in a \"wave form\" with two points of inflection, we can see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$. By symmetry, $f(4)=0$. Now, WLOG let $f(1)=12$, and $f(2)=f(3)=-12$. Then, we can use finite differences to get that the third (constant) difference is $12$, and therefore $f(0)=12+(24+(24+12))=072.",
"Because a cubic must come in a \"wave form\" with two points of inflection, we can see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$. By symmetry, $f(4)=0$. Now, WLOG let $f(1)=12$, and $f(2)=f(3)=-12$. Then, we can use finite differences to get that the third (constant) difference is $12$, and therefore $f(0)=12+(24+(24+12))=072.",
"We can rewrite our function as two different cubics, $f(x)=k(x-a)(x-b)(x-c)\\pm12=k(x-d)(x-e)(x-f)\\mp12$. Note that $k$ is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that $a+b+c=d+e+f$ and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing $a=1$, $b=5$, $c=6$, $d=2$, $e=3$, $f=7$ yields: \\[kx^3-12kx^2+41kx-30k\\pm12=kx^3-12kx^2+41kx-42k\\mp12\\] For the constant terms to have a difference of 24 ($|\\pm12-\\mp12|$), we must have $k=\\pm2$, so the constant term of our polynomial is $\\pm72$, the absolute value of which is $072. -- Solution by eiis1000",
"We can rewrite our function as two different cubics, $f(x)=k(x-a)(x-b)(x-c)\\pm12=k(x-d)(x-e)(x-f)\\mp12$. Note that $k$ is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that $a+b+c=d+e+f$ and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing $a=1$, $b=5$, $c=6$, $d=2$, $e=3$, $f=7$ yields: \\[kx^3-12kx^2+41kx-30k\\pm12=kx^3-12kx^2+41kx-42k\\mp12\\] For the constant terms to have a difference of 24 ($|\\pm12-\\mp12|$), we must have $k=\\pm2$, so the constant term of our polynomial is $\\pm72$, the absolute value of which is $072. -- Solution by eiis1000",
"We can rewrite the function as $f(x) - 12 = z(x-1)(x-5)(x-6)$ and $f(x) + 12 = z(x-2)(x-3)(x-7)$. Since we need to find $|f(0)|$, substitute 0 for x in these two equations. \\[f(0) - 12 = z(0-1)(0-5)(0-6), f(0) + 12 = z(0-2)(0-3)(0-7)\\] \\[f(0) - 12 = z\\cdot -30, f(0) + 12 = z\\cdot -42\\] Isolating $z$ in both of the equations, \\[z = \\frac{f(0)-12}{-30}, z = \\frac{f(0) + 12}{-42}\\] Equating the two and solving for $f(0)$, we see $|f(0)| = 72$. ~YBSuburbanTea",
"We can rewrite the function as $f(x) - 12 = z(x-1)(x-5)(x-6)$ and $f(x) + 12 = z(x-2)(x-3)(x-7)$. Since we need to find $|f(0)|$, substitute 0 for x in these two equations. \\[f(0) - 12 = z(0-1)(0-5)(0-6), f(0) + 12 = z(0-2)(0-3)(0-7)\\] \\[f(0) - 12 = z\\cdot -30, f(0) + 12 = z\\cdot -42\\] Isolating $z$ in both of the equations, \\[z = \\frac{f(0)-12}{-30}, z = \\frac{f(0) + 12}{-42}\\] Equating the two and solving for $f(0)$, we see $|f(0)| = 72$. ~YBSuburbanTea",
"Let the leading coefficient of $f(x)$ be $a$. Then, it is obvious that $(f(x) - 12)(f(x) + 12) = a^2(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7)$. Let us now, let $x = 0$. We then have, after cleaning it up nicely, $[f(0)]^2 = 2^2 * 3^2 * (35a^2 + 4)$. We now take the square root of both sides, to obtain $|f(0)|$. So, $|f(0)| = 6 * \\sqrt{35a^2 + 4}$. Now, this is the cheese part. Since this is aime, we know that the answer must be an integer, so we assume $a = 2$. Thus, we get $|f(0)| = 6 * \\sqrt{144} = 6 * 12 = 072$. ~~triggod",
"Let the leading coefficient of $f(x)$ be $a$. Then, it is obvious that $(f(x) - 12)(f(x) + 12) = a^2(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7)$. Let us now, let $x = 0$. We then have, after cleaning it up nicely, $[f(0)]^2 = 2^2 * 3^2 * (35a^2 + 4)$. We now take the square root of both sides, to obtain $|f(0)|$. So, $|f(0)| = 6 * \\sqrt{35a^2 + 4}$. Now, this is the cheese part. Since this is aime, we know that the answer must be an integer, so we assume $a = 2$. Thus, we get $|f(0)| = 6 * \\sqrt{144} = 6 * 12 = 072$. ~~triggod"
] |
2015-I-11
| 2,015
| 11
|
Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ .
| 108
|
I
|
[
"Let $D$ be the midpoint of $\\overline{BC}$. Then by SAS Congruence, $\\triangle ABD \\cong \\triangle ACD$, so $\\angle ADB = \\angle ADC = 90^o$. Now let $BD=y$, $AB=x$, and $\\angle IBD = \\dfrac{\\angle ABD}{2} = \\theta$. Then $\\mathrm{cos}{(\\theta)} = \\dfrac{y}{8}$ and $\\mathrm{cos}{(2\\theta)} = \\dfrac{y}{x} = 2\\mathrm{cos^2}{(\\theta)} - 1 = \\dfrac{y^2-32}{32}$. Cross-multiplying yields $32y = x(y^2-32)$. Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$. Additionally, since $\\triangle IBD$ has hypotenuse $\\overline{IB}$ of length $8$, $BD=y < 8$. Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$. However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\\dfrac{32(6)}{(6)^2-32}=48$. Thus the perimeter of $\\triangle ABC$ must be $2(x+y) = 108.",
"Let $AB=x$ and the foot of the altitude from $A$ to $BC$ be point $E$ and $BE=y$. Since ABC is isosceles, $I$ is on $AE$. By Pythagorean Theorem, $AE=\\sqrt{x^2-y^2}$. Let $IE=a$ and $IA=b$. By Angle Bisector theorem, $\\frac{y}{a}=\\frac{x}{b}$. Also, $a+b=\\sqrt{x^2-y^2}$. Solving for $a$, we get $a=\\frac{\\sqrt{x^2-y^2}}{1+\\frac{x}{y}}$. Then, using Pythagorean Theorem on $\\triangle BEI$ we have $y^2+\\left(\\frac{\\sqrt{x^2-y^2}}{1+\\frac{x}{y}}\\right)^2=8^2=64$. Simplifying, we have $y^2+y^2\\frac{x^2-y^2}{(x+y)^2}=64$. Factoring out the $y^2$, we have $y^2\\left(1+\\frac{x^2-y^2}{(x+y)^2}\\right)=64$. Adding 1 to the fraction and simplifying, we have $\\frac{y^2x(x+y)}{(x+y)^2}=32$. Crossing out the $x+y$, and solving for $x$ yields $32y = x(y^2-32)$. Then, we continue as Solution 1 does.",
"Let $AB=x$ and the foot of the altitude from $A$ to $BC$ be point $E$ and $BE=y$. Since ABC is isosceles, $I$ is on $AE$. By Pythagorean Theorem, $AE=\\sqrt{x^2-y^2}$. Let $IE=a$ and $IA=b$. By Angle Bisector theorem, $\\frac{y}{a}=\\frac{x}{b}$. Also, $a+b=\\sqrt{x^2-y^2}$. Solving for $a$, we get $a=\\frac{\\sqrt{x^2-y^2}}{1+\\frac{x}{y}}$. Then, using Pythagorean Theorem on $\\triangle BEI$ we have $y^2+\\left(\\frac{\\sqrt{x^2-y^2}}{1+\\frac{x}{y}}\\right)^2=8^2=64$. Simplifying, we have $y^2+y^2\\frac{x^2-y^2}{(x+y)^2}=64$. Factoring out the $y^2$, we have $y^2\\left(1+\\frac{x^2-y^2}{(x+y)^2}\\right)=64$. Adding 1 to the fraction and simplifying, we have $\\frac{y^2x(x+y)}{(x+y)^2}=32$. Crossing out the $x+y$, and solving for $x$ yields $32y = x(y^2-32)$. Then, we continue as Solution 1 does.",
"Let $AB=x$, call the midpoint of $BC$ point $E$, call the point where the incircle meets $AB$ point $D$, and let $BE=y$. We are looking for the minimum value of $2(x+y)$. $AE$ is an altitude because the triangle is isosceles. By Pythagoras on $BEI$, the inradius is $\\sqrt{64-y^2}$ and by Pythagoras on $ABE$, $AE$ is $\\sqrt{x^2-y^2}$. By equal tangents, $BE=BD=y$, so $AD=x-y$. Since $ID$ is an inradius, $ID=IE$ and using pythagoras on $ADI$ yields $AI=$$\\sqrt{x^2-2xy+64}$. $ADI$ is similar to $AEB$ by $AA$, so we can write $\\frac{x-y}{\\sqrt{x^2-2xy+64}}=\\frac{\\sqrt{x^2-y^2}}{x}$. Simplifying, $\\frac{x}{\\sqrt{x^2-2xy+64}}=\\sqrt{\\frac{x+y}{x-y}}$. Squaring, subtracting 1 from both sides, and multiplying everything out, we get $yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2$, which turns into $32y=x(y^2-32)$. Finish as in Solution 1.",
"Angle bisectors motivate trig bash. Define angle $IBC = x$. Foot of perpendicular from $I$ to $BC$ is point $P$. $\\overline{BC} = 2\\overline{BP} = 2(8\\cos(x)) = N$, where $N$ is an integer. Thus, $\\cos(x) = \\frac{N}{16}$. Via double angle, we calculate $\\overline{AB}$ to be $\\frac{8\\cos(x)}{2\\cos(x)^2 - 1} = \\frac{64N}{N^2 - 128}$. This is to be an integer. We can bound $N$ now, as $N > 11$ to avoid negative values and $N < 16$ due to triangle inequality. Testing, $N = 12$ works, giving $\\overline{AB} = 48, \\overline{BC} = 12$. Our answer is $2 * 48 + 12 = 108. - whatRthose",
"Let $M$ be midpoint $BC, BM = x, AB = y, \\angle IBM = \\alpha.$ $BI$ is the bisector of $\\angle ABM$ in $\\triangle ABM.$ $BI = \\frac {2 xy \\cos \\alpha}{x+y} = 8, \\cos \\alpha = \\frac {x}{8} \\implies \\frac {x^2 y}{x+y} = 32.$ \\[y = \\frac {32 x} {x^2 - 32}.\\] $BC = 2x$ is integer, $5.5^2 < 32 \\implies x \\ge 6.$ $BM < BI \\implies x =\\{ 6, 6.5, 7, 7.5 \\}.$ If $x > 6$ then $y$ is not integer. \\[x = 6 \\implies y = 48 \\implies 2(x+y) = \\textbf{108}.\\] [email protected], vvsss"
] |
2015-I-12
| 2,015
| 12
|
Consider all 1000-element subsets of the set $\{ 1, 2, 3, ... , 2015 \}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .
| 431
|
I
|
[
"Let $M$ be the desired mean. Then because $\\dbinom{2015}{1000}$ subsets have 1000 elements and $\\dbinom{2015 - i}{999}$ have $i$ as their least element, \\begin{align*} \\binom{2015}{1000} M &= 1 \\cdot \\binom{2014}{999} + 2 \\cdot \\binom{2013}{999} + \\dots + 1016 \\cdot \\binom{999}{999} \\\\ &= \\binom{2014}{999} + \\binom{2013}{999} + \\dots + \\binom{999}{999} \\\\ & + \\binom{2013}{999} + \\binom{2012}{999} + \\dots + \\binom{999}{999} \\\\ & \\dots \\\\ & + \\binom{999}{999} \\\\ &= \\binom{2015}{1000} + \\binom{2014}{1000} + \\dots + \\binom{1000}{1000} \\\\ &= \\binom{2016}{1001}. \\end{align*} Using the definition of binomial coefficient and the identity $n! = n \\cdot (n-1)!$, we deduce that \\[M = \\frac{2016}{1001} = \\frac{288}{143}.\\] The answer is $431",
"Each 1000-element subset $\\left\\{ a_1, a_2,a_3,...,a_{1000}\\right\\}$ of $\\left\\{1,2,3,...,2015\\right\\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\\left\\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$. There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is equal to the sum. But choosing a set $\\left\\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\\right\\}$ is the same as choosing a 1001-element subset from $\\left\\{1,2,3,...,2016\\right\\}$! Thus, the average is $\\frac{\\binom{2016}{1001}}{\\binom{2015}{1000}}=\\frac{2016}{1001}=\\frac{288}{143}$. Our answer is $p+q=288+143=431 we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.",
"Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$: For $p = 2$ and $q = 2$, the answer is $1$. For $p = 3$ and $q = 2$, the answer is $\\frac43$. For $p = 4$ and $q = 2$, the answer is $\\frac53$. For $p = 3$ and $q = 3$, the answer is $1$. For $p = 4$ and $q = 3$, the answer is $\\frac54$. For $p = 5$ and $q = 3$, the answer is $\\frac32$. At this point, we can see a pattern: our desired answer is always $\\frac{p+1}{q+1}$. Plugging in $p = 2015$ and $q = 1000$, the answer is $\\frac{2016}{1001}=\\frac{288}{143}$, so $288 + 143 = 431 we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.",
"Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$: For $p = 2$ and $q = 2$, the answer is $1$. For $p = 3$ and $q = 2$, the answer is $\\frac43$. For $p = 4$ and $q = 2$, the answer is $\\frac53$. For $p = 3$ and $q = 3$, the answer is $1$. For $p = 4$ and $q = 3$, the answer is $\\frac54$. For $p = 5$ and $q = 3$, the answer is $\\frac32$. At this point, we can see a pattern: our desired answer is always $\\frac{p+1}{q+1}$. Plugging in $p = 2015$ and $q = 1000$, the answer is $\\frac{2016}{1001}=\\frac{288}{143}$, so $288 + 143 = 431 we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.",
"In the \"average case\", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is $\\frac{2016}{1001}$. -tigershark22 VIEWER NOTE: This solution doesn't always work, for example, take $n^2$ on $[0,1]$. The \"average case\" is $n=\\frac{1}{2}\\implies n^2=\\frac{1}{4}$ but integrating $n^2$ from $0$ to $1$ we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.",
"In the \"average case\", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is $\\frac{2016}{1001}$. -tigershark22 VIEWER NOTE: This solution doesn't always work, for example, take $n^2$ on $[0,1]$. The \"average case\" is $n=\\frac{1}{2}\\implies n^2=\\frac{1}{4}$ but integrating $n^2$ from $0$ to $1$ we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations."
] |
2015-I-13
| 2,015
| 13
|
With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ .
| 91
|
I
|
[
"Let $x = \\cos 1^\\circ + i \\sin 1^\\circ$. Then from the identity \\[\\sin 1 = \\frac{x - \\frac{1}{x}}{2i} = \\frac{x^2 - 1}{2 i x},\\] we deduce that (taking absolute values and noticing $|x| = 1$) \\[|2\\sin 1| = |x^2 - 1|.\\] But because $\\csc$ is the reciprocal of $\\sin$ and because $\\sin z = \\sin (180^\\circ - z)$, if we let our product be $M$ then \\[\\frac{1}{M} = \\sin 1^\\circ \\sin 3^\\circ \\sin 5^\\circ \\dots \\sin 177^\\circ \\sin 179^\\circ\\] \\[= \\frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \\dots |x^{354} - 1| |x^{358} - 1|\\] because $\\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \\dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\\dots (z - x^{358}) = z^{90} + 1$, and so \\[\\frac{1}{M} = \\dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \\dots |1 - x^{358}| = \\dfrac{1}{2^{90}} |1^{90} + 1| = \\dfrac{1}{2^{89}}.\\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = 91.",
"Let $p=\\sin1\\sin3\\sin5...\\sin89$ \\[p=\\sqrt{\\sin1\\sin3\\sin5...\\sin177\\sin179}\\] \\[=\\sqrt{\\frac{\\sin1\\sin2\\sin3\\sin4...\\sin177\\sin178\\sin179}{\\sin2\\sin4\\sin6\\sin8...\\sin176\\sin178}}\\] \\[=\\sqrt{\\frac{\\sin1\\sin2\\sin3\\sin4...\\sin177\\sin178\\sin179}{(2\\sin1\\cos1)\\cdot(2\\sin2\\cos2)\\cdot(2\\sin3\\cos3)\\cdot....\\cdot(2\\sin89\\cos89)}}\\] \\[=\\sqrt{\\frac{1}{2^{89}}\\frac{\\sin90\\sin91\\sin92\\sin93...\\sin177\\sin178\\sin179}{\\cos1\\cos2\\cos3\\cos4...\\cos89}}\\] $=\\sqrt{\\frac{1}{2^{89}}}$ because of the identity $\\sin(90+x)=\\cos(x)$ we want $\\frac{1}{p^2}=2^{89}$ Thus the answer is $2+89=091",
"Similar to Solution $2$, so we use $\\sin{2\\theta}=2\\sin\\theta\\cos\\theta$ and we find that: \\begin{align*}\\sin(4)\\sin(8)\\sin(12)\\sin(16)\\cdots\\sin(84)\\sin(88)&=(2\\sin(2)\\cos(2))(2\\sin(4)\\cos(4))(2\\sin(6)\\cos(6))(2\\sin(8)\\cos(8))\\cdots(2\\sin(42)\\cos(42))(2\\sin(44)\\cos(44))\\\\ &=(2\\sin(2)\\sin(88))(2\\sin(4))\\sin(86))(2\\sin(6)\\sin(84))(2\\sin(8)\\sin(82))\\cdots(2\\sin(42)\\sin(48))(2\\sin(44)\\sin(46))\\\\ &=2^{22}(\\sin(2)\\sin(88)\\sin(4)\\sin(86)\\sin(6)\\sin(84)\\sin(8)\\sin(82)\\cdots\\sin(42)\\sin(48)\\sin(44)\\sin(46))\\\\ &=2^{22}(\\sin(2)\\sin(4)\\sin(6)\\sin(8)\\cdots\\sin(82)\\sin(84)\\sin(86)\\sin(88))\\end{align*} Now we can cancel the sines of the multiples of $4$: \\[1=2^{22}(\\sin(2)\\sin(6)\\sin(10)\\sin(14)\\cdots\\sin(82)\\sin(86))\\] So $\\sin(2)\\sin(6)\\sin(10)\\sin(14)\\cdots\\sin(82)\\sin(86)=2^{-22}$ and we can apply the double-angle formula again: \\begin{align*}2^{-22}&=(\\sin(2)\\sin(6)\\sin(10)\\sin(14)\\cdots\\sin(82)\\sin(86)\\\\ &=(2\\sin(1)\\cos(1))(2\\sin(3)\\cos(3))(2\\sin(5)\\cos(5))(2\\sin(7)\\cos(7))\\cdots(2\\sin(41)\\cos(41))(2\\sin(43)\\cos(43))\\\\ &=(2\\sin(1)\\sin(89))(2\\sin(3)\\sin(87))(2\\sin(5)\\sin(85))(2\\sin(7)\\sin(87))\\cdots(2\\sin(41)\\sin(49))(2\\sin(43)\\sin(47))\\\\ &=2^{22}(\\sin(1)\\sin(89)\\sin(3)\\sin(87)\\sin(5)\\sin(85)\\sin(7)\\sin(83)\\cdots\\sin(41)\\sin(49)\\sin(43)\\sin(47))\\\\ &=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(41)\\sin(43))(\\sin(47)\\sin(49)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\end{align*} Of course, $\\sin(45)=2^{-\\frac{1}{2}}$ is missing, so we multiply it to both sides: \\[2^{-22}\\sin(45)=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(41)\\sin(43))(\\sin(45))(\\sin(47)\\sin(49)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\] \\[\\left(2^{-22}\\right)\\left(2^{-\\frac{1}{2}}\\right)=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\] \\[2^{-\\frac{45}{2}}=2^{22}(\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89))\\] Now isolate the product of the sines: \\[\\sin(1)\\sin(3)\\sin(5)\\sin(7)\\cdots\\sin(83)\\sin(85)\\sin(87)\\sin(89)=2^{-\\frac{89}{2}}\\] And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: \\[\\csc^2(1)\\csc^2(3)\\csc^2(5)\\csc^2(7)\\cdots\\csc^2(83)\\csc^2(85)\\csc^2(87)\\csc^2(89)=\\left(\\frac{1}{2^{-\\frac{89}{2}}}\\right)^2=\\left(2^{\\frac{89}{2}}\\right)^2=2^{89}\\] The answer is therefore $m+n=(2)+(89)=091.",
"Let $p=\\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ$. Then, $\\sqrt{\\frac{1}{p}}=\\prod_{k=1}^{45} \\sin(2k-1)^\\circ$. Since $\\sin\\theta=\\cos(90^{\\circ}-\\theta)$, we can multiply both sides by $\\frac{\\sqrt{2}}{2}$ to get $\\sqrt{\\frac{1}{2p}}=\\prod_{k=1}^{23} \\sin(2k-1)^\\circ\\cos(2k-1)^\\circ$. Using the double-angle identity $\\sin2\\theta=2\\sin\\theta\\cos\\theta$, we get $\\sqrt{\\frac{1}{2p}}=\\frac{1}{2^{23}}\\prod_{k=1}^{23} \\sin(4k-2)^\\circ$. Note that the right-hand side is equal to $\\frac{1}{2^{23}}\\prod_{k=1}^{45} \\sin(2k)^\\circ\\div \\prod_{k=1}^{22} \\sin(4k)^\\circ$, which is equal to $\\frac{1}{2^{23}}\\prod_{k=1}^{45} \\sin(2k)^\\circ\\div \\prod_{k=1}^{22} 2\\sin(2k)^\\circ\\cos(2k)^\\circ$, again, from using our double-angle identity. Putting this back into our equation and simplifying gives us $\\sqrt{\\frac{1}{2p}}=\\frac{1}{2^{45}}\\prod_{k=23}^{45} \\sin(2k)^\\circ\\div \\prod_{k=1}^{22} \\cos(2k)^\\circ$. Using the fact that $\\sin\\theta=\\cos(90^{\\circ}-\\theta)$ again, our equation simplifies to $\\sqrt{\\frac{1}{2p}}=\\frac{\\sin90^\\circ}{2^{45}}$, and since $\\sin90^\\circ=1$, it follows that $2p = 2^{90}$, which implies $p=2^{89}$. Thus, $m+n=2+89=091.",
"Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think). Recall that the roots of $x^n+1$ are $e^{\\frac{(2k-1)\\pi i}{n}}, k=1,2,...,n$, we have \\[x^n + 1 = \\prod_{k=1}^{n}(x-e^{\\frac{(2k-1)\\pi i}{n}})\\] Let $x=1$, and take absolute value of both sides, \\[2 = \\prod_{k=1}^{n}|1-e^{\\frac{(2k-1)\\pi i}{n}}|= 2^n\\prod_{k=1}^{n}|\\sin\\frac{(2k-1)\\pi}{2n}|\\] or, \\[\\prod_{k=1}^{n}|\\sin\\frac{(2k-1)\\pi}{2n}| = 2^{-(n-1)}\\] Let $n$ be even, then, \\[\\sin\\frac{(2k-1)\\pi}{2n} = \\sin\\left(\\pi - \\frac{(2k-1)\\pi}{2n}\\right) = \\sin\\left(\\frac{(2(n-k+1)-1)\\pi}{2n}\\right)\\] so, \\[\\prod_{k=1}^{n}\\left|\\sin\\frac{(2k-1)\\pi}{n}\\right| = \\prod_{k=1}^{\\frac{n}{2}}\\sin^2\\frac{(2k-1)\\pi}{2n}\\] Set $n=90$ and we have \\[\\prod_{k=1}^{45}\\sin^2\\frac{(2k-1)\\pi}{180} = 2^{-89}\\], \\[\\prod_{k=1}^{45}\\csc^2\\frac{(2k-1)\\pi}{180} = 2^{89}\\] -Mathdummy",
"Recall that $\\sin\\alpha\\cdot \\sin(60^{\\circ}-\\alpha)\\cdot \\sin(60^{\\circ}+\\alpha)=\\frac{1}{4}\\cdot \\sin3\\alpha$ Since it is in csc, we can write in sin and then take reciprocal. We can group them by threes, $P=(\\sin1^{\\circ}\\cdot \\sin59^{\\circ}\\cdot \\sin61^{\\circ})\\cdots(\\sin29^{\\circ}\\cdot \\sin31^{\\circ}\\cdot \\sin89^{\\circ})$. Thus \\begin{align*} P &=\\frac{1}{4^{15}}\\cdot \\sin3^{\\circ}\\cdot \\sin9^{\\circ}\\cdots\\sin87^{\\circ}\\\\ &=\\frac{1}{4^{20}}\\cdot \\sin9^{\\circ}\\cdot \\sin27^{\\circ}\\cdot \\sin45^{\\circ}\\cdot \\sin63^{\\circ}\\cdot \\sin81^{\\circ}\\\\ &=\\frac{1}{4^{20}}\\cdot \\frac{\\sqrt{2}}{2}\\cdot \\sin9^{\\circ}\\cdot \\cos9^{\\circ}\\cdot \\sin27^{\\circ}\\cdot \\cos27^{\\circ}\\\\ &=\\frac{1}{4^{21}}\\cdot \\frac{\\sqrt{2}}{2}\\cdot \\sin18^{\\circ}\\cdot \\cos36^{\\circ}=\\frac{\\sqrt{2}}{2^{45}} \\end{align*} So we take reciprocal, $\\frac 1P=2^{\\frac{89}{2}}$, the desired answer is $\\frac{1}{P^2}=2^{89}$ leads to answer $091 ~bluesoul",
"We have \\[\\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ = \\left(\\frac{1}{\\sin1^\\circ \\cdot \\sin3^\\circ \\cdots \\sin89^\\circ}\\right)^2.\\] Multiplying by $\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}$ gives \\[\\left(\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\sin1^\\circ \\sin2^\\circ \\cdot \\sin3^\\circ \\cdots \\sin88^\\circ \\cdot \\sin89^\\circ}\\right)^2\\] \\[= \\left(\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\sin1^\\circ \\sin2^\\circ \\cdot \\sin3^\\circ \\cdots \\sin 45^\\circ \\cdot \\cos 44^\\circ \\cdot \\cos 43^\\circ \\cdots \\cos1^\\circ}\\right)^2.\\] Using $\\sin\\alpha \\cos\\alpha = \\frac{1}{2}\\sin{2\\alpha}$ gives \\[\\left(\\frac{\\sin2^\\circ \\cdot \\sin4^\\circ \\cdots \\sin88^\\circ}{\\frac{1}{2} \\sin2^\\circ \\cdot \\frac{1}{2} \\sin4^\\circ \\cdots \\frac{1}{2} \\sin88^\\circ \\cdot \\sin45^\\circ}\\right) ^2\\] \\[= \\left(\\frac{1}{(\\frac{1}{2})^{44} \\cdot \\frac{\\sqrt{2}}{2}}\\right)^2\\] \\[= 2^{89}.\\] Thus, the answer is $2+89 = 091",
"Consider the product $\\prod_{k=1}^{45} \\csc^2(2k-1) = \\prod_{k=45}^{1} \\sec^2(2k-1) = \\prod_{k=1}^{45} \\sec^2(2k-1) = \\prod_{k=1}^{45} (1+\\tan^2(2k-1))$ However, note that the $45$ numbers in the form $\\sqrt{\\tan^2(2k-1)} = \\tan(2k-1)$ for $1\\le{k}\\le{45}$ are precisely the roots of the equation $\\frac{1}{\\tan{(\\tan^{-1}{90x}})} = 0$. Thus, it suffices to find $|P(-1)|$, where $P$ is the polynomial formed by the denominator of $\\tan{(\\tan^{-1}{90\\sqrt{x}})}$. This is true because $\\prod_{k=1}^{45} (x-\\tan^2(2k-1)) = \\prod_{k=1}^{45} -(-x+\\tan^2(2k-1))$ gives us the factored root form of the equation where $P$ is undefined, which corresponds to the equation where the denominator equals $0$. It remains to find the denominator of $P$; fortunately, we may use tangent angle multiplication rules. Specifically, the denominator of $P$ will be $\\sum_{n=0} ^{45} (-1)^n\\binom{90}{2n}\\sqrt{x}^{2n} = \\sum_{n=0} ^{45} (-1)^n\\binom{90}{2n}x^n$. Evaluating at $x = -1$, we may obtain the sum $\\sum_{n=0} ^{45} (-1)^n\\binom{90}{2n}(-1)^{n} = \\sum_{n=0} ^{45} \\binom{90}{2n}$. From here, there are two ways to finish; the first is to recognize the well known sum $\\sum_{n=0} ^{k} \\binom{2k}{2n} = 2^(2k-1)$, from which we may plug in $k = 45$ to see $|P(-1)| = 2^(45*2-1) = 2^{89}$ to obtain the answer of $2+89=091"
] |
2015-I-14
| 2,015
| 14
|
For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.
| 483
|
I
|
[
"Let $n\\ge 2$ and define $a(n) = \\left\\lfloor \\sqrt n \\right\\rfloor$. For $2\\le n \\le 1000$, we have $1\\le a(n)\\le 31$. For $a^2 \\le x < (a+1)^2$ we have $y=ax$. Thus $A(n+1)-A(n)=a(n+\\tfrac 12) = \\Delta_n$ (say), and $\\Delta_n$ is an integer if $a$ is even; otherwise $\\Delta_n$ is an integer plus $\\tfrac 12$. If $a=1$, $n\\in \\{1,2,3\\}$ and $\\Delta_n$ is of the form $k(n)+\\tfrac 12$ so $A(n)$ is an integer when $n$ is even. If $a=2$, $n\\in\\{4,\\ldots , 8\\}$ and $\\Delta_n$ is an integer for all $n$. Since $A(3)$ is not an integer, so $A(n)$ is not an integer for any $n$. If $a=3$, $n\\in\\{9,\\ldots , 15\\}$ and $\\Delta_n$ is of the form $k(n)+\\tfrac 12$. Since $A(8)$ is of the form $k+\\tfrac 12$ so $A(n)$ is an integer only when $n$ is odd. If $a=4$, $n\\in\\{16,\\ldots , 24\\}$ and $\\Delta_n$ is an integer for all $n$. Since $A(15)$ is an integer so $A(n)$ is an integer for all $n$. Now we are back to where we started; i.e., the case $a=5$ will be the same as $a=1$ and so on. Thus, \\begin{align} a(n)\\equiv 1\\pmod 4 \\qquad &\\Longrightarrow \\qquad A(n) \\in \\mathbb{Z} \\textrm{ for even } n, \\\\ a(n)\\equiv 2\\pmod 4 \\qquad &\\Longrightarrow \\qquad A(n) \\not\\in \\mathbb{Z} \\textrm{ for any } n, \\\\ a(n)\\equiv 3\\pmod 4 \\qquad &\\Longrightarrow \\qquad A(n) \\in \\mathbb{Z} \\textrm{ for odd } n, \\\\ a(n)\\equiv 0\\pmod 4 \\qquad &\\Longrightarrow \\qquad A(n) \\in \\mathbb{Z} \\textrm{ for all } n. \\end{align} For each $a$ there are $2a+1$ corresponding values of $n$: i.e., $n\\in \\{a^2, \\ldots , (a+1)^2-1\\}$. Thus, the number of values of $n$ corresponding to $(4)$ (i.e., $a(n)\\equiv 0\\pmod 4$) is given by \\[\\sum_{\\substack{a=4k \\\\ a\\le 31}}(2a+1) = \\sum_{k=1}^7 (8k+1)=231.\\] The cases $(1)$ and $(3)$ combine to account for half the values of $n$ corresponding to odd values of $a(n)$; i.e., \\[\\frac 12 \\cdot \\sum_{\\substack{a=2k+1 \\\\ a\\le 31}} (2a+1) = \\sum_{k=0}^{15} (2k+\\tfrac 32) = 264\\]However, this also includes the odd integers in $\\{1001, \\ldots , 1023\\}$. Subtracting $12$ to account for these, we get the number of values of $n$ corresponding to cases $(1)$ and $(3)$ to be $264-12=252$. Adding the contributions from all cases we get our answer to be $231+252= 483.",
"By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$. The width of each trapezoid is $3, 5, 7$, etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\\frac{1}{2}$. Whenever $a$ is even, the value of $A(n)$ increases by an integer value. Since each trapezoid always has an odd width, every value of $n$ is not an integer when $a \\pmod{4} \\equiv 2$, and is an integer when $a \\pmod{4} \\equiv 0$. Every other value is an integer when $a$ is odd. Therefore, it is simply a matter of determining the number of values of $n$ where $a \\pmod{4} \\equiv 0$ ($(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)$), and adding the number of values of $n$ where $a$ is odd ($\\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}$). Adding the two values gives $231+252=483."
] |
2015-I-15
| 2,015
| 15
|
A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$ , and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\text{o}$ . The block is then sliced in half along the plane that passes through point $A$ , point $B$ , and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$ , where $a$ , $b$ , and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$ . [asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]
| 53
|
I
|
[
"Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $N$ be the center of the bottom face, and $M$ the midpoint of $\\overline{AB}$. Then $ON=4$, $MN=3$ (because of the 120 degree angle), and so $OM=5$. Project $C$ and $D$ onto the bottom face to get $X$ and $Y$, respectively. Then the section $ABCD$ (whose area we need to find), is a stretching of the section $ABXY$ on the bottom face. The ratio of stretching is $\\frac{OM}{MN}=\\frac{5}{3}$, and we do not square this value when finding the area because it is only stretching in one direction. Using 30-60-90 triangles and circular sectors, we find that the area of the section $ABXY$ is $18\\sqrt{3}\\ + 12 \\pi$. Thus, the area of section $ABCD$ is $20\\pi + 30\\sqrt{3}$, and so our answer is $20+30+3=053.",
"Label the points same as in the first sentence above. Consider a view of the cylinder such that height is disregarded, i.e. a top view. From this view, note that Cylinder $O$ has become a circle with $\\overarc{AB}$ = $\\overarc{CD}$ = $120^\\text{o}$. Using 30-60-90 triangles, we get rectangle $ABCD$ to have a horizontal component of $6$. Now, consider a side view, such that $A$ and $B$ coincide at the bottom of the diagram. From this view, consider the right triangle composed of hypotenuse $AD$ and a point along the base of the viewpoint, which will be labeled as $E$. From the top view, $AE = 6$. Because of the height of the cylinder, $DE$ is equal to $8$. This makes $AD$ equal to $10$. Now, the use of simple calculus is required. Conceptualize an infinite number of lines perpendicular to $AE$ intersecting both $AE$ and $AD$. Consider the area between point $A$ and the first vertical line. Label the point where the line intersects AE as E', and the point where the line intersects AD as D'. The area of the part of the initial unpainted face within these two positions approaches a rectangle with length AD' and width $w$. The area of the base within these two positions approaches a rectangle with length AE' and width $w$. The ratio of AD':AE' is 10:6, since the ratio of AD:AE is 10:6. This means that the area of the initial unpainted surface within these two positions to the area of the base within these two positions is equal to 10$w$:6$w$ = 10:6. Through a similar argument, the areas between each set of vertical lines also maintains a ratio of 10:6. Therefore, the ratio of the area we wish to find to the area of the base between AB and CD (from the top perspective) is 10:6. Using 30-60-90 triangles and partial circles, the area of the base between AB and CD is calculated to be $18\\sqrt{3}\\ + 12 \\pi$. The area of the unpainted surface therefore becomes $20\\pi + 30\\sqrt{3}$, and so our answer is $053.",
"This problem can be calculus-bashed (for those like me who never noticed the surface was merely a stretch of its projection). Label points as in the first paragraph of Solution 1 ($A$ and $B$ as given, $M$ the midpoint of $AB$, $O$ the center of the cylinder, $T$ the center of the bottom face of the cylinder). Because of the 120 degrees and right triangle calculations, we find $MT$ = 3, $OT$ = 4, $OM$ = 5). We will be integrating with respect to the y-coordinate which we define as distance downwards from $O$ (in this system, the $y$-coordinate of the bottom face would be 4). We note that by similar triangles, we have that the length from $O$ to the point on the unpainted surface of coordinate $y$ is $\\ell = \\frac{5}{4} y$, and therefore $d\\ell = \\frac{5}{4} dy$. Define the segment $A'B'$ to be the intersection of the painted surface with the circular cross section of the cylinder of coordinate $y$, with endpoints $A'$ and $B'$ and midpoint $M'$, with $T'$ the center of this circular cross section. Then, by similar triangles, $T'M' = \\frac{3}{4} y$ and thus \\[A'B' = 2A' M' = 2 \\sqrt{ 6^2 - \\left( \\frac{3}{4}y\\right)^2 } = \\frac{3}{2} \\sqrt{ 64 - y^2 }\\]. We know that $A'B'$ is perpendicular to $OM$. Now we can set up our integral: we will integrate $y$ from 0 to 4 and multiply by two because the total height is 8. \\[A = 2\\int_0^4 \\left(\\frac{3}{2}\\sqrt{ 64 - y^2 }\\right) \\left( \\frac{5}{4} dy\\right)\\] \\[A = \\frac{15}{4} \\int_0^4 \\sqrt{ 64 - y^2 }dy\\] Then we substitute $y = 8\\sin \\theta$ with $dy = 8 \\cos \\theta d \\theta$, changing the bounds to 0 to $\\frac{\\pi}{6}$ as appropriate. \\[A = \\frac{15}{4} \\int_0^\\frac{\\pi}{6} \\sqrt{ 64 - 64\\sin^2 \\theta }\\cdot 8\\cos\\theta d\\theta\\] \\[A = 240 \\int_0^\\frac{\\pi}{6} \\cos^2 \\theta d\\theta\\] \\[A = 240 \\left[ \\frac{\\theta}{2} + \\frac{\\sin 2\\theta}{4} \\right]_0^\\frac{\\pi}{6} = 240 \\left[ \\frac{\\pi}{12} + \\frac{\\sqrt{3}}{8} \\right] = 20{\\pi} + 30\\sqrt{3}\\] Therefore, $a + b + c = 20 + 30 + 3 = 053.",
"Extend the cylinder such that the cylinder has a height of $16$ and the same radius of $6$. Let $Q$ be a point on the circumference of the top face of the cylinder. Let $R$ be a point on the bottom face of the cylinder, and the farthest point from A on the cylinder. Let a plane $P$ pass through points $Q$ and $R$. The cross section formed by the intersection of the plane and this stretched cylinder, is an ellipse. Notice that the cross section area we want to find is a part the part of this ellipse contained in the original cylinder. The semi-major axis length of the ellipse is \\[\\sqrt{6^2 + 8^2} = 10\\]. The semiminor axis length is just the radius of the cylinder, which is $6$. If we take the plane of this ellipse to be the $2$ dimensional Cartesian plane and take the center of this ellipse as $(0, 0)$, this ellipse has the equation : \\[\\frac{x^2}{10^2} + \\frac{y^2}{6^2} = 1\\]. Notice that the cross section area we want to find is the area bounded by this ellipse from $x = -5$ to $x = 5$. Thus we must integrate the equation of the ellipse from $-5$ to $5$. That will give us the positive area bounded by the ellipse and the $x$ axis. We want to double it, as we want both the positive and negative area as that is our cross section. Rewrite the ellispe equation as \\[y = \\frac{3}{5} \\sqrt{100 - x^2}\\]. We want to find \\[2\\int_{-5}^{5} \\left(\\frac{3}{5}\\sqrt{100 - x^2 }\\right) \\left( dx\\right)\\]. Evaluate this integral : It comes out to be \\[20\\pi + 30\\sqrt{3}\\]. Finally, the answer we give, is \\[20 + 30 + 3 = 053\\]. Motivation : Notice that a cross section of any diagonal plane not going through the top or bottom face of the cylinder is an ellipse. When the plane does cross the bottom and top faces, the resulting cross section will be a portion of an ellipse. When the plane is perpendicular to its top and bottom faces, the resulting cross section is a rectangle. So, from the moment the cross section starts cutting both circular faces, to the moment it is perpendicular to the circular bases, it goes from a full ellipse, to a rectangle. Hence, the cross section we want is a part of the ellipse."
] |
2015-II-1
| 2,015
| 1
|
Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ .
| 131
|
II
|
[
"If $N$ is $22$ percent less than one integer $k$, then $N=\\frac{78}{100}k=\\frac{39}{50}k$. In addition, $N$ is $16$ percent greater than another integer $m$, so $N=\\frac{116}{100}m=\\frac{29}{25}m$. Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$. Setting these two equal, we have $\\frac{39}{50}k=\\frac{29}{25}m$. Multiplying by $50$ on both sides, we get $39k=58m$. The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$, so $N=1131$. The answer is $131.",
"Continuing from Solution 1, we have $N=\\frac{39}{50}k$ and $N=\\frac{29}{25}m$. It follows that $k=\\frac{50}{39}N$ and $m=\\frac{25}{29}N$. Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\\frac{50}{39}$ and $\\frac{25}{29}$. In other words, $N$ is a multiple of both $29$ and $39$. That makes $N=\\operatorname{lcm}(29,39)=29\\cdot39=1131$. The answer is $131.",
"We can express $N$ as $0.78a$ and $1.16b$, where $a$ and $b$ are some positive integers. $N=0.78a=1.16b\\implies100N=78a=116b.$ Let us try to find the smallest possible value of $100N$, first ignoring the integral constraint. Obviously, we are just trying to find the LCM of $78$ and $116.$ They have no common factors but $2$, so we multiply $78$ and $116$ and divide by $2$ to get $4524.$ This is obviously not divisible by $100$, so this is not possible, as it would imply that $N=\\dfrac{4524}{100},$ which is not an integer. This can be simplified to $\\dfrac{1131}{25}$. We know that any possible value of $N$ will be an integral multiple of this value; the smallest such $N$ is achieved by multiplying this value by $25.$ We arrive at $1131$, which is $131 ~Technodoggo"
] |
2015-II-2
| 2,015
| 2
|
In a new school, $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 25
|
II
|
[
"We see that $40\\% \\cdot 100\\% + 30\\% \\cdot 80\\% + 20\\% \\cdot 50\\% + 10\\% \\cdot 20\\% = 76\\%$ of students are learning Latin. In addition, $30\\% \\cdot 80\\% = 24\\%$ of students are sophomores learning Latin. Thus, our desired probability is $\\dfrac{24}{76}=\\dfrac{6}{19}$ and our answer is $6+19=025.",
"Assume that there are $100$ students in the school. There are $40$ freshmen taking Latin, $24$ sophomores taking Latin, $10$ juniors taking Latin, and $2$ seniors taking Latin. We get the probability to be the number of sophomores taking Latin over the total number of students taking Latin, or $\\dfrac{24}{76}$. Simplifying, we get $\\dfrac{6}{19}$. Adding, we get $025."
] |
2015-II-3
| 2,015
| 3
|
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ .
| 476
|
II
|
[
"The three-digit integers divisible by $17$, and their digit sum: \\[\\begin{array}{c|c} m & s(m)\\\\ \\hline 102 & 3 \\\\ 119 & 11\\\\ 136 & 10\\\\ 153 & 9\\\\ 170 & 8\\\\ 187 & 16\\\\ 204 & 6\\\\ 221 & 5\\\\ 238 & 13\\\\ 255 & 12\\\\ 272 & 11\\\\ 289 & 19\\\\ 306 & 9\\\\ 323 & 8\\\\ 340 & 7\\\\ 357 & 15\\\\ 374 & 14\\\\ 391 & 13\\\\ 408 & 12\\\\ 425 & 11\\\\ 442 & 10\\\\ 459 & 18\\\\ 476 & 17 \\end{array}\\] Thus the answer is $476.",
"We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$. Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$, shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds digit is less than $9$. Unfortunately, $68$ does not fit the criteria, but $374$ does, meaning that, instead of continually adding multiples of $17$, we can stop here and simply add $102$ to reach our final answer of $476. ~Tiblis (Comment from another person: Actually, this doesn't work because you can't be sure there are no numbers between 374 and 476 that work. This solution just lucks out.)",
"We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$. Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$, shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds digit is less than $9$. Unfortunately, $68$ does not fit the criteria, but $374$ does, meaning that, instead of continually adding multiples of $17$, we can stop here and simply add $102$ to reach our final answer of $476. ~Tiblis (Comment from another person: Actually, this doesn't work because you can't be sure there are no numbers between 374 and 476 that work. This solution just lucks out.)",
"The digit sum of a base $10$ integer $m$ is just $m\\pmod{9}$. In this problem, we know $17\\mid m$, or $m=17k$ for a positive integer $k$. Also, we know that $m\\equiv 17\\equiv -1\\pmod{9}$, or $17k\\equiv -k\\equiv -1\\pmod{9}$. Obviously $k=1$ is a solution. This means in general, $k=9x+1$ is a solution for non-negative integer $x$. Checking the first few possible solutions, we find that $m=476, and we're done.",
"Since the sum of the digits in the base-10 representation of $m$ is $17$, we must have $m\\equiv 17 \\pmod{9}$ or $m\\equiv -1\\pmod{9}$. We also know that since $m$ is divisible by 17, $m\\equiv 0 \\pmod{17}$. To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set $m\\equiv (-1)(17)(8)\\pmod {153}$, we find that $m\\equiv 0\\pmod{17}$ and $m\\equiv -1\\pmod{9}$, because $17\\cdot 8\\equiv 136 \\equiv 1\\pmod{9}$. The trick to getting here was to find the number $x$ such that $17x\\equiv 1\\pmod{9}$, so that when we take things $\\pmod{9}$, the $17$ goes away. We can do this using the Extended Euclidean Algorithm or by guess and check to find that $x\\equiv 8\\pmod{9}$. Finally, since $m\\equiv 17\\pmod{153}$, we repeatedly add multiples of $153$ until we get a number in which its digits sum to 17, which first happens when $m=476.",
"We proceed by casework on the number of digits. Clearly the answer must have at least two digits, seeing as the maximum digit sum for a one-digit number is 9. The answer must also have less than 4 digits, because this is the AIME. Case 1: The answer is a 2-digit number. Represent the number as $10a + b$, where $0 < a \\leq 9$ and $0 \\leq b \\leq 9$. The conditions of the problem translated into algebra are: \\[17|10a+b\\] \\[a+b=17\\] By the Euclidean Algorithm, this is equivalent to: \\[17|9a\\] 9 is not a factor of 17, so $17|a$. So $a$ must be a multiple of 17, but this is impossible because of the conditions we placed on $a$ and $b$. (Alternatively, note that the only possible options are 89 and 98, and neither works.) Case 2: The answer is a 3-digit number. Represent the number as $100a+10b+c$, where $0 < a \\leq 9$ and $0 \\leq b,c \\leq 9$. Translating the conditions again: \\[17|100a+10b+c\\] \\[a+b+c=17\\] \\[17|99a+9b\\] \\[17|9(11a+b)\\] \\[17|11a+b\\] Testing multiples of 17 yields $(4, 7, 6)$ as the minimal solution for $(a, b, c)$ and thus the answer is $476. - gting"
] |
2015-II-4
| 2,015
| 4
|
In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ .
| 18
|
II
|
[
"Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$, where $E$ is closer to $D$. Subtract the two bases and divide to find that $ED$ is $\\log 8$. The altitude can be expressed as $\\frac{4}{3} \\log 8$. Therefore, the two legs are $\\frac{5}{3} \\log 8$, or $\\log 32$. The perimeter is thus $\\log 32 + \\log 32 + \\log 192 + \\log 3$ which is $\\log 2^{16} 3^2$. So $p + q = 018",
"Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$, with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\\log 192 - \\log 3 = \\log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $\\log 64$ being 6. Then the 2 legs of the trapezoid is $\\sqrt{3^2+4^2}=5=\\log 32$. And we have the answer: $\\log 192 + \\log 32 + \\log 32 + \\log 3 = \\log(192 \\cdot 32 \\cdot 32 \\cdot 3) = \\log(2^6 \\cdot 3 \\cdot 2^5 \\cdot 2^5 \\cdot 3) = \\log(2^{16} \\cdot 3^2) \\Rightarrow 16+2 = 18 -dragoon",
"Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$, with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\\log 192 - \\log 3 = \\log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $\\log 64$ being 6. Then the 2 legs of the trapezoid is $\\sqrt{3^2+4^2}=5=\\log 32$. And we have the answer: $\\log 192 + \\log 32 + \\log 32 + \\log 3 = \\log(192 \\cdot 32 \\cdot 32 \\cdot 3) = \\log(2^6 \\cdot 3 \\cdot 2^5 \\cdot 2^5 \\cdot 3) = \\log(2^{16} \\cdot 3^2) \\Rightarrow 16+2 = 18 -dragoon",
"Let $ABCD$ be the trapezoid, where $\\overline{AB} || \\overline{CD}$ and $AB = \\log 3$ and $CD = \\log 192$. Draw altitudes from $A$ and $B$ to $\\overline{CD}$ with feet at $E$ and $F$, respectively. $AB = \\log 3$, so $EF = \\log 3$. Now, we attempt to find $DE + FC$, or what's left of $CD$ after we take out $EF$. We make use of the two logarithmic rules: \\[\\log(xy) = \\log x + \\log y\\] \\[\\log(x^a) = a\\log(x)\\] \\[CD = \\log 192 = \\log (3 \\cdot 2^6) = \\log 3 + \\log(2^6) = \\log 3 + 6\\log 2\\] Thus, since $CD = DE + EF + FC = \\log 3 + 6\\log 2$, $CD - EF = \\log 3 + 6\\log 2 - \\log 3 = 6\\log 2 = DE + FC$. Now, why was finding $DE + FC$ important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially \"glue\" triangles $\\triangle DAE$ and $\\triangle BFC$ together to get $\\triangle XC'D'$, where $X$ is the point where $A$ and $B$ became one. Note we can do this because $\\triangle DAE$ and $\\triangle BFC$ are both right triangles with a common leg length (the altitude of trapezoid $ABCD$). Triangle $XC'D'$ has a base of $C'D'$, which is just equal to $DE + FC = 6\\log 2$. It is equal to $DE + FC$ because when we brought triangles $\\triangle DAE$ and $\\triangle BFC$ together, the length of $CD$ was not changed except for taking out $EF$. $XC' = XD'$ since $AD = BC$ because the problem tells us we have an isosceles trapezoid. Drop and altitude from $X$ to $C'D'$ The altitude has length $\\log 16 = 4\\log 2$. The altitude also bisects $C'D'$ since $\\triangle XC'D'$ is isosceles. Let the foot of the altitude be $M$. Then $MD' = 3\\log 2$ (Remember that C'D' was $6\\log 2$, and then it got bisected by the altitude). Thus, the hypotenuse, $XD'$ must be $5\\log 2$ from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of $\\log 2$. Since $XD' = XC' = BC = AD$, $BC = AD = 5\\log 2 = \\log 2^5$. Now, we have $CD = \\log (3 \\cdot 2^6)$, $AB = \\log 3$, and $BC = AD = \\log 2^5$. Thus, their sum is \\[\\log (3 \\cdot 2^6) + \\log 3 + \\log 2^5 + \\log 2^5 = \\log (2^16 \\cdot 3^2)\\] Thus, $p + q = 16 + 2 = 18. ~Extremelysupercooldude",
"Let $a=\\log2$ and $b=\\log3$ so that the base lengths are $\\log3=b$ and $\\log192=\\log(3\\cdot64)=\\log3+\\log\\left(2^6\\right)=6a+b$ and the altitudes are $\\log16=4a$. Then we have the following picture: [asy] import graph; unitsize(1cm); draw((0,0)--(3,4)--(9,4)--(12,0)--cycle); draw((3,4)--(3,0));draw((9,4)--(9,0)); label(\"$b$\",(3,4)--(9,4),N); label(\"$6a+b$\",(0,0)--(12,0),S); label(\"$b$\",(0,0)--(12,0),N); label(\"$4a$\", (3,0)--(3,4),W); label(\"$4a$\", (9,0)--(9,4),E); [/asy] Note that we have the two right triangles to the side; one of each of their bases is an altitude, which we know the length to be $4a$. The length of the other base can be calculated as $\\dfrac{(6a+b)-b}2=3a$ via simple isosceles trapezoid geometry; it is clear that each right triangle is actually a $3-4-5$ triangle, so we know their hypotenuses (and the remaining unknown sides of our original trapezoid) have length $5a$ (because of the $3a$ and $4a$ bases). Our answer is therefore $5a+5a+b+6a+b=16a+2b=16\\log2+2\\log3=\\log\\left(2^{16}3^2\\right)$; $p+q=16+2=018. QED. ~Technodoggo"
] |
2015-II-5
| 2,015
| 5
|
Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ .
| 90
|
II
|
[
"Call the given grid \"Grid A\". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \\times (n-1)$. There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid B is $2n(n-1)$, and the number of ways to pick two squares out of Grid A is $\\dbinom{n^2}{2}$. So, the probability that the two chosen squares are adjacent is $\\frac{2n(n-1)}{\\binom{n^2}{2}} = \\frac{2n(n-1)}{\\frac{n^2(n^2-1)}{2}} = \\frac{4}{n(n+1)}$. We wish to find the smallest positive integer $n$ such that $\\frac{4}{n(n+1)} < \\frac{1}{2015}$, and by inspection the first such $n$ is $090.",
"Consider a $3 \\times 3$ grid, where there are $4$ corner squares, $4$ edge squares, and $1$ center square. A $4 \\times 4$ grid has $4$ corner squares, $8$ edge squares, and $4$ center squares. By examining simple cases, we can conclude that for a grid that is $n \\times n$, there are always $4$ corners squares, $4(n-2)$ edge squares, and $n^2-4n+4$ center squares. Each corner square is adjacent to $2$ other squares, edge squares to $3$ other squares, and center squares to $4$ other squares. In the problem, the second square can be any square that is not the first, which means there are $n^2-1$ to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is $\\frac{2}{n^2-1}(\\frac{4}{n^2}) +\\frac{3}{n^2-1}(\\frac{4(n-2)}{n^2}) +\\frac{4}{n^2-1}(\\frac{n^2-4n+4}{n^2})$. Simplifying, we get $\\frac{4}{n(n+1)}$ which we can set to be less than $\\frac{1}{2015}$. By inspection, we find that the first such $n$ is $090.",
"There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is $\\frac{(n-2)(n-2)}{n^2}$ multiplied by $\\frac{4}{n^2 -1}$. Add that to the probability of an edge and an adjacent square( $\\frac{4n-8}{n^2}$ multiplied by $\\frac{3}{n^2-1}$) and the probability of a corner and an adjacent square( $\\frac{4}{n^2}$ multiplied by $\\frac{2}{n^2-1}$) to get $\\frac{4n^2-4n}{n^4-n^2}$. Simplify to get $\\frac{4}{n^2+n}$. With some experimentation, we realize that the smallest value of n is $090.",
"Notice how a chosen unit square on the grid has 4 vertically & horizontally adjacent squares around it (not counting corners or sides.) That's $\\frac{4}{n^2}$. Using this, we rewrite $\\frac{1}{2015}$ as $\\frac{4}{8060}$. Notice that the denominator $8060$ is really close to $90^2$, and the problem is asking for the least positive integer less than $\\frac{1}{2015}$. Therefore, the closest possible estimation is $090 as the correct answer. ~orenbad",
"Notice how a chosen unit square on the grid has 4 vertically & horizontally adjacent squares around it (not counting corners or sides.) That's $\\frac{4}{n^2}$. Using this, we rewrite $\\frac{1}{2015}$ as $\\frac{4}{8060}$. Notice that the denominator $8060$ is really close to $90^2$, and the problem is asking for the least positive integer less than $\\frac{1}{2015}$. Therefore, the closest possible estimation is $090 as the correct answer. ~orenbad"
] |
2015-II-6
| 2,015
| 6
|
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?" After some calculations, Jon says, "There is more than one such polynomial." Steve says, "You're right. Here is the value of $a$ ." He writes down a positive integer and asks, "Can you tell me the value of $c$ ?" Jon says, "There are still two possible values of $c$ ." Find the sum of the two possible values of $c$ .
| 440
|
II
|
[
"We call the three roots (some may be equal to one another) $x_1$, $x_2$, and $x_3$. Using Vieta's formulas, we get $x_1+x_2+x_3 = a$, $x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}$, and $x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2}$. Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \\cdot x_1 \\cdot x_2+2 \\cdot x_1 \\cdot x_3+2 \\cdot x_2 \\cdot x_3 = a^2$. We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \\cdot \\frac{a^2-81}{2}$. Simplifying the right side: \\begin{align*} a^2-2 \\cdot \\frac{a^2-81}{2} &= a^2-a^2+81\\\\ &= 81.\\\\ \\end{align*} So, we know $x_1^2+x_2^2+x_3^2 = 81$. We can then list out all the triples of positive integers whose squares sum to $81$: We get $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$. These triples give $a$ values of $13$, $15$, and $15$, respectively, and $c$ values of $64$, $216$, and $224$, respectively. We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$. Thus, the two $c$ values are $216$ and $224$, which sum to $\\text{440}. ~BealsConjecture~",
"We call the three roots (some may be equal to one another) $x_1$, $x_2$, and $x_3$. Using Vieta's formulas, we get $x_1+x_2+x_3 = a$, $x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}$, and $x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2}$. Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \\cdot x_1 \\cdot x_2+2 \\cdot x_1 \\cdot x_3+2 \\cdot x_2 \\cdot x_3 = a^2$. We can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \\cdot \\frac{a^2-81}{2}$. Simplifying the right side: \\begin{align*} a^2-2 \\cdot \\frac{a^2-81}{2} &= a^2-a^2+81\\\\ &= 81.\\\\ \\end{align*} So, we know $x_1^2+x_2^2+x_3^2 = 81$. We can then list out all the triples of positive integers whose squares sum to $81$: We get $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$. These triples give $a$ values of $13$, $15$, and $15$, respectively, and $c$ values of $64$, $216$, and $224$, respectively. We know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$. Thus, the two $c$ values are $216$ and $224$, which sum to $\\text{440}. ~BealsConjecture~",
"First things first. Vietas gives us the following: \\begin{align} x_1+x_2+x_3 = a\\\\ x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}\\\\ x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2} \\end{align} From $(2)$, $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$, which implies that both sides of $(2)$ are even. Then, from $(1)$, we see that an odd number of $x_1$, $x_2$, and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers. Now, the theoretical maximum value of the left side of $(2)$ is $3 \\cdot \\biggl(\\frac{a}{3}\\biggr)^2=\\frac{a^2}{3}$. That means that the maximum bound of $a$ is where \\[\\frac{a^2}{3} > \\frac{a^2-81}{2},\\] which simplifies to $\\sqrt{243} > a$, meaning \\[16 > a.\\] So now we have that $9<a$ from $(2)$, $a<16$, and $a$ is odd from $(2)$. This means that $a$ could equal $11$, $13$, or $15$. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$, of which the last two return equal $a$ values. Then, $2(3 \\cdot 6 \\cdot 6+4 \\cdot 4 \\cdot 7)=440 AWD.",
"First things first. Vietas gives us the following: \\begin{align} x_1+x_2+x_3 = a\\\\ x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}\\\\ x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2} \\end{align} From $(2)$, $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$, which implies that both sides of $(2)$ are even. Then, from $(1)$, we see that an odd number of $x_1$, $x_2$, and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers. Now, the theoretical maximum value of the left side of $(2)$ is $3 \\cdot \\biggl(\\frac{a}{3}\\biggr)^2=\\frac{a^2}{3}$. That means that the maximum bound of $a$ is where \\[\\frac{a^2}{3} > \\frac{a^2-81}{2},\\] which simplifies to $\\sqrt{243} > a$, meaning \\[16 > a.\\] So now we have that $9<a$ from $(2)$, $a<16$, and $a$ is odd from $(2)$. This means that $a$ could equal $11$, $13$, or $15$. At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$, $(3, 6, 6)$, and $(4, 4, 7)$, of which the last two return equal $a$ values. Then, $2(3 \\cdot 6 \\cdot 6+4 \\cdot 4 \\cdot 7)=440 AWD.",
"Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\\frac{4x\\pm\\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\\pm\\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because the value of $a$ alone does not determine the polynomial, $a$, $a$ must equal $15$. Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root is less than (or equal to) the $x$ value at the local maximum (picture the graph of a cubic equation), it suffices to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has positive integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$, and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$. Thus, $c=2\\cdot 3\\cdot 6\\cdot 6=216$, or $c=2\\cdot 4\\cdot 4\\cdot 7=224$. Our answer is $216+224=440 ~bad_at_mathcounts",
"Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic formula gives $a=\\frac{4x\\pm\\sqrt{324-8x^2}}{2}$, which simplifies to $a=2x\\pm\\sqrt{81-2x^2}$. We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$, and $a=15$ or $a=9$ when $x=6$. Because the value of $a$ alone does not determine the polynomial, $a$, $a$ must equal $15$. Now our polynomial equals $2x^3-30x^2+144x-c$. Because one root is less than (or equal to) the $x$ value at the local maximum (picture the graph of a cubic equation), it suffices to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has positive integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$, and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$. Thus, $c=2\\cdot 3\\cdot 6\\cdot 6=216$, or $c=2\\cdot 4\\cdot 4\\cdot 7=224$. Our answer is $216+224=440 ~bad_at_mathcounts"
] |
2015-II-7
| 2,015
| 7
|
Triangle $ABC$ has side lengths $AB = 12$ , $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = w$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[\text{Area}(PQRS) = \alpha w - \beta \cdot w^2.\] Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 161
|
II
|
[
"If $\\omega = 25$, the area of rectangle $PQRS$ is $0$, so \\[\\alpha\\omega - \\beta\\omega^2 = 25\\alpha - 625\\beta = 0\\] and $\\alpha = 25\\beta$. If $\\omega = \\frac{25}{2}$, we can reflect $APQ$ over $PQ$, $PBS$ over $PS$, and $QCR$ over $QR$ to completely cover rectangle $PQRS$, so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \\frac{12 + 17 + 25}{2} = 27$, \\[[ABC] = \\sqrt{27 \\cdot 15 \\cdot 10 \\cdot 2} = 90\\] so \\[45 = \\alpha\\omega - \\beta\\omega^2 = \\frac{625}{2} \\beta - \\beta\\frac{625}{4} = \\beta\\frac{625}{4}\\] and \\[\\beta = \\frac{180}{625} = \\frac{36}{125}\\] so the answer is $m + n = 36 + 125 = 161.",
"[asy] unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label(\"$A$\",A,N); label(\"$B$\",B,SW); label(\"$C$\",C,SE); label(\"$E$\",E,SE); label(\"$F$\",F,NE); label(\"$P$\",P,NW); label(\"$Q$\",Q,NE); label(\"$R$\",R,SE); label(\"$S$\",S,SW); draw(rightanglemark(B,E,A,12)); dot(E); dot(F); [/asy] Similar triangles can also solve the problem. First, solve for the area of the triangle. $[ABC] = 90$. This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $AC$ and solving. (The $8$ side would be collinear with line $AB$) After finding the area, solve for the altitude to $BC$. Let $E$ be the intersection of the altitude from $A$ and side $BC$. Then $AE = \\frac{36}{5}$. Solving for $BE$ using the Pythagorean Formula, we get $BE = \\frac{48}{5}$. We then know that $CE = \\frac{77}{5}$. Now consider the rectangle $PQRS$. Since $SR$ is collinear with $BC$ and parallel to $PQ$, $PQ$ is parallel to $BC$ meaning $\\Delta APQ$ is similar to $\\Delta ABC$. Let $F$ be the intersection between $AE$ and $PQ$. By the similar triangles, we know that $\\frac{PF}{FQ}=\\frac{BE}{EC} = \\frac{48}{77}$. Since $PF+FQ=PQ=\\omega$. We can solve for $PF$ and $FQ$ in terms of $\\omega$. We get that $PF=\\frac{48}{125} \\omega$ and $FQ=\\frac{77}{125} \\omega$. Let's work with $PF$. We know that $PQ$ is parallel to $BC$ so $\\Delta APF$ is similar to $\\Delta ABE$. We can set up the proportion: $\\frac{AF}{PF}=\\frac{AE}{BE}=\\frac{3}{4}$. Solving for $AF$, $AF = \\frac{3}{4} PF = \\frac{3}{4} \\cdot \\frac{48}{125} \\omega = \\frac{36}{125} \\omega$. We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \\frac{36}{5} - \\frac{36}{125} \\omega$. Therefore, $[PQRS] = PQ \\cdot PS = \\omega (\\frac{36}{5} - \\frac{36}{125} \\omega) = \\frac{36}{5}\\omega - \\frac{36}{125} \\omega^2$. This means that $\\beta = \\frac{36}{125} \\Rightarrow (m,n) = (36,125) \\Rightarrow m+n = 161.",
"Heron's Formula gives $[ABC] = \\sqrt{27 \\cdot 15 \\cdot 10 \\cdot 2} = 90,$ so the altitude from $A$ to $BC$ has length $\\dfrac{2[ABC]}{BC} = \\dfrac{36}{5}.$ Now, draw a parallel to $AB$ from $Q$, intersecting $BC$ at $T$. Then $BT = w$ in parallelogram $QPBT$, and so $CT = 25 - w$. Clearly, $CQT$ and $CAB$ are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so \\[\\frac{QR}{\\frac{36}{5}} = \\frac{25 - w}{25}.\\] Solving gives $[PQRS] = \\dfrac{36}{5} \\cdot \\dfrac{25 - w}{25} \\cdot w = \\dfrac{36w}{5} - \\dfrac{36w^2}{125}$, so the answer is $36 + 125 = 161$.",
"Using the diagram from Solution 2 above, label $AF$ to be $h$. Through Heron's formula, the area of $\\triangle ABC$ turns out to be $90$, so using $AE$ as the height and $BC$ as the base yields $AE=\\frac{36}{5}$. Now, through the use of similarity between $\\triangle APQ$ and $\\triangle ABC$, you find $\\frac{w}{25}=\\frac{h}{36/5}$. Thus, $h=\\frac{36w}{125}$. To find the height of the rectangle, subtract $h$ from $\\frac{36}{5}$ to get $\\left(\\frac{36}{5}-\\frac{36w}{125}\\right)$, and multiply this by the other given side $w$ to get $\\frac{36w}{5}-\\frac{36w^2}{125}$ for the area of the rectangle. Finally, $36+125=161.",
"Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$. Hence, $h=\\frac{36}{5}$ Now, we see that $\\sin{B}=\\frac{PS}{12-L}\\implies PS=\\sin{B}(12-L)$ We easily find that $\\sin{B}=\\frac{3}{5}$ Hence, $PS=\\frac{3}{5}(12-L)$ Now, we see that $[PQRS]=\\frac{3}{5}(12-L)(w)$ Now, it is obvious that we want to find $L$ in terms of $W$. Looking at the diagram, we see that because $PQRS$ is a rectangle, $\\triangle{APQ}\\sim{\\triangle{ABC}}$ Hence.. we can now set up similar triangles. We have that $\\frac{AP}{AB}=\\frac{PQ}{BC}\\implies \\frac{L}{12}=\\frac{W}{25}\\implies 25L=12W\\implies L=\\frac{12W}{25}$. Plugging back in.. $[PQRS]=\\frac{3w}{5}(12-(\\frac{12W}{25}))\\implies \\frac{3w}{5}(\\frac{300-12W}{25})\\implies \\frac{900W-36W^2}{125}$ Simplifying, we get $\\frac{36W}{5}-\\frac{36W^2}{125}$ Hence, $125+36=161",
"Proceed as in solution 1. When $\\omega$ is equal to zero, $\\alpha - \\beta\\omega=\\alpha$ is equal to the altitude. This means that $25\\beta$ is equal to $\\frac{36}{5}$, so $\\beta = \\frac{36}{125}$, yielding $161.",
"Using the diagram in Solution 2, the area of $PQRS$ is ($w$)($PR$), we only need to find $PR$. Because $ABC$ and $APQ$ are similar in a ratio of 25:w. Because of $AB$:$AC$=12:15 and $AP$:$AB$= $w$:25, we can derive that $BP$=frac{(25-w)(12)}{25}. By using LoC on $ABC$, it is obvious that Cos(B)=\\frac{4}{5} and Sin($B$)=\\frac{3}{5}. $PR$=\\frac{3bp}{5}=\\frac{(25-w)(36)}{125}. Multiply $w$, we get $\\frac{36W}{5}-\\frac{36W^2}{125}$ which means the answer is161",
"Using the diagram in Solution 2, the area of $PQRS$ is ($w$)($PR$), we only need to find $PR$. Because $ABC$ and $APQ$ are similar in a ratio of 25:w. Because of $AB$:$AC$=12:15 and $AP$:$AB$= $w$:25, we can derive that $BP$=frac{(25-w)(12)}{25}. By using LoC on $ABC$, it is obvious that Cos(B)=\\frac{4}{5} and Sin($B$)=\\frac{3}{5}. $PR$=\\frac{3bp}{5}=\\frac{(25-w)(36)}{125}. Multiply $w$, we get $\\frac{36W}{5}-\\frac{36W^2}{125}$ which means the answer is161"
] |
2015-II-8
| 2,015
| 8
|
Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 36
|
II
|
[
"Let us call the quantity $\\frac{a^3b^3+1}{a^3+b^3}$ as $N$ for convenience. Knowing that $a$ and $b$ are positive integers, we can legitimately rearrange the given inequality so that $a$ is by itself, which makes it easier to determine the pairs of $(a, b)$ that work. Doing so, we have \\[\\frac{ab+1}{a+b} < \\frac{3}{2}\\] \\[\\implies 2ab + 2 < 3a + 3b \\implies 2ab - 3a < 3b - 2\\] \\[\\implies a < \\frac{3b - 2}{2b - 3}.\\] Now, observe that if $b = 1$ we have that $N = \\frac{a^3 + 1}{a^3 + 1} = 1$, regardless of the value of $a$. If $a = 1$, we have the same result: that $N = \\frac{b^3 + 1}{b^3 + 1} = 1$, regardless of the value of $b$. Hence, we want to find pairs of positive integers $(a, b)$ existing such that neither $a$ nor $b$ is equal to $1$, and that the conditions given in the problem are satisfied in order to check that the maximum value for $N$ is not $1$. To avoid the possibility that $a = 1$, we want to find values of $b$ such that $\\frac{3b - 2}{2b - 3} > 2$. If we do this, we will have that $a < \\frac{3b - 2}{2b - 3} = k$, where $k$ is greater than $2$, and this allows us to choose values of $a$ greater than $1$. Again, since $b$ is a positive integer, and we want $b > 1$, we can legitimately multiply both sides of $\\frac{3b - 2}{2b - 3} > 2$ by $2b - 3$ to get $3b - 2 > 4b - 6 \\implies b < 4$. For $b = 3$, we have that $a < \\frac{7}{3}$, so the only possibility for $a$ greater than $1$ is obviously $2$. Plugging these values into $N$, we have that $N = \\frac{8(27) + 1}{8 + 27} = \\frac{217}{35} = \\frac{31}{5}$. For $b = 2$, we have that $a < \\frac{4}{1} = 4$. Plugging $a = 3$ and $b = 2$ in for $N$ yields the same result of $\\frac{31}{5}$, but plugging $a = 2$ and $b = 2$ into $N$ yields that $N = \\frac{8(8) + 1}{8 + 8} = \\frac{65}{16}$. Clearly, $\\frac{31}{5}$ is the largest value we can have for $N$, so our answer is $31 + 5 = 036. (Technically, we would have to find that b > 1 before dividing both sides of the inequality by 2b - 3, but otherwise this solution is correct)",
"\\[\\frac{ab + 1}{a + b} < \\frac{3}{2} \\rightarrow 2ab + 2 < 3a + 3b \\rightarrow 4ab - 6a - 6b + 4 < 0 \\rightarrow (2a - 3)(2b - 3) < 5.\\] \\[2a - 3, 2b - 3 \\in \\{x \\neq 2k, k \\in Z \\} \\rightarrow\\] $(2a-3)$ and $(2b-3)$ each cannot be even or else $a$ and $b$ will not be integers \\[(2a - 3)(2b - 3) = 1, 3 \\rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).\\] \\[(a, b) = (2, 2), (2, 3), (3, 2).\\] \\[\\frac{a^3 b^3 + 1}{a^3 + b^3} = \\frac{65}{16}, \\frac{31}{5}.\\] \\[\\frac{31}{5} \\rightarrow 036).",
"\\[\\frac{ab + 1}{a + b} < \\frac{3}{2} \\rightarrow 2ab + 2 < 3a + 3b \\rightarrow 4ab - 6a - 6b + 4 < 0 \\rightarrow (2a - 3)(2b - 3) < 5.\\] \\[2a - 3, 2b - 3 \\in \\{x \\neq 2k, k \\in Z \\} \\rightarrow\\] $(2a-3)$ and $(2b-3)$ each cannot be even or else $a$ and $b$ will not be integers \\[(2a - 3)(2b - 3) = 1, 3 \\rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).\\] \\[(a, b) = (2, 2), (2, 3), (3, 2).\\] \\[\\frac{a^3 b^3 + 1}{a^3 + b^3} = \\frac{65}{16}, \\frac{31}{5}.\\] \\[\\frac{31}{5} \\rightarrow 036).",
"Notice that for $\\frac{a^3b^3+1}{a^3+b^3}$ to be maximized, $\\frac{ab+1}{a+b}$ has to be maximized. We simplify as above to $2ab + 2 < 3a + 3b$, which is $(a-\\frac{3}{2})(b-\\frac{3}{2}) < \\frac{5}{4}$. To maximize, $a$ has to be as close to $b$ as possible, making $a$ close to $\\frac{3+\\sqrt{5}}{2}$. Because $a$ and $b$ are positive integers, $a = 3$, and checking back gives $b = 2$ as the maximum or the other way around, which the answer is thus $\\frac{216+1}{27+8} = \\frac{217}{35} = \\frac{31}{5} \\rightarrow 036.",
"Guess and check a few values of $a$ and $b$ you will find two things. One, that the highest values of $a$ and $b$ are the closest to $\\frac{3}{2}$. Two, that the pair $(2,3)$ is the highest possible value of $a, b$. So plugging in $a=2$ and $b=3$ we get $((2*3)^3)+1/(8+27)$ = $\\frac{217}{35}$ = $\\frac{31}{5}$, and $31+5 = 36$. Solution by Helloitsaaryan"
] |
2015-II-9
| 2,015
| 9
|
A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$ . [asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]
| 384
|
II
|
[
"Our aim is to find the volume of the part of the cube submerged in the cylinder. In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is $4$, by the Law of Cosines, the side length s of the equilateral triangle is \\[s^2 = 2\\cdot(4^2) - 2\\cdot(4^2)\\cos(120^{\\circ}) = 3(4^2)\\] so $s = 4\\sqrt{3}$.* Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are $\\frac{4\\sqrt{3}}{\\sqrt{2}} = 2\\sqrt{6}$ (the three triangular faces touching the submerged vertex are all $45-45-90$ triangles) so \\[v = \\frac{1}{3}(2\\sqrt{6})\\left(\\frac{1}{2} \\cdot (2\\sqrt{6})^2\\right) = \\frac{1}{6} \\cdot 48\\sqrt{6} = 8\\sqrt{6}\\] so \\[v^2 = 64 \\cdot 6 = 384 relationship for 30-60-90 triangles.",
"Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equilateral triangle, using the fact that the circle of radius $4$ is the circumcircle of the equilateral triangle. Using equilateral triangle properties, you can find that the height of the triangle is $6$, and the side length is $\\frac{6}{\\sqrt{3}}=2\\sqrt{3} \\cdot 2=4\\sqrt{3}$. As the other faces are right isosceles triangles, they are $\\frac{4\\sqrt{3}}{\\sqrt{2}}=2\\sqrt{6}$. Therefore the volume of this tetrahedron is \\[\\left(\\frac{2\\sqrt{6}}{2}\\right)^2=12 \\ \\cdot \\ (2\\sqrt{6})=24\\sqrt{6} \\implies \\frac{24\\sqrt{6}}{3}=8\\sqrt{6} \\implies (8\\sqrt{6})^2=384\\] -dragoon -lpieleanu (minor latex changes) Note: The height of the cylinder and the side length of the cube had no effect on the result of the problem. Note 2: If this doesn’t make sense, just imagine slicing a corner off a cube.",
"Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equilateral triangle, using the fact that the circle of radius $4$ is the circumcircle of the equilateral triangle. Using equilateral triangle properties, you can find that the height of the triangle is $6$, and the side length is $\\frac{6}{\\sqrt{3}}=2\\sqrt{3} \\cdot 2=4\\sqrt{3}$. As the other faces are right isosceles triangles, they are $\\frac{4\\sqrt{3}}{\\sqrt{2}}=2\\sqrt{6}$. Therefore the volume of this tetrahedron is \\[\\left(\\frac{2\\sqrt{6}}{2}\\right)^2=12 \\ \\cdot \\ (2\\sqrt{6})=24\\sqrt{6} \\implies \\frac{24\\sqrt{6}}{3}=8\\sqrt{6} \\implies (8\\sqrt{6})^2=384\\] -dragoon -lpieleanu (minor latex changes) Note: The height of the cylinder and the side length of the cube had no effect on the result of the problem. Note 2: If this doesn’t make sense, just imagine slicing a corner off a cube.",
"We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\\sqrt3$. From here, its area is \\[\\dfrac{\\bigl(4\\sqrt3\\bigr)^2\\sqrt3}4=12\\sqrt3.\\] The leg of the isosceles right triangle is $\\dfrac{4\\sqrt3}{\\sqrt2}=2\\sqrt6$, and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram. [asy] import olympiad; pair V, T, B; V = (-4, 0); B = origin; T = (0, 2*sqrt(2)); draw(V--B--T--cycle); draw(rightanglemark(V, B, T)); label(\"Vertex\", V, W); label(\"Tip\", T, N); label(\"Base\", B, SE); label(\"$4$\", V--B, S); label(\"$2\\sqrt6$\", V--T, NW); [/asy] The height from the tip to the base is $2\\sqrt2$, so the volume is $\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $384. -integralarefun",
"We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\\sqrt3$. From here, its area is \\[\\dfrac{\\bigl(4\\sqrt3\\bigr)^2\\sqrt3}4=12\\sqrt3.\\] The leg of the isosceles right triangle is $\\dfrac{4\\sqrt3}{\\sqrt2}=2\\sqrt6$, and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram. [asy] import olympiad; pair V, T, B; V = (-4, 0); B = origin; T = (0, 2*sqrt(2)); draw(V--B--T--cycle); draw(rightanglemark(V, B, T)); label(\"Vertex\", V, W); label(\"Tip\", T, N); label(\"Base\", B, SE); label(\"$4$\", V--B, S); label(\"$2\\sqrt6$\", V--T, NW); [/asy] The height from the tip to the base is $2\\sqrt2$, so the volume is $\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $384. -integralarefun",
"Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically. The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$. This means the medians of the triangle have length $\\frac{3}{2} \\cdot 4=6$, because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$. Using $30-60-90$ triangles, the side length of the triangle is $4\\sqrt{3}$, and its area is $\\frac{(4\\sqrt{3})^2\\sqrt{3}}{4}=12\\sqrt{3}$. Next, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$. The area of each is then $\\frac{x^2}{2}$. By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\\left( \\frac{x^2}{2} \\right) ^2=(12\\sqrt{3})^2$. Solving yields $x=2\\sqrt{6}$. The height of the pyramid is then $\\sqrt{(2\\sqrt{6})^2-4^2}=2\\sqrt{2}$, by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\\frac{1}{3}\\cdot 12\\sqrt{3} \\cdot 2\\sqrt{2}=8\\sqrt{6}$, and the requested answer is $v^2=(8\\sqrt{6})^2=384. ~bad_at_mathcounts",
"Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically. The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$. This means the medians of the triangle have length $\\frac{3}{2} \\cdot 4=6$, because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$. Using $30-60-90$ triangles, the side length of the triangle is $4\\sqrt{3}$, and its area is $\\frac{(4\\sqrt{3})^2\\sqrt{3}}{4}=12\\sqrt{3}$. Next, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$. The area of each is then $\\frac{x^2}{2}$. By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\\left( \\frac{x^2}{2} \\right) ^2=(12\\sqrt{3})^2$. Solving yields $x=2\\sqrt{6}$. The height of the pyramid is then $\\sqrt{(2\\sqrt{6})^2-4^2}=2\\sqrt{2}$, by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\\frac{1}{3}\\cdot 12\\sqrt{3} \\cdot 2\\sqrt{2}=8\\sqrt{6}$, and the requested answer is $v^2=(8\\sqrt{6})^2=384. ~bad_at_mathcounts"
] |
2015-II-10
| 2,015
| 10
|
Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, $53421$ and $14253$ are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but $45123$ is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$ .
| 486
|
II
|
[
"The simple recurrence can be found. When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$, before $n - 2$, and at the very end. Ex. Inserting 4 into the string 123: 4 can go before the 2 (1423), before the 3 (1243), and at the very end (1234). Only the addition of the next number, $n$, will change anything. Thus the number of permutations with $n$ elements is three times the number of permutations with $n-1$ elements. Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$. Thus for $n=7$ there are $2*3^5=486 permutations. When you are faced with a brain-fazing equation and combinatorics is part of the problem, use recursion! This same idea appeared on another AIME with an 8-box problem."
] |
2015-II-11
| 2,015
| 11
|
The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ , $BC=4$ , $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 23
|
II
|
[
"Solution 1 Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\\triangle{OMB} \\sim \\triangle{QOB}$ because both are right triangles, and $\\angle{OBQ} \\cong \\angle{OBM}$. By $\\frac{MB}{BO}=\\frac{BO}{BQ}$, $MB = r\\left(\\frac{r}{4.5}\\right) = \\frac{r^2}{4.5}$. However, since $O$ is the circumcenter of triangle $ABC$, $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\\frac{r^2}{4.5} = 2 \\implies r = 3$. Since we know $BN=\\frac{5}{2}$ and $\\triangle BOP \\sim \\triangle BNO$, we have $\\frac{BP}{3} = \\frac{3}{\\frac{5}{2}}$. Thus, $BP = \\frac{18}{5}$. $m + n=023. ~awang11's sol",
"Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\\triangle{OMB} \\sim \\triangle{QOB}$ because both are right triangles, and $\\angle{OBQ} \\cong \\angle{OBM}$. By $\\frac{MB}{BO}=\\frac{BO}{BQ}$, $MB = r\\left(\\frac{r}{4.5}\\right) = \\frac{r^2}{4.5}$. However, since $O$ is the circumcenter of triangle $ABC$, $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\\frac{r^2}{4.5} = 2 \\implies r = 3$. Since we know $BN=\\frac{5}{2}$ and $\\triangle BOP \\sim \\triangle BNO$, we have $\\frac{BP}{3} = \\frac{3}{\\frac{5}{2}}$. Thus, $BP = \\frac{18}{5}$. $m + n=023. ~awang11's sol",
"Minor arc $BC = 2A$ so $\\angle{BOC}=2A$. Since $\\triangle{BOC}$ is isosceles ($BO$ and $OC$ are radii), $\\angle{CBO}=(180-2A)/2=90-A$. $\\angle{CBO}=90-A$, so $\\angle{BQO}=A$. From this we get that $\\triangle{BPQ}\\sim \\triangle{BCA}$. So $\\dfrac{BP}{BC}=\\dfrac{BQ}{BA}$, plugging in the given values we get $\\dfrac{BP}{4}=\\dfrac{4.5}{5}$, so $BP=\\dfrac{18}{5}$, and $m+n=023. ~awang11's sol",
"Minor arc $BC = 2A$ so $\\angle{BOC}=2A$. Since $\\triangle{BOC}$ is isosceles ($BO$ and $OC$ are radii), $\\angle{CBO}=(180-2A)/2=90-A$. $\\angle{CBO}=90-A$, so $\\angle{BQO}=A$. From this we get that $\\triangle{BPQ}\\sim \\triangle{BCA}$. So $\\dfrac{BP}{BC}=\\dfrac{BQ}{BA}$, plugging in the given values we get $\\dfrac{BP}{4}=\\dfrac{4.5}{5}$, so $BP=\\dfrac{18}{5}$, and $m+n=023. ~awang11's sol",
"Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, \\[OM=\\sqrt{r^2-4}\\] Thus, \\[OQ=\\frac{\\sqrt{4r^2+9}}{2}\\] Using $\\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\\triangle{BON}$ ~ $\\triangle{OPN}$, ${NP=11/10}$. Therefore, \\[BP=\\frac{5}{2}+\\frac{11}{10}=18/5\\] $18+5=023. ~awang11's sol",
"Let $\\angle{BQO}=\\alpha$. Extend $OB$ to touch the circumcircle at a point $K$. Then, note that $\\angle{KAC}=\\angle{CBK}=\\angle{QBO}=90^\\circ-\\alpha$. But since $BK$ is a diameter, $\\angle{KAB}=90^\\circ$, implying $\\angle{CAB}=\\alpha$. It follows that $APCQ$ is a cyclic quadrilateral. Let $BP=x$. By Power of a Point, \\[5x=4\\cdot\\frac 9 2\\implies x=\\frac{18}{5}.\\]The answer is $18+5=023. ~awang11's sol",
"$\\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$ Denote the circumradius of $ABC$ to be $R$, the circumcircle of $ABC$ to be $O$, and the shortest distance from $Q$ to circle $O$ to be $x$. Using Power of a Point on $Q$ relative to circle $O$, we get that $x(x+2r) = 0.5 \\cdot 4.5 = \\frac{9}{4}$. Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \\frac{81}{4}$. Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$. Now, set $\\cos{ABC} = y$. Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\\frac{\\sqrt{4^2 + 5^2 - 2 \\cdot 4 \\cdot 5 x}}{\\sqrt{1 - x^2}} = 2R = 6$. Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$. Now, we get $x = \\frac{10 \\pm \\sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \\frac{10 + \\sqrt{55}}{18}$(You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\\frac{1}{\\sin{OPB}} = \\frac{3}{\\cos{\\angle ABC - \\angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$, we get that $BP = \\frac{162}{2\\cos{B} + \\sqrt{5}\\sin{B}}$. Now, the hard part is to find what $\\sin{B}$ is. We therefore want $\\frac{\\sqrt{324 - (10 + \\sqrt{55})^2}}{18} = \\frac{\\sqrt{169 - 20\\sqrt{55}}}{18}$. For the numerator, by inspection $(a + b\\sqrt{55})^2$ will not work for integers $a$ and $b$. The other case is if there is $(a\\sqrt{5} + b\\sqrt{11})^2$. By inspection, $5\\sqrt{5} - 2\\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\\frac{18}{5} \\rightarrow 23. ~awang11's sol",
"[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */ draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */ draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.82,0.87),dotstyle); label(\"$A$\", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); dot((3.178984115621537,0.7692140299269852),dotstyle); label(\"$B$\", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); dot((2.6857306099246263,4.738685150758791),dotstyle); label(\"$C$\", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); label(\"$O$\", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); label(\"$P$\", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); label(\"$Q$\", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); label(\"$A'$\", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); label(\"$P'$\", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); label(\"$D$\", (2.603644633462422,5.802794720137042), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] Reflect $A$, $P$ across $OB$ to points $A'$ and $P'$, respectively with $A'$ on the circle and $P, O, P'$ collinear. Now, $\\angle A'CQ = 180^{\\circ} - \\angle A'CB = \\angle A'AB = \\angle P'PB$ by parallel lines. From here, $\\angle P'PB = \\angle PP'B = \\angle A'P'Q$ as $P, P', Q$ collinear. From here, $A'P'QC$ is cyclic, and by power of a point we obtain $\\frac{18}{5} \\implies 023. ~awang11's sol"
] |
2015-II-12
| 2,015
| 12
|
There are $2^{10} = 1024$ possible $10$ -letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than $3$ adjacent letters that are identical.
| 548
|
II
|
[
"Let $a_{n}$ be the number of ways to form $n$-letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical. Note that, at the end of each $n$-letter string, there are $3$ possibilities for the last letter chain: it must be either $1$, $2$, or $3$ letters long. Removing this last chain will make a new string that is $n-1$, $n-2$, or $n-3$ letters long, respectively. Therefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$. We can see that \\[a_{1}=2\\] \\[a_{2}=2^{2}=4\\] \\[a_{3}=2^{3}=8\\] so using our recursive relation we find \\[a_{4}=14\\] \\[a_{5}=26\\] \\[a_{6}=48\\] \\[a_{7}=88\\] \\[a_{8}=162\\] \\[a_{9}=298\\] \\[a_{10}=548\\]",
"The solution is a simple recursion: We have three cases for the ending of a string: three in a row, two in a row, and a single: ...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$ (Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.) For case $(1)$, we could only add a B to the end, making it a case $(3)$. For case $(2)$, we could add an A or a B to the end, making it a case $(1)$ if you add an A, or a case $(3)$ if you add a B. For case $(3)$, we could add an A or a B to the end, making it a case $(2)$ or a case $(3)$. Let us create three series to represent the number of permutations for each case: $\\{a\\}$, $\\{b\\}$, and $\\{c\\}$ representing case $(1)$, $(2)$, and $(3)$ respectively. The series have the following relationship: $a_n=b_{n-1}$; $b_n=c_{n-1}$; $c_n=c_{n-1}+a_{n-1}+b_{n-1}$ For $n=3$: $a_3$ and $b_3$ both equal $2$, $c_3=4$. With some simple math, we have: $a_{10}=88$, $b_{10}=162$, and $c_{10}=298$. Summing the three up we have our solution: $88+162+298=548.",
"This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$. Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$. Note that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\\ge4$. Similarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}$ for all $n\\ge4$. Here is why: every valid strings of $n$ letters $(n\\ge4)$ where the first letter is $A$ must begin with one of the following: $AAAB$ - and the number of valid ways is $b_{n-3}$. $AAB$ - and the number of valid ways is $b_{n-2}$. $AB$ - and there are $b_{n-1}$ ways. We know that $a_1=1$, $a_2=2$, and $a_3=4$. Similarly, we have $b_1=1$, $b_2=2$, and $b_3=4$. We can quickly check our recursion to see if our recursive formula works. By the formula, $a_4=b_3+b_2+b_1=7$, and listing out all $a_4$, we can quickly verify our formula. Therefore, we have the following: $\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\\\\hline a & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\\\b & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\end{tabular}$ The total number of valid $10$ letter strings is equal to $a_{10}+b_{10}=274+274=548. Therefore, we didn't really need to list out both recursion formulas, which could save us some time.",
"Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this! We can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along]. $4 \\rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$ $5 \\rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$ $6 \\rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$ $7 \\rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$ $8 \\rightarrow (3,1...1) = 8, (2,2,1...1) = 28$ $9 \\rightarrow (2,1...1) = 9$ $10 \\rightarrow (1,1....1) =1$ Adding them all up gives you 274; multiplying by 2 gives $548.",
"Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this! We can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along]. $4 \\rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$ $5 \\rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$ $6 \\rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$ $7 \\rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$ $8 \\rightarrow (3,1...1) = 8, (2,2,1...1) = 28$ $9 \\rightarrow (2,1...1) = 9$ $10 \\rightarrow (1,1....1) =1$ Adding them all up gives you 274; multiplying by 2 gives $548.",
"We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string. Let $a_n$ be the number of strings of length n that satisfy the problem statement and also has a \"streak\" of length 1 at the end. ABABBA has a streak of length 1. Let $b_n$ be the number of strings of length n that satisfy the problem statement and also has of length 2 at the end. ABABBAA has a streak of length 2. There are 2 \"A\" s at the end of the string. Let $c_n$ be the number of string of length n that that satisfy the problem statement and also has a \"streak\" of length 3 at the end. ABABBAAA has a streak of length 3. There are 3 \"A\" s at the end of the string. Let's establish a recursive relationship. $a_{n+1} = a_n+b_n+c_n$, since you can simply break the streak. $b_{n+1} = a_n$, and $c_{n+1} = b_n$ Since you can just add to the streak. We can log everything using a table. $\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\\\\hline a & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 & 298\\\\b & 0 & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 \\\\c & 0 & 0 & 2 & 2 &4 & 8 & 14 & 26 & 48 & 88\\end{tabular}$ Adding $a_{10}$, $b_{10}$, $c_{10}$ gets the total number of numbers that doesn't have more than 3 concecutive letters. That gets a total of $298+162+88 = 548 -AlexLikeMath",
"Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$. For $n \\ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choices for the fifth character. However we have to be careful if the last three characters of $S_4$ are all the same, as in that case we would be overcounting, (i.e if we have a string $\\dots aaa$, we can only add $b$ to the end so that the problem criteria is satisfied. Additionally strings such as $\\dots aaa$ or $\\dots bbb$ should only be counted once as we only have one choice for the $n$th character to add ($b$ and $a$ respectively). Thus to compute $S_n$ we start with $S_{n-1}$ then take out the strings ending in $\\dots bbb$ or $\\dots aaa$. There are $S_{n-4}$ remaining valid strings (i.e if we pick any of the $S_{n-4}$ valid strings, examine the last character, if it is a $b$ then we append $aaa$ to it, and if if it an $a$ we append $bbb$ to it, hence the number of $S_{n-1}$ strings are in one-to-one correspondence with the number of strings in $S_{n-4}$. Thus we have the recursion $S_n = 2(S_{n-1} - S_{n-4}) + S_{n-4}$ (where we are first taking away $S_{n-4}$, doubling the result, then adding $S_{n-4}$ back in signifying that we only count it once, as described above). The recursion simplifies to $S_n = 2S_{n-1} - S_{n-4}$ and we can now quickly compute the remaining values: \\[S_5 = 2S_4 - S_1 = 26, S_6 = 2(26)-4 = 48, S_7 = 2(48) - 8 = 88, S_8 = 2(88) - 14 = 162, S_9 = 2(162) - 26 = 298, S_{10} = 2S_9 - 48 = 548\\] ~afroromanian",
"Let us define a \"run\" as a set of consecutive letters that are all the same and let a \"maximal run\" be a run that is not the proper subset of any other run, in other words, it cannot be expanded. From now on, we consider all runs to be maximal. Note that the minimum number of runs in the $10$-letter string is $4$ (or else the Pigeonhole Principle tells us that at least $1$ run has $4$ or more letters in it, contradiction), and the maximum number of runs is clearly $10.$ Since an arbitrary run can have anywhere from $1$ to $3$ letters in it, it follows that the number of $10$-letter strings with $n$ runs (where $4 \\leq n \\leq 10$) is\\[2[x^{10}] (x+x^2+x^3)^n,\\]i.e. twice the coefficient of $x^{10}$ in the expansion of $(x+x^2+x^3)^n.$ (Note that we multiplied by $2$ because there are two choices for which letter the first run has and then the rest are fixed). Hence, we wish to find\\[2\\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n.\\] First, we can rewrite this as \\begin{align*} \\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n &= \\sum_{n=4}^{10} [x^{10}] \\ x^n \\cdot (1+x+x^2)^n \\\\ &= \\sum_{n=4}^{10} [x^{10-n}] \\left(\\frac{1-x^3}{1-x}\\right)^n \\\\ &= \\sum_{n=4}^{10} [x^{10-n}] \\frac{(1-x^3)^n}{(1-x)^n}. \\\\ \\end{align*} Now, we proceed case by case, utilizing the Binomial Theorem for the numerator in all of the cases: For $n=4,$ we have \\begin{align*} [x^6] \\frac{(1-x^3)^4}{(1-x)^4} &= \\binom{4}{0} [x^6] (1-x)^{-4} - \\binom{4}{1} [x^3] (1-x)^{-4} + \\binom{4}{2} [x^0] (1-x)^{-4} \\\\ &= 1\\cdot\\binom{9}{3} - 4\\cdot\\binom{6}{3} + 6\\cdot\\binom{3}{3} \\\\ &= 1\\cdot84-4\\cdot20+6\\cdot1 \\\\ &= 84-80+6 \\\\ &= \\underline{10}. \\end{align*} For $n=5,$ we have \\begin{align*} [x^5] \\frac{(1-x^3)^5}{(1-x)^5} &= \\binom{5}{0} [x^5] (1-x)^{-5} - \\binom{5}{1}[x^2] (1-x)^{-5} \\\\ &= 1\\cdot\\binom{9}{4}-5\\cdot\\binom{6}{4} \\\\ &= 1\\cdot126-5\\cdot15 \\\\ &= 126-75 \\\\ &= \\underline{51}. \\end{align*}For $n=6,$ we have \\begin{align*} [x^4] \\frac{(1-x^3)^6}{(1-x)^6} &= \\binom{6}{0} [x^4](1-x)^{-6} - \\binom{6}{1}[x^1](1-x)^{-6} \\\\ &= 1 \\cdot \\binom{9}{5} - 6 \\cdot \\binom{6}{5} \\\\ &= 1 \\cdot 126 - 6 \\cdot 6 \\\\ &= 126-36 \\\\ &= \\underline{90}. \\end{align*}For $n=7,$ we have \\begin{align*} [x^3] \\frac{(1-x^3)^7}{(1-x)^7} &= \\binom{7}{0}[x^3](1-x)^{-7} - \\binom{7}{1}[x^0](1-x)^{-7} \\\\ &= 1\\cdot \\binom{9}{6} - 7\\cdot\\binom{6}{6} \\\\ &= 1\\cdot84-7\\cdot1\\\\ &= 84-7 \\\\ &= \\underline{77}. \\end{align*}For $n=8,$ we have \\begin{align*} [x^2] \\frac{(1-x^3)^8}{(1-x)^8} &=\\binom{8}{0} [x^2](1-x)^{-8} \\\\ &= 1\\cdot\\binom{9}{7} \\\\ &= 1\\cdot36 \\\\ &= \\underline{36}. \\end{align*}For $n=9,$ we have \\begin{align*} [x^1] \\frac{(1-x^3)^9}{(1-x)^9} &= \\binom{9}{0} [x^1] (1-x)^{-9}\\\\ &= 1\\cdot\\binom{9}{8} \\\\ &= 1\\cdot9 \\\\ &= \\underline{9}. \\end{align*}For $n=10,$ we have \\begin{align*} [x^0] \\frac{(1-x^3)^{10}}{(1-x)^{10}} &= [x^0] (1-x)^{-10} \\\\ &= 1\\cdot\\binom{9}{9} \\\\ &= 1\\cdot1 \\\\ &= \\underline{1}. \\end{align*}Hence, the answer is \\begin{align*} 2\\sum_{n=4}^{10} [x^{10-n}] \\frac{(1-x^3)^n}{(1-x)^n} &= 2(10+51+90+77+36+9+1) \\\\ &= 2\\cdot274 \\\\ &= 548. \\end{align*} -lpieleanu",
"Let us define a \"run\" as a set of consecutive letters that are all the same and let a \"maximal run\" be a run that is not the proper subset of any other run, in other words, it cannot be expanded. From now on, we consider all runs to be maximal. Note that the minimum number of runs in the $10$-letter string is $4$ (or else the Pigeonhole Principle tells us that at least $1$ run has $4$ or more letters in it, contradiction), and the maximum number of runs is clearly $10.$ Since an arbitrary run can have anywhere from $1$ to $3$ letters in it, it follows that the number of $10$-letter strings with $n$ runs (where $4 \\leq n \\leq 10$) is\\[2[x^{10}] (x+x^2+x^3)^n,\\]i.e. twice the coefficient of $x^{10}$ in the expansion of $(x+x^2+x^3)^n.$ (Note that we multiplied by $2$ because there are two choices for which letter the first run has and then the rest are fixed). Hence, we wish to find\\[2\\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n.\\] First, we can rewrite this as \\begin{align*} \\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n &= \\sum_{n=4}^{10} [x^{10}] \\ x^n \\cdot (1+x+x^2)^n \\\\ &= \\sum_{n=4}^{10} [x^{10-n}] \\left(\\frac{1-x^3}{1-x}\\right)^n \\\\ &= \\sum_{n=4}^{10} [x^{10-n}] \\frac{(1-x^3)^n}{(1-x)^n}. \\\\ \\end{align*} Now, we proceed case by case, utilizing the Binomial Theorem for the numerator in all of the cases: For $n=4,$ we have \\begin{align*} [x^6] \\frac{(1-x^3)^4}{(1-x)^4} &= \\binom{4}{0} [x^6] (1-x)^{-4} - \\binom{4}{1} [x^3] (1-x)^{-4} + \\binom{4}{2} [x^0] (1-x)^{-4} \\\\ &= 1\\cdot\\binom{9}{3} - 4\\cdot\\binom{6}{3} + 6\\cdot\\binom{3}{3} \\\\ &= 1\\cdot84-4\\cdot20+6\\cdot1 \\\\ &= 84-80+6 \\\\ &= \\underline{10}. \\end{align*} For $n=5,$ we have \\begin{align*} [x^5] \\frac{(1-x^3)^5}{(1-x)^5} &= \\binom{5}{0} [x^5] (1-x)^{-5} - \\binom{5}{1}[x^2] (1-x)^{-5} \\\\ &= 1\\cdot\\binom{9}{4}-5\\cdot\\binom{6}{4} \\\\ &= 1\\cdot126-5\\cdot15 \\\\ &= 126-75 \\\\ &= \\underline{51}. \\end{align*}For $n=6,$ we have \\begin{align*} [x^4] \\frac{(1-x^3)^6}{(1-x)^6} &= \\binom{6}{0} [x^4](1-x)^{-6} - \\binom{6}{1}[x^1](1-x)^{-6} \\\\ &= 1 \\cdot \\binom{9}{5} - 6 \\cdot \\binom{6}{5} \\\\ &= 1 \\cdot 126 - 6 \\cdot 6 \\\\ &= 126-36 \\\\ &= \\underline{90}. \\end{align*}For $n=7,$ we have \\begin{align*} [x^3] \\frac{(1-x^3)^7}{(1-x)^7} &= \\binom{7}{0}[x^3](1-x)^{-7} - \\binom{7}{1}[x^0](1-x)^{-7} \\\\ &= 1\\cdot \\binom{9}{6} - 7\\cdot\\binom{6}{6} \\\\ &= 1\\cdot84-7\\cdot1\\\\ &= 84-7 \\\\ &= \\underline{77}. \\end{align*}For $n=8,$ we have \\begin{align*} [x^2] \\frac{(1-x^3)^8}{(1-x)^8} &=\\binom{8}{0} [x^2](1-x)^{-8} \\\\ &= 1\\cdot\\binom{9}{7} \\\\ &= 1\\cdot36 \\\\ &= \\underline{36}. \\end{align*}For $n=9,$ we have \\begin{align*} [x^1] \\frac{(1-x^3)^9}{(1-x)^9} &= \\binom{9}{0} [x^1] (1-x)^{-9}\\\\ &= 1\\cdot\\binom{9}{8} \\\\ &= 1\\cdot9 \\\\ &= \\underline{9}. \\end{align*}For $n=10,$ we have \\begin{align*} [x^0] \\frac{(1-x^3)^{10}}{(1-x)^{10}} &= [x^0] (1-x)^{-10} \\\\ &= 1\\cdot\\binom{9}{9} \\\\ &= 1\\cdot1 \\\\ &= \\underline{1}. \\end{align*}Hence, the answer is \\begin{align*} 2\\sum_{n=4}^{10} [x^{10-n}] \\frac{(1-x^3)^n}{(1-x)^n} &= 2(10+51+90+77+36+9+1) \\\\ &= 2\\cdot274 \\\\ &= 548. \\end{align*} -lpieleanu"
] |
2015-II-13
| 2,015
| 13
|
Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$ .
| 628
|
II
|
[
"If $n = 1$, $a_n = \\sin(1) > 0$. Then if $n$ satisfies $a_n < 0$, $n \\ge 2$, and \\[a_n = \\sum_{k=1}^n \\sin(k) = \\cfrac{1}{\\sin{1}} \\sum_{k=1}^n\\sin(1)\\sin(k) = \\cfrac{1}{2\\sin{1}} \\sum_{k=1}^n\\cos(k - 1) - \\cos(k + 1) = \\cfrac{1}{2\\sin(1)} [\\cos(0) + \\cos(1) - \\cos(n) - \\cos(n + 1)].\\] Since $2\\sin 1$ is positive, it does not affect the sign of $a_n$. Let $b_n = \\cos(0) + \\cos(1) - \\cos(n) - \\cos(n + 1)$. Now since $\\cos(0) + \\cos(1) = 2\\cos\\left(\\cfrac{1}{2}\\right)\\cos\\left(\\cfrac{1}{2}\\right)$ and $\\cos(n) + \\cos(n + 1) = 2\\cos\\left(n + \\cfrac{1}{2}\\right)\\cos\\left(\\cfrac{1}{2}\\right)$, $b_n$ is negative if and only if $\\cos\\left(\\cfrac{1}{2}\\right) < \\cos\\left(n + \\cfrac{1}{2}\\right)$, or when $n \\in [2k\\pi - 1, 2k\\pi]$. Since $\\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\\pi, 4\\pi, \\cdots$. Then the hundredth such value will be when $k = 100$ and $n = \\lfloor 200\\pi \\rfloor = \\lfloor 628.318 \\rfloor = 628 So n = 628. ~mathboy282",
"Notice that $a_n$ is the imaginary part of $\\sum_{k=1}^n e^{ik}$, by Euler's formula. Using the geometric series formula, we find that this sum is equal to \\[\\frac{e^{i(n+1)}-e^i}{e^i-1} = \\frac{\\cos (n+1) - \\cos 1 + i (\\sin (n+1) - \\sin 1) }{\\cos 1 - 1 + i \\sin 1}\\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \\[\\frac{(\\cos 1 - 1)(\\cos(n+1)-\\cos 1) + (\\sin 1)(\\sin(n+1)-\\sin 1) + i((\\sin(n+1) - \\sin 1)(\\cos 1 - 1) - (\\sin 1)(\\cos(n+1)-\\cos 1))}{\\cos^2 1 - 2 \\cos 1 + 1 + \\sin^2 1}\\] We only need to look at the imaginary part, which is \\[\\frac{(\\sin(n+1) \\cos 1 - \\cos(n+1) \\sin 1) - \\sin 1 \\cos 1 + \\sin 1 - \\sin (n+1) + \\sin 1 \\cos 1}{2-2 \\cos 1} = \\frac{\\sin n - \\sin(n+1) + \\sin 1}{2-2 \\cos 1}\\] Since $\\cos 1 < 1$, $2-2 \\cos 1 > 0$, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\\sin (n+1) - \\sin n > \\sin 1 \\implies 2 \\cos \\left(n + \\frac{1}{2} \\right) \\sin \\frac{1}{2} > \\sin 1 \\implies \\cos \\left( n + \\frac{1}{2} \\right) > \\frac{\\sin 1}{2 \\sin{\\frac{1}{2}}} = \\cos \\left(\\frac{1}{2} \\right),$ by sum to product. This only holds when $n$ is between $2\\pi k - 1$ and $2\\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$th such $n$ is $\\lfloor 200\\pi \\rfloor = 628.",
"Similar to solution 2, we set a complex number $z=\\cos 1+i\\sin 1$. We start from $z$ instead of $1$ because $k$ starts from $1$: be careful. The sum of $z+z^2+z^3+z^4+z^5\\dots=\\frac{z-z^{n+1}}{1-z}=\\frac{z}{z-1}\\left(z^n-1\\right)$. We are trying to make $n$ so that the imaginary part of this expression is negative. The argument of $z$ is $1$. The argument of $z-1$, however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$. The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$-axis is $half$ the change, due to Circle Theorems (this intercepted arc is the argument of $z$), because the $y$- axis is tangent to the circle at the origin. So $\\text{arg}(z-1)=\\frac{\\pi+1}{2}$. Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$-axis is $\\frac{1-\\pi}{2}$. We want the argument of the whole expression $-\\pi<\\theta<0$. This translates into $\\frac{-\\pi-1}{2}<\\text{arg}\\left(z^n-1\\right)<\\frac{\\pi-1}{2}$. $z^n-1$ also consists of points on the circle centered at $(-1,0)$, so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$-axis. Since $z^n-1$ is always negative, $\\frac{-3\\pi}{2}<\\text{arg}\\left(z^n-1\\right)<\\frac{-\\pi}{2}$, and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\\frac{\\pi}{2}-\\frac{\\pi-1}{2}=\\frac{-1}{2}$ angle with the $y$-axis both ways. So the argument of $z^n$ must be in the bound $-1<\\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category per rotation because $\\pi$ is irrational, $n_{100}\\equiv z^{628}$ and the answer is $628.",
"By the product-to-sum formula, \\[\\sin \\frac{1}{2} \\sin k = \\frac{1}{2} \\left[ \\cos \\left( k - \\frac{1}{2} \\right) - \\cos \\left( k + \\frac{1}{2} \\right) \\right].\\]Thus, we can make the sum in the problem telescope: \\begin{align*} a_n &= \\sum_{k = 1}^n \\sin k \\\\ &= \\sum_{k = 1}^n \\frac{\\sin \\frac{1}{2} \\sin k}{\\sin \\frac{1}{2}} \\\\ &= \\sum_{k = 1}^n \\frac{\\cos (k - \\frac{1}{2}) - \\cos (k + \\frac{1}{2})}{2 \\sin \\frac{1}{2}} \\\\ &= \\frac{(\\cos \\frac{1}{2} - \\cos \\frac{3}{2}) + (\\cos \\frac{3}{2} - \\cos \\frac{5}{2}) + \\dots + (\\cos \\frac{2n - 1}{2} - \\cos \\frac{2n + 1}{2})}{2 \\sin \\frac{1}{2}} \\\\ &= \\frac{\\cos \\frac{1}{2} - \\cos \\frac{2n + 1}{2}}{2 \\sin \\frac{1}{2}}. \\end{align*}Then $a_n < 0$ when $\\cos \\frac{1}{2} < \\cos \\frac{2n + 1}{2}.$ This occurs if and only if \\[2 \\pi k - \\frac{1}{2} < \\frac{2n + 1}{2} < 2 \\pi k + \\frac{1}{2}\\]for some integer $k.$ Equivalently, \\[2 \\pi k - 1 < n < 2 \\pi k.\\]In other words, $n = \\lfloor 2 \\pi k \\rfloor.$ The 100th index of this form is then $\\lfloor 2 \\pi \\cdot 100 \\rfloor = 628"
] |
2015-II-14
| 2,015
| 14
|
Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ .
| 89
|
II
|
[
"The expression we want to find is $2(x^3+y^3) + x^3y^3$. Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$, respectively. Dividing the latter by the former equation yields $\\frac{x^2-xy+y^2}{xy} = \\frac{945}{810}$. Adding 3 to both sides and simplifying yields $\\frac{(x+y)^2}{xy} = \\frac{25}{6}$. Solving for $x+y$ and substituting this expression into the first equation yields $\\frac{5\\sqrt{6}}{6}(xy)^{\\frac{9}{2}} = 810$. Solving for $xy$, we find that $xy = 3\\sqrt[3]{2}$, so $x^3y^3 = 54$. Substituting this into the second equation and solving for $x^3+y^3$ yields $x^3+y^3=\\frac{35}{2}$. So, the expression to evaluate is equal to $2 \\times \\frac{35}{2} + 54 = 089 we substitute and find the answer to be 89.",
"Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x+y)(x^2-xy+y^2)=945$, respectively. By the first equation, $x+y=\\frac{810}{x^4y^4}$. Plugging this in to the second equation and simplifying yields $(\\frac{x}{y}-1+\\frac{y}{x})=\\frac{7}{6}$. Now substitute $\\frac{x}{y}=a$. Solving the quadratic in $a$, we get $a=\\frac{x}{y}=\\frac{2}{3}$ or $\\frac{3}{2}$ As both of the original equations were symmetric in $x$ and $y$, WLOG, let $\\frac{x}{y}=\\frac{2}{3}$, so $x=\\frac{2}{3}y$. Now plugging this in to either one of the equations, we get the solutions $y=\\frac{3(2^{\\frac{2}{3}})}{2}$, $x=2^{\\frac{2}{3}}$. Now plugging into what we want, we get $8+54+27=089",
"Add three times the first equation to the second equation and factor to get $(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375$. Taking the cube root yields $xy(x+y)=15$. Noting that the first equation is $(xy)^3\\cdot(xy(x+y))=810$, we find that $(xy)^3=\\frac{810}{15}=54$. Plugging this into the second equation and dividing yields $x^3+y^3 = \\frac{945}{54} = \\frac{35}{2}$. Thus the sum required, as noted in Solution 1, is $54+\\frac{35}{2}\\cdot2 = 089.",
"As with the other solutions, factor. But this time, let $a=xy$ and $b=x+y$. Then $a^4b=810$. Notice that $x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)$, so we are looking for $2b(b^2-3a)+a^3$. Now, if we divide the second equation by the first one, we get $7/6 = \\frac{b^2-3a}{a}$; then $\\frac{b^2}{a}=\\frac{25}{6}$. Therefore, $a = \\frac{6}{25}b^2$. Substituting $\\frac{6}{25}b^2$ for $a$ in equation 1, simplifying, and then taking the cube root gives us $b^3 = \\frac{5^3}{2}.$ Finding $a^3$ by cubing $a = \\frac{6}{25}b^2$ on both sides and simplifying using our previous substitution, we get $a^3 = 54$. Substituting this into the first equation and then dividing by $27$, we get $2b(b^2 - 3a) = 35$. Our final answer is $35+54=089.",
"Factor the given equations as: \\[x^4y^4(x+y)=810\\] \\[x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945\\] We note that these expressions (as well as the desired expression) can be written exclusively in terms of $x+y$ and $xy$. We make the substitution $s=x+y$ and $p=xy$ (for sum and product, respectively). \\[x^4y^4(x+y)=p^4s=810\\] \\[x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945\\] We see that $p^4s$ shows up in both equations, so we can eliminate it and find $sp$, after which we can get $p^3$ from the first equation. If you rewrite the desired expression using $s$ and $p$, it becomes clear that you don't need to actually find the values of $s$ and $p$, but I will do so for the sake of completion. \\[s^3p^3=945+3p^4s\\] \\[s^3p^3=945+3(810)=3375\\] \\[sp=15\\] \\[p^3=\\frac{810}{sp}=54\\] \\[p=3\\cdot2^{1/3}\\] \\[s=\\frac{15}{p}=5\\cdot2^{-1/3}\\] The desired expression can be written as: \\[2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3\\] \\[2(s)(s^2-3p)+p^3=2s^3-6sp+p^3\\] Plugging in $s$ and $p$, we get: \\[2(5\\cdot2^{-1/3})^3-6(15)+54=125-90+54=089\\] - gting",
"Factor the first and second equations as $(xy)^4(x+y)=810$ and $(xy)^3(x+y)(x^2-xy+y^2)=945$, respectively. Dividing them (allowed, since neither are $0$), we have \\[\\frac{xy}{x^2-xy+y^2}=\\frac67\\] or \\[x^2-\\frac{13}{6}xy+y^2=0.\\] Plugging into the quadratic formula and solving for $x$ in terms of $y,$ we have \\[x=\\frac{\\frac{13y}{6}\\pm \\sqrt{\\frac{169y^2}{36}-4y^2}}{2}=\\frac{2y}3 , \\frac{3y}2 .\\] WLOG, let $x=\\frac{3y}2 .$ Plugging into our first equation, we have \\[\\left(\\frac{3}{2}y\\right)^4\\left(\\frac52 y\\right)=810 \\implies y^3 = 4 .\\] Plugging this result (and the one for $x$ in terms of $y$) into our desired expression, we have \\begin{align*} 2x^3+(xy)^3+2y^3 &= \\frac{27}{4}y^3 + \\left(\\frac{3}{2} y^2\\right)^3 +2y^3 \\\\ &= \\frac{35}{4}y^3 +\\frac{27}{8}(y^3)^2 \\\\ &= 35+54 \\\\ &= 089 \\end{align*} ~ASAB",
"Take $w=x+y$ and $z=xy$. Remark that \\begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\\\ =~&2(x^{3}+y^{3})+(xy)^{3} \\\\ =~&2\\left[(x+y)^{3}-3xy(x+y)\\right]+(xy)^{3} \\\\ =~&2(w^{3}-3wz)+z^{3} \\\\ =~& 2w^{3}-6wz+z^{3}.\\end{align*} The given equations imply that \\[wz^{4}=810~~~\\text{and}~~~(wz)^{3}-3wz^{4}=945.\\] Substituting the first equation into the second, we have that $(wz)^{3}=945+3\\cdot 810=3375$, thus $wz=\\sqrt[3]{3375}=15$. Now \\[z^{3}=\\frac{wz^{4}}{wz}=\\frac{810}{15}=54\\] and \\[w^{3}=\\frac{3375}{54}=\\frac{125}{2}.\\] Thus \\begin{align*}2w^{3}-6wz+z^{3}&=2\\left(\\frac{125}{2}\\right)-6(15)+54 \\\\ &=125-90+54 \\\\ &=89.\\end{align*}",
"$x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \\frac{x^2-xy+y^2}{xy}=\\frac{7}{6}, \\frac{x^2+y^2}{xy}=\\frac{13}{6}$ Let $x^2+y^2=13k; xy=6k$, then we can see $(x+y)^2-12k=13k, x+y=5\\sqrt{k}$, now, we see $x^4y^4\\cdot (x+y)=1296k^4\\cdot 5\\sqrt{k}, k=\\frac{1}{\\sqrt[3]{4}}$. The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\\cdot 35k^{\\frac{3}{2}}=89 ~bluesoul",
"Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \\cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$. Now, we will subtract and add the equations to gather information on $x$ and $y$. When we subtract the equations and clean it up via factoring, we yield $(xy)^3 \\cdot (x + y) \\cdot (x - y)^2 = 115$, and when we add them, we yield $(xy)^3 \\cdot (x + y) \\cdot (x^2 + y^2) = 1755$. Now with some intuition, you should divide the equations to obtain $\\frac{(x^2 + y^2)}{(x - y)^2} = 13$. Now, we clean this up to obtain the following factoring of $0 = 2 \\cdot (2x - 3y) \\cdot (3x - 2y)$. This implies that $x = \\frac{3y}{2}$. We plug that into the target expression to reduce it down to one variable, and get that target expression is $2x^3 + (xy)^3 + 2y^3 = \\frac{27}{4} \\cdot y^3 + \\frac{27}{8} \\cdot y^6 + 2y^3$. This means that if we can find a way to get $y^3$, then the rest is trivial. We get $y^3$ by plugging in $x = \\frac{3y}{2}$ into $x^3 \\cdot y^6 + y^3 \\cdot x^6 = 945$. However, this time we only factor as $(xy)^3 \\cdot (x^3 + y^3)$ because we particularly want a cubic degree on $y$. Plugging in $x = \\frac{3y}{2}$ we get $y^3 = 4$. Now lets plug this into our target expression to get $\\frac{27}{4} \\cdot 4 + \\frac{27}{8} \\cdot 4^2 + 2 \\cdot 4 = 89 by MinecraftPlayer404",
"Let $x, y$ be roots of some polynomial $P(\\alpha)$. Then we have that $P(\\alpha) = (\\alpha - x)(\\alpha - y) \\implies P(\\alpha) = \\alpha^2 - (x + y) \\alpha + xy$. Now from the first equation we have that $x + y = 810/(xy)^4$. For convenience, denote $z = xy$. Define $P_k = x^k + y^k$, where $x, y$ are roots of the above polynomial. Then we have that by Newton's formulas $P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}$, where $P_0 = 2$ and $P_1 = x + y = 810/z^4$. We desire $2P_3 + z^3$. Now building up this recurrence; \\[P_2 = \\frac{810}{z^4}P_1 - zP_0 = \\frac{810^2}{z^8} - 2z.\\] Then we have that \\[P_3 = \\frac{810}{z^4} \\left(\\frac{810^2}{z^8} - 3z \\right).\\] Plugging this value of $P_3$ into the second equation $z^3P_3 = 810$, yields $z^9 = 164025$. Now $z^3 = 54$. To compute $P_3$ is now trivial, as returning this new value of $z$ into the second equation yields $P_3 = 35/2$. Hence $z^3 + 2P_3 = 54 + 35 = 089. -th1nq3r",
"Let $x, y$ be roots of some polynomial $P(\\alpha)$. Then we have that $P(\\alpha) = (\\alpha - x)(\\alpha - y) \\implies P(\\alpha) = \\alpha^2 - (x + y) \\alpha + xy$. Now from the first equation we have that $x + y = 810/(xy)^4$. For convenience, denote $z = xy$. Define $P_k = x^k + y^k$, where $x, y$ are roots of the above polynomial. Then we have that by Newton's formulas $P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}$, where $P_0 = 2$ and $P_1 = x + y = 810/z^4$. We desire $2P_3 + z^3$. Now building up this recurrence; \\[P_2 = \\frac{810}{z^4}P_1 - zP_0 = \\frac{810^2}{z^8} - 2z.\\] Then we have that \\[P_3 = \\frac{810}{z^4} \\left(\\frac{810^2}{z^8} - 3z \\right).\\] Plugging this value of $P_3$ into the second equation $z^3P_3 = 810$, yields $z^9 = 164025$. Now $z^3 = 54$. To compute $P_3$ is now trivial, as returning this new value of $z$ into the second equation yields $P_3 = 35/2$. Hence $z^3 + 2P_3 = 54 + 35 = 089. -th1nq3r"
] |
2015-II-15
| 2,015
| 15
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Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ such that $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
| 129
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II
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[
"Let $M$ be the intersection of $\\overline{BC}$ and the common internal tangent of $\\mathcal P$ and $\\mathcal Q.$ We claim that $M$ is the circumcenter of right $\\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\\overline{BC},$ implying that $\\angle{BAC} = 90.$ Now draw $\\overline{PA}, \\overline{PB}, \\overline{PM},$ where $P$ is the center of circle $\\mathcal P.$ Quadrilateral $PAMB$ is cyclic, and by Pythagorean Theorem $PM = \\sqrt{5},$ so by Ptolemy on $PAMB$ we have \\[AB \\sqrt{5} = 2 \\cdot 1 + 2 \\cdot 1 = 4 \\iff AB = \\dfrac{4 \\sqrt{5}}{5}.\\] Do the same thing on cyclic quadrilateral $QAMC$ (where $Q$ is the center of circle $\\mathcal Q$ and get $AC = \\frac{8 \\sqrt{5}}{5}.$ Let $\\angle A = \\angle{DAB}.$ By Law of Sines, $BD = 2R \\sin A = 2 \\sin A.$ Note that $\\angle{D} = \\angle{ABC}$ from inscribed angles, so \\begin{align*} [ABD] &= \\dfrac{1}{2} BD \\cdot AB \\cdot \\sin{\\angle B} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{4 \\sqrt{5}}{5} \\cdot 2 \\sin A \\sin{\\left(180 - \\angle A - \\angle D\\right)} \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\sin{\\left(\\angle A + \\angle D\\right)} \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\left(\\sin A \\cos D + \\cos A \\sin D\\right) \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\left(\\sin A \\cos{\\angle{ABC}} + \\cos A \\sin{\\angle{ABC}}\\right) \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\left(\\dfrac{\\sqrt{5} \\sin A}{5} + \\dfrac{2 \\sqrt{5} \\cos A}{5}\\right) \\\\ &= \\dfrac{4}{5} \\cdot \\sin A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Similarly, $\\angle{EAC} = 90 - \\angle A,$ and by Law of Sines $CE = 8 \\sin{\\angle{EAC}} = 8 \\cos A.$ Note that $\\angle{E} = \\angle{ACB}$ from inscribed angles, so \\begin{align*} [ACE] &= \\dfrac{1}{2} AC \\cdot CE \\sin{\\angle C} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{8 \\sqrt{5}}{5} \\cdot 8 \\cos A \\sin{\\left[180 - \\left(90 - \\angle A\\right) - \\angle E\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\sin{\\left[\\left(90 - \\angle A\\right) + \\angle{ACB}\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\left(\\dfrac{2 \\sqrt{5} \\cos A}{5} + \\dfrac{\\sqrt{5} \\sin A}{5}\\right) \\\\ &= \\dfrac{32}{5} \\cdot \\cos A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Setting the two areas equal, we get \\[\\tan A = \\frac{\\sin A}{\\cos A} = 8 \\iff \\sin A = \\frac{8}{\\sqrt{65}}, \\cos A = \\frac{1}{\\sqrt{65}}\\] after Pythagorean Identity. Now plug back in and the common area is $\\frac{64}{65} \\iff 129",
"Let $M$ be the intersection of $\\overline{BC}$ and the common internal tangent of $\\mathcal P$ and $\\mathcal Q.$ We claim that $M$ is the circumcenter of right $\\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\\overline{BC},$ implying that $\\angle{BAC} = 90.$ Now draw $\\overline{PA}, \\overline{PB}, \\overline{PM},$ where $P$ is the center of circle $\\mathcal P.$ Quadrilateral $PAMB$ is cyclic, and by Pythagorean Theorem $PM = \\sqrt{5},$ so by Ptolemy on $PAMB$ we have \\[AB \\sqrt{5} = 2 \\cdot 1 + 2 \\cdot 1 = 4 \\iff AB = \\dfrac{4 \\sqrt{5}}{5}.\\] Do the same thing on cyclic quadrilateral $QAMC$ (where $Q$ is the center of circle $\\mathcal Q$ and get $AC = \\frac{8 \\sqrt{5}}{5}.$ Let $\\angle A = \\angle{DAB}.$ By Law of Sines, $BD = 2R \\sin A = 2 \\sin A.$ Note that $\\angle{D} = \\angle{ABC}$ from inscribed angles, so \\begin{align*} [ABD] &= \\dfrac{1}{2} BD \\cdot AB \\cdot \\sin{\\angle B} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{4 \\sqrt{5}}{5} \\cdot 2 \\sin A \\sin{\\left(180 - \\angle A - \\angle D\\right)} \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\sin{\\left(\\angle A + \\angle D\\right)} \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\left(\\sin A \\cos D + \\cos A \\sin D\\right) \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\left(\\sin A \\cos{\\angle{ABC}} + \\cos A \\sin{\\angle{ABC}}\\right) \\\\ &= \\dfrac{4 \\sqrt{5}}{5} \\cdot \\sin A \\cdot \\left(\\dfrac{\\sqrt{5} \\sin A}{5} + \\dfrac{2 \\sqrt{5} \\cos A}{5}\\right) \\\\ &= \\dfrac{4}{5} \\cdot \\sin A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Similarly, $\\angle{EAC} = 90 - \\angle A,$ and by Law of Sines $CE = 8 \\sin{\\angle{EAC}} = 8 \\cos A.$ Note that $\\angle{E} = \\angle{ACB}$ from inscribed angles, so \\begin{align*} [ACE] &= \\dfrac{1}{2} AC \\cdot CE \\sin{\\angle C} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{8 \\sqrt{5}}{5} \\cdot 8 \\cos A \\sin{\\left[180 - \\left(90 - \\angle A\\right) - \\angle E\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\sin{\\left[\\left(90 - \\angle A\\right) + \\angle{ACB}\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\left(\\dfrac{2 \\sqrt{5} \\cos A}{5} + \\dfrac{\\sqrt{5} \\sin A}{5}\\right) \\\\ &= \\dfrac{32}{5} \\cdot \\cos A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Setting the two areas equal, we get \\[\\tan A = \\frac{\\sin A}{\\cos A} = 8 \\iff \\sin A = \\frac{8}{\\sqrt{65}}, \\cos A = \\frac{1}{\\sqrt{65}}\\] after Pythagorean Identity. Now plug back in and the common area is $\\frac{64}{65} \\iff 129",
"[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"$A$\",A,NE);label(\"$O_1$\",O_1,NE);label(\"$O_2$\",O_2,NE);label(\"$B$\",B,SW);label(\"$C$\",C,SW);label(\"$D$\",D,NE);label(\"$E$\",E,NE);label(\"$N$\",N,W);label(\"$K$\",(-24/15,0.2));label(\"$L$\",(24/15,0.2));label(\"$n$\",(-0.8,-0.12));label(\"$p$\",((29/15,-48/15)));label(\"$\\mathcal{P}$\",(-1.6,1.1));label(\"$\\mathcal{Q}$\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy] Call $O_1$ and $O_2$ the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\\triangle{O_1BN} \\sim \\triangle{O_2CN}$ and setting $NO_1 = k$, we have that $\\frac{k+5}{k} = \\frac{4}{1} \\implies k=\\frac{5}{3}$. We can do some more length chasing using triangles similar to $O_1BN$ to get that $AK = AL = \\frac{24}{15}$, $BK = \\frac{12}{15}$, and $CL = \\frac{48}{15}$. Now, consider the circles $\\mathcal{P}$ and $\\mathcal{Q}$ on the coordinate plane, where $A$ is the origin. If the line $\\ell$ through $A$ intersects $\\mathcal{P}$ at $D$ and $\\mathcal{Q}$ at $E$ then $4 \\cdot DA = AE$. To verify this, notice that $\\triangle{AO_1D} \\sim \\triangle{EO_2A}$ from the fact that both triangles are isosceles with $\\angle{O_1AD} \\cong \\angle{O_2AE}$, which are corresponding angles. Since $O_2A = 4\\cdot O_1A$, we can conclude that $4 \\cdot DA = AE$. Hence, we need to find the slope $m$ of line $\\ell$ such that the perpendicular distance $n$ from $B$ to $AD$ is four times the perpendicular distance $p$ from $C$ to $AE$. This will mean that the product of the bases and heights of triangles $ACE$ and $DBA$ will be equal, which in turn means that their areas will be equal. Let the line $\\ell$ have the equation $y = -mx \\implies mx + y = 0$, and let $m$ be a positive real number so that the negative slope of $\\ell$ is preserved. Setting $A = (0,0)$, the coordinates of $B$ are $(x_B, y_B) = \\left(\\frac{-24}{15}, \\frac{-12}{15}\\right)$, and the coordinates of $C$ are $(x_C, y_C) = \\left(\\frac{24}{15}, \\frac{-48}{15}\\right)$. Using the point-to-line distance formula and the condition $n = 4p$, we have \\[\\frac{|mx_B + 1(y_B) + 0|}{\\sqrt{m^2 + 1}} = \\frac{4|mx_C + 1(y_C) + 0|}{\\sqrt{m^2 + 1}}\\] \\[\\implies |mx_B + y_B| = 4|mx_C + y_C| \\implies \\left|\\frac{-24m}{15} + \\frac{-12}{15}\\right| = 4\\left|\\frac{24m}{15} + \\frac{-48}{15}\\right|.\\] If $m > 2$, then clearly $B$ and $C$ would not lie on the same side of $\\ell$. Thus since $m > 0$, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have \\[\\frac{6m}{15} + \\frac{3}{15} = \\frac{48}{15} - \\frac{24m}{15} \\implies 2m = 3 \\implies m = \\frac{3}{2}.\\] Thus, the equation of $\\ell$ is $y = -\\frac{3}{2}x$. Then we can find the coordinates of $D$ by finding the point $(x,y)$ other than $A = (0,0)$ where the circle $\\mathcal{P}$ intersects $\\ell$. $\\mathcal{P}$ can be represented with the equation $(x + 1)^2 + y^2 = 1$, and substituting $y = -\\frac{3}{2}x$ into this equation yields $x = 0, -\\frac{8}{13}$ as solutions. Discarding $x = 0$, the $y$-coordinate of $D$ is $-\\frac{3}{2} \\cdot -\\frac{8}{13} = \\frac{12}{13}$. The distance from $D$ to $A$ is then $\\frac{4}{\\sqrt{13}}.$ The perpendicular distance from $B$ to $AD$ or the height of $\\triangle{DBA}$ is $\\frac{|\\frac{3}{2}\\cdot\\frac{-24}{15} + \\frac{-12}{15} + 0|}{\\sqrt{\\frac{3}{2}^2 + 1}} = \\frac{\\frac{48}{15}}{\\frac{\\sqrt{13}}{2}} = \\frac{32}{5\\sqrt{13}}.$ Finally, the common area is $\\frac{1}{2}\\left(\\frac{32}{5\\sqrt{13}} \\cdot \\frac{4}{\\sqrt{13}}\\right) = \\frac{64}{65}$, and $m + n = 64 + 65 = 129.",
"By homothety, we deduce that $AE = 4 AD$. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of $P$ and $Q$ to $l$.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from $B$ to $l$ is four times that from $C$ to $l$. Let the distance from $C$ be $x$ and the distance from $B$ be $4x$. Let $P$ and $Q$ be the centers of their respective circles. Then dropping a perpendicular from $P$ to $Q$ creates a $3-4-5$ right triangle, from which $BC = 4$ and, if $\\alpha = \\angle{AQC}$, that $\\cos \\alpha = \\dfrac{3}{5}$. Then $\\angle{BPA} = 180^\\circ - \\alpha$, and the Law of Cosines on triangles $APB$ and $AQC$ gives $AB = \\dfrac{4}{\\sqrt{5}}$ and $AC = \\dfrac{8}{\\sqrt{5}}.$ Now, using the Pythagorean Theorem to express the length of the projection of $BC$ onto line $l$ gives \\[\\sqrt{\\frac{16}{5} - 16x^2} + \\sqrt{\\frac{64}{5} - x^2} = \\sqrt{16 - 9x^2}.\\] Squaring and simplifying gives \\[\\sqrt{\\left(\\frac{1}{5} - x^2\\right)\\left(\\frac{64}{5} - x^2\\right)} = x^2,\\] and squaring and solving gives $x = \\dfrac{8}{5\\sqrt{13}}.$ By the Law of Sines on triangle $ABD$, we have \\[\\frac{BD}{\\sin A} = 2.\\] But we know $\\sin A = \\dfrac{4x}{AB}$, and so a small computation gives $BD = \\dfrac{16}{\\sqrt{65}}.$ The Pythagorean Theorem now gives \\[AD = \\sqrt{BD^2 - (4x)^2} + \\sqrt{AB^2 - (4x)^2} = \\frac{4}{\\sqrt{13}},\\] and so the common area is $\\dfrac{1}{2} \\cdot \\frac{4}{\\sqrt{13}} \\cdot \\frac{32}{5\\sqrt{13}} = \\frac{64}{65}.$ The answer is $129",
"$DE$ goes through $A$, the point of tangency of both circles. So $DE$ intercepts equal arcs in circle $P$ and $Q$: homothety. Hence, $AE=4AD$. We will use such similarity later. The diagonal distance between the centers of the circles is $4+1=5$. The difference in heights is $4-1=3$. So $BC=\\sqrt{5^2-3^2}=4$. The triangle connecting the centers with a side parallel to $BC$ is a $3-4-5$ right triangle. Since $O_PA=1$, the height of $A$ is $1+3/5=8/5$. Drop an altitude from $A$ to $BC$ and call it $I$: $IB=4/5$ and $IC=4-4/5=16/5$. Since right $\\triangle AIB\\sim\\triangle CIB$, $ABC$ is a right triangle also; $IB:IA:IC$ form a geometric progression $\\times 2$. Extend $BA$ through $A$ to a point $G$ on the other side of $\\circ Q$. By homothety, $\\triangle DAB\\sim\\triangle EAG$. By angle chasing $\\triangle DAB$ through right triangle $ABC$, we deduce that $\\angle CEG$ is a right angle. Since $ACEG$ is cyclic, $\\angle GAC$ is also right. So $CG$ is a diameter of $\\circ G$. Because of this, $CG \\perp BC$, the tangent line. $\\triangle BCG$ is right and $\\triangle BCG\\sim\\triangle ABC\\sim\\triangle CAG$. $AC=\\sqrt{(8/5)^2+(16/5)^2}=8\\sqrt{5}/5$ so $AG=2AC=16\\sqrt{5}/5$ and $[\\triangle CAG]=64/5$. Since $[\\triangle DAB]=[\\triangle ACE]$, the common area is $[ACEG]/17$. $16[\\triangle DAB]=[\\triangle GAE]$ because the triangles are similar with a ratio of $1:4$. So we only need to find $[\\triangle CEG]$ now. Extend $DE$ through $E$ to intersect the tangent at $F$. Because $4DA=AE$, the altitude from $B$ to $AD$ is $4$ times the height from $C$ to $EA$. So $BC=3/4BF$ and $BF=16/3$. We look at right triangle $\\triangle AIF$. $IF=68/15$ and $AI=8/5$. $\\triangle AIF$ is a $17-6-5\\sqrt{13}$ right triangle. Hypotenuse $AF$ intersects $CG$ at a point, we call it $H$. $CH=4/3\\div 68/15\\cdot 8/5=8/17$. So $HG=8-8/17=128/17$. By Power of a Point, $CH\\cdot HG=AH\\cdot HE$. $AH=16/5\\cdot 5\\sqrt{13}/17=16\\sqrt{13}/17.$ So $HE=1024/289\\cdot 17/(16\\sqrt{13})=64/(17\\sqrt{13})$. The height from $E$ to $CG$ is $17/(5\\sqrt{13})\\cdot 64/(17\\sqrt{13})=64/65$. Thus, $[\\triangle CEG]=64/65\\cdot 8\\div 2=256/65$. The area of the whole cyclic quadrilateral is $64/5+256/65=(832+256)/65=1088/65$. Lastly, the common area is $1/17$ the area of the quadrilateral, or $64/65$. So $64+65=129.",
"[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"$A$\",A,NE);label(\"$O_1$\",O_1,NE);label(\"$O_2$\",O_2,NE);label(\"$B$\",B,SW);label(\"$C$\",C,SW);label(\"$D$\",D,NE);label(\"$E$\",E,NE);label(\"$N$\",N,W);label(\"$K$\",(-24/15,0.2));label(\"$L$\",(24/15,0.2));label(\"$n$\",(-0.8,-0.12));label(\"$p$\",((29/15,-48/15)));label(\"$\\mathcal{P}$\",(-1.6,1.1));label(\"$\\mathcal{Q}$\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5)); //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); path circle2 = Circle((4,0),4); N = (-8/3,0); pair X =rotate(180,O_2)*E; pair Y = (8,0); draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y); dot(\"$X$\", X, NE);dot(\"$Y$\", Y, NE); [/asy] Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because $AE=XY$ and $AE \\parallel XY$, $XYE$ is right angle. First, $\\frac{NO_1}{NO_1+5} = \\frac{1}{4}$, so $NO_1=\\frac{5}{3}$. And, \\[\\cos{\\angle{AO_2C}}=\\cos{2\\angle{AYC}} = \\frac{O_2C}{NO_1+5} = \\frac{3}{5}\\] \\[\\sin{\\angle{AYC}} = \\sqrt{\\dfrac{1-\\cos{2\\angle{AYC}}}{2}}=\\frac{1}{\\sqrt{5}}\\] \\[\\cos{\\angle{AYC}} = \\frac{2}{\\sqrt{5}}\\] Then, \\[[AEC] = \\frac{1}{2}AE*CE*\\sin{\\angle{ACE}}=\\frac{1}{2}AE*8\\sin{\\angle{CYE}}*\\frac{1}{\\sqrt{5}}\\] \\[[CXY] = \\frac{1}{2}CX*XY*\\sin{\\angle{CXY}}=\\frac{1}{2}*8\\sin{\\angle{XYC}}*XY*\\sin{\\angle{CAY}}\\] Since $\\angle{CAY} = 90 - \\angle{AYC}$, $\\angle{XYC} = 90 - \\angle{CYE}$, $XY = AE$, we have \\[[CXY] = \\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\cos{\\angle{AYC}}=\\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\frac{2}{\\sqrt{5}}\\] Since $\\triangle{CXY}$ is four times in scale to $\\triangle{AEC}$, their area ratio is 16. Divide the two equations for the two areas, we have \\[\\tan{\\angle{CYE}} = \\frac{1}{8}\\] With this angle found, everything else just follows. \\[\\sin{\\angle{CYE}} = \\dfrac{\\tan{\\angle{CYE}}}{\\sqrt{1+\\tan^2{\\angle{CYE}}}}=\\dfrac{1}{\\sqrt{65}}\\] \\[\\cos{\\angle{CYE}} = \\dfrac{8}{\\sqrt{65}}\\] \\[\\sin{\\angle{AYE}} = \\sin({\\angle{AYC}+\\angle{CYE}} )= \\frac{1}{\\sqrt{5}}*\\frac{8}{\\sqrt{65}} + \\frac{2}{\\sqrt{5}}*\\frac{1}{\\sqrt{65}} = \\frac{2}{\\sqrt{13}}\\] \\[AE = 8\\sin{\\angle{AYE}} = \\frac{16}{\\sqrt{13}}\\] \\[[AEC] = \\frac{1}{2}*8*\\frac{16}{\\sqrt{13}}*\\dfrac{1}{\\sqrt{65}}*\\frac{1}{\\sqrt{5}}=\\frac{64}{65}\\] Thus, our answer is $64+65=129.",
"[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"$A$\",A,NE);label(\"$O_1$\",O_1,NE);label(\"$O_2$\",O_2,NE);label(\"$B$\",B,SW);label(\"$C$\",C,SW);label(\"$D$\",D,NE);label(\"$E$\",E,NE);label(\"$N$\",N,W);label(\"$K$\",(-24/15,0.2));label(\"$L$\",(24/15,0.2));label(\"$n$\",(-0.8,-0.12));label(\"$p$\",((29/15,-48/15)));label(\"$\\mathcal{P}$\",(-1.6,1.1));label(\"$\\mathcal{Q}$\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5)); //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); path circle2 = Circle((4,0),4); N = (-8/3,0); pair X =rotate(180,O_2)*E; pair Y = (8,0); draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y); dot(\"$X$\", X, NE);dot(\"$Y$\", Y, NE); [/asy] Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because $AE=XY$ and $AE \\parallel XY$, $XYE$ is right angle. First, $\\frac{NO_1}{NO_1+5} = \\frac{1}{4}$, so $NO_1=\\frac{5}{3}$. And, \\[\\cos{\\angle{AO_2C}}=\\cos{2\\angle{AYC}} = \\frac{O_2C}{NO_1+5} = \\frac{3}{5}\\] \\[\\sin{\\angle{AYC}} = \\sqrt{\\dfrac{1-\\cos{2\\angle{AYC}}}{2}}=\\frac{1}{\\sqrt{5}}\\] \\[\\cos{\\angle{AYC}} = \\frac{2}{\\sqrt{5}}\\] Then, \\[[AEC] = \\frac{1}{2}AE*CE*\\sin{\\angle{ACE}}=\\frac{1}{2}AE*8\\sin{\\angle{CYE}}*\\frac{1}{\\sqrt{5}}\\] \\[[CXY] = \\frac{1}{2}CX*XY*\\sin{\\angle{CXY}}=\\frac{1}{2}*8\\sin{\\angle{XYC}}*XY*\\sin{\\angle{CAY}}\\] Since $\\angle{CAY} = 90 - \\angle{AYC}$, $\\angle{XYC} = 90 - \\angle{CYE}$, $XY = AE$, we have \\[[CXY] = \\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\cos{\\angle{AYC}}=\\frac{1}{2}*8AE\\cos{\\angle{CYE}}*\\frac{2}{\\sqrt{5}}\\] Since $\\triangle{CXY}$ is four times in scale to $\\triangle{AEC}$, their area ratio is 16. Divide the two equations for the two areas, we have \\[\\tan{\\angle{CYE}} = \\frac{1}{8}\\] With this angle found, everything else just follows. \\[\\sin{\\angle{CYE}} = \\dfrac{\\tan{\\angle{CYE}}}{\\sqrt{1+\\tan^2{\\angle{CYE}}}}=\\dfrac{1}{\\sqrt{65}}\\] \\[\\cos{\\angle{CYE}} = \\dfrac{8}{\\sqrt{65}}\\] \\[\\sin{\\angle{AYE}} = \\sin({\\angle{AYC}+\\angle{CYE}} )= \\frac{1}{\\sqrt{5}}*\\frac{8}{\\sqrt{65}} + \\frac{2}{\\sqrt{5}}*\\frac{1}{\\sqrt{65}} = \\frac{2}{\\sqrt{13}}\\] \\[AE = 8\\sin{\\angle{AYE}} = \\frac{16}{\\sqrt{13}}\\] \\[[AEC] = \\frac{1}{2}*8*\\frac{16}{\\sqrt{13}}*\\dfrac{1}{\\sqrt{65}}*\\frac{1}{\\sqrt{5}}=\\frac{64}{65}\\] Thus, our answer is $64+65=129.",
"Let $K$ be the intersection of $BC$ and $AE$. Since the radii of the two circles are 1:4, so we have $AD:AE=1:4$, and the distance from $B$ to line $l$ and the distance from $C$ to line $l$ are in a ratio of 4:1, so $BK:CK=4:1$. We can easily calculate the length of $BC$ to be 4, so $CK=\\frac{4}{3}$. Let $J$ be the foot of perpendicular line from $A$ to $BC$, we can know that $BJ:CJ=1:4$, so $BJ = 0.8$, $CJ=3.2$, $AJ=1.6$, and $AK=\\sqrt{1.6^2+\\left(3.2+\\frac{4}{3}\\right)^2}=\\frac{4}{3}\\sqrt{13}$. Since $CK^2 = EK\\cdot AK$, so $EK=\\frac{4}{39}\\sqrt{13}$, and $AE = \\frac{4}{3}\\sqrt{13} - \\frac{4}{39}\\sqrt{13}= \\frac{16}{13}\\sqrt{13}$. $\\sin\\angle AKB=\\frac{AJ}{AK} = \\frac{1.6}{\\frac{4}{3}\\sqrt{13}}=\\frac{1.2}{\\sqrt{13}}$, so the distance from $C$ to line $l$ is $d=CK\\cdot \\sin\\angle AKB = \\frac{4}{3}\\cdot \\frac{1.2}{\\sqrt{13}}=\\frac{1.6}{\\sqrt{13}}$. so the area is \\[[ACE] = \\frac{1}{2}\\cdot AE\\cdot d = \\frac{1}{2}\\cdot\\frac{16}{13}\\sqrt{13}\\frac{1.6}{\\sqrt{13}} = \\frac{64}{65}\\] The final answer is $129. --- by Dan Li",
"Let $K$ be the intersection of $BC$ and $AE$. Since the radii of the two circles are 1:4, so we have $AD:AE=1:4$, and the distance from $B$ to line $l$ and the distance from $C$ to line $l$ are in a ratio of 4:1, so $BK:CK=4:1$. We can easily calculate the length of $BC$ to be 4, so $CK=\\frac{4}{3}$. Let $J$ be the foot of perpendicular line from $A$ to $BC$, we can know that $BJ:CJ=1:4$, so $BJ = 0.8$, $CJ=3.2$, $AJ=1.6$, and $AK=\\sqrt{1.6^2+\\left(3.2+\\frac{4}{3}\\right)^2}=\\frac{4}{3}\\sqrt{13}$. Since $CK^2 = EK\\cdot AK$, so $EK=\\frac{4}{39}\\sqrt{13}$, and $AE = \\frac{4}{3}\\sqrt{13} - \\frac{4}{39}\\sqrt{13}= \\frac{16}{13}\\sqrt{13}$. $\\sin\\angle AKB=\\frac{AJ}{AK} = \\frac{1.6}{\\frac{4}{3}\\sqrt{13}}=\\frac{1.2}{\\sqrt{13}}$, so the distance from $C$ to line $l$ is $d=CK\\cdot \\sin\\angle AKB = \\frac{4}{3}\\cdot \\frac{1.2}{\\sqrt{13}}=\\frac{1.6}{\\sqrt{13}}$. so the area is \\[[ACE] = \\frac{1}{2}\\cdot AE\\cdot d = \\frac{1}{2}\\cdot\\frac{16}{13}\\sqrt{13}\\frac{1.6}{\\sqrt{13}} = \\frac{64}{65}\\] The final answer is $129. --- by Dan Li",
"Consider the common tangent from $A$ to both circles. Let this intersect $BC$ at point $K$. From equal tangents, we have $BK=AK=CK$, which implies that $\\angle BAC = 90^\\circ$. Let the center of $\\mathcal{P}$ be $O_1$, and the center of $\\mathcal{Q}$ be $O_2$. Angle chasing, we find that $\\triangle O_1DA \\sim \\triangle O_2EA$ with a ratio of $1:4$. Hence $4AD = AE$. We can easily deduce that $BC=4$ by dropping an altitude from $O_1$ to $O_2C$. Let $\\angle ABC = \\theta$. By some simple angle chasing, we obtain that $\\angle BO_1A = 2\\angle BDA = 2\\angle ABC = 2\\theta,$ and similarly $\\angle CO_2A = 180 - 2\\theta$. Using LoC, we get that $AB = \\sqrt{2-2\\cos2\\theta}$ and $AC = \\sqrt{32+32\\cos2\\theta}$. From Pythagorean theorem, we have \\[AB^2 + AC^2 = BC^2 \\implies \\cos 2\\theta = -\\frac{3}{5} \\implies \\cos \\theta = \\frac{1}{\\sqrt5}, \\sin \\theta = \\frac{2}{\\sqrt 5}\\] In other words, $AB = \\frac{4}{\\sqrt5}, AC = \\frac{8}{\\sqrt5}$. Using the area condition, we have: \\begin{align*} \\frac12 AD*AB \\sin \\angle DAB &= \\frac 12 AE*AC \\sin(90-\\angle DAB) \\\\ AD*\\frac{4}{\\sqrt5} \\sin \\angle DAB &= 4AD*\\frac{8}{\\sqrt5} \\cos \\angle DAB \\\\ \\sin \\angle DAB &= 8 \\cos \\angle DAB \\\\ \\implies \\sin \\angle DAB &= \\frac{8}{\\sqrt{65}} \\end{align*} Now, for brevity, let $\\angle D = \\angle ADB$ and $\\angle A = \\angle DAB$. From Law of Sines on $\\triangle ABD$, we have \\begin{align*} \\frac{AB}{\\sin \\angle D} &= \\frac{AD}{\\sin (180-\\angle A - \\angle D)} \\\\ \\frac{\\frac{4}{\\sqrt5}}{\\frac{2}{\\sqrt5}} &= \\frac{AD}{\\sin \\angle A \\cos\\angle D + \\sin \\angle D\\cos\\angle A} \\\\ 2 &= \\frac{AD}{\\frac{2}{\\sqrt{13}}} \\\\ AD &= \\frac{4}{\\sqrt{13}} \\end{align*} It remains to find the area of $\\triangle ABD$. This is just \\[\\frac12 AD*AB*\\sin \\angle A = \\frac12 * \\frac{4}{\\sqrt{13}}*\\frac{4}{\\sqrt5}*\\frac{8}{\\sqrt{65}} = \\frac{64}{65}\\] for an answer of $129 This solution was brought to you by Leonard_my_dude.",
"Add in the line $k$ as the internal tangent between the two circles. Let $M$ be the midpoint of $BC$; It is well-known that $M$ is on $k$ and because $k$ is the radical axis of the two circles, $AM=BM=CM$. Therefore because $M$ is the circumcenter of $\\triangle{BAC}$, $\\angle{BAC}=90^{\\circ}$. Let $O_P$ be the center of circle $\\mathcal{P}$ and likewise let $O_Q$ be the center of circle $\\mathcal{Q}$. It is well known that by homothety $O_P, A,$ and $O_Q$ are collinear. It is well-known that $\\angle{ABC}=\\angle{ADB}=b$, and likewise $\\angle{ACB}=\\angle{AEC}=c$. By homothety, $AD=4AE$, therefore since the two triangles mentioned in the problem, the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$. Using the Pythagorean Theorem, $BC = 4$. By angle-chasing, $O_{P}AMB$ is cyclic, and likewise $O_{Q}CMA$ is cyclic. Use the Pythagorean Theorem for $\\triangle{O_{P}BM}$ to get $O_{P}M=\\sqrt{5}$. Then by Ptolemy's Theorem $AB=\\frac{4\\sqrt{5}}{5} \\implies AC=\\frac{8\\sqrt{5}}{5}$. Now to compute the area, using what we know about the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$, letting $x$ be the length of the altitude from $C$ to $AE$, $x=\\frac{8}{5\\sqrt{13}}$. From the Law of Sines, $\\frac{BD}{\\sin{A}}=2 \\implies \\sin{A}=\\frac{4x}{AB} \\implies BD=\\frac{8x}{AB}=\\frac{16\\sqrt{65}}{65}$. Then use the Pythagorean Theorem twice and add up the lengths to get $AD=\\frac{4}{\\sqrt{13}}$. Use the formula $\\frac{b \\times h}{2}$ to get $\\frac{64}{65} = 129 as the answer. ~First",
"Add in the line $k$ as the internal tangent between the two circles. Let $M$ be the midpoint of $BC$; It is well-known that $M$ is on $k$ and because $k$ is the radical axis of the two circles, $AM=BM=CM$. Therefore because $M$ is the circumcenter of $\\triangle{BAC}$, $\\angle{BAC}=90^{\\circ}$. Let $O_P$ be the center of circle $\\mathcal{P}$ and likewise let $O_Q$ be the center of circle $\\mathcal{Q}$. It is well known that by homothety $O_P, A,$ and $O_Q$ are collinear. It is well-known that $\\angle{ABC}=\\angle{ADB}=b$, and likewise $\\angle{ACB}=\\angle{AEC}=c$. By homothety, $AD=4AE$, therefore since the two triangles mentioned in the problem, the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$. Using the Pythagorean Theorem, $BC = 4$. By angle-chasing, $O_{P}AMB$ is cyclic, and likewise $O_{Q}CMA$ is cyclic. Use the Pythagorean Theorem for $\\triangle{O_{P}BM}$ to get $O_{P}M=\\sqrt{5}$. Then by Ptolemy's Theorem $AB=\\frac{4\\sqrt{5}}{5} \\implies AC=\\frac{8\\sqrt{5}}{5}$. Now to compute the area, using what we know about the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$, letting $x$ be the length of the altitude from $C$ to $AE$, $x=\\frac{8}{5\\sqrt{13}}$. From the Law of Sines, $\\frac{BD}{\\sin{A}}=2 \\implies \\sin{A}=\\frac{4x}{AB} \\implies BD=\\frac{8x}{AB}=\\frac{16\\sqrt{65}}{65}$. Then use the Pythagorean Theorem twice and add up the lengths to get $AD=\\frac{4}{\\sqrt{13}}$. Use the formula $\\frac{b \\times h}{2}$ to get $\\frac{64}{65} = 129 as the answer. ~First",
"Let $P$ and $Q$ be the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$ , respectively. Let $M$ be midpoint $BC, \\beta = \\angle ACB.$ Upper diagram shows that $\\sin 2\\beta = \\frac {4}{5}$ and $AC = 2 AB.$ Therefore $\\cos 2\\beta = \\frac {3}{5}.$ Let $CH\\perp l, BH'\\perp l.$ Lower diagram shows that $\\angle CAE = \\angle ABH' = \\alpha$ (perpendicular sides) and $\\angle CQE = 2\\alpha$ (the same intersept $\\overset{\\Large\\frown} {CE}).$ \\[\\tan\\alpha = \\frac {1}{8}, \\sin2\\alpha = \\frac{2 \\tan \\alpha}{1 + \\tan^2 \\alpha} = \\frac {16}{65}, \\cos2\\alpha = \\frac{1 - \\tan^2 \\alpha}{1 + \\tan^2 \\alpha} = \\frac {63}{65}.\\] The area \\[[ACE] = [AQC]+[CQE]– [AQE].\\] Hence \\[[ACE] =\\frac{AQ^2}{2} \\left(\\sin 2\\alpha + \\sin 2\\beta - \\sin(2\\alpha + 2\\beta)\\right),\\] \\[[ACE] = 8\\left( \\frac{16}{65}+\\frac{4}{5} - \\frac{4}{5}\\cdot \\frac{63}{65} - \\frac{3}{5}\\cdot \\frac{16}{65}\\right) = \\frac{64}{65}\\implies 129.\\] [email protected], vvsss",
"Let $P$ and $Q$ be the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$ , respectively. Let $M$ be midpoint $BC, \\beta = \\angle ACB.$ Upper diagram shows that $\\sin 2\\beta = \\frac {4}{5}$ and $AC = 2 AB.$ Therefore $\\cos 2\\beta = \\frac {3}{5}.$ Let $CH\\perp l, BH'\\perp l.$ Lower diagram shows that $\\angle CAE = \\angle ABH' = \\alpha$ (perpendicular sides) and $\\angle CQE = 2\\alpha$ (the same intersept $\\overset{\\Large\\frown} {CE}).$ \\[\\tan\\alpha = \\frac {1}{8}, \\sin2\\alpha = \\frac{2 \\tan \\alpha}{1 + \\tan^2 \\alpha} = \\frac {16}{65}, \\cos2\\alpha = \\frac{1 - \\tan^2 \\alpha}{1 + \\tan^2 \\alpha} = \\frac {63}{65}.\\] The area \\[[ACE] = [AQC]+[CQE]– [AQE].\\] Hence \\[[ACE] =\\frac{AQ^2}{2} \\left(\\sin 2\\alpha + \\sin 2\\beta - \\sin(2\\alpha + 2\\beta)\\right),\\] \\[[ACE] = 8\\left( \\frac{16}{65}+\\frac{4}{5} - \\frac{4}{5}\\cdot \\frac{63}{65} - \\frac{3}{5}\\cdot \\frac{16}{65}\\right) = \\frac{64}{65}\\implies 129.\\] [email protected], vvsss",
"We begin by extending $\\overline{AB}$ upwards until it intersects Circle $\\mathcal{Q}$. We can call this point of intersection $F$. Connect $F$ with $E$, $C$, and $A$ for future use. Create a trapezoid with points $B$, $C$, and the origins of Circles $\\mathcal{P}$ and $\\mathcal{Q}$. After quick inspection, we can conclude that the distance between the origins is 5 and that $\\overline{BC}$ is 4. (Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is $2\\cdot \\sqrt{a\\cdot b}$, where $a$ and $b$ are the two respective radii of the circles. In our case, we get $2\\cdot \\sqrt{4} = 4$. Using similar triangles or homotheties, $AE=4\\cdot AD$ and $AF=4\\cdot BA$. $BC^2 = BA\\cdot BF$. \\[16 = BA\\cdot (5\\cdot BA)\\] \\[BA = \\frac{4}{\\sqrt{5}}\\] \\[AF = \\frac{16}{\\sqrt{5}}\\] \\[BF = 4\\cdot \\sqrt{5}\\] Inspecting $\\triangle{BFC}$, we recognize that it is a right triangle ($\\angle{BCF} = 90$) as the final length ($\\overline{FC}$) being 8 would allow for an $x-2x-x\\sqrt{5}$ triangle. Hence, the diameter of circle $\\mathcal{Q}$ = $CF$. This also means that $\\angle{CAF} = \\angle{CEF} = 90$. From the fact that $\\triangle{ABC}$ is a right triangle: \\[AB^2 + AC^2 = BC^2\\] \\[AC = \\frac{8}{\\sqrt{5}}\\] (Note: We could have also used $\\triangle{FAC}$.) Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between $\\overline{FC}$ and $\\overline{DE}$ as $M$. Label $\\angle{CAE} = \\angle{CFE} = \\theta$. Since $\\angle{BAC} = 90$, $\\angle{EAF} = \\angle{DAB} = 90-\\theta$. Now, use the sine formula and the fact that the areas of $\\triangle{DBA}$ and $\\triangle{ACE}$ are equal to get: \\[\\frac{1}{2}\\cdot AC\\cdot AE\\cdot \\sin{\\theta} = \\frac{1}{2}\\cdot AD\\cdot AB\\cdot \\sin{(90-\\theta)}\\] Since $\\sin{(90-\\theta)} = \\cos{\\theta}$: \\[\\tan{\\theta} = \\frac{AD}{2\\cdot AE}\\] Using right $\\triangle{FCE}$, since $\\angle{ACM} = \\angle{CFE}$, $\\tan{\\theta} = \\frac{CE}{FE}$. Hence, plugging into the previous equation: \\[\\frac{CE}{FE} = \\frac{AD}{2\\cdot AE}\\] Using the Pythagorean theorem on $\\triangle{FCE}$, $FE = \\sqrt{(64 - CE^2)}$. We also know that $AE=4\\cdot AD$ Plugging back in: \\[\\frac{CE}{\\sqrt{(64 - CE^2)}} = \\frac{AD}{2\\cdot (4\\cdot AD)}\\] \\[\\frac{CE}{\\sqrt{(64 - CE^2)}} = \\frac{1}{8}\\]. From here, we can square both sides and bring everything to one side to get: \\[EF^2 + 64EF - 64 = 0\\]. \\[EF = \\frac{64}{\\sqrt{65}}\\] \\[CE = \\frac{8}{\\sqrt{65}}\\] We should also return to the fact that $\\sin{\\theta} = \\frac{CE}{CF}$ from $\\triangle{FCE}$, so \\[\\sin{\\theta} = \\frac{1}{\\sqrt{65}}\\] From the fact that $\\angle{CAF} = \\angle{CEF} = 90$, we can use Ptolemy's Theorem on quadrilateral $ACEF$. $AC\\cdot EF + CE\\cdot FA = CF\\cdot AE$. Plugging in and solving, we get that $AE = \\frac{16}{\\sqrt{13}}$. We now have all of our pieces to use the Sine Formula on $\\triangle{ACE}$. \\[\\frac{1}{2}\\cdot AC\\cdot AF\\cdot \\sin{\\theta}\\] \\[\\frac{1}{2}\\cdot \\frac{8}{\\sqrt{5}}\\cdot \\frac{16}{\\sqrt{13}}\\cdot \\frac{1}{\\sqrt{65}} = \\frac{64}{65} = 129\\] ~Solution by: armang32324",
"We begin by extending $\\overline{AB}$ upwards until it intersects Circle $\\mathcal{Q}$. We can call this point of intersection $F$. Connect $F$ with $E$, $C$, and $A$ for future use. Create a trapezoid with points $B$, $C$, and the origins of Circles $\\mathcal{P}$ and $\\mathcal{Q}$. After quick inspection, we can conclude that the distance between the origins is 5 and that $\\overline{BC}$ is 4. (Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is $2\\cdot \\sqrt{a\\cdot b}$, where $a$ and $b$ are the two respective radii of the circles. In our case, we get $2\\cdot \\sqrt{4} = 4$. Using similar triangles or homotheties, $AE=4\\cdot AD$ and $AF=4\\cdot BA$. $BC^2 = BA\\cdot BF$. \\[16 = BA\\cdot (5\\cdot BA)\\] \\[BA = \\frac{4}{\\sqrt{5}}\\] \\[AF = \\frac{16}{\\sqrt{5}}\\] \\[BF = 4\\cdot \\sqrt{5}\\] Inspecting $\\triangle{BFC}$, we recognize that it is a right triangle ($\\angle{BCF} = 90$) as the final length ($\\overline{FC}$) being 8 would allow for an $x-2x-x\\sqrt{5}$ triangle. Hence, the diameter of circle $\\mathcal{Q}$ = $CF$. This also means that $\\angle{CAF} = \\angle{CEF} = 90$. From the fact that $\\triangle{ABC}$ is a right triangle: \\[AB^2 + AC^2 = BC^2\\] \\[AC = \\frac{8}{\\sqrt{5}}\\] (Note: We could have also used $\\triangle{FAC}$.) Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between $\\overline{FC}$ and $\\overline{DE}$ as $M$. Label $\\angle{CAE} = \\angle{CFE} = \\theta$. Since $\\angle{BAC} = 90$, $\\angle{EAF} = \\angle{DAB} = 90-\\theta$. Now, use the sine formula and the fact that the areas of $\\triangle{DBA}$ and $\\triangle{ACE}$ are equal to get: \\[\\frac{1}{2}\\cdot AC\\cdot AE\\cdot \\sin{\\theta} = \\frac{1}{2}\\cdot AD\\cdot AB\\cdot \\sin{(90-\\theta)}\\] Since $\\sin{(90-\\theta)} = \\cos{\\theta}$: \\[\\tan{\\theta} = \\frac{AD}{2\\cdot AE}\\] Using right $\\triangle{FCE}$, since $\\angle{ACM} = \\angle{CFE}$, $\\tan{\\theta} = \\frac{CE}{FE}$. Hence, plugging into the previous equation: \\[\\frac{CE}{FE} = \\frac{AD}{2\\cdot AE}\\] Using the Pythagorean theorem on $\\triangle{FCE}$, $FE = \\sqrt{(64 - CE^2)}$. We also know that $AE=4\\cdot AD$ Plugging back in: \\[\\frac{CE}{\\sqrt{(64 - CE^2)}} = \\frac{AD}{2\\cdot (4\\cdot AD)}\\] \\[\\frac{CE}{\\sqrt{(64 - CE^2)}} = \\frac{1}{8}\\]. From here, we can square both sides and bring everything to one side to get: \\[EF^2 + 64EF - 64 = 0\\]. \\[EF = \\frac{64}{\\sqrt{65}}\\] \\[CE = \\frac{8}{\\sqrt{65}}\\] We should also return to the fact that $\\sin{\\theta} = \\frac{CE}{CF}$ from $\\triangle{FCE}$, so \\[\\sin{\\theta} = \\frac{1}{\\sqrt{65}}\\] From the fact that $\\angle{CAF} = \\angle{CEF} = 90$, we can use Ptolemy's Theorem on quadrilateral $ACEF$. $AC\\cdot EF + CE\\cdot FA = CF\\cdot AE$. Plugging in and solving, we get that $AE = \\frac{16}{\\sqrt{13}}$. We now have all of our pieces to use the Sine Formula on $\\triangle{ACE}$. \\[\\frac{1}{2}\\cdot AC\\cdot AF\\cdot \\sin{\\theta}\\] \\[\\frac{1}{2}\\cdot \\frac{8}{\\sqrt{5}}\\cdot \\frac{16}{\\sqrt{13}}\\cdot \\frac{1}{\\sqrt{65}} = \\frac{64}{65} = 129\\] ~Solution by: armang32324",
"Let the center of the larger circle be $O,$ and the center of the smaller circle be $P.$ It is not hard to find the areas of $ACO$ and $ABP$ using pythagorean theorem, which are $\\frac{32}{5}$ and $\\frac{2}{5}$ respectively. Assign $\\angle AOC=a,\\angle COE=c,\\angle DPB=d,\\angle BPA=b.$ We can figure out that $\\angle AOE=\\angle DOA=\\theta$ using vertical angles and isosceles triangles. Now, using $[ABC]=\\frac{1}{2}ab\\sin C$ \\[[ACE]=[AOC]+[COE]-[AOC]=\\dfrac{32}{5}-8\\sin c-8\\sin \\theta,\\] \\[[DAB]=[APD]+[APB]+[BPD]=\\dfrac{2}{5}+\\dfrac{1}{2}\\sin \\theta+\\dfrac{1}{2}\\sin d.\\] We can also figure out that $\\sin a=\\dfrac{4}{5},\\cos a=\\dfrac{3}{5},\\sin b=\\dfrac{4}{5}, \\cos b=-\\dfrac{3}{5}.$ Also, $c=\\theta-a$ and $d=360-\\theta-b.$ Using sum and difference identities: \\[\\sin c=\\dfrac{3}{5}\\sin \\theta-\\dfrac{4}{5}\\cos \\theta,\\] \\[\\sin d=\\dfrac{3}{5}\\sin\\theta-\\dfrac{4}{5}\\cos\\theta.\\] (We can also notice that $c+d=360-\\theta-b+\\theta-a=360-(a+b)=180$ which means that $\\sin c=\\sin d.$) Substituting in the equations for $\\sin c$ and $\\sin d$ into the equations for $[ACE]$ and $[DAB],$ setting them equal, and simplifying: \\[3=2\\sin\\theta+3\\cos\\theta.\\] Solving this equation we get that $\\sin\\theta=\\frac{12}{13}$ and $\\cos\\theta=\\frac{5}{13}.$ Doing a lot of substitution gives us \\[[ACE]=[DAB]=\\dfrac{64}{65},\\] which means the answer is $64+65=129 ~BS2012",
"Let the center of the larger circle be $O,$ and the center of the smaller circle be $P.$ It is not hard to find the areas of $ACO$ and $ABP$ using pythagorean theorem, which are $\\frac{32}{5}$ and $\\frac{2}{5}$ respectively. Assign $\\angle AOC=a,\\angle COE=c,\\angle DPB=d,\\angle BPA=b.$ We can figure out that $\\angle AOE=\\angle DOA=\\theta$ using vertical angles and isosceles triangles. Now, using $[ABC]=\\frac{1}{2}ab\\sin C$ \\[[ACE]=[AOC]+[COE]-[AOC]=\\dfrac{32}{5}-8\\sin c-8\\sin \\theta,\\] \\[[DAB]=[APD]+[APB]+[BPD]=\\dfrac{2}{5}+\\dfrac{1}{2}\\sin \\theta+\\dfrac{1}{2}\\sin d.\\] We can also figure out that $\\sin a=\\dfrac{4}{5},\\cos a=\\dfrac{3}{5},\\sin b=\\dfrac{4}{5}, \\cos b=-\\dfrac{3}{5}.$ Also, $c=\\theta-a$ and $d=360-\\theta-b.$ Using sum and difference identities: \\[\\sin c=\\dfrac{3}{5}\\sin \\theta-\\dfrac{4}{5}\\cos \\theta,\\] \\[\\sin d=\\dfrac{3}{5}\\sin\\theta-\\dfrac{4}{5}\\cos\\theta.\\] (We can also notice that $c+d=360-\\theta-b+\\theta-a=360-(a+b)=180$ which means that $\\sin c=\\sin d.$) Substituting in the equations for $\\sin c$ and $\\sin d$ into the equations for $[ACE]$ and $[DAB],$ setting them equal, and simplifying: \\[3=2\\sin\\theta+3\\cos\\theta.\\] Solving this equation we get that $\\sin\\theta=\\frac{12}{13}$ and $\\cos\\theta=\\frac{5}{13}.$ Doing a lot of substitution gives us \\[[ACE]=[DAB]=\\dfrac{64}{65},\\] which means the answer is $64+65=129 ~BS2012"
] |
2016-I-1
| 2,016
| 1
|
For $-1<r<1$ , let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$ . Find $S(a)+S(-a)$ .
| 336
|
I
|
[
"The sum of an infinite geometric series is $\\frac{a}{1-r}\\rightarrow \\frac{12}{1\\mp a}$. The product $S(a)S(-a)=\\frac{144}{1-a^2}=2016$. $\\frac{12}{1-a}+\\frac{12}{1+a}=\\frac{24}{1-a^2}$, so the answer is $\\frac{2016}{6}=336."
] |
2016-I-2
| 2,016
| 2
|
Two dice appear to be normal dice with their faces numbered from $1$ to $6$ , but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$ . The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 71
|
I
|
[
"It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\\cdot(1\\cdot 6+2\\cdot 5+3\\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\\frac{56}{441}=\\frac{8}{63}$. The answer is $8+63=071 See also 2006 AMC 12B Problems/Problem 17",
"Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling $n$ is $\\frac{n}{21}$ because $21=\\frac{6 \\cdot 7}{2}$ Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice. The probability that 7 is rolled is now $2(\\frac{1}{21} \\cdot \\frac{6}{21}+\\frac{2}{21} \\cdot \\frac{5}{21} + \\frac{3}{21} \\cdot \\frac{4}{21})$ which is equal to $\\frac{56}{441}=\\frac{8}{63}$. Therefore the answer is $8+63=071 ~PEKKA",
"Since the probability of rolling a $1$ is $\\frac{1}{21}$, the probability of rolling a $2$ is $\\frac{2}{21}$ the probability of rolling a $3$ is $\\frac{3}{21}$ and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of $1$ and $6$, $2$ and $5$, and $3$ and $4$, and the rest is symmetry and the others are irrelevant. We have: $2 \\cdot (\\frac{1}{21} \\cdot \\frac{2}{7}$ $+$ $\\frac{2}{21} \\cdot \\frac{5}{21}$ $+$ $\\frac{1}{7} \\cdot \\frac{4}{21})$ $=$ $2 \\cdot (\\frac{2}{147} + \\frac{10}{441} + \\frac{4}{147})$ = $2 \\cdot \\frac{4}{63} = \\frac{8}{63}$. Therefore, the answer is $8 + 63$ = $071 ~Arcticturn"
] |
2016-I-3
| 2,016
| 3
|
A regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated. [asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy]
| 810
|
I
|
[
"Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place. Then, there are 2 paths from that point down to the third plane. Within the third plane, there are 9 paths as well (consider the logic from the second plane). Finally, there is only one way down to the bottom vertex. Therefore there are $5 \\cdot 9 \\cdot 2 \\cdot 9 \\cdot 1=810 paths.",
"Assume an ant is on the top of this icosahedron. Note that the icosahedron has two pentagon planes and two points where the ant starts and ends. Also note that when the ant hits a vertex of the pentagon, there is only two ways to go down. When the ant ends up at the last vertex and is about to head down, there is only one way to go down. Case $1$: The ant move down to the first pentagon, then the next pentagon, and finally to the last point in a total of four moves; there are a total of $5 \\cdot 2 \\cdot 1 = 10$ ways to achieve this. Case $2$: The ant goes to the first pentagon and moves to another vertex in the same pentagon. The ant has a total of $8$ paths doing this since it may go through $1, 2, 3,$ or $4$ edges from either two directions when going across the edges of the pentagon (recall the ant may not repeat a move to another vertex) . Now the ant moves to the second pentagon below. The ant again has a total of $8$ moves if he wanders around the pentagon. Finally, the ant moves down after wandering to the last vertex. There is a total of $5 \\cdot 8 \\cdot 2 \\cdot 8 = 640$ ways. Case $3$: Assume the ant goes to the first pentagon and wanders around. Then the ant goes the next pentagon and then heads directly to the last vertex. There are $5 \\cdot 8 \\cdot 2 = 80$ ways. Case $4$: Now let the ant go to the first pentagon and then go directly down to the next pentagon. The ant wanders around the second pentagon before heading to the last vertex. Like case $3$ above, there are $5 \\cdot 2 \\cdot 8 = 80$ ways. Adding up all four cases, we get $10 + 640 + 80 + 80 =810 total paths, as desired. ~skyscraper",
"Go to 2020 AMC 10A #19, and connect all of the centers of the faces on the dodecahedron to get the icosahedron. The answer is $810."
] |
2016-I-4
| 2,016
| 4
|
A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60^\circ$ . Find $h^2$ .
| 108
|
I
|
[
"Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$, and let $D$ be the last vertex of the triangular pyramid. Then $\\angle CAB = 120^\\circ$. Let $X$ be the foot of the altitude from $A$ to $\\overline{BC}$. Then since $\\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$. Since the dihedral angle between $\\triangle ABC$ and $\\triangle BCD$ is $60^\\circ$, $\\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\\sqrt{3} = h$. Thus $h^2 = 108. ~gundraja",
"Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$, and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$. Then, we have $AM=6$ ($30-60-90$) triangles and $DM=\\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\\angle AMD=60^{\\circ}$, we get \\[h^2=36+h^2+36-\\frac12 \\cdot 12 \\sqrt{36+h^2} \\implies h^2=108,\\] as desired. -A1001"
] |
2016-I-5
| 2,016
| 5
|
Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ .
| 53
|
I
|
[
"Let $d$ be the number of days Anh reads for. Because the difference between the number of pages and minutes Anh reads for each day stays constant and is an integer, $d$ must be a factor of the total difference, which is $374-319=55$. Also note that the number of pages Anh reads is $dn+\\frac{d(d-1)}{2}$. Similarly, the number of minutes she reads for is $dt+\\frac{d(d-1)}{2}$. When $d$ is odd (which it must be), both of these numbers are multiples of $d$. Therefore, $d$ must be a factor of $55$, $319$, and $374$. The only such numbers are $1$ and $11$. We know that Anh reads for at least $2$ days. Therefore, $d=11$. Using this, we find that she reads $55$ \"additional\" pages and $55$ \"additional\" minutes. Therefore, $n=\\frac{374-55}{11}=29$, while $t=\\frac{319-55}{11}=24$. The answer is therefore $29+24=053.",
"We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$, the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\\left \\lfloor{\\frac{11}{2}}\\right \\rfloor=5$ from each to get $(n,t)=(29,24)$ and $29+24=053.",
"If we let $k$ be equal to the number of days it took to read the book, the sum of $n$ through $n+k$ is equal to $(2n+k)(k+1)=748$ Similarly, $(2t+k)(k+1)=638$ We know that both factors must be integers and we see that the only common multiple of $748$ and $638$ not equal to $1$ that will get us positive integer solutions for $n$ and $t$ is $11$. We set $k+1=11$ so $k=10$. We then solve for $n$ and $t$ in their respective equations, getting $2n+10=68$. $n=29$ We also get $2t+10=58$. $t=24$. Our final answer is $29+24=053",
"Notice $374=34\\cdot 11$ and $319=29\\cdot 11$. Also, note the sum of an arithmetic series is $\\frac{2n+k}{2} \\cdot b$, where $n$ is our first term, $n+k$ is our final term, and $b$ is the number of terms. Since we know both sequences of $n$ and $t$ have the same length, and since $11$ is prime and shared by both $319$ and $374$, we deduce that $b=11$. Thus from here we know $2n+k=68$ and $2t+k=58$ by using our other factors $34$ and $29$. Finally, we add the two systems up and we get $2t+2k+2n=126$. But, notice that $k=b-1$, since the first term has $k=0$, and our last term has $k=b-1$. Plugging this back into our equation we get $2n+2t=106 \\implies n+t=053",
"We list two equations: \\begin{align*} n+(n+1)+...+(n+k)&=374\\\\ t+(t+1)+...+(t+k)&=319. \\end{align*} Subtracting the two, we get: \\[(n-t)(k+1)=374-319=55.\\] Manipulating the first and second equation, we get: \\begin{align*} n(k+1)+\\frac{k(k+1)}{2}&=374 \\\\ t(k+1)+\\frac{k(k+1)}{2}&=319. \\end{align*} We factor out the common factor $k+1$: \\begin{align*} (k+1)\\left(n+\\frac{k}{2}\\right)&=374 \\\\ (k+1)\\left(t+\\frac{k}{2}\\right)&=319. \\end{align*} Note that $374$ and $319$ have a GCD of $11,$ now combining this with our equation that $(n-t)(k+1)=55,$ we see that $k+1$ has to equal $11.$ Thus, we get: \\[(n,t)=(29,24) \\Rightarrow n+t=29+24=053.\\] Note: We see that because n-t=5, it becomes impossible for n+k/2 and t+k/2 to both be multiples of 11. Thus, this satisfies our condition. Thus k+1 must be 11 to satisfy the common factor 11 constraint. mathboy282"
] |
2016-I-6
| 2,016
| 6
|
In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $\overline{AB}$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC= \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 13
|
I
|
[
"Suppose we label the angles as shown below. [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label(\"$A$\",A,SW,fontsize(8)); label(\"$B$\",B,SE,fontsize(8)); label(\"$C$\",C,N,fontsize(8)); label(\"$D$\",D,S,fontsize(8)); label(\"$I$\",I,NE,fontsize(8)); label(\"$L$\",L,SW,fontsize(8)); label(\"$\\alpha$\",A,5*dir(midangle(C,A,I)),fontsize(8)); label(\"$\\alpha$\",A,5*dir(midangle(I,A,B)),fontsize(8)); label(\"$\\beta$\",B,12*dir(midangle(A,B,I)),fontsize(8)); label(\"$\\beta$\",B,12*dir(midangle(I,B,C)),fontsize(8)); label(\"$\\gamma$\",C,5*dir(midangle(A,C,I)),fontsize(8)); label(\"$\\gamma$\",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] As $\\angle BCD$ and $\\angle BAD$ intercept the same arc, we know that $\\angle BAD=\\gamma$. Similarly, $\\angle ABD=\\gamma$. Also, using $\\triangle ICA$, we find $\\angle CIA=180-\\alpha-\\gamma$. Therefore, $\\angle AID=\\alpha+\\gamma=\\angle DAI$, so $\\triangle AID$ must be isosceles with $AD=ID=5$. Similarly, $BD=ID=5$. Then $\\triangle DLB \\sim \\triangle ALC$, hence $\\frac{AL}{AC} = \\frac{3}{5}$. Also, $AI$ bisects $\\angle LAC$, so by the Angle Bisector Theorem $\\frac{CI}{IL} =\\frac{AC}{AL}= \\frac{5}{3}$. Thus $CI = \\frac{10}{3}$, and the answer is $013.",
"See the diagram in Solution 1. Let $\\Delta ABC$'s $C$-excenter be $J_C$. Since $CI$ is an angle bisector, $\\angle ACD = \\angle BCD$, meaning that $D$ is the midpoint of arc $\\overarc{BC}$. By the Incenter/Excenter Lemma, $DA=DI=DB=DJ_C=2+3=5$, and applying Power of a Point on circle $D$ gives $AL\\cdot AB = LI\\cdot LJ_C = 2(3+5) = 16$. Applying Power of a Point again on $\\Delta ABC$'s circumcircle gives $AL\\cdot LB = LC\\cdot LD = 16$, and since $LD = 3$, $LC = \\frac{16}{3}$. Thus $IC = LC-LI = \\frac{16}{3}-2 = \\frac{10}{3}$. We submit $10+3=013. - NamelyOrange",
"See the diagram in Solution 1. Let $\\Delta ABC$'s $C$-excenter be $J_C$. Since $CI$ is an angle bisector, $\\angle ACD = \\angle BCD$, meaning that $D$ is the midpoint of arc $\\overarc{BC}$. By the Incenter/Excenter Lemma, $DA=DI=DB=DJ_C=2+3=5$, and applying Power of a Point on circle $D$ gives $AL\\cdot AB = LI\\cdot LJ_C = 2(3+5) = 16$. Applying Power of a Point again on $\\Delta ABC$'s circumcircle gives $AL\\cdot LB = LC\\cdot LD = 16$, and since $LD = 3$, $LC = \\frac{16}{3}$. Thus $IC = LC-LI = \\frac{16}{3}-2 = \\frac{10}{3}$. We submit $10+3=013. - NamelyOrange",
"WLOG assume $\\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$, and $\\angle CLB=\\angle CLA=90^\\circ$. Draw the perpendicular from $I$ to $CB$, and let it meet $CB$ at $E$. Since $IL=2$, $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$. Then, triangles $BLD$ and $CEI$ are similar, and $\\tfrac{2}{3}=\\tfrac{CI}{x}$. Thus, $CI=\\tfrac{2x}{3}$. $\\triangle CBD \\sim \\triangle CEI$, so $\\tfrac{IE}{DB}=\\tfrac{CI}{CD}$. Thus $\\tfrac{2}{x}=\\tfrac{(2x/3)}{(2x/3+5)}$. Solving for $x$, we have: $x^2-2x-15=0$, or $x=5, -3$. $x$ is positive, so $x=5$. As a result, $CI=\\tfrac{2x}{3}=\\tfrac{10}{3}$ and the answer is $013",
"WLOG assume $\\triangle ABC$ is isosceles (with vertex $C$). Let $O$ be the center of the circumcircle, $R$ the circumradius, and $r$ the inradius. A simple sketch will reveal that $\\triangle ABC$ must be obtuse (as an acute triangle will result in $LI$ being greater than $DL$) and that $O$ and $I$ are collinear. Next, if $OI=d$, $DO+OI=R+d$ and $R+d=DL+LI=5$. Euler gives us that $d^{2}=R(R-2r)$, and in this case, $r=LI=2$. Thus, $d=\\sqrt{R^{2}-4R}$. Solving for $d$, we have $R+\\sqrt{R^{2}-4R}=5$, then $R^{2}-4R=25-10R+R^{2}$, yielding $R=\\frac{25}{6}$. Next, $R+d=5$ so $d=\\frac{5}{6}$. Finally, $OC=OI+IC$ gives us $R=d+IC$, and $IC=\\frac{25}{6}-\\frac{5}{6}=\\frac{10}{3}$. Our answer is then $013.",
"Since $\\angle{LAD} = \\angle{BDC}$ and $\\angle{DLA}=\\angle{DCB}$, $\\triangle{DLA}\\sim\\triangle{DBC}$. Also, $\\angle{DAC}=\\angle{BLC}$ and $\\angle{ACD}=\\angle{LCB}$ so $\\triangle{DAC}\\sim\\triangle{BLC}$. Now we can call $AC$, $b$ and $BC$, $a$. By angle bisector theorem, $\\frac{AD}{DB}=\\frac{AC}{BC}$. So let $AD=bk$ and $DB=ak$ for some value of $k$. Now call $IC=x$. By the similar triangles we found earlier, $\\frac{3}{ak}=\\frac{bk}{x+2}$ and $\\frac{b}{x+5}=\\frac{x+2}{a}$. We can simplify this to $abk^2=3x+6$ and $ab=(x+5)(x+2)$. So we can plug the $ab$ into the first equation and get $(x+5)(x+2)k^2=3(x+2) \\rightarrow k^2(x+5)=3$. We can now draw a line through $A$ and $I$ that intersects $BC$ at $E$. By mass points, we can assign a mass of $a$ to $A$, $b$ to $B$, and $a+b$ to $D$. We can also assign a mass of $(a+b)k$ to $C$ by angle bisector theorem. So the ratio of $\\frac{DI}{IC}=\\frac{(a+b)k}{a+b}=k=\\frac{2}{x}$. So since $k=\\frac{2}{x}$, we can plug this back into the original equation to get $\\left(\\frac{2}{x}\\right)^2(x+5)=3$. This means that $\\frac{3x^2}{4}-x-5=0$ which has roots -2 and $\\frac{10}{3}$ which means our $CI=\\frac{10}{3}$ and our answer is $013.",
"Since $\\angle BCD$ and $\\angle BAD$ both intercept arc $BD$, it follows that $\\angle BAD=\\gamma$. Note that $\\angle AID=\\alpha+\\gamma$ by the external angle theorem. It follows that $\\angle DAI=\\angle AID=\\alpha+\\gamma$, so we must have that $\\triangle AID$ is isosceles, yielding $AD=ID=5$. Note that $\\triangle DLA \\sim \\triangle DAC$, so $\\frac{DA}{DL} = \\frac{DC}{DA}$. This yields $DC = \\frac{25}{3}$. It follows that $CI = DC - DI = \\frac{10}{3}$, giving a final answer of $013.",
"Let $I_C$ be the excenter opposite to $C$ in $ABC$. By the incenter-excenter lemma $DI=DC \\therefore$ $LI_C=8,LI=2,II_C=10$. Its well known that $(I_C,I,L,C)=-1 \\implies \\dfrac{LI_C}{LI}.\\dfrac{CI}{CI_C}=-1 \\implies \\dfrac{CI}{CI+10}=\\dfrac{1}{4} \\implies CI=\\dfrac{10}{3}. (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)",
"We can just say that quadrilateral $ADBC$ is a right kite with right angles at $A$ and $B$. Let us construct another similar right kite with the points of tangency on $AC$ and $BC$ called $E$ and $F$ respectively, point $I$, and point $C$. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call $CI$ $x$ for simplicity's sake. Based on the fact that $\\triangle BCD$ is similar to $\\triangle FCI$ we can use triangle proportionality to say that $BD$ is $2\\frac{x+5}{x}$. Using geometric mean theorem we can show that $BL$ must be $\\sqrt{3x+6}$. With Pythagorean Theorem we can say that $3x+6+9=4{(\\frac{x+5}{x})}^2$. Multiplying both sides by $x^2$ and moving everything to LHS will give you $3{x}^3+11{x}^2-40x-100=0$ Since $x$ must be in the form $\\frac{p}{q}$ we can assume that $x$ is most likely a positive fraction in the form $\\frac{p}{3}$ where $p$ is a factor of $100$. Testing the factors in synthetic division would lead $x = \\frac{10}{3}$, giving us our desired answer $013. ~Lopkiloinm",
"[asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label(\"$A$\",A,SW,fontsize(8)); label(\"$B$\",B,SE,fontsize(8)); label(\"$C$\",C,N,fontsize(8)); label(\"$D$\",D,S,fontsize(8)); label(\"$I$\",I,NE,fontsize(8)); label(\"$L$\",L,SW,fontsize(8)); label(\"$\\alpha$\",A,5*dir(midangle(C,A,I)),fontsize(8)); label(\"$\\alpha$\",A,5*dir(midangle(I,A,B)),fontsize(8)); label(\"$\\beta$\",B,12*dir(midangle(A,B,I)),fontsize(8)); label(\"$\\beta$\",B,12*dir(midangle(I,B,C)),fontsize(8)); label(\"$\\gamma$\",C,5*dir(midangle(A,C,I)),fontsize(8)); label(\"$\\gamma$\",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$. Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral. Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\\angle ABD$ and $\\angle DAB$ are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at $D$), and the lengths of $BC$, $AC$, $AB$, and $CI$ as $a,b,c, x$, respectively. After applying Ptolemy's, one will get that: \\[z(a+b)=c(x+5)\\] Next, since $ACBD$ is cyclic, triangles $ALD$ and $CLB$ are similar, yielding the following equation once simplifications are made to the equation $\\frac{AD}{CB}=\\frac{AL}{BL}$, with the length of $BL$ written in terms of $a,b,c$ using the angle bisector theorem on triangle $ABC$: \\[zc=3(a+b)\\] Next, drawing in the bisector of $\\angle BAC$ to the incenter $I$, and applying the angle bisector theorem, we have that: \\[cx=2(a+b)\\] Now, solving for $z$ in the second equation, and $x$ in the third equation and plugging them both back into the first equation, and making the substitution $w=\\frac{a+b}{c}$, we get the quadratic equation: \\[3w^2-2w-5=0\\] Solving, we get $w=5/3$, which gives $z=5$ and $x=10/3$, when we rewrite the above equations in terms of $w$. Thus, our answer is $013 and we're done. -mathislife52",
"[asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label(\"$A$\",A,SW,fontsize(8)); label(\"$B$\",B,SE,fontsize(8)); label(\"$C$\",C,N,fontsize(8)); label(\"$D$\",D,S,fontsize(8)); label(\"$I$\",I,NE,fontsize(8)); label(\"$L$\",L,SW,fontsize(8)); label(\"$\\alpha$\",A,5*dir(midangle(C,A,I)),fontsize(8)); label(\"$\\alpha$\",A,5*dir(midangle(I,A,B)),fontsize(8)); label(\"$\\beta$\",B,12*dir(midangle(A,B,I)),fontsize(8)); label(\"$\\beta$\",B,12*dir(midangle(I,B,C)),fontsize(8)); label(\"$\\gamma$\",C,5*dir(midangle(A,C,I)),fontsize(8)); label(\"$\\gamma$\",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$. Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral. Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\\angle ABD$ and $\\angle DAB$ are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at $D$), and the lengths of $BC$, $AC$, $AB$, and $CI$ as $a,b,c, x$, respectively. After applying Ptolemy's, one will get that: \\[z(a+b)=c(x+5)\\] Next, since $ACBD$ is cyclic, triangles $ALD$ and $CLB$ are similar, yielding the following equation once simplifications are made to the equation $\\frac{AD}{CB}=\\frac{AL}{BL}$, with the length of $BL$ written in terms of $a,b,c$ using the angle bisector theorem on triangle $ABC$: \\[zc=3(a+b)\\] Next, drawing in the bisector of $\\angle BAC$ to the incenter $I$, and applying the angle bisector theorem, we have that: \\[cx=2(a+b)\\] Now, solving for $z$ in the second equation, and $x$ in the third equation and plugging them both back into the first equation, and making the substitution $w=\\frac{a+b}{c}$, we get the quadratic equation: \\[3w^2-2w-5=0\\] Solving, we get $w=5/3$, which gives $z=5$ and $x=10/3$, when we rewrite the above equations in terms of $w$. Thus, our answer is $013 and we're done. -mathislife52",
"[email protected], vvsss",
"[email protected], vvsss",
"Let $AB=c,BC=a,CA=b$, and $x=\\tfrac{a+b}{c}$. Then, notice that $\\tfrac{CI}{IL}=\\tfrac{a+b}{c}=x$, so $CI=IL\\cdot{}x=2x$. Also, by the incenter-excenter lemma, $AD=BD=ID=IL+LD=5$. Therefore, by Ptolemy's Theorem on cyclic quadrilateral $ABCD$, $5a+5b=c(2x+5)$, so $5\\left(\\tfrac{a+b}{c}\\right)=2x+5$, so $5x=2x+5$. Solving, we get that $x=\\tfrac{5}{3}$, so $CI=\\tfrac{10}{3}$ and the answer is $10+3=013.",
"Perform a $\\sqrt{bc}$ Inversion followed by a reflection along the angle bisector of $\\angle BCA$. It's well known that \\[AB \\leftrightarrow \\odot CBA \\implies L \\leftrightarrow D\\] \\[I \\leftrightarrow I_A\\] where $I_A$ is the $A-$excenter. Also by Fact 5, $DI_A = 5$. So, \\[CL \\cdot CD = CI \\cdot CI_A\\] \\[\\implies (CI + IL) \\cdot (CI + ID) = (CI) \\cdot (CI + II_A)\\] \\[\\implies (CI + 2) \\cdot (CI + 5) = (CI) \\cdot (CI + 10)\\] \\[\\implies 7CI +10= 10CI\\] \\[\\implies CI = \\dfrac{10}{3}.\\blacksquare\\] ~kamatadu",
"Without loss of generality, let $\\triangle ABC$ be isosceles. Note that by the incenter-excenter lemma, $DI = DA = DB.$ Hence, $DA=DB=5.$ Let the point of tangency of the incircle and $\\overline{BC}$ be $F$ and the point of tangency of the incircle and $\\overline{AC}$ be $E.$ We note that $\\angle ALC = \\angle BLC = 90^\\circ$ and $LA=LB=4,$ which immediately gives $AE=BF=4.$ Applying the Pythagorean Theorem on $\\triangle ALC$ and $\\triangle IEC$ gives $2^2+x^2=y^2$ and $4^2+(2+y)^2 = (4+x)^2.$ Solving for $y$ gives us $y=\\frac{10}{3}.$ Therefore, $IC = \\frac{10}{3}$ so the answer is $13 ~peelybonehead",
"Let $C_1\\in AB$ be the point such that $IC_1\\perp AB$, and let $D_1\\in AB$ be defined similarly for $D$. We know that $\\triangle IC_1L\\sim\\triangle DD_1L$, so by triangle ratios $DD_1=\\frac{3}{2}r$, where $r$ is the inradius. Additionally, by cyclic quadrilaterals, we know that $\\angle BAD=\\angle DAB=\\frac{\\gamma}{2}$, where $\\gamma$ is equivalent to $\\angle ACB$. Thus $\\triangle ADB$ is isosceles and $DD_1$ is the perpendicular bisector of the triangle, so $AD_1=\\frac{c}{2}$. Since $\\tan\\left(\\frac{\\gamma}{2}\\right)=\\sqrt{\\frac{(s-a)(s-b)}{s(s-c)}}$ from formulas (where $s$ is half the perimeter of $\\triangle ABC$) and since $\\tan\\left(\\frac{\\gamma}{2}\\right)=\\frac{\\frac{3}{2}r}{\\frac{1}{2}c}$ from $\\triangle ADB$, we can set up an equation: $\\sqrt{\\frac{(s-a)(s-b)}{s(s-c)}}=\\frac{\\frac{3}{2}r}{\\frac{1}{2}c}\\implies\\frac{\\sqrt{s(s-a)(s-b)(s-c)}}{s(s-c)}=\\frac{3rs}{cs}\\implies\\frac{[ABC]}{s(s-c)}=\\frac{3[ABC]}{cs}$ $\\implies\\frac{1}{s-c}=\\frac{3}{c}\\implies \\frac{s}{c}=\\frac{4}{3}$ Let $C_2\\in AB$ such that $CC_2\\perp AB$. Then $CC_2=\\frac{2[ABC]}{c}$. Using the area formula $[ABC]=rs$ and our fact from above yields $CC_2=\\frac{8}{3}r$. We then notice that $\\triangle CC_2L\\sim\\triangle IC_1L$, so if we let $x=CI$, by triangle ratios we find that $\\frac{\\frac{8}{3}r}{x+2}=\\frac{r}{2}$, leading to $x=\\frac{10}{3}$. Thus the answer is $10+3=013. ~eevee9406",
"Let $C_1\\in AB$ be the point such that $IC_1\\perp AB$, and let $D_1\\in AB$ be defined similarly for $D$. We know that $\\triangle IC_1L\\sim\\triangle DD_1L$, so by triangle ratios $DD_1=\\frac{3}{2}r$, where $r$ is the inradius. Additionally, by cyclic quadrilaterals, we know that $\\angle BAD=\\angle DAB=\\frac{\\gamma}{2}$, where $\\gamma$ is equivalent to $\\angle ACB$. Thus $\\triangle ADB$ is isosceles and $DD_1$ is the perpendicular bisector of the triangle, so $AD_1=\\frac{c}{2}$. Since $\\tan\\left(\\frac{\\gamma}{2}\\right)=\\sqrt{\\frac{(s-a)(s-b)}{s(s-c)}}$ from formulas (where $s$ is half the perimeter of $\\triangle ABC$) and since $\\tan\\left(\\frac{\\gamma}{2}\\right)=\\frac{\\frac{3}{2}r}{\\frac{1}{2}c}$ from $\\triangle ADB$, we can set up an equation: $\\sqrt{\\frac{(s-a)(s-b)}{s(s-c)}}=\\frac{\\frac{3}{2}r}{\\frac{1}{2}c}\\implies\\frac{\\sqrt{s(s-a)(s-b)(s-c)}}{s(s-c)}=\\frac{3rs}{cs}\\implies\\frac{[ABC]}{s(s-c)}=\\frac{3[ABC]}{cs}$ $\\implies\\frac{1}{s-c}=\\frac{3}{c}\\implies \\frac{s}{c}=\\frac{4}{3}$ Let $C_2\\in AB$ such that $CC_2\\perp AB$. Then $CC_2=\\frac{2[ABC]}{c}$. Using the area formula $[ABC]=rs$ and our fact from above yields $CC_2=\\frac{8}{3}r$. We then notice that $\\triangle CC_2L\\sim\\triangle IC_1L$, so if we let $x=CI$, by triangle ratios we find that $\\frac{\\frac{8}{3}r}{x+2}=\\frac{r}{2}$, leading to $x=\\frac{10}{3}$. Thus the answer is $10+3=013. ~eevee9406"
] |
2016-I-7
| 2,016
| 7
|
For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left(\frac{\sqrt{|a+b|}}{ab+100}\right)i.\] Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number.
| 103
|
I
|
[
"We consider two cases: Case 1: $ab \\ge -2016$. In this case, if \\[0 = \\text{Im}\\left({\\frac{\\sqrt{ab+2016}}{ab+100}-\\left({\\frac{\\sqrt{|a+b|}}{ab+100}}\\right)i}\\right) = -\\frac{\\sqrt{|a+b|}}{ab+100}\\] then $ab \\ne -100$ and $|a + b| = 0 = a + b$. Thus $ab = -a^2$ so $a^2 < 2016$. Thus $a = -44,-43, ... , -1, 0, 1, ..., 43, 44$, yielding $89$ values. However since $ab = -a^2 \\ne -100$, we have $a \\ne \\pm 10$. Thus there are $87$ allowed tuples $(a,b)$ in this case. Case 2: $ab < -2016$. In this case, we want \\[0 = \\text{Im}\\left({\\frac{\\sqrt{ab+2016}}{ab+100}-\\left({\\frac{\\sqrt{|a+b|}}{ab+100}}\\right)i}\\right) = \\frac{\\sqrt{-ab-2016} - \\sqrt{|a+b|}}{ab+100}\\] Squaring, we have the equations $ab \\ne -100$ (which always holds in this case) and \\[-(ab + 2016)= |a + b|.\\] Then if $a > 0$ and $b < 0$, let $c = -b$. If $c > a$, \\[ac - 2016 = c - a \\Rightarrow (a - 1)(c + 1) = 2015 \\Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).\\] Note that $ab < -2016$ for every one of these solutions. If $c < a$, then \\[ac - 2016 = a - c \\Rightarrow (a + 1)(c - 1) = 2015 \\Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).\\] Again, $ab < -2016$ for every one of the above solutions. This yields $8$ solutions. Similarly, if $a < 0$ and $b > 0$, there are $8$ solutions. Thus, there are a total of $16$ solutions in this case. Thus, the answer is $87 + 16 = 103."
] |
2016-I-8
| 2,016
| 8
|
For a permutation $p = (a_1,a_2,\ldots,a_9)$ of the digits $1,2,\ldots,9$ , let $s(p)$ denote the sum of the three $3$ -digit numbers $a_1a_2a_3$ , $a_4a_5a_6$ , and $a_7a_8a_9$ . Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is $0$ . Let $n$ denote the number of permutations $p$ with $s(p) = m$ . Find $|m - n|$ .
| 162
|
I
|
[
"To minimize $s(p)$, the numbers $1$, $2$, and $3$ (which sum to $6$) must be in the hundreds places. For the units digit of $s(p)$ to be $0$, the numbers in the ones places must have a sum of either $10$ or $20$. However, since the tens digit contributes more to the final sum $s(p)$ than the ones digit, and we are looking for the minimum value of $s(p)$, we take the sum's units digit to be $20$. We know that the sum of the numbers in the tens digits is $\\sum\\limits_{i=1}^9 (i) -6-20=45-6-20=19$. Therefore, $m=100\\times6+10\\times19+20=810$. To find $n$, realize that there are $3!=6$ ways of ordering the numbers in each of the places. Additionally, there are three possibilities for the numbers in the ones place: $(4,7,9)$, $(5,7,8)$, and $(5,6,9)$. Therefore there are $6^3\\times3=648$ ways in total. $|m-n|=|810-648|=162."
] |
2016-I-9
| 2,016
| 9
|
Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$ . This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$ . Find the maximum possible area of $AQRS$ .
| 744
|
I
|
[
"Note that if angle $BAC$ is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where $A$ is obtuse. Therefore, angle A is acute. Let angle $CAS=n$ and angle $BAQ=m$. Then, $\\overline{AS}=31\\cos(n)$ and $\\overline{AQ}=40\\cos(m)$. Then the area of rectangle $AQRS$ is $1240\\cos(m)\\cos(n)$. By product-to-sum, $\\cos(m)\\cos(n)=\\frac{1}{2}(\\cos(m+n)+\\cos(m-n))$. $\\cos(m+n)=\\sin(90-m-n)=\\sin(BAC)=\\frac{1}{5}$. The maximum possible value of $\\cos(m-n)$ is 1, which occurs when $m=n$. Thus the maximum possible value of $\\cos(m)\\cos(n)$ is $\\frac{1}{2}(\\frac{1}{5}+1)=\\frac{3}{5}$ so the maximum possible area of $AQRS$ is $1240\\times{\\frac{3}{5}}=744.",
"We start by drawing a diagram; [asy] size(400); import olympiad; import geometry; pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0); draw(A--B--C--cycle); draw(A--Q); draw(Q--R); draw(R--S); draw(S--A); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$C$\", C, N); label(\"$Q$\", Q, E); label(\"$R$\", R, E); label(\"$S$\", S, W); label(\"$w$\", (-1,10)); label(\"$l$\", (15,21)); label(\"$y$\", (7.5,-1)); label(\"$x$\", (31,15)); label(\"$31$\",(7.5,10), E); label(\"$40$\",(15,15), N); markangle(Label(\"$\\alpha$\", Relative(0.5)), n=1, C, A, B); markangle(Label(\"$\\beta$\", Relative(0.5)), n=1, B, A, Q); markangle(Label(\"$\\gamma$\", Relative(0.5)), n=1, S, A, C); [/asy] We know that $\\sin \\alpha = \\frac{1}{5}$. Since $\\sin \\alpha = \\cos (90- \\alpha)$, \\[\\cos (90- \\alpha) = \\frac{1}{5} \\implies \\cos (\\beta + \\gamma) = \\frac{1}{5}\\] Using our angle sum identities, we expand this to $\\cos \\beta \\cdot \\cos \\gamma - \\sin \\beta \\cdot \\sin \\gamma = \\frac{1}{5}$. We can now use the right triangle definition of cosine and sine to rewrite this equation as; \\[\\frac{l}{40} \\cdot \\frac{w}{31} - \\frac{x}{40} \\cdot \\frac{y}{31} = \\frac{1}{5} \\implies lw- xy = 8 \\cdot 31 \\implies lw = xy + 31 \\cdot 8\\] Hang on; $lw$ is the area we want to maximize! Therefore, to maximize this area we must maximize $xy = 40 \\sin \\beta \\cdot 31 \\sin \\gamma = 31 \\cdot 40 \\cdot \\frac{1}{2}( \\cos (\\beta - \\gamma) - \\cos ( \\beta + \\gamma)) = 31 \\cdot 20 \\cdot (\\cos(\\beta-\\gamma)-\\frac{1}{5})$. Since $\\cos(\\beta-\\gamma)$ is the only variable component of this expression, to maximize the expression we must maximize $\\cos(\\beta-\\gamma)$. The cosine function has a maximum value of 1, so our equation evaluates to $xy = 31 \\cdot 20 \\cdot (1-\\frac{1}{5}) = 31 \\cdot 20 \\cdot \\frac{4}{5} = 31 \\cdot 16$ (Note that at this max value, since $\\beta$ and $\\gamma$ are both acute, $\\beta-\\gamma=0 \\implies \\beta=\\gamma$). Finally, $lw = xy + 31 \\cdot 8 = 31 \\cdot 16 + 31\\cdot 8 = 31 \\cdot 24 = 744 ~KingRavi",
"As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\\text{cis} A$, and let $z$ be a complex number with $|z|=1$, $\\text{Arg}(z)\\ge 0^\\circ$ and $\\text{Arg}(zw)\\le90^\\circ$. Then we represent $B$ by $40z$ and $C$ by $31zw$. The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$. Thus \\[[AQRS]=\\Re(40z)\\cdot \\Im(31zw)=1240\\left(\\frac{z+\\overline{z}}{2}\\right)\\left(\\frac{zw-\\overline{zw}}{2i}\\right).\\] We can expand this, using the fact that $z\\overline{z}=|z|^2$, finding \\[[AQRS]=620\\left(\\frac{z^2w-\\overline{z^2w}+w-\\overline{w}}{2i}\\right)=620(\\Im(z^2w)+\\Im(w)).\\] Now as $w=\\text{cis}A$, we know that $\\Im(w)=\\frac15$. Also, $|z^2w|=1$, so the maximum possible imaginary part of $z^2w$ is $1$. This is clearly achievable under our conditions on $z$. Therefore, the maximum possible area of $AQRS$ is $620(1+\\tfrac15)=744.",
"Let $\\theta$ be the angle $\\angle BAQ$. The height of the rectangle then can be expressed as $h = 31 \\sin (A+\\theta)$, and the length of the rectangle can be expressed as $l = 40\\cos \\theta$. The area of the rectangle can then be written as a function of $\\theta$, $[AQRS] = a(\\theta) = 31\\sin (A+\\theta)\\cdot 40 \\cos \\theta = 1240 \\sin (A+\\theta) \\cos \\theta$. For now, we will ignore the $1240$ and focus on the function $f(\\theta) = \\sin (A+\\theta) \\cos \\theta = (\\sin A \\cos \\theta + \\cos A \\sin \\theta)(\\cos \\theta) = \\sin A \\cos^2 \\theta + \\cos A \\sin \\theta \\cos \\theta = \\sin A \\cos^2 \\theta + \\frac{1}{2} \\cos A \\sin 2\\theta$. Taking the derivative, $f'(\\theta) = \\sin A \\cdot -2\\cos \\theta \\sin \\theta + \\cos A \\cos 2\\theta = \\cos A \\cos 2\\theta - \\sin A \\sin 2\\theta = \\cos(2\\theta + A)$. Setting this equal to $0$, we get $\\cos(2 \\theta + A) = 0 \\Rightarrow 2\\theta +A = 90, 270 ^\\circ$. Since we know that $A+ \\theta < 90$, the $270^\\circ$ solution is extraneous. Thus, we get that $\\theta = \\frac{90 - A}{2} = 45 - \\frac{A}{2}$. Plugging this value into the original area equation, $a(45 - \\frac{A}{2}) = 1240 \\sin (45 - \\frac{A}{2} + A) \\cos (45 - \\frac{A}{2}) = 1240\\sin( 45+ \\frac{A}{2})\\cos(45 - \\frac{A}{2})$. Using a product-to-sum formula, we get that: \\[1240\\sin( 45+ \\frac{A}{2})\\cos(45 - \\frac{A}{2}) =\\] \\[1240\\cdot \\frac{1}{2}\\cdot(\\sin((45 + \\frac{A}{2}) + (45 -\\frac{A}{2}))+\\sin((45 +\\frac{A}{2})-(45 - \\frac{A}{2})))=\\] \\[620 (\\sin 90^\\circ + \\sin A) = 620 \\cdot \\frac{6}{5} = 744\\].",
"Let $\\theta$ be the angle $\\angle BAQ$. The height of the rectangle then can be expressed as $h = 31 \\sin (A+\\theta)$, and the length of the rectangle can be expressed as $l = 40\\cos \\theta$. The area of the rectangle can then be written as a function of $\\theta$, $[AQRS] = a(\\theta) = 31\\sin (A+\\theta)\\cdot 40 \\cos \\theta = 1240 \\sin (A+\\theta) \\cos \\theta$. For now, we will ignore the $1240$ and focus on the function $f(\\theta) = \\sin (A+\\theta) \\cos \\theta = (\\sin A \\cos \\theta + \\cos A \\sin \\theta)(\\cos \\theta) = \\sin A \\cos^2 \\theta + \\cos A \\sin \\theta \\cos \\theta = \\sin A \\cos^2 \\theta + \\frac{1}{2} \\cos A \\sin 2\\theta$. Taking the derivative, $f'(\\theta) = \\sin A \\cdot -2\\cos \\theta \\sin \\theta + \\cos A \\cos 2\\theta = \\cos A \\cos 2\\theta - \\sin A \\sin 2\\theta = \\cos(2\\theta + A)$. Setting this equal to $0$, we get $\\cos(2 \\theta + A) = 0 \\Rightarrow 2\\theta +A = 90, 270 ^\\circ$. Since we know that $A+ \\theta < 90$, the $270^\\circ$ solution is extraneous. Thus, we get that $\\theta = \\frac{90 - A}{2} = 45 - \\frac{A}{2}$. Plugging this value into the original area equation, $a(45 - \\frac{A}{2}) = 1240 \\sin (45 - \\frac{A}{2} + A) \\cos (45 - \\frac{A}{2}) = 1240\\sin( 45+ \\frac{A}{2})\\cos(45 - \\frac{A}{2})$. Using a product-to-sum formula, we get that: \\[1240\\sin( 45+ \\frac{A}{2})\\cos(45 - \\frac{A}{2}) =\\] \\[1240\\cdot \\frac{1}{2}\\cdot(\\sin((45 + \\frac{A}{2}) + (45 -\\frac{A}{2}))+\\sin((45 +\\frac{A}{2})-(45 - \\frac{A}{2})))=\\] \\[620 (\\sin 90^\\circ + \\sin A) = 620 \\cdot \\frac{6}{5} = 744\\].",
"Let $\\alpha$ be the angle $\\angle CAS$ and $\\beta$ be the angle $\\angle BAQ$. Then \\[\\alpha + \\beta + \\angle A = 90^\\circ \\Rightarrow \\alpha + \\beta = 90^\\circ - \\angle A\\] \\[\\cos(\\alpha + \\beta) = \\cos(90^\\circ - \\angle A)\\] \\[\\cos(\\alpha + \\beta) = \\sin(\\angle A) = \\frac{1}{5}\\] \\[\\cos\\alpha\\cos\\beta - \\sin\\alpha\\sin\\beta = \\frac{1}{5}\\] \\[\\cos\\alpha\\cos\\beta - \\sqrt{(1-\\cos^2\\alpha)(1-\\cos^2\\beta)} = \\frac{1}{5}\\] \\[\\cos\\alpha\\cos\\beta - \\sqrt{1-\\cos^2\\alpha-\\cos^2\\beta+\\cos^2\\alpha\\cos^2\\beta} = \\frac{1}{5}\\] However, by AM-GM: \\[\\cos^2\\alpha+\\cos^2\\beta \\ge 2\\cos\\alpha\\beta\\] Therefore, \\[1-\\cos^2\\alpha-\\cos^2\\beta+\\cos^2\\alpha\\cos^2\\beta \\le 1-2\\cos\\alpha\\beta+\\cos^2\\alpha\\cos^2\\beta = (1-\\cos\\alpha\\cos\\beta)^2\\] \\[\\sqrt{1-\\cos^2\\alpha-\\cos^2\\beta+\\cos^2\\alpha\\cos^2\\beta} \\le 1-\\cos\\alpha\\cos\\beta\\] So, \\[\\frac{1}{5} \\ge \\cos\\alpha\\cos\\beta - (1-\\cos\\alpha\\cos\\beta) = 2\\cos\\alpha\\cos\\beta-1\\] \\[\\frac{3}{5} \\ge \\cos\\alpha\\cos\\beta\\]. However, the area of the rectangle is just $AS \\cdot AQ = 31\\cos\\alpha \\cdot 40\\cos\\beta \\le 31 \\cdot 40 \\cdot \\frac{3}{5} = 744."
] |
2016-I-10
| 2,016
| 10
|
A strictly increasing sequence of positive integers $a_1$ , $a_2$ , $a_3$ , $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ , $a_{2k}$ , $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ , $a_{2k+1}$ , $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ .
| 504
|
I
|
[
"We first create a similar sequence where $a_1=1$ and $a_2=2$. Continuing the sequence, \\[1, 2,4,6,9,12,16,20,25,30,36,42,49,\\cdots\\] Here we can see a pattern; every second term (starting from the first) is a square, and every second term (starting from the third) is the end of a geometric sequence. This can be proven by induction. Similarly, $a_{13}$ would also need to be the end of a geometric sequence (divisible by a square). We see that $2016$ is $2^5 \\cdot 3^2 \\cdot 7$, so the squares that would fit in $2016$ are $1^2=1$, $2^2=4$, $3^2=9$, $2^4=16$, $2^2 \\cdot 3^2 = 36$, and $2^4 \\cdot 3^2 = 144$. By simple inspection $144$ is the only plausible square, since the other squares in the sequence don't have enough elements before them to go all the way back to $a_1$ while still staying as positive integers. $a_{13}=2016=14\\cdot 144$, so $a_1=14\\cdot 36=504. ~IYN~",
"Setting $a_1 = a$ and $a_2 = ka$, the sequence becomes: \\[a, ka, k^2a, k(2k-1)a, (2k-1)^2a, (2k-1)(3k-2)a, (3k-2)^2a, \\cdots\\] and so forth, with $a_{2n+1} = (nk-(n-1))^2a$. Then, $a_{13} = (6k-5)^2a = 2016$. Keep in mind, $k$ need not be an integer, only $k^2a, (k+1)^2a,$ etc. does. $2016 = 2^5*3^2*7$, so only the squares $1, 4, 9, 16, 36,$ and $144$ are plausible for $(6k-5)^2$. But when that is anything other than $2$, $k^2a$ is not an integer. Therefore, $a = 2016/2^2 = 504$. Thanks for reading, Rowechen Zhong.",
"Instead of setting $a_1$ equal to something and $a_2$ equal to something, note that it is rather easier to set $a_1=x^2$ and $a_3=y^2$ so that $a_2=xy,a_4=y(2y-x),a_5=(2y-x)^2$ and so on until you reached $a_{13}=(6y-5x)^2$ (Or simply notice the pattern), so $6y-5x=\\sqrt{2016}=12\\sqrt{14}$. Note that since each of the terms has degree 2 so if you multiply $x$ and $y$ by $\\sqrt{14}$ you multiply each term by $14$ so each term is still a integer if the terms are already integers before you multiply $x$ and $y$ by $\\sqrt{14}$, so let $w=\\frac{x}{\\sqrt{14}}$ and $z=\\frac{y}{\\sqrt{14}}$ so $6z-5w=12$. Then, for the sequence to be strictly increasing positive integers we have $(w,z)=(6,7)$ so $x=6\\sqrt{14}$ and $a_1=x^2=6^2 \\cdot 14=504~Ddk001",
"The thirteenth term of the sequence is $2016$, which makes that fourteenth term of the sequence $2016+r$ and the $15^{\\text{th}}$ term $\\frac{(2016+r)^2}{2016}$. We note that $r$ is an integer so that means $\\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$, which is $168$. We bash all the way back to the first term and get our answer of $504. -Pleaseletmewin",
"The thirteenth term of the sequence is $2016$, which makes that fourteenth term of the sequence $2016+r$ and the $15^{\\text{th}}$ term $\\frac{(2016+r)^2}{2016}$. We note that $r$ is an integer so that means $\\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$, which is $168$. We bash all the way back to the first term and get our answer of $504. -Pleaseletmewin",
"Let $a_{2k-1}=s$ where $k=1$. Then, $a_{2k}=sr \\Longrightarrow a_{2k+1}=sr^2$. Continuing on, we get $a_{2k+2}=sr(r-1)+sr^2=sr(2r-1) \\Longrightarrow a_{2k+3}=sr^2(\\frac{2r-1}{r})^2=s(2r-1)^2$. Moreover, $a_{2k+4}=s(2r-1)(r-1)+s(2r-1)^2=s(2r-1)(3r-2) \\Longrightarrow a_{2k+5}=s(2r-1)^2(\\frac{3r-2}{2r-1})^2=s(3r-2)^2$. It is clear now that $a_{2k+2c}=s(cr-(c-1))((c+1)r-c)$ and $a_{2k+2c-1}=s(cr-(c-1))^2$. Plugging in $c=6$, $a_{13}=s(6r-5)^2=2016$. The prime factorization of $2016=2^5\\cdot3^2\\cdot7$ so we look for perfect squares. $6r-5\\equiv (6r-5)^2\\equiv 1\\pmod{6}$ if $r$ is an integer, and $\\frac{\\omega+5}{6}=r \\Longrightarrow 6\\mid{s}$ if $r$ is not an integer and $\\omega$ is rational, so $6\\mid{s}$. This forces $s=2\\cdot3^2\\cdot7\\cdot{N}$. Assuming $(6r-5)$ is an integer, it can only be $2^x$, $x\\in{1,2}$. If $6r-5=2^1$, $r=\\frac{7}{6}$. If $6r-5=2^2$, $r=\\frac{3}{2}$. Note that the latter cannot work since $a_{2k+1}=s(\\frac{9}{4}) \\Longrightarrow 4\\mid{s}$ but $N=1 \\Longrightarrow s=2\\cdot3^2\\cdot7$ in this scenario. Therefore, $r=\\frac{7}{6} \\Longrightarrow s=\\frac{2016}{2^2}=504$. Plugging back $k=1$, $a_1=s=504 ~Magnetoninja",
"Bash all the way from $a_1$ and $a_2$ to $a_{13}=2016$ to get to the equation $a_1\\left(\\frac{6a_2}{a_1}-5\\right)^2=2016$ and then $a_2=2\\sqrt{14a_1}+\\frac{5}{6}a_1$ (direct consequence). Notice that from the two equations we can deduce that: 1) $a_1\\mid2016$, 2) $6\\mid a_1$, and 3) $\\sqrt{\\frac{a_1}{14}}\\in\\mathbb{Z}$. Since $2016=2^5\\cdot3^2\\cdot7=14(2^4\\cdot 3^2)$, the only possible values of $a_1$ that satisfy the conditions are $126$, $504$, and $2016$ itself. Obviously $2016$ cannot be a correct answer choice (from AIME rules), and if $a_1=126$, $a_3\\not\\in\\mathbb{Z}$, so $504 is the solution. ~eevee9406",
"Bash all the way from $a_1$ and $a_2$ to $a_{13}=2016$ to get to the equation $a_1\\left(\\frac{6a_2}{a_1}-5\\right)^2=2016$ and then $a_2=2\\sqrt{14a_1}+\\frac{5}{6}a_1$ (direct consequence). Notice that from the two equations we can deduce that: 1) $a_1\\mid2016$, 2) $6\\mid a_1$, and 3) $\\sqrt{\\frac{a_1}{14}}\\in\\mathbb{Z}$. Since $2016=2^5\\cdot3^2\\cdot7=14(2^4\\cdot 3^2)$, the only possible values of $a_1$ that satisfy the conditions are $126$, $504$, and $2016$ itself. Obviously $2016$ cannot be a correct answer choice (from AIME rules), and if $a_1=126$, $a_3\\not\\in\\mathbb{Z}$, so $504 is the solution. ~eevee9406"
] |
2016-I-11
| 2,016
| 11
|
Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 109
|
I
|
[
"Plug in $x=1$ to get $(1-1)P(1+1) = 0 = (1+2)P(1) \\Rightarrow P(1) = 0$. Plug in $x=0$ to get $(0-1)P(0+1) = (0+2)P(0)\\Rightarrow P(0) = -\\frac{1}{2}P(1) = 0$. Plug in $x=-1$ to get $(-1-1)P(-1+1) = (-1+2)P(-1)\\Rightarrow (-2)P(0)=P(-1)\\Rightarrow P(-1) = 0$. So $P(x) = x(x-1)(x+1)Q(x)$ for some polynomial $Q(x)$. Using the initial equation, once again, \\[(x-1)P(x+1) = (x+2)P(x)\\] \\[(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))\\] \\[(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)\\] \\[Q(x+1) = Q(x)\\] From here, we know that $Q(x) = C$ for a constant $C$ ($Q(x)$ cannot be periodic since it is a polynomial), so $P(x) = Cx(x-1)(x+1)$. We know that $\\left(P(2)\\right)^2 = P(3)$. Plugging those into our definition of $P(x)$: $(C \\cdot 2 \\cdot (2-1) \\cdot (2+1))^2 = C \\cdot 3 \\cdot (3-1) \\cdot (3+1) \\Rightarrow (6C)^2 = 24C \\Rightarrow 36C^2 - 24C = 0 \\Rightarrow C = 0$ or $\\frac{2}{3}$. So we know that $P(x) = \\frac{2}{3}x(x-1)(x+1)$. So $P(\\frac{7}{2}) = \\frac{2}{3} \\cdot \\frac{7}{2} \\cdot (\\frac{7}{2} - 1) \\cdot (\\frac{7}{2} + 1) = \\frac{105}{4}$. Thus, the answer is $105 + 4 = 109.",
"From the equation we see that $x-1$ divides $P(x)$ and $(x+2)$ divides $P(x+1)$ so we can conclude that $x-1$ and $x+1$ divide $P(x)$ (if we shift the function right by 1, we get $(x-2)P(x) = (x+1)P(x-1)$, and from here we can see that $x+1$ divides $P(x)$). This means that $1$ and $-1$ are roots of $P(x)$. Plug in $x = 0$ and we see that $P(0) = 0$ so $0$ is also a root. Suppose we had another root that is not one of those $3$. Notice that the equation above indicates that if $r$ is a root then $r+1$ and $r-1$ is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three. That means $P(x) = cx(x-1)(x+1)$. We can use $P(2)^2 = P(3)$ to get $c = \\frac{2}{3}$. Plugging in $\\frac{7}{2}$ is now trivial and we see that it is $\\frac{105}{4}$ so our answer is $109",
"Although this may not be the most mathematically rigorous answer, we see that $\\frac{P(x+1)}{P(x)}=\\frac{x+2}{x-1}$. Using a bit of logic, we can make a guess that $P(x+1)$ has a factor of $x+2$, telling us $P(x)$ has a factor of $x+1$. Similarly, we guess that $P(x)$ has a factor of $x-1$, which means $P(x+1)$ has a factor of $x$. Now, since $P(x)$ and $P(x+1)$ have so many factors that are off by one, we may surmise that when you plug $x+1$ into $P(x)$, the factors \"shift over,\" i.e. $P(x)=(A)(A+1)(A+2)...(A+n)$, which goes to $P(x+1)=(A+1)(A+2)(A+3)...(A+n+1)$. This is useful because these, when divided, result in $\\frac{P(x+1)}{P(x)}=\\frac{A+n+1}{A}$. If $\\frac{A+n+1}{A}=\\frac{x+2}{x-1}$, then we get $A=x-1$ and $A+n+1=x+2$, $n=2$. This gives us $P(x)=(x-1)x(x+1)$ and $P(x+1)=x(x+1)(x+2)$, and at this point we realize that there has to be some constant $a$ multiplied in front of the factors, which won't affect our fraction $\\frac{P(x+1)}{P(x)}=\\frac{x+2}{x-1}$ but will give us the correct values of $P(2)$ and $P(3)$. Thus $P(x)=a(x-1)x(x+1)$, and we utilize $P(2)^2=P(3)$ to find $a=\\frac{2}{3}$. Evaluating $P \\left ( \\frac{7}{2} \\right )$ is then easy, and we see it equals $\\frac{105}{4}$, so the answer is $109",
"Substituting $x=2$ into the given equation, we find that $P(3)=4P(2)=P(2)^2$. Therefore, either $P(2)=0$ or $P(2)=4$. Now for integers $n\\ge 2$, we know that \\[P(n+1)=\\frac{n+2}{n-1}P(n).\\] Applying this repeatedly, we find that \\[P(n+1)=\\frac{(n+2)!/3!}{(n-1)!}P(2).\\] If $P(2)=0$, this shows that $P(x)$ has infinitely many roots, meaning that $P(x)$ is identically equal to zero. But this contradicts the problem statement. Therefore, $P(2)=4$, and we find $P(n+1)=\\frac{2}{3}(n+2)(n+1)n$ for all positive integers $n\\ge2$. This cubic polynomial matches the values $P(n+1)$ for infinitely many numbers, hence the two polynomials are identically equal. In particular, $P\\left(\\frac72\\right)=\\frac23\\cdot\\frac92\\cdot\\frac72\\cdot\\frac52=\\frac{105}{4}$, and the answer is $109.",
"We can find zeroes of the polynomial by making the first given equation $0 = 0$. Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$, respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \\cdot P(2) = 0$, which means $P(2)$ is not necessarily $0$. If $P(2) = 0$, then plugging in $2$ to the equation yields $P(3) = 0$, plugging in $3$ to the equation yields $P(4) = 0$, and so on, a contradiction of \"nonzero polynomial\". So $2$ is not a zero. Note that plugging in $x = 0$ to the equation does not yield any additional zeros. Thus, the only zeroes of $P(x)$ are $-1, 0,$ and $1$, so $P(x) = a(x + 1)x(x - 1)$ for some nonzero constant $a$. We can plug in $2$ and $3$ into the polynomial and use the second given equation to find an equation for $a$. $P(2) = 6a$ and $P(3) = 24a$, so: \\[(6a)^2 = 24a \\implies 36a^2 = 24a \\implies a = \\frac23\\] Plugging in $\\frac72$ into the polynomial $\\frac23(x + 1)x(x - 1)$ yields $\\frac{105}{4}$. $105 + 4 = 109.",
"Plug in $x=2$ yields $P(3)=4P(2)$. Since also $(P(2))^2=P(3)$, we have $P(2)=4$ and $P(3)=16$. Plug in $x=3$ yields $2P(4)=5P(3)$ so $P(4)=40$. Repeat the action gives $P(2)=4$, $P(3)=16$, $P(4)=40$, $P(5)=80$, and $P(6)=140$. Since $P(x)$ is a polynomial, the $k$th difference is constant, where $k=\\deg(P(x))$. Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant. \\[4,16,40,80,140\\] \\[12,24,40,60\\] \\[12,16,20\\] \\[4,4,4\\] Since the 3rd difference of $P(x)$ is constant, we can conclude that $\\deg(P(x))=3$. Let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$. Plug in the values for $x$ and solve the system of 4 equations gives $(a_3,a_2,a_1,a_0)=(\\frac{2}{3},0,-\\frac{2}{3},0)$ Thus $P(x)=\\frac{2}{3}x^3-\\frac{2}{3}x$ and $P(\\frac{7}{2})=\\frac{105}{4}\\Longrightarrow m+n=109, the 3rd difference of P(x). ~inaccessibles"
] |
2016-I-12
| 2,016
| 12
|
Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.
| 132
|
I
|
[
"$m(m-1)$ is the product of two consecutive integers, so it is always even. Thus $m(m-1)+11$ is odd and never divisible by $2$. Thus any prime $p$ that divides $m^2-m+11$ must divide $4m^2-4m+44=(2m-1)^2+43$. We see that $(2m-1)^2\\equiv -43\\pmod{p}$. We can verify that $-43$ is not a perfect square mod $p$ for each of $p=3,5,7$. Therefore, all prime factors of $m^2-m+11$ are $\\ge 11$. Let $m^2 - m + 11 = pqrs$ for primes $11\\le p \\le q \\le r \\le s$. From here, we could go a few different ways: Solution 1a Suppose $p=11$; then $m^2-m+11=11qrs$. Reducing modulo 11, we get $m\\equiv 1,0 \\pmod{11}$ so $k(11k\\pm 1)+1 = qrs$. Suppose $q=11$. Then we must have $11k^2\\pm k + 1 = 11rs$, which leads to $k\\equiv \\pm 1 \\pmod{11}$, i.e., $k\\in \\{1,10,12,21,23,\\ldots\\}$. $k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=101$, a prime (impossible). Finally, for $k=12$ we get $rs=143=11\\cdot 13$. Thus our answer is $m=11k= 132 and we are done.",
"Suppose $p=11$; then $m^2-m+11=11qrs$. Reducing modulo 11, we get $m\\equiv 1,0 \\pmod{11}$ so $k(11k\\pm 1)+1 = qrs$. Suppose $q=11$. Then we must have $11k^2\\pm k + 1 = 11rs$, which leads to $k\\equiv \\pm 1 \\pmod{11}$, i.e., $k\\in \\{1,10,12,21,23,\\ldots\\}$. $k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=101$, a prime (impossible). Finally, for $k=12$ we get $rs=143=11\\cdot 13$. Thus our answer is $m=11k= 132 and we are done.",
"Let $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\\ge11$. If $p, q, r, s = 11$, then $m^2-m+11=11^4$. We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\\cdot 11^4-43$. But \\[(2\\cdot 11^2-1)^2=4\\cdot 11^4-4\\cdot 11^2+1 <4\\cdot 11^4-43<(2\\cdot 11^2)^2,\\] hence we have pinned a perfect square $(2m-1)^2=4\\cdot 11^4-43$ strictly between two consecutive perfect squares, a contradiction. Hence $pqrs \\ge 11^3 \\cdot 13$. Thus $m^2-m+11\\ge 11^3\\cdot 13$, or $(m-132)(m+131)\\ge0$. From the inequality, we see that $m \\ge 132$. $132^2 - 132 + 11 = 11^3 \\cdot 13$, so $m = 132 and we are done.",
"First, we can show that $m^2 - m + 11 \\not |$ $2,3,5,7$. This can be done by just testing all residue classes. For example, we can test $m \\equiv 0 \\mod 2$ or $m \\equiv 1 \\mod 2$ to show that $m^2 - m + 11$ is not divisible by 2. Case 1: m = 2k $m^2 - m + 11 \\equiv 2(2 \\cdot k^2 - k + 5) +1 \\equiv 1 \\mod 2$ Case 2: m = 2k+1 $m^2 - m + 11 \\equiv 2(2 \\cdot k^2 + k + 5) +1 \\equiv 1 \\mod 2$ Now, we can test $m^2 - m + 11 = 11^4$, which fails, so we test $m^2 - m + 11 = 11^3 \\cdot 13$, and we get m = $132$. -AlexLikeMath"
] |
2016-I-13
| 2,016
| 13
|
Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$ . A fence is located at the horizontal line $y = 0$ . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0$ , with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y < 0$ . Freddy starts his search at the point $(0, 21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river.
| 273
|
I
|
[
"Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and \\[E(y)=1+\\frac{E(y+1)+E(y-1)+2E(y)}{4}\\] for all $y$ such that $1\\le y\\le 23$. Also note that $E(0)=1+\\frac{2E(0)+E(1)}{3}$. This gives $E(0)=E(1)+3$. Plugging this into the equation for $E(1)$ gives that \\[E(1)=1+\\frac{E(2)+3E(1)+3}{4},\\] or $E(1)=E(2)+7$. Iteratively plugging this in gives that $E(n)=E(n+1)+4n+3$. Thus $E(23)=E(24)+95$, $E(22)=E(23)+91=E(24)+186$, and $E(21)=E(22)+87=E(24)+273=273."
] |
2016-I-14
| 2,016
| 14
|
Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ .
| 574
|
I
|
[
"First note that $1001 = 143 \\cdot 7$ and $429 = 143 \\cdot 3$ so every point of the form $(7k, 3k)$ is on the line. Then consider the line $l$ from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$. Translate the line $l$ so that $(7k, 3k)$ is now the origin. There is one square and one circle that intersect the line around $(0,0)$. Then the points on $l$ with an integral $x$-coordinate are, since $l$ has the equation $y = \\frac{3x}{7}$: \\[(0,0), \\left(1, \\frac{3}{7}\\right), \\left(2, \\frac{6}{7}\\right), \\left(3, 1 + \\frac{2}{7}\\right), \\left(4, 1 + \\frac{5}{7}\\right), \\left(5, 2 + \\frac{1}{7}\\right), \\left(6, 2 + \\frac{4}{7}\\right), (7,3).\\] We claim that the lower right vertex of the square centered at $(2,1)$ lies on $l$. Since the square has side length $\\frac{1}{5}$, the lower right vertex of this square has coordinates $\\left(2 + \\frac{1}{10}, 1 - \\frac{1}{10}\\right) = \\left(\\frac{21}{10}, \\frac{9}{10}\\right)$. Because $\\frac{9}{10} = \\frac{3}{7} \\cdot \\frac{21}{10}$, $\\left(\\frac{21}{10}, \\frac{9}{10}\\right)$ lies on $l$. Since the circle centered at $(2,1)$ is contained inside the square, this circle does not intersect $l$. Similarly the upper left vertex of the square centered at $(5,2)$ is on $l$. Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between $(0,0)$ and $(7,3)$ that intersect $l$. Since there are $\\frac{1001}{7} = \\frac{429}{3} = 143$ segments from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$, the above count is yields $143 \\cdot 2 = 286$ squares. Since every lattice point on $l$ is of the form $(3k, 7k)$ where $0 \\le k \\le 143$, there are $144$ lattice points on $l$. Centered at each lattice point, there is one square and one circle, hence this counts $288$ squares and circles. Thus $m + n = 286 + 288 = 574. (Solution by gundraja)",
"See if you can solve the problem with the following. [asy]size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted); for(int i=0;i<8;++i)for(int j=0; j<4; ++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));} [/asy] Solution to Solution 2 This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$, which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \\cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=574. -Patrick4President",
"This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$, which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \\cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=574. -Patrick4President",
"This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$, the slope of our line is $\\frac{3}{7}$, and we only need to consider the line in the rectangle from the origin to $(7,3)$, and we can iterate the line $11*13=143$ times. First, we consider how to figure out if the line intersects a square. Given a lattice point $(x_1, y_1)$, we can think of representing a square centered at that lattice point as all points equal to $(x_1 \\pm a, y_1 \\pm b)$ s.t. $0 \\leq a,b \\leq \\frac{1}{10}$. If the line $y = \\frac{3}{7}x$ intersects the square, then we must have $\\frac{y_1 + b}{x_1 + a} = \\frac{3}{7}$. The line with the least slope that intersects the square intersects at the bottom right corner and the line with the greatest slope that intersects the square intersects at the top left corner; thus we must have that $\\frac{3}{7}$ lies in between these slopes, or that $\\frac{y_1-\\frac{1}{10}}{x_1+\\frac{1}{10}} \\leq \\frac{3}{7} \\leq \\frac{y_1+\\frac{1}{10}}{x_1-\\frac{1}{10}}$. Simplifying, $3x_1 - 1 \\leq 7y_1 \\leq 3x_1 + 1$. Because $y$ can only equal $0, 1, 2, 3$, we just do casework based on the values of $y$ and find that the points $(2, 1)$ and $(5, 2)$ are intersected just at the corner of the square and $(0, 0), (7, 3)$ are intersected through the center of the square. However, we disregard one of $(0, 0)$ and $(7, 3)$, WLOG $(0, 0)$, since we just use it in our count for the next of the 143 segments. Therefore, in one of our \"segments\", 3 squares are intersected and 1 circle is intersected giving 4 total. Thus our answer is $143*4 = 572$. HOWEVER, we cannot forget that we ignored $(0, 0)$, which contributes another square and circle to our count, making the final answer $572 + 2 = 574. -Patrick4President"
] |
2016-I-15
| 2,016
| 15
|
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ , $Y$ , $D$ are collinear, $XC = 67$ , $XY = 47$ , and $XD = 37$ . Find $AB^2$ .
| 270
|
I
|
[
"Using the radical axis theorem, the lines $\\overline{AD}, \\overline{BC}, \\overline{XY}$ are all concurrent at one point, call it $F$. Now recall by Miquel's theorem in $\\triangle FDC$ the fact that quadrilaterals $DAXY$ and $CBXY$ are cyclic implies $FAXB$ is cyclic as well. Denote $\\omega_{3}\\equiv(FAXB)$ and $Z\\equiv\\ell\\cap\\overline{FXY}$. Since point $Z$ lies on the radical axis of $\\omega_{1},\\omega_{2}$, it has equal power with respect to both circles, thus \\[AZ^{2}=\\text{Pow}_{\\omega_{1}}(Z)=ZX\\cdot ZY=\\text{Pow}_{\\omega_{2}}(Z)=ZB^{2}\\implies AZ=ZB.\\] Also, notice that \\[AZ\\cdot ZB=\\text{Pow}_{\\omega_{3}}(Z)=ZX\\cdot ZF\\implies ZY=ZF.\\] The diagonals of quadrilateral $FAYB$ bisect each other at $Z$, so we conclude that $FAYB$ is a parallelogram. Let $u:=ZX$, so that $ZY=ZF=u+47$. Because $FAYB$ is a parallelogram and quadrilaterals $DAXY, CBXY$ are cyclic, \\[\\angle DFX=\\angle AFX=\\angle BYX=\\angle BCX=\\angle FCX~~\\text{and}~~\\angle XDF=\\angle XDA=\\angle XYA=\\angle XFB=\\angle XFC\\] so we have the pair of similar triangles $\\triangle DFX~\\sim~\\triangle FCX$. Thus \\[\\dfrac{37}{2u+47}=\\dfrac{2u+47}{67}\\implies 2u+47=\\sqrt{37\\cdot 67}\\implies u=\\dfrac{1}{2}\\left(\\sqrt{37\\cdot 67}-47\\right).\\] Now compute \\[AB^{2}=4AZ^{2}=4\\cdot ZX\\cdot ZY=4u(u+47)=37\\cdot 67-47^{2}=\\textbf{270}.\\]",
"Let $Z = XY \\cap AB$. By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$. Moreover, $\\triangle DXP \\sim \\triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$. Then \\[\\frac{y}{37} = \\frac{67}{y} \\qquad \\implies \\qquad y^2 = 37 \\cdot 67.\\] Now, $AZ^2 = \\tfrac 14 AB^2$, and by power of a point, \\begin{align*} x(y-x) &= \\tfrac 14 AB^2, \\quad \\text{and} \\\\ x(47+x) &= \\tfrac 14 AB^2 \\end{align*} Solving, we get \\[\\tfrac 14 AB^2 = \\tfrac 12 (y-47)\\cdot \\tfrac 12 (y+47) \\qquad \\implies\\] \\[\\qquad AB ^ 2 = 37\\cdot 67 - 47^2 = 270\\]",
"By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$. Let $AB$ and $EY$ intersect at $S$. Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel's Theorem $AXBE$ is cyclic as well. Thus \\[\\angle AEX = \\angle ABX = \\angle XCB = \\angle XYB\\]and \\[\\angle XEB = \\angle XAB = \\angle XDA = \\angle XYA.\\]Thus $AY \\parallel EB$ and $YB \\parallel EA$, so $AEBY$ is a parallelogram. Hence $AS = SB$ and $SE = SY$. But notice that $DXE$ and $EXC$ are similar by $AA$ Similarity, so $XE^2 = XD \\cdot XC = 37 \\cdot 67$. But \\[XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \\cdot 2XS = 2SY \\cdot 2SX = 4SA^2 = AB^2.\\]Hence $AB^2 = 37 \\cdot 67 - 47^2 = 270",
"First, we note that as $\\triangle XDY$ and $\\triangle XYC$ have bases along the same line, $\\frac{[\\triangle XDY]}{[\\triangle XYC]}=\\frac{DY}{YC}$. We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\\omega_1$ and if $R_2$ is the radius of $\\omega_2$, then \\[\\frac{[\\triangle XDY]}{[\\triangle XYC]}=\\frac{(37\\cdot 47\\cdot DY)/(4R_1)}{(47\\cdot 67\\cdot YC)/(4R_2)}=\\frac{37\\cdot DY\\cdot R_2}{67\\cdot YC\\cdot R_1}.\\] Since we showed this to be $\\frac{DY}{YC}$, we see that $\\frac{R_2}{R_1}=\\frac{67}{37}$. We extend $AD$ and $BC$ to meet at point $P$, and we extend $AB$ and $CD$ to meet at point $Q$ as shown below. [asy] size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label(\"$A$\",a,NW,fontsize(8)); label(\"$B$\",b,NE,fontsize(8)); label(\"$C$\",c,SE,fontsize(8)); label(\"$D$\",d,SW,fontsize(8)); label(\"$X$\",x,2*WNW,fontsize(8)); label(\"$Y$\",y,3*S,fontsize(8)); label(\"$P$\",p,N,fontsize(8)); label(\"$Q$\",q,W,fontsize(8)); [/asy] As $ABCD$ is cyclic, we know that $\\angle BCD=180-\\angle DAB=\\angle BAP$. But then as $AB$ is tangent to $\\omega_2$ at $B$, we see that $\\angle BCD=\\angle ABY$. Therefore, $\\angle ABY=\\angle BAP$, and $BY\\parallel PD$. A similar argument shows $AY\\parallel PC$. These parallel lines show $\\triangle PDC\\sim\\triangle ADY\\sim\\triangle BYC$. Also, we showed that $\\frac{R_2}{R_1}=\\frac{67}{37}$, so the ratio of similarity between $\\triangle ADY$ and $\\triangle BYC$ is $\\frac{37}{67}$, or rather \\[\\frac{AD}{BY}=\\frac{DY}{YC}=\\frac{YA}{CB}=\\frac{37}{67}.\\] We can now use the parallel lines to find more similar triangles. As $\\triangle AQD\\sim \\triangle BQY$, we know that \\[\\frac{QA}{QB}=\\frac{QD}{QY}=\\frac{AD}{BY}=\\frac{37}{67}.\\] Setting $QA=37x$, we see that $QB=67x$, hence $AB=30x$, and the problem simplifies to finding $30^2x^2$. Setting $QD=37^2y$, we also see that $QY=37\\cdot 67y$, hence $DY=37\\cdot 30y$. Also, as $\\triangle AQY\\sim \\triangle BQC$, we find that \\[\\frac{QY}{QC}=\\frac{YA}{CB}=\\frac{37}{67}.\\] As $QY=37\\cdot 67y$, we see that $QC=67^2y$, hence $YC=67\\cdot30y$. Applying Power of a Point to point $Q$ with respect to $\\omega_2$, we find \\[67^2x^2=37\\cdot 67^3 y^2,\\] or $x^2=37\\cdot 67 y^2$. We wish to find $AB^2=30^2x^2=30^2\\cdot 37\\cdot 67y^2$. Applying Stewart's Theorem to $\\triangle XDC$, we find \\[37^2\\cdot (67\\cdot 30y)+67^2\\cdot(37\\cdot 30y)=(67\\cdot 30y)\\cdot (37\\cdot 30y)\\cdot (104\\cdot 30y)+47^2\\cdot (104\\cdot 30y).\\] We can cancel $30\\cdot 104\\cdot y$ from both sides, finding $37\\cdot 67=30^2\\cdot 67\\cdot 37y^2+47^2$. Therefore, \\[AB^2=30^2\\cdot 37\\cdot 67y^2=37\\cdot 67-47^2=270.\\]",
"[asy] size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label(\"$A$\",a,NW,fontsize(9)); label(\"$B$\",b,NE,fontsize(9)); label(\"$C$\",c,SE,fontsize(9)); label(\"$D$\",d,SW,fontsize(9)); label(\"$X$\",x,2*N,fontsize(9)); label(\"$Y$\",y,3*S,fontsize(9)); [/asy] First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\\angle DAX = \\angle XYC = \\theta$, and $\\angle XYD = \\angle CBX = 180 - \\theta$, due to the properties of cyclic quadrilaterals. In addition, let $\\angle BAX = x$ and $\\angle ABX = y$. Thus, $\\angle ADX = \\angle AYX = x$ and $\\angle XYB = \\angle XCB = y$. Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums: \\[\\theta + x +\\angle XCY + y = 180^{\\circ}\\] \\[180^{\\circ} - \\theta + y + \\angle XDY + x = 180^{\\circ}\\] Cancelling out $180^{\\circ}$ in the second equation and isolating $\\theta$ yields $\\theta = y + \\angle XDY + x$. Substituting $\\theta$ back into the first equation, we obtain \\[2x + 2y + \\angle XCY + \\angle XDY = 180^{\\circ}\\] Since \\[x + y +\\angle XAY + \\angle XCY + \\angle DAY = 180^{\\circ}\\] \\[x + y + \\angle XDY + \\angle XCY + \\angle DAY = 180^{\\circ}\\] we can then imply that $\\angle DAY = x + y$. Similarly, $\\angle YBC = x + y$. So then $\\angle DXY = \\angle YXC = x + y$, so since we know that $XY$ bisects $\\angle DXC$, we can solve for $DY$ and $YC$ with Stewart’s Theorem. Let $DY = 37n$ and $YC = 67n$. Then \\[37n \\cdot 67n \\cdot 104n + 47^2 \\cdot 104n = 37^2 \\cdot 67n + 67^2 \\cdot 37n\\] \\[37n \\cdot 67n + 47^2 = 37 \\cdot 67\\] \\[n^2 = \\frac{270}{2479}\\] Now, since $\\angle AYX = x$ and $\\angle BYX = y$, $\\angle AYB = x + y$. From there, let $\\angle AYD = \\alpha$ and $\\angle BYC = \\beta$. From angle chasing we can derive that $\\angle YDX = \\angle YAX = \\beta - x$ and $\\angle YCX = \\angle YBX = \\alpha - y$. From there, since $\\angle ADX = x$, it is quite clear that $\\angle ADY = \\beta$, and $\\angle YAB = \\beta$ can be found similarly. From there, since $\\angle ADY = \\angle YAB = \\angle BYC = \\beta$ and $\\angle DAY = \\angle AYB = \\angle YBC = x + y$, we have $AA$ similarity between $\\triangle DAY$, $\\triangle AYB$, and $\\triangle YBC$. Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\\triangle DAY \\sim \\triangle AYB$). However, $\\triangle DAY \\sim \\triangle AYB \\sim \\triangle YBC$ yields the proportion $\\frac{AD}{DY} = \\frac{YA}{AB} = \\frac{BY}{YC}$; hence, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$. We can now simply use arithmetic to calculate $AB^2$. \\[AB^2 = DY \\cdot YC\\] \\[AB^2 = 37 \\cdot 67 \\cdot \\frac{270}{2479}\\] \\[AB^2 = 270\\] -Solution by TheBoomBox77",
"Let $E = DA \\cap CB$. By Radical Axes, $E$ lies on $XY$. Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\\triangle EDC$ in this configuration. Claim. $\\triangle DXE \\sim \\triangle EXC$ Proof. We angle chase. \\[\\measuredangle XEC = \\measuredangle XEB = \\measuredangle XAB = \\measuredangle XDA = \\measuredangle XDE\\]and\\[\\measuredangle XCE = \\measuredangle XCB = \\measuredangle XBA = \\measuredangle XEA = \\measuredangle XED. \\square\\] Let $F = EX \\cap AB$. Note \\[FA^2 = FX \\cdot FY = FB^2\\]and\\[EF \\cdot FX = AF \\cdot FB = FA^2 = FX \\cdot FY \\implies EF = FY\\]By our claim, \\[\\frac{DX}{XE} = \\frac{EX}{XC} \\implies EX^2 = DX \\cdot XC = 67 \\cdot 37 \\implies FY = \\frac{EY}{2} = \\frac{EX+XY}{2} = \\frac{\\sqrt{67 \\cdot 37}+47}{2}\\]and\\[FX=FY-47=\\frac{\\sqrt{67 \\cdot 37}-47}{2}\\]Finally, \\[AB^2 = (2 \\cdot FA)^2 = 4 \\cdot FX \\cdot FY = 4 \\cdot \\frac{(67 \\cdot 37) - 47^2}{4} = 270. \\blacksquare\\]~Mathscienceclass",
"Let $E = DA \\cap CB$. By Radical Axes, $E$ lies on $XY$. Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\\triangle EDC$ in this configuration. Claim. $\\triangle DXE \\sim \\triangle EXC$ Proof. We angle chase. \\[\\measuredangle XEC = \\measuredangle XEB = \\measuredangle XAB = \\measuredangle XDA = \\measuredangle XDE\\]and\\[\\measuredangle XCE = \\measuredangle XCB = \\measuredangle XBA = \\measuredangle XEA = \\measuredangle XED. \\square\\] Let $F = EX \\cap AB$. Note \\[FA^2 = FX \\cdot FY = FB^2\\]and\\[EF \\cdot FX = AF \\cdot FB = FA^2 = FX \\cdot FY \\implies EF = FY\\]By our claim, \\[\\frac{DX}{XE} = \\frac{EX}{XC} \\implies EX^2 = DX \\cdot XC = 67 \\cdot 37 \\implies FY = \\frac{EY}{2} = \\frac{EX+XY}{2} = \\frac{\\sqrt{67 \\cdot 37}+47}{2}\\]and\\[FX=FY-47=\\frac{\\sqrt{67 \\cdot 37}-47}{2}\\]Finally, \\[AB^2 = (2 \\cdot FA)^2 = 4 \\cdot FX \\cdot FY = 4 \\cdot \\frac{(67 \\cdot 37) - 47^2}{4} = 270. \\blacksquare\\]~Mathscienceclass",
"$AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \\cdot XD - XY^2 = 67 \\cdot 37 - 47^2 = 270 [email protected], vvsss",
"$AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \\cdot XD - XY^2 = 67 \\cdot 37 - 47^2 = 270 [email protected], vvsss",
"Extend $\\overline{AD}$ and $\\overline{BC}$ to meet at point $P$. Let $M$ be the midpoint of segment $AB$. Then by radical axis on $(ADY)$, $(BCY)$ and $(ABCD)$, $P$ lies on $XY$. By the bisector lemma, $M$ lies on $XY$. It is well-known that $P$, $A$, $X$, and $B$ are concyclic. By Power of a point on $M$ with respect to $(PAXB)$ and $(ADY)$, \\[|\\text{Pow}(M, (PAXB))| = MX \\cdot MP = MA^2 = |\\text{Pow}(M, (ADY))| = MX \\cdot MY,\\] so $MP=MY$. Thus $AB$ and $PY$ bisect each other, so $PAYB$ is a parallelogram. This implies that \\[\\angle DAY = \\angle YBC,\\] so by the inscribed angle theorem $\\overline{XY}$ bisects $\\angle DXC$. Claim: $AB^2 = DY \\cdot YC$. Proof. Define the linear function $f(\\bullet) := \\text{Pow}(\\bullet, (ADY)) - \\text{Pow}(\\bullet, (ABCD))$. Since $\\overline{BY}$ is parallel to the radical axis $\\overline{AD}$ of $(ADY)$ and $(ABCD)$ by our previous parallelism, $f(B)=f(Y)$. Note that $f(B)=AB^2$ while $f(Y)=DY \\cdot YC$, so we conclude. $\\square$ By Stewart's theorem on $\\triangle DXC$, $DY \\cdot YC=37 \\cdot 67 - 47^2 = 270$, so $AB^2=270. ~ Leo.Euler",
"Extend $\\overline{AD}$ and $\\overline{BC}$ to meet at point $P$. Let $M$ be the midpoint of segment $AB$. Then by radical axis on $(ADY)$, $(BCY)$ and $(ABCD)$, $P$ lies on $XY$. By the bisector lemma, $M$ lies on $XY$. It is well-known that $P$, $A$, $X$, and $B$ are concyclic. By Power of a point on $M$ with respect to $(PAXB)$ and $(ADY)$, \\[|\\text{Pow}(M, (PAXB))| = MX \\cdot MP = MA^2 = |\\text{Pow}(M, (ADY))| = MX \\cdot MY,\\] so $MP=MY$. Thus $AB$ and $PY$ bisect each other, so $PAYB$ is a parallelogram. This implies that \\[\\angle DAY = \\angle YBC,\\] so by the inscribed angle theorem $\\overline{XY}$ bisects $\\angle DXC$. Claim: $AB^2 = DY \\cdot YC$. Proof. Define the linear function $f(\\bullet) := \\text{Pow}(\\bullet, (ADY)) - \\text{Pow}(\\bullet, (ABCD))$. Since $\\overline{BY}$ is parallel to the radical axis $\\overline{AD}$ of $(ADY)$ and $(ABCD)$ by our previous parallelism, $f(B)=f(Y)$. Note that $f(B)=AB^2$ while $f(Y)=DY \\cdot YC$, so we conclude. $\\square$ By Stewart's theorem on $\\triangle DXC$, $DY \\cdot YC=37 \\cdot 67 - 47^2 = 270$, so $AB^2=270. ~ Leo.Euler"
] |
2016-II-1
| 2,016
| 1
|
Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
| 108
|
II
|
[
"Let $r$ be the common ratio, where $r>1$. We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$. We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$, so we have $3ar=432,$ or $ar=144$, which is how much Betty had. Now we have $144+\\dfrac{144}{r}+144r=444$, or $144(r+\\dfrac{1}{r})=300$, or $r+\\dfrac{1}{r}=\\dfrac{25}{12}$, which solving for $r$ gives $r=\\dfrac{4}{3}$, since $r>1$, so Alex had $\\dfrac{3}{4} \\cdot 144=108 peanuts.",
"Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$. Adding $k$ to both sides and factoring, \\[\\frac{444}{a}+k=(k+1)^2\\] For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$. Simplifying, $k^2-2k+1=\\frac{12}{a}$. Factoring, \\[(k-1)^2=\\frac{12}{a}\\] \\[(k+1)^2-(k-1)^2=4k \\implies 4k=\\frac{444-12}{a}+k \\implies k=\\frac{144}{a}\\] Substitute $k$ in the second equation to get $(\\frac{144-a}{a})^2=\\frac{12}{a}$. Expanding and applying the quadratic formula, \\[a=150\\pm\\frac{\\sqrt{300^2-4(144^2)}}{2}\\] Taking out $4^2\\cdot3^2$ from under the radical leaves \\[a=150\\pm6\\sqrt{625-576}=108, 192\\] Since Alex's peanut number was the lowest of the trio, and $3*192>444$, Alex initially had $108 peanuts.",
"Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$. Adding $k$ to both sides and factoring, \\[\\frac{444}{a}+k=(k+1)^2\\] For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$. Simplifying, $k^2-2k+1=\\frac{12}{a}$. Factoring, \\[(k-1)^2=\\frac{12}{a}\\] \\[(k+1)^2-(k-1)^2=4k \\implies 4k=\\frac{444-12}{a}+k \\implies k=\\frac{144}{a}\\] Substitute $k$ in the second equation to get $(\\frac{144-a}{a})^2=\\frac{12}{a}$. Expanding and applying the quadratic formula, \\[a=150\\pm\\frac{\\sqrt{300^2-4(144^2)}}{2}\\] Taking out $4^2\\cdot3^2$ from under the radical leaves \\[a=150\\pm6\\sqrt{625-576}=108, 192\\] Since Alex's peanut number was the lowest of the trio, and $3*192>444$, Alex initially had $108 peanuts.",
"Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$, $b$, and $c$ respectively. Let the final numbers of peanuts, after eating, be $a'$, $b'$, and $c'$. We are given that $a + b + c = 444$. Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$. Since $a'$, $b'$, and $c'$ form an arithmetic progression, we have that $a' = b' - x$ and $c' = b' + x$ for some integer $x$. Substituting yields $3b' = 405$ and so $b' = 135$. Since Betty ate $9$ peanuts, it follows that $b = b' + 9 = 144$. Since $a$, $b$, and $c$ form a geometric progression, we have that $a = \\frac{b}{r}$ and $c = br$. Multiplying yields $ac = b^2 = 144^2$. Since $a + c = 444 - b = 300$, it follows that $a = 150 - \\lambda$ and $c = 150 + \\lambda$ for some integer $\\lambda$. Substituting yields $(150-\\lambda)(150+\\lambda) = 144^2$, which expands and rearranges to $\\lambda^2 = 150^2-144^2 = 42^2$. Since $\\lambda > 0$, we must have $\\lambda = 42$, and so $a = 150 - \\lambda = 108.",
"Bashing is not difficult. All we have to consider is the first equation. We can write it as $x*(1+r+r^2) = 444$. The variable $x$ must be an integer, and after trying all the factors of $444$, it's clear that $r$ is a fraction smaller than $10$. When calculating the coefficient of $x$, we must consider that the fraction produced will very likely have a numerator that divides $444$. Trying a couple will make it easier to find the fraction, and soon you will find that $\\frac{4}{3}$ gives a numerator of $37$, a rather specific factor of $444$. Solving for the rest will give you an integer value of $108. This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals.",
"Let $b$ be the finish number of Betty's peanuts. Then \\[3b = 444-(5 + 9 + 25) = 405 = 3 \\cdot 135 \\implies b = 135, b+ 9 = 144.\\] Let $k > 1$ be the common ratio. Then \\[\\frac{144}{k} + 144 + k \\cdot 144 = 444 \\implies \\frac{144}{k} + k \\cdot 144 = 300\\implies \\frac{12}{k} + k \\cdot 12 = 25\\implies k = \\frac{4}{3} \\implies \\frac {144\\cdot 3}{4} = 108.\\] [email protected], vvsss"
] |
2016-II-2
| 2,016
| 2
|
There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a+b$ .
| 107
|
II
|
[
"Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\\dfrac{3}{5}x+\\dfrac{2}{5}2x = \\dfrac{3}{10} \\implies \\dfrac{7}{5}x=\\dfrac{3}{10}$ $\\implies x=\\dfrac{3}{14}$. Therefore, the probability that it doesn't rain on either day is $\\left(1-\\dfrac{3}{14}\\right)\\left(\\dfrac{3}{5}\\right)=\\dfrac{33}{70}$. Therefore, the probability that rains on at least one of the days is $1-\\dfrac{33}{70}=\\dfrac{37}{70}$, so adding up the $2$ numbers, we have $37+70=107."
] |
2016-II-3
| 2,016
| 3
|
Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4. \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$ .
| 265
|
II
|
[
"First, we get rid of logs by taking powers: $xyz-3+\\log_5 x=2^{5}=32$, $xyz-3+\\log_5 y=3^{4}=81$, and $(xyz-3+\\log_5 z)=4^{4}=256$. Adding all the equations up and using the $\\log {xy}=\\log {x}+\\log{y}$ property, we have $3xyz+\\log_5{xyz} = 378$, so we have $xyz=125$. Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\\log_5 x=-90, \\log_5 y=-41, \\log_5 z=134$, so adding all the absolute values we have $90+41+134=265."
] |
2016-II-4
| 2,016
| 4
|
An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.
| 180
|
II
|
[
"By counting the number of green cubes $2$ different ways, we have $12a=20b$, or $a=\\dfrac{5}{3} b$. Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \\times b \\times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \\times a \\times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$. We check a few small values of $a,b$ and solve for $c$, checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$, $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$, and $(a,b)=(15,9)$ gives $c=4$, with a volume of $540$. Any higher $(a,b)$ will $ab>180$, so therefore, the minimum volume is $180.",
"The total number of green cubes is given by $12a=20b\\Longrightarrow a=\\frac{5}{3}b$. Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$. Substitute $a=\\frac{5}{3}b$ gives $r=15$. Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$. Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$. Multiplying yields $abc^2=2160$. Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$, $b$ are minimized (since each of $ac$, $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\\Longrightarrow abc=180 ~ Nafer"
] |
2016-II-5
| 2,016
| 5
|
Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ .
| 182
|
II
|
[
"Do note that by counting the area in 2 ways, the first altitude is $x = \\frac{ab}{c}$. By similar triangles, the common ratio is $\\rho = \\frac{a}{c}$ for each height, so by the geometric series formula, we have \\begin{align} 6p=\\frac{x}{1-\\rho} = \\frac{ab}{c-a}. \\end{align} Writing $p=a+b+c$ and clearing denominators, we get \\[13a=6p .\\]Thus $p=13q$, $a=6q$, and $b+c=7q$, i.e. $c=7q-b$. Plugging these into $(1)$, we get $78q(q-b)=6bq$, i.e., $14b=13q$. Thus $q=14r$ and $p=182r$, $b=13r$, $a=84r$, $c=85r$. Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=182. Note this sulotion seems to have some serious errors",
"Note that by counting the area in 2 ways, the first altitude is $\\dfrac{ab}{c}$. By similar triangles, the common ratio is $\\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\\dfrac{\\dfrac{ab}{c}}{1-\\dfrac{a}{c}}$. Multiplying by the denominator and expanding, the equation becomes $\\dfrac{ab}{c}=6a+6b+6c-\\dfrac{6a^2}{c}-\\dfrac{6ab}{c}-6a$. Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$, so $7ab = 6bc+6b^2$ and $7a=6b+6c$. Checking for Pythagorean triples gives $13,84,$ and $85$, so $p=13+84+85=182 Solution modified/fixed from Shaddoll's solution.",
"Note that by counting the area in 2 ways, the first altitude is $\\dfrac{ab}{c}$. By similar triangles, the common ratio is $\\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\\dfrac{\\dfrac{ab}{c}}{1-\\dfrac{a}{c}}$. Multiplying by the denominator and expanding, the equation becomes $\\dfrac{ab}{c}=6a+6b+6c-\\dfrac{6a^2}{c}-\\dfrac{6ab}{c}-6a$. Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$, so $7ab = 6bc+6b^2$ and $7a=6b+6c$. Checking for Pythagorean triples gives $13,84,$ and $85$, so $p=13+84+85=182 Solution modified/fixed from Shaddoll's solution.",
"We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even. First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\\geq2$. The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\\left(\\frac{C_{n-1}C_{n}}{C_{0}B}\\right)$. Thus, $p\\sum_{n=1}^{\\infty}\\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$. Simplifying, $\\sum_{n=1}^{\\infty}C_{2n-1}C_{2n}=6C_{0}B + \\frac{(C_{0}B)^2}{p}=C_{0}B\\left(6+\\frac{C_{0}B}{p}\\right)$. (1) Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even, $p\\sum_{n=1}^{\\infty}\\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$. Simplifying, $\\sum_{n=1}^{\\infty}C_{2n-2}C_{2n-1}=C_{0}B\\left(7-\\frac{C_{0}B}{p}\\right)$. (2) Adding (1) and (2) together, we find that $6p=13C_{0}B \\Rightarrow p=\\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \\Rightarrow \\frac{7C_{0}B}{6}=C_{0}A+AB \\Rightarrow 7C_{0}B=6C_{0}A + 6AB$. Setting $a=C_{0}B$, $b=C_{0}A$, and $c=AB$, we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=182.",
"[asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label(\"$A$\", A, SE); label(\"$B$\", B, NW); label(\"$C_0$\", C0, SW); label(\"$C_1$\", C1, NE); label(\"$C_2$\", C2, W); label(\"$C_3$\", C3, NE); label(\"$C_4$\", C4, W); label(\"$a$\", (B+C0)/2, W); label(\"$b$\", (A+C0)/2, S); label(\"$c$\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy] Let $a = BC_0$, $b = AC_0$, and $c = AB$. Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\\frac{a+c}{b}$. The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \\cdots$, shown in red in the figure below. Since each of the triangles $\\triangle C_0 C_1 C_2, \\triangle C_2 C_3 C_4, \\dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\\frac{a+c}{b}$. In other words, we have that $a\\left(\\frac{a+c}{b}\\right) = 6p$. Guessing and checking Pythagorean triples reveals that $a = 84$, $b=13$, $c = 85$, and $p = a + b + c = 182\", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); draw(C0--B, blue); [/asy]",
"This solution proceeds from $\\frac{\\frac{ab}{c}}{1-\\frac{a}{c}} = \\frac{\\frac{ab}{c}}{\\frac{c-a}{c}} = \\frac{ab}{c-a} = 6(a+b+c)$. Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$. Thus $a = 168n^2, b = 26n^2, c = 170n^2$. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = 182.",
"For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$, we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$. Let $b=2xy$, then this generalization gives $6(a+b+c)(c-b)=ab$ \\[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\\] \\[(x+y)xy=6(x^2+xy)(x-y)\\] \\[xy=6x(x-y)\\] \\[7xy=6x^2\\] \\[y=\\frac{6}{7}x\\] Now this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $182."
] |
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