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2009-II-8
| 2,009
| 8
|
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m+n$ .
| 41
|
II
|
[
"Solution 1 There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at least two more throws than Dave, and our answer will therefore be $1-2p$. How to compute $p$? Suppose that Linda made exactly $t$ throws. The probability that this happens is $(5/6)^{t-1}\\cdot (1/6)$, as she must make $t-1$ unsuccessful throws followed by a successful one. In this case, we need Dave to make at least $t+2$ throws. This happens if his first $t+1$ throws are unsuccessful, hence the probability is $(5/6)^{t+1}$. Thus for a fixed $t$ the probability that Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \\cdot (1/6)$. Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$. Hence: \\begin{align*} p & = \\sum_{t=1}^\\infty \\left(\\frac 56\\right)^{2t} \\cdot \\frac 16 \\\\ & = \\frac 16 \\cdot \\left(\\frac 56\\right)^2 \\cdot \\sum_{x=0}^\\infty \\left(\\frac{25}{36}\\right)^x \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac 1{1 - \\dfrac{25}{36}} \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac{36}{11} \\\\ & = \\frac {25}{66} \\end{align*} Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\\cdot \\frac{25}{66} = 1 - \\frac{25}{33} = \\frac 8{33}$, and the answer is $8+33 = 041",
"There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at least two more throws than Dave, and our answer will therefore be $1-2p$. How to compute $p$? Suppose that Linda made exactly $t$ throws. The probability that this happens is $(5/6)^{t-1}\\cdot (1/6)$, as she must make $t-1$ unsuccessful throws followed by a successful one. In this case, we need Dave to make at least $t+2$ throws. This happens if his first $t+1$ throws are unsuccessful, hence the probability is $(5/6)^{t+1}$. Thus for a fixed $t$ the probability that Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \\cdot (1/6)$. Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$. Hence: \\begin{align*} p & = \\sum_{t=1}^\\infty \\left(\\frac 56\\right)^{2t} \\cdot \\frac 16 \\\\ & = \\frac 16 \\cdot \\left(\\frac 56\\right)^2 \\cdot \\sum_{x=0}^\\infty \\left(\\frac{25}{36}\\right)^x \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac 1{1 - \\dfrac{25}{36}} \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac{36}{11} \\\\ & = \\frac {25}{66} \\end{align*} Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\\cdot \\frac{25}{66} = 1 - \\frac{25}{33} = \\frac 8{33}$, and the answer is $8+33 = 041",
"Let $p$ be the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. (We will call this event \"a win\", and the opposite event will be \"a loss\".) Let both players roll their first die. With probability $\\frac 1{36}$, both throw a six and we win. With probability $\\frac{10}{36}$ exactly one of them throws a six. In this case, we win if the remaining player throws a six in their next throw, which happens with probability $\\frac 16$. Finally, with probability $\\frac{25}{36}$ none of them throws a six. Now comes the crucial observation: At this moment, we are in exactly the same situation as in the beginning. Hence in this case we will win with probability $p$. We just derived the following linear equation: \\[p = \\frac 1{36} + \\frac{10}{36} \\cdot \\frac 16 + \\frac{25}{36} \\cdot p\\] Solving for $p$, we get $p=\\frac 8{33}$, hence the answer is $8+33 = 041",
"Let's write out the probabilities with a set number of throws that Linda rolls before getting a 6. The probability of Linda rolling once and gets 6 right away is $\\frac{1}{6}$. The probability that Dave will get a six in the same, one less, or one more throw is $\\frac{1}{6} + \\frac{5}{6} * \\frac{5}{6}$. Thus the combined probability is $\\frac{11}{216}$. Let's do the same with the probability that Linda rolls twice and getting a six. This time it is $\\frac{5}{6} * \\frac{1}{6}$. The probability that Dave meets the requirements set is $\\frac{1}{6} + \\frac{5}{6} * \\frac{1}{6} + \\frac{5}{6} * \\frac{5}{6} * \\frac{1}{6}$. Combine the probabilities again to get $\\frac{455}{7776}$. (or not, because you can simplify without calculating later) It's clear that as the number of rolls before getting a six increases, the probability that Dave meets the requirements is multiplied by $\\frac{5}{6} * \\frac{5}{6}$. We can use this pattern to solve for the sum of an infinite geometric series. First, set the case where Linda rolls only once aside. It doesn't fit the same pattern as the rest, so we'll add it separately at the end. Next, let $a = (\\frac{5}{6} * \\frac{1}{6}) * (\\frac{1}{6} + \\frac{5}{6} * \\frac{1}{6} + \\frac{5}{6} * \\frac{5}{6} * \\frac{1}{6}) = \\frac{455}{7776}$ as written above. Each probability where the number of tosses Linda makes increases by one will be $a * (\\frac{25}{36})^{n+1}$. Let $S$ be the sum of all these probabilities. $S = a + a * \\frac{25}{36} + a * (\\frac{25}{36})^2...$ $S * \\frac{25}{36} = a * \\frac{25}{36} + a * (\\frac{25}{36})^2 + a * (\\frac{25}{36})^3...$ Subtract the second equation from the first to get $S * \\frac{11}{36} = a$ $S = a * \\frac{36}{11}$ $S = \\frac{455}{2376}$ Don't forget to add the first case where Linda rolls once. $\\frac{455}{2376} + \\frac{11}{216} = \\frac{8}{33}$ $8 + 33 = 41"
] |
2009-II-9
| 2,009
| 9
|
Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$ , and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$ . Find the remainder when $m-n$ is divided by $1000$ .
| 0
|
II
|
[
"Solution 1 It is actually reasonably easy to compute $m$ and $n$ exactly. First, note that if $4x+3y+2z=2009$, then $y$ must be odd. Let $y=2y'-1$. We get $4x + 6y' - 3 + 2z = 2009$, which simplifies to $2x + 3y' + z = 1006$. For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly one $z$ such that the equality holds. Hence we need to count the pairs $(x,y')$. For a fixed $y'$, $x$ can be at most $\\left\\lfloor \\dfrac{1005-3y'}2 \\right\\rfloor$. Hence the number of solutions is \\begin{align*} m & = \\sum_{y'=1}^{334} \\left\\lfloor \\dfrac{1005-3y'}2 \\right\\rfloor \\\\ & = 501 + 499 + 498 + 496 + \\cdots + 6 + 4 + 3 + 1 \\\\ & = 1000 + 994 + \\cdots + 10 + 4 \\\\ & = 83834 \\end{align*} Similarly, we can compute that $n=82834$, hence $m-n = 1000 \\equiv 000.",
"It is actually reasonably easy to compute $m$ and $n$ exactly. First, note that if $4x+3y+2z=2009$, then $y$ must be odd. Let $y=2y'-1$. We get $4x + 6y' - 3 + 2z = 2009$, which simplifies to $2x + 3y' + z = 1006$. For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly one $z$ such that the equality holds. Hence we need to count the pairs $(x,y')$. For a fixed $y'$, $x$ can be at most $\\left\\lfloor \\dfrac{1005-3y'}2 \\right\\rfloor$. Hence the number of solutions is \\begin{align*} m & = \\sum_{y'=1}^{334} \\left\\lfloor \\dfrac{1005-3y'}2 \\right\\rfloor \\\\ & = 501 + 499 + 498 + 496 + \\cdots + 6 + 4 + 3 + 1 \\\\ & = 1000 + 994 + \\cdots + 10 + 4 \\\\ & = 83834 \\end{align*} Similarly, we can compute that $n=82834$, hence $m-n = 1000 \\equiv 000.",
"We can avoid computing $m$ and $n$, instead we will compute $m-n$ directly. Note that $4x+3y+2z=2009$ if and only if $4(x-1)+3(y-1)+2(z-1)=2000$. Hence there is an almost 1-to-1 correspondence between the positive integer solutions of the two equations. The only exceptions are the solutions of the first equation in which at least one of the variables is equal to $1$. The value $m-n$ is the number of such solutions. If $x=1$, we get the equation $3y+2z=2005$. The variable $y$ must be odd, and it must be between $1$ and $667$, inclusive. For each such $y$ there is exactly one valid $z$. Hence in this case there are $334$ valid solutions. If $y=1$, we get the equation $4x+2z=2006$, or equivalently $2x+z=1003$. The variable $x$ must be between $1$ and $501$, inclusive, and for each such $x$ there is exactly one valid $z$. Hence in this case there are $501$ valid solutions. If $z=1$, we get the equation $4x+3y=2007$. The variable $y$ must be odd, thus let $y=2u-1$. We get $4x+6u=2010$, or equivalently, $2x+3u=1005$. Again, we see that $u$ must be odd, thus let $u=2v-1$. We get $2x+6v=1008$, which simplifies to $x+3v=504$. Now, we see that $v$ must be between $1$ and $167$, inclusive, and for each such $v$ we have exactly one valid $x$. Hence in this case there are $167$ valid solutions. Finally, we must note that there are two special solutions: one with $x=y=1$, and one with $y=z=1$. We counted each of them twice, hence we have to subtract two from the total. Therefore $m-n = 334 + 501 + 167 - 2 = 1000$, and the answer is $1000\\bmod 1000 = 000.",
"In this solution we will perform a similar operation as in Solution 2, but only on $y$: $4x+3y+2z=2009$ if and only if $4x+3(y-3)+2z=2000$. There is a one-to-one correspondence between the solutions of these two equations. Let $y'=y-3$ and require $y'$ to be positive as well. Then the second equation becomes $4x+3y'+2z=2000$. Notice that there are several \"extra\" solutions in the first equation that cannot be included in the second equation (since that would make $y'$ non-positive). The value $m-n$ is therefore the number of \"extra\" solutions. Since $y'=y-3$, in order for $y'$ to be non-positive $1 \\leq y \\leq 3$. However, equation (1) requires y to be odd, so we have two cases to consider: $y=1$ and $y=3$. This results in the two equations $4x+3+2z=2009$ and $4x+9+2z=2009$. $4x+3+2z=2009$ simplifies to $2x+z=1003$. There is exactly one valid $z$ for each $x$; $x$ must be between $1$ and $501$ (inclusive) to obtain positive integer solutions. Therefore, there are $501$ solutions in this case. $4x+9+2z=2009$ simplifies to $2x+z=1000$. There is exactly one valid $z$ for each $x$; $x$ must be between $1$ and $499$ (inclusive) to obtain positive integer solutions. Therefore, there are $499$ solutions in this case. Thus, $m-n = 501 + 499 = 1000; 1000 \\equiv 000."
] |
2009-II-10
| 2,009
| 10
|
Four lighthouses are located at points $A$ , $B$ , $C$ , and $D$ . The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$ , the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$ , and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$ . To an observer at $A$ , the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$ , the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac{p\sqrt{r}}{q}$ , where $p$ , $q$ , and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p+q+r$ .
| 96
|
II
|
[
"Let $O$ be the intersection of $BC$ and $AD$. By the Angle Bisector Theorem, $\\frac {5}{BO}$ = $\\frac {13}{CO}$, so $BO$ = $5x$ and $CO$ = $13x$, and $BO$ + $OC$ = $BC$ = $12$, so $x$ = $\\frac {2}{3}$, and $OC$ = $\\frac {26}{3}$. Let $P$ be the foot of the altitude from $D$ to $OC$. It can be seen that triangle $DOP$ is similar to triangle $AOB$, and triangle $DPC$ is similar to triangle $ABC$. If $DP$ = $15y$, then $CP$ = $36y$, $OP$ = $10y$, and $OD$ = $5y\\sqrt {13}$. Since $OP$ + $CP$ = $46y$ = $\\frac {26}{3}$, $y$ = $\\frac {13}{69}$, and $AD$ = $\\frac {60\\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$). The answer is $60$ + $13$ + $23$ = $096.",
"Extend $AB$ and $CD$ to intersect at $P$. Note that since $\\angle ACB=\\angle PCB$ and $\\angle ABC=\\angle PBC=90^{\\circ}$ by ASA congruency we have $\\triangle ABC\\cong \\triangle PBC$. Therefore $AC=PC=13$. By the angle bisector theorem, $PD=\\dfrac{130}{23}$ and $CD=\\dfrac{169}{23}$. Now we apply Stewart's theorem to find $AD$: \\begin{align*}13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=13\\cdot 13\\cdot \\dfrac{130}{23}+10\\cdot 10\\cdot \\dfrac{169}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=\\dfrac{169\\cdot 130+169\\cdot 100}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=1690\\\\ AD^2&=130-\\dfrac{130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130\\cdot 23^2-130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130(23^2-169)}{23^2}\\\\ AD^2&=\\dfrac{130(360)}{23^2}\\\\ AD&=\\dfrac{60\\sqrt{13}}{23}\\\\ \\end{align*} and our final answer is $60+13+23=096.",
"Notice that by extending $AB$ and $CD$ to meet at a point $E$, $\\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$, $B=(12,0)$, $E=(12,-5)$ and $A=(12,5)$, and the equation of line $CD$ is $y=-\\dfrac{5}{12}x$. Let F be the intersection point of $AD$ and $BC$, and by using the Angle Bisector Theorem: $\\dfrac{BF}{AB}=\\dfrac{FC}{AC}$ we have $FC=\\dfrac{26}{3}$. Then the equation of the line $AF$ through the points $(12,5)$ and $\\left(\\frac{26}{3},0\\right)$ is $y=\\frac32 x-13$. Hence the intersection point of $AF$ and $CD$ is the point $D$ at the coordinates $\\left(\\dfrac{156}{23},-\\dfrac{65}{23}\\right)$. Using the distance formula, $AD=\\sqrt{\\left(12-\\dfrac{156}{23}\\right)^2+\\left(5+\\dfrac{65}{23}\\right)^2}=\\dfrac{60\\sqrt{13}}{23}$ for an answer of $60+13+23=096.",
"After drawing a good diagram, we reflect $ABC$ over the line $BC$, forming a new point that we'll call $A'$. Also, let the intersection of $AD$ and $BC$ be point $E$. Point $D$ lies on line $A'C$. Since line $AD$ bisects $\\angle{CAB}$, we can use the Angle Bisector Theorem. $AA'=10$ and $AC=13$, so $\\frac{CD}{A'D}=\\frac{13}{10}$. Letting the segments be $13x$ and $10x$ respectively, we now have $13x+10x=13$. Therefore, $x=\\frac{13}{23}$. By the Pythagorean Theorem, $AE=\\frac{5\\sqrt{13}}{3}$. Using the Angle Bisector Theorem on $\\angle{ACD}$, we have that $ED=\\frac{5x\\sqrt{13}}{3}$. Substituting in $x=\\frac{13}{23}$, we have that $AD=\\left(\\frac{5\\sqrt{13}}{3}\\right)(1+x)=\\frac{60\\sqrt{13}}{23}$, so the answer is $60+13+23=096. -RootThreeOverTwo",
"Let $CD$ and $AB$ meet at $A'$; then $\\overline{AD}$ is an angle bisector of isosceles $\\triangle AA'C$. Then by the Angle Bisector Theorem, $A'D=\\frac {10}{10+13} \\cdot 13=\\frac {130}{23}$, and $\\cos \\angle DA'A=\\frac {5}{13}$. By the Law of Cosines on $\\triangle AA'D$, we have \\[AD^2=10^2+\\left(\\frac {130}{23}\\right)^2-2 \\cdot 10 \\cdot \\frac {130}{23} \\cdot \\frac {5}{13}=\\frac {2^4 \\cdot 3^2 \\cdot 5^2 \\cdot 13}{23^2} \\Longrightarrow AD=\\frac {60\\sqrt {13}}{23}\\] and the answer is $60+13+23=096. [asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype(\"4 4\")); MP(\"5\", (A+B)/2, f); MP(\"13\", (A+C)/2, NE,f); MP(\"A\",D(A),f); MP(\"B\",D(B),f); MP(\"C\",D(C),N,f); MP(\"A'\",D(E),f); MP(\"D\",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy] -azjps",
"Let $CD$ and $AB$ meet at $A'$; then $\\overline{AD}$ is an angle bisector of isosceles $\\triangle AA'C$. Then by the Angle Bisector Theorem, $A'D=\\frac {10}{10+13} \\cdot 13=\\frac {130}{23}$, and $\\cos \\angle DA'A=\\frac {5}{13}$. By the Law of Cosines on $\\triangle AA'D$, we have \\[AD^2=10^2+\\left(\\frac {130}{23}\\right)^2-2 \\cdot 10 \\cdot \\frac {130}{23} \\cdot \\frac {5}{13}=\\frac {2^4 \\cdot 3^2 \\cdot 5^2 \\cdot 13}{23^2} \\Longrightarrow AD=\\frac {60\\sqrt {13}}{23}\\] and the answer is $60+13+23=096. [asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype(\"4 4\")); MP(\"5\", (A+B)/2, f); MP(\"13\", (A+C)/2, NE,f); MP(\"A\",D(A),f); MP(\"B\",D(B),f); MP(\"C\",D(C),N,f); MP(\"A'\",D(E),f); MP(\"D\",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy] -azjps",
"[asy]pathpen = linewidth(0.7); pen f = fontsize(10); size(144); pair B=(0,0); pair A=(5,0); pair C=(0,12); pair E=(-5,0); pair abA = A+unit(C-A)+unit(B-A); pair D = extension(A,abA,C,E); D(A--C); D(C--B); D(A--B); D(C--D); D(A--D); D(D--E,dashed); D(B--E,dashed); D(rightanglemark(C,B,A,20),red); MP(\"A\",D(A),plain.SE,f); MP(\"B\",D(B),plain.S,f); MP(\"C\",D(C),plain.N,f); MP(\"D\",D(D),plain.NW,f); MP(\"E\",D(E),plain.SW,f); MP(\"5\",(A+B)/2,plain.S,f); MP(\"12\",(B+C)/2,plain.E,f); MP(\"13\",(A+C)/2,plain.NE,f); MP(\"\\alpha\",A,4*(unit(D-A)+unit(C-A))/2,f); MP(\"\\alpha\",A,4*(unit(B-A)+unit(D-A))/2,f); MP(\"2\\alpha\",E,4*(unit(C-E)+unit(B-E))/2,f); MP(\"\\beta\",C,6*(unit(A-C)+unit(B-C))/2,f); MP(\"\\beta\",C,6*(unit(E-C)+unit(B-C))/2,f);[/asy] Using the law of sines on $\\bigtriangleup ADE$, \\begin{align*} AD &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(180^\\circ-3\\alpha)}\\\\ &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha+2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha)\\cos(2\\alpha)+\\cos(\\alpha)\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sqrt{\\dfrac{1-\\cos(2\\alpha)}{2}}\\cdot\\cos(2\\alpha)+\\sqrt{\\dfrac{1+\\cos(2\\alpha)}{2}}\\cdot\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\dfrac{12}{13}}{\\sqrt{\\dfrac{8}{26}}\\cdot\\dfrac{5}{13}+\\sqrt{\\dfrac{18}{26}}\\cdot\\dfrac{12}{13}}\\\\ &= 10 \\cdot \\dfrac{12\\sqrt{13}}{10+36} = \\dfrac{60\\sqrt{13}}{23} \\end{align*} $\\therefore$ the answer is $60+13+23=096. -m1sterzer0",
"[asy]pathpen = linewidth(0.7); pen f = fontsize(10); size(144); pair B=(0,0); pair A=(5,0); pair C=(0,12); pair E=(-5,0); pair abA = A+unit(C-A)+unit(B-A); pair D = extension(A,abA,C,E); D(A--C); D(C--B); D(A--B); D(C--D); D(A--D); D(D--E,dashed); D(B--E,dashed); D(rightanglemark(C,B,A,20),red); MP(\"A\",D(A),plain.SE,f); MP(\"B\",D(B),plain.S,f); MP(\"C\",D(C),plain.N,f); MP(\"D\",D(D),plain.NW,f); MP(\"E\",D(E),plain.SW,f); MP(\"5\",(A+B)/2,plain.S,f); MP(\"12\",(B+C)/2,plain.E,f); MP(\"13\",(A+C)/2,plain.NE,f); MP(\"\\alpha\",A,4*(unit(D-A)+unit(C-A))/2,f); MP(\"\\alpha\",A,4*(unit(B-A)+unit(D-A))/2,f); MP(\"2\\alpha\",E,4*(unit(C-E)+unit(B-E))/2,f); MP(\"\\beta\",C,6*(unit(A-C)+unit(B-C))/2,f); MP(\"\\beta\",C,6*(unit(E-C)+unit(B-C))/2,f);[/asy] Using the law of sines on $\\bigtriangleup ADE$, \\begin{align*} AD &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(180^\\circ-3\\alpha)}\\\\ &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha+2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha)\\cos(2\\alpha)+\\cos(\\alpha)\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sqrt{\\dfrac{1-\\cos(2\\alpha)}{2}}\\cdot\\cos(2\\alpha)+\\sqrt{\\dfrac{1+\\cos(2\\alpha)}{2}}\\cdot\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\dfrac{12}{13}}{\\sqrt{\\dfrac{8}{26}}\\cdot\\dfrac{5}{13}+\\sqrt{\\dfrac{18}{26}}\\cdot\\dfrac{12}{13}}\\\\ &= 10 \\cdot \\dfrac{12\\sqrt{13}}{10+36} = \\dfrac{60\\sqrt{13}}{23} \\end{align*} $\\therefore$ the answer is $60+13+23=096. -m1sterzer0"
] |
2009-II-11
| 2,009
| 11
|
For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$ . Find the sum of all possible values of the product $m \cdot n$ .
| 125
|
II
|
[
"We have $\\log m - \\log k = \\log \\left( \\frac mk \\right)$, hence we can rewrite the inequality as follows: \\[- \\log n < \\log \\left( \\frac mk \\right) < \\log n\\] We can now get rid of the logarithms, obtaining: \\[\\frac 1n < \\frac mk < n\\] And this can be rewritten in terms of $k$ as \\[\\frac mn < k < mn\\] From $k<mn$ it follows that the $50$ solutions for $k$ must be the integers $mn-1, mn-2, \\dots, mn-50$. This will happen if and only if the lower bound on $k$ is in a suitable range -- we must have $mn-51 \\leq \\frac mn < mn-50$. Obviously there is no solution for $n=1$. For $n>1$ the left inequality can be rewritten as $m\\leq\\dfrac{51n}{n^2-1}$, and the right one as $m > \\dfrac{50n}{n^2-1}$. Remember that we must have $m\\geq n$. However, for $n\\geq 8$ we have $\\dfrac{51n}{n^2-1} < n$, and hence $m<n$, which is a contradiction. This only leaves us with the cases $n\\in\\{2,3,4,5,6,7\\}$. For $n=2$ we have $\\dfrac{100}3 < m \\leq \\dfrac{102}3$ with a single integer solution $m=\\dfrac{102}3=34$. For $n=3$ we have $\\dfrac{150}8 < m \\leq \\dfrac{153}8$ with a single integer solution $m=\\dfrac{152}8=19$. For $n=4,5,6,7$ our inequality has no integer solutions for $m$. Therefore the answer is $34\\cdot 2 + 19\\cdot 3 = 68 + 57 = 125.",
"Realize that the question has the number \"50\" in it, where a \"5\" can be taken out. Then look at the year, which is 2009; subtract 2000 from 2009 to get 9, whose square root is 3. Then you get 5 to the power of 3, which is $125.",
"Realize that the question has the number \"50\" in it, where a \"5\" can be taken out. Then look at the year, which is 2009; subtract 2000 from 2009 to get 9, whose square root is 3. Then you get 5 to the power of 3, which is $125."
] |
2009-II-12
| 2,009
| 12
|
From the set of integers $\{1,2,3,\dots,2009\}$ , choose $k$ pairs $\{a_i,b_i\}$ with $a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$ . Find the maximum possible value of $k$ .
| 803
|
II
|
[
"Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\\cdots+2k=k(2k+1)$. On the other hand, as the sum of each pair is distinct and at most equal to $2009$, the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \\cdots + (2009-(k-1)) = \\frac{k(4019-k)}{2}$. Hence we get a necessary condition on $k$: For a solution to exist, we must have $\\frac{k(4019-k)}{2} \\geq k(2k+1)$. As $k$ is positive, this simplifies to $\\frac{4019-k}{2} \\geq 2k+1$, whence $5k\\leq 4017$, and as $k$ is an integer, we have $k\\leq \\lfloor 4017/5\\rfloor = 803$. If we now find a solution with $k=803$, we can be sure that it is optimal. From the proof it is clear that we don't have much \"maneuvering space\", if we want to construct a solution with $k=803$. We can try to use the $2k$ smallest numbers: $1$ to $2\\cdot 803 = 1606$. When using these numbers, the average sum will be $1607$. Hence we can try looking for a nice systematic solution that achieves all sums between $1607-401=1206$ and $1607+401=2008$, inclusive. Such a solution indeed does exist, here is one: Partition the numbers $1$ to $1606$ into four sequences: $A=1,3,5,7,\\dots,801,803$ $B=1205,1204,1203,1202,\\dots,805,804$ $C=2,4,6,8,\\dots,800,802$ $D=1606,1605,1604,1603,\\dots,1207,1206$ Sequences $A$ and $B$ have $402$ elements each, and the sums of their corresponding elements are $1206,1207,1208,1209,\\dots,1606,1607$. Sequences $C$ and $D$ have $401$ elements each, and the sums of their corresponding elements are $1608,1609,1610,1611,\\dots,2007,2008$. Thus we have shown that there is a solution for $k=803 no solution exists."
] |
2009-II-13
| 2,009
| 13
|
Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$ . The arc is divided into seven congruent arcs by six equally spaced points $C_1,C_2,\dots,C_6$ . All chords of the form $\overline{AC_i}$ or $\overline{BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the remainder when $n$ is divided by $1000$ .
| 672
|
II
|
[
"Solution 1 Let the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then $C_1,\\ldots, C_6$ are 6 of the 14th roots of unity. Let $\\omega=\\text{cis}\\frac{360^{\\circ}}{14}$; then $C_1,\\ldots, C_6$ correspond to $\\omega,\\ldots, \\omega^6$. Let $C_1',\\ldots, C_6'$ be their reflections across the diameter. These points correspond to $\\omega^8\\ldots, \\omega^{13}$. Then the lengths of the segments are $|1-\\omega|,\\ldots, |1-\\omega^6|,|1-\\omega^8|,\\ldots |1-\\omega^{13}|$. Noting that $B$ represents 1 in the complex plane, the desired product is \\begin{align*} BC_1\\cdots BC_6 \\cdot AC_1\\cdots AC_6&= BC_1\\cdots BC_6 \\cdot BC_1'\\cdots BC_6'\\\\ &= |(x-\\omega^1)\\ldots(x-\\omega^6)(x-\\omega^8)\\ldots(x-\\omega^{13})| \\end{align*} for $x=1$. However, the polynomial $(x-\\omega^1)\\ldots(x-\\omega^6)(x-\\omega^8)\\ldots(x-\\omega^{13})$ has as its zeros all 14th roots of unity except for $-1$ and $1$. Hence \\[(x-\\omega^1)\\ldots(x-\\omega^6)(x-\\omega^8)\\ldots(x-\\omega^{13})=\\frac{x^{14}-1}{(x-1)(x+1)}=x^{12}+x^{10}+\\cdots +x^2+1.\\] Thus the product is $|x^{12}+\\cdots +x^2+1|=7$ when the radius is 1, and the product is $2^{12}\\cdot 7=28672$. Thus the answer is $672 ~Solution by sml1809",
"Let the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then $C_1,\\ldots, C_6$ are 6 of the 14th roots of unity. Let $\\omega=\\text{cis}\\frac{360^{\\circ}}{14}$; then $C_1,\\ldots, C_6$ correspond to $\\omega,\\ldots, \\omega^6$. Let $C_1',\\ldots, C_6'$ be their reflections across the diameter. These points correspond to $\\omega^8\\ldots, \\omega^{13}$. Then the lengths of the segments are $|1-\\omega|,\\ldots, |1-\\omega^6|,|1-\\omega^8|,\\ldots |1-\\omega^{13}|$. Noting that $B$ represents 1 in the complex plane, the desired product is \\begin{align*} BC_1\\cdots BC_6 \\cdot AC_1\\cdots AC_6&= BC_1\\cdots BC_6 \\cdot BC_1'\\cdots BC_6'\\\\ &= |(x-\\omega^1)\\ldots(x-\\omega^6)(x-\\omega^8)\\ldots(x-\\omega^{13})| \\end{align*} for $x=1$. However, the polynomial $(x-\\omega^1)\\ldots(x-\\omega^6)(x-\\omega^8)\\ldots(x-\\omega^{13})$ has as its zeros all 14th roots of unity except for $-1$ and $1$. Hence \\[(x-\\omega^1)\\ldots(x-\\omega^6)(x-\\omega^8)\\ldots(x-\\omega^{13})=\\frac{x^{14}-1}{(x-1)(x+1)}=x^{12}+x^{10}+\\cdots +x^2+1.\\] Thus the product is $|x^{12}+\\cdots +x^2+1|=7$ when the radius is 1, and the product is $2^{12}\\cdot 7=28672$. Thus the answer is $672 ~Solution by sml1809",
"Let $O$ be the midpoint of $A$ and $B$. Assume $C_1$ is closer to $A$ instead of $B$. $\\angle AOC_1$ = $\\frac {\\pi}{7}$. Using the Law of Cosines, $\\overline {AC_1}^2$ = $8 - 8 \\cos \\frac {\\pi}{7}$, $\\overline {AC_2}^2$ = $8 - 8 \\cos \\frac {2\\pi}{7}$, . . . $\\overline {AC_6}^2$ = $8 - 8 \\cos \\frac {6\\pi}{7}$ So $n$ = $(8^6)(1 - \\cos \\frac {\\pi}{7})(1 - \\cos \\frac {2\\pi}{7})\\dots(1 - \\cos \\frac{6\\pi}{7})$. It can be rearranged to form $n$ = $(8^6)(1 - \\cos \\frac {\\pi}{7})(1 - \\cos \\frac {6\\pi}{7})\\dots(1 - \\cos \\frac {3\\pi}{7})(1 - \\cos \\frac {4\\pi}{7})$. Since $\\cos a = - \\cos (\\pi - a)$, we have $n$ = $(8^6)(1 - \\cos \\frac {\\pi}{7})(1 + \\cos \\frac {\\pi}{7}) \\dots (1 - \\cos \\frac {3\\pi}{7})(1 + \\cos \\frac {3\\pi}{7})$ = $(8^6)(1 - \\cos^2 \\frac {\\pi}{7})(1 - \\cos^2 \\frac {2\\pi}{7})(1 - \\cos^2 \\frac {3\\pi}{7})$ = $(8^6)(\\sin^2 \\frac {\\pi}{7})(\\sin^2 \\frac {2\\pi}{7})(\\sin^2 \\frac {3\\pi}{7})$ It can be shown that $\\sin \\frac {\\pi}{7} \\sin \\frac {2\\pi}{7} \\sin \\frac {3\\pi}{7}$ = $\\frac {\\sqrt {7}}{8}$, so $n$ = $8^6(\\frac {\\sqrt {7}}{8})^2$ = $7(8^4)$ = $28672$, so the answer is $672 ~Solution by sml1809",
"Note that for each $k$ the triangle $ABC_k$ is a right triangle. Hence the product $AC_k \\cdot BC_k$ is twice the area of the triangle $ABC_k$. Knowing that $AB=4$, the area of $ABC_k$ can also be expressed as $2c_k$, where $c_k$ is the length of the altitude from $C_k$ onto $AB$. Hence we have $AC_k \\cdot BC_k = 4c_k$. By the definition of $C_k$ we obviously have $c_k = 2\\sin\\frac{k\\pi}7$. From these two observations we get that the product we should compute is equal to $8^6 \\cdot \\prod_{k=1}^6 \\sin \\frac{k\\pi}7$, which is the same identity as in Solution 2. Computing the product of sines In this section we show one way how to evaluate the product $\\prod_{k=1}^6 \\sin \\frac{k\\pi}7 = \\prod_{k=1}^3 (\\sin \\frac{k\\pi}7)^2$. Let $\\omega_k = \\cos \\frac{2k\\pi}7 + i\\sin \\frac{2k\\pi}7$. The numbers $1,\\omega_1,\\omega_2,\\dots,\\omega_6$ are the $7$-th complex roots of unity. In other words, these are the roots of the polynomial $x^7-1$. Then the numbers $\\omega_1,\\omega_2,\\dots,\\omega_6$ are the roots of the polynomial $\\frac{x^7-1}{x-1} = x^6+x^5+\\cdots+x+1$. We just proved the identity $\\prod_{k=1}^6 (x - \\omega_k) = x^6+x^5+\\cdots+x+1$. Substitute $x=1$. The right hand side is obviously equal to $7$. Let's now examine the left hand side. We have: \\begin{align*} (1-\\omega_k)(1-\\omega_{7-k})=|1-\\omega_k|^2 & = \\left( 1-\\cos \\frac{2k\\pi}7 \\right)^2 + \\left( \\sin \\frac{2k\\pi}7 \\right)^2 \\\\ & = 2-2\\cos \\frac{2k\\pi}7 \\\\ & = 2-2 \\left( 1 - 2 \\left( \\sin \\frac{k\\pi}7 \\right)^2 \\right) \\\\ & = 4\\left( \\sin \\frac{k\\pi}7 \\right)^2 \\end{align*} Therefore the size of the left hand side in our equation is $\\prod_{k=1}^3 4 (\\sin \\frac{k\\pi}7)^2 = 2^6 \\prod_{k=1}^3 (\\sin \\frac{k\\pi}7)^2$. As the right hand side is $7$, we get that $\\prod_{k=1}^3 (\\sin \\frac{k\\pi}7)^2 = \\frac{7}{2^6}$. Solution 4 (Product of Sines) Lemma 1: A chord $ab$ of a circle with center $O$ and radius $r$ has length $2r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$. Proof: Denote $H$ as the projection from $O$ to line $AB$. Then, by definition, $HA=HB=r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$. Thus, $AB = 2r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$, which concludes the proof. Lemma 2: $\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n} = \\dfrac{n}{2^{n-1}}$ Proof: Let $w=\\text{cis}\\;\\dfrac{\\pi}{n}$. Thus, \\[\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n} = \\prod_{k=1}^{n-1} \\dfrac{w^k-w^{-k}}{2i} = \\dfrac{w^{\\frac{n(n-1)}{2}}}{(2i)^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k}) = \\dfrac{1}{2^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k})\\] Since, $w^{-2k}$ are just the $n$th roots of unity excluding $1$, by Vieta's, $\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n}=\\dfrac{1}{2^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k}) = \\dfrac{n}{2^{n-1}}$, thus completing the proof. By Lemma 1, the length $AC_k=2r\\sin\\dfrac{k\\pi}{14}$ and similar lengths apply for $BC_k$. Now, the problem asks for $\\left(\\prod_{k=1}^6 \\left(4\\sin\\dfrac{k\\pi}{14}\\right)\\right)^2$. This can be rewritten, due to $\\sin \\theta = \\sin (\\pi-\\theta)$, as $\\prod_{k=1}^6 \\left(4\\sin\\dfrac{k\\pi}{14}\\right) \\cdot \\prod_{k=8}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right) = \\dfrac{1}{\\sin \\dfrac{7\\pi}{14}}\\cdot \\prod_{k=1}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right) = \\prod_{k=1}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right).$ By Lemma 2, this furtherly boils down to $4^{12}\\cdot \\dfrac{14}{2^{13}} = 7\\cdot 2^{12} = 672 ~Solution by sml1809",
"Lemma 1: A chord $ab$ of a circle with center $O$ and radius $r$ has length $2r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$. Proof: Denote $H$ as the projection from $O$ to line $AB$. Then, by definition, $HA=HB=r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$. Thus, $AB = 2r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$, which concludes the proof. Lemma 2: $\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n} = \\dfrac{n}{2^{n-1}}$ Proof: Let $w=\\text{cis}\\;\\dfrac{\\pi}{n}$. Thus, \\[\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n} = \\prod_{k=1}^{n-1} \\dfrac{w^k-w^{-k}}{2i} = \\dfrac{w^{\\frac{n(n-1)}{2}}}{(2i)^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k}) = \\dfrac{1}{2^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k})\\] Since, $w^{-2k}$ are just the $n$th roots of unity excluding $1$, by Vieta's, $\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n}=\\dfrac{1}{2^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k}) = \\dfrac{n}{2^{n-1}}$, thus completing the proof. By Lemma 1, the length $AC_k=2r\\sin\\dfrac{k\\pi}{14}$ and similar lengths apply for $BC_k$. Now, the problem asks for $\\left(\\prod_{k=1}^6 \\left(4\\sin\\dfrac{k\\pi}{14}\\right)\\right)^2$. This can be rewritten, due to $\\sin \\theta = \\sin (\\pi-\\theta)$, as $\\prod_{k=1}^6 \\left(4\\sin\\dfrac{k\\pi}{14}\\right) \\cdot \\prod_{k=8}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right) = \\dfrac{1}{\\sin \\dfrac{7\\pi}{14}}\\cdot \\prod_{k=1}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right) = \\prod_{k=1}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right).$ By Lemma 2, this furtherly boils down to $4^{12}\\cdot \\dfrac{14}{2^{13}} = 7\\cdot 2^{12} = 672 ~Solution by sml1809",
"Lemma 1: A chord $ab$ of a circle with center $O$ and radius $r$ has length $2r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$. Proof: Denote $H$ as the projection from $O$ to line $AB$. Then, by definition, $HA=HB=r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$. Thus, $AB = 2r\\sin\\left(\\dfrac{\\angle AOB}{2}\\right)$, which concludes the proof. Lemma 2: $\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n} = \\dfrac{n}{2^{n-1}}$ Proof: Let $w=\\text{cis}\\;\\dfrac{\\pi}{n}$. Thus, \\[\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n} = \\prod_{k=1}^{n-1} \\dfrac{w^k-w^{-k}}{2i} = \\dfrac{w^{\\frac{n(n-1)}{2}}}{(2i)^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k}) = \\dfrac{1}{2^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k})\\] Since, $w^{-2k}$ are just the $n$th roots of unity excluding $1$, by Vieta's, $\\prod_{k=1}^{n-1} \\sin \\dfrac{k\\pi}{n}=\\dfrac{1}{2^{n-1}}\\prod_{k=1}^{n-1} (1-w^{-2k}) = \\dfrac{n}{2^{n-1}}$, thus completing the proof. By Lemma 1, the length $AC_k=2r\\sin\\dfrac{k\\pi}{14}$ and similar lengths apply for $BC_k$. Now, the problem asks for $\\left(\\prod_{k=1}^6 \\left(4\\sin\\dfrac{k\\pi}{14}\\right)\\right)^2$. This can be rewritten, due to $\\sin \\theta = \\sin (\\pi-\\theta)$, as $\\prod_{k=1}^6 \\left(4\\sin\\dfrac{k\\pi}{14}\\right) \\cdot \\prod_{k=8}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right) = \\dfrac{1}{\\sin \\dfrac{7\\pi}{14}}\\cdot \\prod_{k=1}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right) = \\prod_{k=1}^{13} \\left(4\\sin\\dfrac{k\\pi}{14}\\right).$ By Lemma 2, this furtherly boils down to $4^{12}\\cdot \\dfrac{14}{2^{13}} = 7\\cdot 2^{12} = 672 ~Solution by sml1809"
] |
2009-II-14
| 2,009
| 14
|
The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}$ for $n \geq 0$ . Find the greatest integer less than or equal to $a_{10}$ .
| 983
|
II
|
[
"The \"obvious\" substitution An obvious way how to get the $4^n$ from under the square root is to use the substitution $a_n = 2^n b_n$. Then the square root simplifies as follows: $\\sqrt{4^n - a_n^2} = \\sqrt{4^n - (2^n b_n)^2} = \\sqrt{4^n - 4^n b_n^2} = 2^n \\sqrt{1 - b_n^2}$. The new recurrence then becomes $b_0=0$ and $b_{n+1} = \\frac45 b_n + \\frac 35\\sqrt{1 - b_n^2}$. Solution 1 We can now simply start to compute the values $b_i$ by hand: \\begin{align*} b_1 & = \\frac 35 \\\\ b_2 & = \\frac 45\\cdot \\frac 35 + \\frac 35 \\sqrt{1 - \\left(\\frac 35\\right)^2} = \\frac{24}{25} \\\\ b_3 & = \\frac 45\\cdot \\frac {24}{25} + \\frac 35 \\sqrt{1 - \\left(\\frac {24}{25}\\right)^2} = \\frac{96}{125} + \\frac 35\\cdot\\frac 7{25} = \\frac{117}{125} \\\\ b_4 & = \\frac 45\\cdot \\frac {117}{125} + \\frac 35 \\sqrt{1 - \\left(\\frac {117}{125}\\right)^2} = \\frac{468}{625} + \\frac 35\\cdot\\frac {44}{125} = \\frac{600}{625} = \\frac{24}{25} \\end{align*} We now discovered that $b_4=b_2$. And as each $b_{i+1}$ is uniquely determined by $b_i$, the sequence becomes periodic. In other words, we have $b_3=b_5=b_7=\\cdots=\\frac{117}{125}$, and $b_2=b_4=\\cdots=b_{10}=\\cdots=\\frac{24}{25}$. Therefore the answer is \\begin{align*} \\lfloor a_{10} \\rfloor & = \\left\\lfloor 2^{10} b_{10} \\right\\rfloor = \\left\\lfloor \\dfrac{1024\\cdot 24}{25} \\right\\rfloor = \\left\\lfloor \\dfrac{1025\\cdot 24}{25} - \\dfrac{24}{25} \\right\\rfloor \\\\ & = \\left\\lfloor 41\\cdot 24 - \\dfrac{24}{25} \\right\\rfloor = 41\\cdot 24 - 1 = 983.",
"We can now simply start to compute the values $b_i$ by hand: \\begin{align*} b_1 & = \\frac 35 \\\\ b_2 & = \\frac 45\\cdot \\frac 35 + \\frac 35 \\sqrt{1 - \\left(\\frac 35\\right)^2} = \\frac{24}{25} \\\\ b_3 & = \\frac 45\\cdot \\frac {24}{25} + \\frac 35 \\sqrt{1 - \\left(\\frac {24}{25}\\right)^2} = \\frac{96}{125} + \\frac 35\\cdot\\frac 7{25} = \\frac{117}{125} \\\\ b_4 & = \\frac 45\\cdot \\frac {117}{125} + \\frac 35 \\sqrt{1 - \\left(\\frac {117}{125}\\right)^2} = \\frac{468}{625} + \\frac 35\\cdot\\frac {44}{125} = \\frac{600}{625} = \\frac{24}{25} \\end{align*} We now discovered that $b_4=b_2$. And as each $b_{i+1}$ is uniquely determined by $b_i$, the sequence becomes periodic. In other words, we have $b_3=b_5=b_7=\\cdots=\\frac{117}{125}$, and $b_2=b_4=\\cdots=b_{10}=\\cdots=\\frac{24}{25}$. Therefore the answer is \\begin{align*} \\lfloor a_{10} \\rfloor & = \\left\\lfloor 2^{10} b_{10} \\right\\rfloor = \\left\\lfloor \\dfrac{1024\\cdot 24}{25} \\right\\rfloor = \\left\\lfloor \\dfrac{1025\\cdot 24}{25} - \\dfrac{24}{25} \\right\\rfloor \\\\ & = \\left\\lfloor 41\\cdot 24 - \\dfrac{24}{25} \\right\\rfloor = 41\\cdot 24 - 1 = 983.",
"After we do the substitution, we can notice the fact that $\\left( \\frac 35 \\right)^2 + \\left( \\frac 45 \\right)^2 = 1$, which may suggest that the formula may have something to do with the unit circle. Also, the expression $\\sqrt{1-x^2}$ often appears in trigonometry, for example in the relationship between the sine and the cosine. Both observations suggest that the formula may have a neat geometric interpretation. Consider the equation: \\[y = \\frac45 x + \\frac 35\\sqrt{1 - x^2}\\] Note that for $t=\\sin^{-1} \\frac 35$ we have $\\sin t=\\frac 35$ and $\\cos t = \\frac 45$. Now suppose that we have $x=\\sin s$ for some $s$. Then our equation becomes: \\[y=\\cos t \\cdot \\sin s + \\sin t \\cdot |\\cos s|\\] Depending on the sign of $\\cos s$, this is either the angle addition, or the angle subtraction formula for sine. In other words, if $\\cos s \\geq 0$, then $y=\\sin(s+t)$, otherwise $y=\\sin(s-t)$. We have $b_0=0=\\sin 0$. Therefore $b_1 = \\sin(0+t) = \\sin t$, $b_2 = \\sin(t+t) = \\sin (2t)$, and so on. (Remember that $t$ is the constant defined as $t=\\sin^{-1} \\frac 35$.) This process stops at the first $b_k = \\sin (kt)$, where $kt$ exceeds $\\frac{\\pi}2$. Then we'll have $b_{k+1} = \\sin(kt - t) = \\sin ((k-1)t) = b_{k-1}$ and the sequence will start to oscillate. Note that $\\sin \\frac{\\pi}6 = \\frac 12 < \\frac 35$, and $\\sin \\frac{\\pi}4 = \\frac{\\sqrt 2}2 > \\frac 35$, hence $t$ is strictly between $\\frac{\\pi}6$ and $\\frac{\\pi}4$. Then $2t\\in\\left(\\frac{\\pi}3,\\frac{\\pi}2 \\right)$, and $3t\\in\\left( \\frac{\\pi}2, \\frac{3\\pi}4 \\right)$. Therefore surely $2t < \\frac{\\pi}2 < 3t$. Hence the process stops with $b_3 = \\sin (3t)$, we then have $b_4 = \\sin (2t) = b_2$. As in the previous solution, we conclude that $b_{10}=b_2$, and that the answer is $\\lfloor a_{10} \\rfloor = \\left\\lfloor 2^{10} b_{10} \\right\\rfloor = 983."
] |
2009-II-15
| 2,009
| 15
|
Let $\overline{MN}$ be a diameter of a circle with diameter $1$ . Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\dfrac 35$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with the chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt t$ , where $r$ , $s$ , and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$ .
| 14
|
II
|
[
"Solution 1 (Quick Calculus) Let $V = \\overline{NM} \\cap \\overline{AC}$ and $W = \\overline{NM} \\cap \\overline{BC}$. Further more let $\\angle NMC = \\alpha$ and $\\angle MNC = 90^\\circ - \\alpha$. Angle chasing reveals $\\angle NBC = \\angle NAC = \\alpha$ and $\\angle MBC = \\angle MAC = 90^\\circ - \\alpha$. Additionally $NB = \\frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem. By the Angle Bisector Formula, \\[\\frac{NV}{MV} = \\frac{\\sin (\\alpha)}{\\sin (90^\\circ - \\alpha)} = \\tan (\\alpha)\\] \\[\\frac{MW}{NW} = \\frac{3\\sin (90^\\circ - \\alpha)}{4\\sin (\\alpha)} = \\frac{3}{4} \\cot (\\alpha)\\] As $NV + MV =MW + NW = 1$ we compute $NW = \\frac{1}{1+\\frac{3}{4}\\cot(\\alpha)}$ and $MV = \\frac{1}{1+\\tan (\\alpha)}$, and finally $VW = NW + MV - 1 = \\frac{1}{1+\\frac{3}{4}\\cot(\\alpha)} + \\frac{1}{1+\\tan (\\alpha)} - 1$. Taking the derivative of $VW$ with respect to $\\alpha$, we arrive at \\[VW' = \\frac{7\\cos^2 (\\alpha) - 4}{(\\sin(\\alpha) + \\cos(\\alpha))^2(4\\sin(\\alpha)+3\\cos(\\alpha))^2}\\] Clearly the maximum occurs when $\\alpha = \\cos^{-1}\\left(\\frac{2}{\\sqrt{7}}\\right)$. Plugging this back in, using the fact that $\\tan(\\cos^{-1}(x)) = \\frac{\\sqrt{1-x^2}}{x}$ and $\\cot(\\cos^{-1}(x)) = \\frac{x}{\\sqrt{1-x^2}}$, we get $VW = 7 - 4\\sqrt{3}$ with $7 + 4 + 3 = 014 is the answer. ~Generic_Username",
"Let $V = \\overline{NM} \\cap \\overline{AC}$ and $W = \\overline{NM} \\cap \\overline{BC}$. Further more let $\\angle NMC = \\alpha$ and $\\angle MNC = 90^\\circ - \\alpha$. Angle chasing reveals $\\angle NBC = \\angle NAC = \\alpha$ and $\\angle MBC = \\angle MAC = 90^\\circ - \\alpha$. Additionally $NB = \\frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem. By the Angle Bisector Formula, \\[\\frac{NV}{MV} = \\frac{\\sin (\\alpha)}{\\sin (90^\\circ - \\alpha)} = \\tan (\\alpha)\\] \\[\\frac{MW}{NW} = \\frac{3\\sin (90^\\circ - \\alpha)}{4\\sin (\\alpha)} = \\frac{3}{4} \\cot (\\alpha)\\] As $NV + MV =MW + NW = 1$ we compute $NW = \\frac{1}{1+\\frac{3}{4}\\cot(\\alpha)}$ and $MV = \\frac{1}{1+\\tan (\\alpha)}$, and finally $VW = NW + MV - 1 = \\frac{1}{1+\\frac{3}{4}\\cot(\\alpha)} + \\frac{1}{1+\\tan (\\alpha)} - 1$. Taking the derivative of $VW$ with respect to $\\alpha$, we arrive at \\[VW' = \\frac{7\\cos^2 (\\alpha) - 4}{(\\sin(\\alpha) + \\cos(\\alpha))^2(4\\sin(\\alpha)+3\\cos(\\alpha))^2}\\] Clearly the maximum occurs when $\\alpha = \\cos^{-1}\\left(\\frac{2}{\\sqrt{7}}\\right)$. Plugging this back in, using the fact that $\\tan(\\cos^{-1}(x)) = \\frac{\\sqrt{1-x^2}}{x}$ and $\\cot(\\cos^{-1}(x)) = \\frac{x}{\\sqrt{1-x^2}}$, we get $VW = 7 - 4\\sqrt{3}$ with $7 + 4 + 3 = 014 is the answer. ~Generic_Username",
"Let $V = \\overline{NM} \\cap \\overline{AC}$ and $W = \\overline{NM} \\cap \\overline{BC}$. Further more let $\\angle NMC = \\alpha$ and $\\angle MNC = 90^\\circ - \\alpha$. Angle chasing reveals $\\angle NBC = \\angle NAC = \\alpha$ and $\\angle MBC = \\angle MAC = 90^\\circ - \\alpha$. Additionally $NB = \\frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem. By the Angle Bisector Formula, \\[\\frac{NV}{MV} = \\frac{\\sin (\\alpha)}{\\sin (90^\\circ - \\alpha)} = \\tan (\\alpha)\\] \\[\\frac{MW}{NW} = \\frac{3\\sin (90^\\circ - \\alpha)}{4\\sin (\\alpha)} = \\frac{3}{4} \\cot (\\alpha)\\] As $NV + MV =MW + NW = 1$ we compute $NW = \\frac{1}{1+\\frac{3}{4}\\cot(\\alpha)}$ and $MV = \\frac{1}{1+\\tan (\\alpha)}$, and finally $VW = NW + MV - 1 = \\frac{1}{1+\\frac{3}{4}\\cot(\\alpha)} + \\frac{1}{1+\\tan (\\alpha)} - 1$. Taking the derivative of $VW$ with respect to $\\alpha$, we arrive at \\[VW' = \\frac{7\\cos^2 (\\alpha) - 4}{(\\sin(\\alpha) + \\cos(\\alpha))^2(4\\sin(\\alpha)+3\\cos(\\alpha))^2}\\] Clearly the maximum occurs when $\\alpha = \\cos^{-1}\\left(\\frac{2}{\\sqrt{7}}\\right)$. Plugging this back in, using the fact that $\\tan(\\cos^{-1}(x)) = \\frac{\\sqrt{1-x^2}}{x}$ and $\\cot(\\cos^{-1}(x)) = \\frac{x}{\\sqrt{1-x^2}}$, we get $VW = 7 - 4\\sqrt{3}$ with $7 + 4 + 3 = 014 is the answer. ~Generic_Username",
"Since $MA = \\frac{\\sqrt{2}}{2} \\approx 0.707 > \\frac{3}{5}$, point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\\overline{AB}$. By the law of cosines, $\\cos \\angle MOB = \\frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \\frac{7}{25}$, so $\\cos \\angle AOB = \\sin \\angle MOB = \\frac{24}{25}$. Then $AB^2 = 2\\left(\\frac{1}{2}\\right)^2 - 2\\left(\\frac{1}{2}\\right)^2 \\cdot \\frac{24}{25} = \\frac{1}{50}$, so $AB = \\frac{1}{5\\sqrt{2}}$. Let $P = AC \\cap MN$ and $Q = BC \\cap MN$, and let $MQ = x$, $PQ = d$, $PN = y$. Note that\\[(M, P; Q, N) \\stackrel{C}{=} (M, A; B, N),\\]that is,\\[\\frac{QP}{QM} \\div \\frac{NP}{NM} = \\frac{BA}{BM} \\div \\frac{NA}{NM}\\]or\\[\\frac{d}{xy} = \\frac{1/(5\\sqrt{2})}{(3/5) \\cdot (\\sqrt{2}/2)} = \\frac{1}{3}.\\]Hence $d = \\frac{1}{3}xy$, and we also know $d+x+y=1$. Now AM-GM gives\\[\\frac{x+y}{2} \\ge \\sqrt{xy} \\implies \\frac{1-d}{2} \\ge \\sqrt{3d}.\\]This gives the quadratic inequality $d^2 - 14d + 1 \\ge 0$, which solves as\\[d \\in \\left(-\\infty, 7-4\\sqrt3\\right] \\cup \\left[7+4\\sqrt3, \\infty\\right).\\]But $d \\le 1$, so the greatest possible value of $d$ is $7-4\\sqrt3$. The answer is $7+4+3=014 is the answer. ~Generic_Username",
"Since $MA = \\frac{\\sqrt{2}}{2} \\approx 0.707 > \\frac{3}{5}$, point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\\overline{AB}$. By the law of cosines, $\\cos \\angle MOB = \\frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \\frac{7}{25}$, so $\\cos \\angle AOB = \\sin \\angle MOB = \\frac{24}{25}$. Then $AB^2 = 2\\left(\\frac{1}{2}\\right)^2 - 2\\left(\\frac{1}{2}\\right)^2 \\cdot \\frac{24}{25} = \\frac{1}{50}$, so $AB = \\frac{1}{5\\sqrt{2}}$. Let $P = AC \\cap MN$ and $Q = BC \\cap MN$, and let $MQ = x$, $PQ = d$, $PN = y$. Note that\\[(M, P; Q, N) \\stackrel{C}{=} (M, A; B, N),\\]that is,\\[\\frac{QP}{QM} \\div \\frac{NP}{NM} = \\frac{BA}{BM} \\div \\frac{NA}{NM}\\]or\\[\\frac{d}{xy} = \\frac{1/(5\\sqrt{2})}{(3/5) \\cdot (\\sqrt{2}/2)} = \\frac{1}{3}.\\]Hence $d = \\frac{1}{3}xy$, and we also know $d+x+y=1$. Now AM-GM gives\\[\\frac{x+y}{2} \\ge \\sqrt{xy} \\implies \\frac{1-d}{2} \\ge \\sqrt{3d}.\\]This gives the quadratic inequality $d^2 - 14d + 1 \\ge 0$, which solves as\\[d \\in \\left(-\\infty, 7-4\\sqrt3\\right] \\cup \\left[7+4\\sqrt3, \\infty\\right).\\]But $d \\le 1$, so the greatest possible value of $d$ is $7-4\\sqrt3$. The answer is $7+4+3=014 is the answer. ~Generic_Username",
"Let $O$ be the center of the circle. Define $\\angle{MOC}=t$, $\\angle{BOA}=2a$, and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$, respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \\pi]$. Let $C'$ be the foot of the perpendicular from $C$ to $MN$. We compute $XY$ as follows. (a) By the Extended Law of Sines in triangle $ABC$, we have \\[CA\\] \\[= \\sin\\angle{ABC}\\] \\[= \\sin\\left(\\frac{\\widehat{AN} + \\widehat{NC}}{2}\\right)\\] \\[= \\sin\\left(\\frac{\\frac{\\pi}{2} + (\\pi-t)}{2}\\right)\\] \\[= \\sin\\left(\\frac{3\\pi}{4} - \\frac{t}{2}\\right)\\] \\[= \\sin\\left(\\frac{\\pi}{4} + \\frac{t}{2}\\right)\\] (b) Note that $CC' = CO\\sin(t) = \\left(\\frac{1}{2}\\right)\\sin(t)$ and $AO = \\frac{1}{2}$. Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \\sin(t)$, and hence, \\[CY/CA\\] \\[= \\frac{CY}{CY + AY}\\] \\[= \\frac{\\sin(t)}{1 + \\sin(t)}\\] \\[= \\frac{\\sin(t)}{\\sin\\left(\\frac{\\pi}{2}\\right) + \\sin(t)}\\] \\[= \\frac{\\sin(t)}{2\\sin\\left(\\frac{\\pi}{4} + \\frac{t}{2}\\right)\\cos\\left(\\frac{\\pi}{4} - \\frac{t}{2}\\right)}\\] (c) We have $\\angle{XCY} = \\frac{\\widehat{AB}}{2}=a$ and $\\angle{CXY} = \\frac{\\widehat{MB}+\\widehat{CN}}{2} = \\frac{\\left(\\frac{\\pi}{2} - 2a\\right) + (\\pi - t)}{2} = \\frac{3\\pi}{4} - a - \\frac{t}{2}$, and hence by the Law of Sines, \\[XY/CY\\] \\[= \\frac{\\sin\\angle{XCY}}{\\sin\\angle{CXY}}\\] \\[= \\frac{\\sin(a)}{\\sin\\left(\\frac{3\\pi}{4} - a - \\frac{t}{2}\\right)}\\] \\[= \\frac{\\sin(a)}{\\sin\\left(\\frac{\\pi}{4} + a + \\frac{t}{2}\\right)}\\] (d) Multiplying (a), (b), and (c), we have \\[XY\\] \\[= CA * (CY/CA) * (XY/CY)\\] \\[= \\frac{\\sin(t)\\sin(a)}{2\\cos\\left(\\frac{\\pi}{4} - \\frac{t}{2}\\right)\\sin\\left(\\frac{\\pi}{4} + a + \\frac{t}{2}\\right)}\\] \\[= \\frac{\\sin(t)\\sin(a)}{\\sin\\left(\\frac{\\pi}{2} + a\\right) + \\sin(a + t)}\\] \\[= \\sin(a)\\times\\frac{\\sin(t)}{\\sin(t + a) + \\cos(a)}\\], which is a function of $t$ (and the constant $a$). Differentiating this with respect to $t$ yields \\[\\sin(a)\\times\\frac{\\cos(t)(\\sin(t + a) + \\cos(a)) - \\sin(t)\\cos(t + a)}{(\\sin(t + a) + \\cos(a))^2}\\], and the numerator of this is \\[\\sin(a) \\times(\\sin(t + a)\\cos(t) - \\cos(t + a)\\sin(t) + \\cos(a)\\cos(t))\\] \\[= \\sin(a) \\times (\\sin(a) + \\cos(a)\\cos(t))\\], which vanishes when $\\sin(a) + \\cos(a)\\cos(t) = 0$. Therefore, the length of $XY$ is maximized when $t=t'$, where $t'$ is the value in $[0, \\pi]$ that satisfies $\\cos(t') = -\\tan(a)$. Note that \\[\\frac{1 - \\tan(a)}{1 + \\tan(a)} = \\tan\\left(\\frac{\\pi}{4} - a\\right) = \\tan((\\widehat{MB})/2) = \\tan\\angle{MNB} = \\frac{3}{4}\\], so $\\tan(a) = \\frac{1}{7}$. We compute \\[\\sin(a) = \\frac{\\sqrt{2}}{10}\\] \\[\\cos(a) = \\frac{7\\sqrt{2}}{10}\\] \\[\\cos(t') = -\\tan(a) = -\\frac{1}{7}\\] \\[\\sin(t') = \\frac{4\\sqrt{3}}{7}\\] \\[\\sin(t' + a)=\\sin(t')\\cos(a) + \\cos(t')\\sin(a) = \\frac{28\\sqrt{6} - \\sqrt{2}}{70}\\], so the maximum length of $XY$ is $\\sin(a)\\times\\frac{\\sin(t')}{\\sin(t' + a) + \\cos(a)} = 7 - 4\\sqrt{3}$, and the answer is $7 + 4 + 3 = 014 is the answer. ~Generic_Username",
"Let $O$ be the center of the circle. Define $\\angle{MOC}=t$, $\\angle{BOA}=2a$, and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$, respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \\pi]$. Let $C'$ be the foot of the perpendicular from $C$ to $MN$. We compute $XY$ as follows. (a) By the Extended Law of Sines in triangle $ABC$, we have \\[CA\\] \\[= \\sin\\angle{ABC}\\] \\[= \\sin\\left(\\frac{\\widehat{AN} + \\widehat{NC}}{2}\\right)\\] \\[= \\sin\\left(\\frac{\\frac{\\pi}{2} + (\\pi-t)}{2}\\right)\\] \\[= \\sin\\left(\\frac{3\\pi}{4} - \\frac{t}{2}\\right)\\] \\[= \\sin\\left(\\frac{\\pi}{4} + \\frac{t}{2}\\right)\\] (b) Note that $CC' = CO\\sin(t) = \\left(\\frac{1}{2}\\right)\\sin(t)$ and $AO = \\frac{1}{2}$. Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \\sin(t)$, and hence, \\[CY/CA\\] \\[= \\frac{CY}{CY + AY}\\] \\[= \\frac{\\sin(t)}{1 + \\sin(t)}\\] \\[= \\frac{\\sin(t)}{\\sin\\left(\\frac{\\pi}{2}\\right) + \\sin(t)}\\] \\[= \\frac{\\sin(t)}{2\\sin\\left(\\frac{\\pi}{4} + \\frac{t}{2}\\right)\\cos\\left(\\frac{\\pi}{4} - \\frac{t}{2}\\right)}\\] (c) We have $\\angle{XCY} = \\frac{\\widehat{AB}}{2}=a$ and $\\angle{CXY} = \\frac{\\widehat{MB}+\\widehat{CN}}{2} = \\frac{\\left(\\frac{\\pi}{2} - 2a\\right) + (\\pi - t)}{2} = \\frac{3\\pi}{4} - a - \\frac{t}{2}$, and hence by the Law of Sines, \\[XY/CY\\] \\[= \\frac{\\sin\\angle{XCY}}{\\sin\\angle{CXY}}\\] \\[= \\frac{\\sin(a)}{\\sin\\left(\\frac{3\\pi}{4} - a - \\frac{t}{2}\\right)}\\] \\[= \\frac{\\sin(a)}{\\sin\\left(\\frac{\\pi}{4} + a + \\frac{t}{2}\\right)}\\] (d) Multiplying (a), (b), and (c), we have \\[XY\\] \\[= CA * (CY/CA) * (XY/CY)\\] \\[= \\frac{\\sin(t)\\sin(a)}{2\\cos\\left(\\frac{\\pi}{4} - \\frac{t}{2}\\right)\\sin\\left(\\frac{\\pi}{4} + a + \\frac{t}{2}\\right)}\\] \\[= \\frac{\\sin(t)\\sin(a)}{\\sin\\left(\\frac{\\pi}{2} + a\\right) + \\sin(a + t)}\\] \\[= \\sin(a)\\times\\frac{\\sin(t)}{\\sin(t + a) + \\cos(a)}\\], which is a function of $t$ (and the constant $a$). Differentiating this with respect to $t$ yields \\[\\sin(a)\\times\\frac{\\cos(t)(\\sin(t + a) + \\cos(a)) - \\sin(t)\\cos(t + a)}{(\\sin(t + a) + \\cos(a))^2}\\], and the numerator of this is \\[\\sin(a) \\times(\\sin(t + a)\\cos(t) - \\cos(t + a)\\sin(t) + \\cos(a)\\cos(t))\\] \\[= \\sin(a) \\times (\\sin(a) + \\cos(a)\\cos(t))\\], which vanishes when $\\sin(a) + \\cos(a)\\cos(t) = 0$. Therefore, the length of $XY$ is maximized when $t=t'$, where $t'$ is the value in $[0, \\pi]$ that satisfies $\\cos(t') = -\\tan(a)$. Note that \\[\\frac{1 - \\tan(a)}{1 + \\tan(a)} = \\tan\\left(\\frac{\\pi}{4} - a\\right) = \\tan((\\widehat{MB})/2) = \\tan\\angle{MNB} = \\frac{3}{4}\\], so $\\tan(a) = \\frac{1}{7}$. We compute \\[\\sin(a) = \\frac{\\sqrt{2}}{10}\\] \\[\\cos(a) = \\frac{7\\sqrt{2}}{10}\\] \\[\\cos(t') = -\\tan(a) = -\\frac{1}{7}\\] \\[\\sin(t') = \\frac{4\\sqrt{3}}{7}\\] \\[\\sin(t' + a)=\\sin(t')\\cos(a) + \\cos(t')\\sin(a) = \\frac{28\\sqrt{6} - \\sqrt{2}}{70}\\], so the maximum length of $XY$ is $\\sin(a)\\times\\frac{\\sin(t')}{\\sin(t' + a) + \\cos(a)} = 7 - 4\\sqrt{3}$, and the answer is $7 + 4 + 3 = 014 is the answer. ~Generic_Username",
"[asy] unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label(\"$A$\",A,N); label(\"$B$\",B,NNW); label(\"$M$\",M,W); label(\"$C$\",C,SSE); label(\"$N$\",n,E); label(\"$D$\",D[0],SE); label(\"$E$\",e[0],SW); label(\"$x$\",(M+C)/2,SW); label(\"$y$\",(n+C)/2,SE); [/asy] Suppose $\\overline{AC}$ and $\\overline{BC}$ intersect $\\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\\overline{CA}$ bisects $\\angle MCN$, and we get \\[\\frac{MC}{MD} = \\frac{NC}{ND}\\Rightarrow MD = \\frac{x}{x + y}.\\] To find $ME$, we note that $\\triangle BNE\\sim\\triangle MCE$ and $\\triangle BME\\sim\\triangle NCE$, so \\begin{align*} \\frac{BN}{NE} &= \\frac{MC}{CE} \\\\ \\frac{ME}{BM} &= \\frac{CE}{NC}. \\end{align*} Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get \\[\\frac{4(ME)}{3 - 3(ME)} = \\frac{x}{y}\\Rightarrow ME = \\frac{3x}{3x + 4y}.\\] The value we wish to maximize is \\begin{align*} DE &= MD - ME \\\\ &= \\frac{x}{x + y} - \\frac{3x}{3x + 4y} \\\\ &= \\frac{xy}{3x^2 + 7xy + 4y^2} \\\\ &= \\frac{1}{3(x/y) + 4(y/x) + 7}. \\end{align*} By the AM-GM inequality, $3(x/y) + 4(y/x)\\geq 2\\sqrt{12} = 4\\sqrt{3}$, so \\[DE\\leq \\frac{1}{4\\sqrt{3} + 7} = 7 - 4\\sqrt{3},\\] giving the answer of $7 + 4 + 3 = 014 is the answer. ~Generic_Username",
"By Pythagoras in $\\triangle BMN,$ we get $BN=\\dfrac{4}{5}.$ Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \\[\\dfrac{XM}{NY}:\\dfrac{NM}{NY}=\\dfrac{AM}{AB}:\\dfrac{MN}{NB}.\\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\\dfrac{(a+b)(b+c)}{b}.$ In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\\cdot MN+AN\\cdot BM=AM\\cdot BN\\implies AB+\\dfrac{3}{5\\sqrt{2}}=\\dfrac{4}{5\\sqrt{2}}\\implies AB=\\dfrac{1}{5\\sqrt{2}}.$ Substituting, we get $\\dfrac{(a+b)(b+c)}{b}=4\\implies b(a+b+c)+ac=4b, b=\\dfrac{ac}{3}$ where we use $a+b+c=1.$ Again using $a+b+c=1,$ we have $a+b+c=1\\implies a+\\dfrac{ac}{3}+c=1\\implies a=3\\dfrac{1-c}{c+3}.$ Then $b=\\dfrac{ac}{3}=\\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \\[(-2c+1)(c+3)-(c-c^2)=0 \\implies c^2+6c-3,c=-3+2\\sqrt{3}.\\] Substituting back yields $b=7-4\\sqrt{3},$ so $7+4+3=014 is the answer. ~Generic_Username",
"By Pythagoras in $\\triangle BMN,$ we get $BN=\\dfrac{4}{5}.$ Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \\[\\dfrac{XM}{NY}:\\dfrac{NM}{NY}=\\dfrac{AM}{AB}:\\dfrac{MN}{NB}.\\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\\dfrac{(a+b)(b+c)}{b}.$ In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\\cdot MN+AN\\cdot BM=AM\\cdot BN\\implies AB+\\dfrac{3}{5\\sqrt{2}}=\\dfrac{4}{5\\sqrt{2}}\\implies AB=\\dfrac{1}{5\\sqrt{2}}.$ Substituting, we get $\\dfrac{(a+b)(b+c)}{b}=4\\implies b(a+b+c)+ac=4b, b=\\dfrac{ac}{3}$ where we use $a+b+c=1.$ Again using $a+b+c=1,$ we have $a+b+c=1\\implies a+\\dfrac{ac}{3}+c=1\\implies a=3\\dfrac{1-c}{c+3}.$ Then $b=\\dfrac{ac}{3}=\\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \\[(-2c+1)(c+3)-(c-c^2)=0 \\implies c^2+6c-3,c=-3+2\\sqrt{3}.\\] Substituting back yields $b=7-4\\sqrt{3},$ so $7+4+3=014 is the answer. ~Generic_Username"
] |
2010-I-1
| 2,010
| 1
|
Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 107
|
I
|
[
"$2010^2 = 2^2\\cdot3^2\\cdot5^2\\cdot67^2$. Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$). Therefore the probability is \\[\\frac {2\\cdot2^4\\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \\frac {26}{81} \\Longrightarrow 26+ 81 = 107.\\]",
"The prime factorization of $2010^2$ is $67^2\\cdot3^2\\cdot2^2\\cdot5^2$. Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$, $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\\binom{16}{1}\\binom{81-16}{1}$. The total number of ways to pick two divisors is $\\binom{81}{2}$ Thus, the probability is \\[\\frac {\\binom{16}{1}\\binom{81-16}{1}}{\\binom{81}{2}} = \\frac {16\\cdot65}{81\\cdot40} = \\frac {26}{81} \\Longrightarrow 26+ 81 = 107.\\]",
"The prime factorization of $2010^2$ is $67^2\\cdot3^2\\cdot2^2\\cdot5^2$. Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$, $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\\binom{16}{1}\\binom{81-16}{1}$. The total number of ways to pick two divisors is $\\binom{81}{2}$ Thus, the probability is \\[\\frac {\\binom{16}{1}\\binom{81-16}{1}}{\\binom{81}{2}} = \\frac {16\\cdot65}{81\\cdot40} = \\frac {26}{81} \\Longrightarrow 26+ 81 = 107.\\]"
] |
2010-I-2
| 2,010
| 2
|
Find the remainder when $9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}$ is divided by $1000$ .
| 109
|
I
|
[
"Note that $999\\equiv 9999\\equiv\\dots \\equiv\\underbrace{99\\cdots9}_{\\text{999 9's}}\\equiv -1 \\pmod{1000}$ (see modular arithmetic). That is a total of $999 - 3 + 1 = 997$ integers, so all those integers multiplied out are congruent to $- 1\\pmod{1000}$. Thus, the entire expression is congruent to $- 1\\times9\\times99 = - 891\\equiv109.",
"The expression also equals $(10-1)(100-1)\\dots({10^{999}}-1)$. To find its modular 1,000, remove all terms from 1,000 and after. Then the expression becomes $(10-1)(100-1)(-1) \\pmod{1000} \\equiv -891 \\pmod{1000} \\equiv 109 By maxamc"
] |
2010-I-3
| 2,010
| 3
|
Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ .
| 529
|
I
|
[
"Substitute $y = \\frac34x$ into $x^y = y^x$ and solve. \\[x^{\\frac34x} = \\left(\\frac34x\\right)^x\\] \\[x^{\\frac34x} = \\left(\\frac34\\right)^x \\cdot x^x\\] \\[x^{-\\frac14x} = \\left(\\frac34\\right)^x\\] \\[x^{-\\frac14} = \\frac34\\] \\[x = \\frac{256}{81}\\] \\[y = \\frac34x = \\frac{192}{81}\\] \\[x + y = \\frac{448}{81}\\] \\[448 + 81 = 529\\]",
"We solve in general using $c$ instead of $3/4$. Substituting $y = cx$, we have: \\[x^{cx} = (cx)^x \\Longrightarrow (x^x)^c = c^x\\cdot x^x\\] Dividing by $x^x$, we get $(x^x)^{c - 1} = c^x$. Taking the $x$th root, $x^{c - 1} = c$, or $x = c^{1/(c - 1)}$. In the case $c = \\frac34$, $x = \\Bigg(\\frac34\\Bigg)^{ - 4} = \\Bigg(\\frac43\\Bigg)^4 = \\frac {256}{81}$, $y = \\frac {64}{27}$, $x + y = \\frac {256 + 192}{81} = \\frac {448}{81}$, yielding an answer of $448 + 81 = 529.",
"Taking the logarithm base $x$ of both sides, we arrive with: \\[y = \\log_x y^x \\Longrightarrow \\frac{y}{x} = \\log_{x} y = \\log_x \\frac{3}{4}x = \\frac{3}{4}\\] Where the last two simplifications were made since $y = \\frac{3}{4}x$. Then, \\[x^{\\frac{3}{4}} = \\frac{3}{4}x \\Longrightarrow x^{\\frac{1}{4}} = \\frac{4}{3} \\Longrightarrow x = \\left(\\frac{4}{3}\\right)^4\\] Then, $y = \\left(\\frac{4}{3}\\right)^3$, and thus: \\[x+y = \\left(\\frac{4}{3}\\right)^3 \\left(\\frac{4}{3} + 1 \\right) = \\frac{448}{81} \\Longrightarrow 448 + 81 = 529\\]",
"Taking the logarithm base $x$ of both sides, we arrive with: \\[y = \\log_x y^x \\Longrightarrow \\frac{y}{x} = \\log_{x} y = \\log_x \\left(\\frac{3}{4}x\\right) = \\frac{3}{4}\\] Now we proceed by the logarithm rule $\\log(ab)=\\log a + \\log b$. The equation becomes: \\[\\log_x \\frac{3}{4} + \\log_x x = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\log_x \\frac{3}{4} + 1 = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\log_x \\frac{3}{4} = -\\frac{1}{4}\\] \\[\\Longleftrightarrow x^{-\\frac{1}{4}} = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\frac{1}{x^{\\frac{1}{4}}} = \\frac{3}{4}\\] \\[\\Longleftrightarrow x^{\\frac{1}{4}} = \\frac{4}{3}\\] \\[\\Longleftrightarrow \\sqrt[4]{x} = \\frac{4}{3}\\] \\[\\Longleftrightarrow x = \\left(\\frac{4}{3}\\right)^4=\\frac{256}{81}\\] Then find $y$ as in solution 3, and we get $529.",
"Taking the logarithm base $x$ of both sides, we arrive with: \\[y = \\log_x y^x \\Longrightarrow \\frac{y}{x} = \\log_{x} y = \\log_x \\left(\\frac{3}{4}x\\right) = \\frac{3}{4}\\] Now we proceed by the logarithm rule $\\log(ab)=\\log a + \\log b$. The equation becomes: \\[\\log_x \\frac{3}{4} + \\log_x x = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\log_x \\frac{3}{4} + 1 = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\log_x \\frac{3}{4} = -\\frac{1}{4}\\] \\[\\Longleftrightarrow x^{-\\frac{1}{4}} = \\frac{3}{4}\\] \\[\\Longleftrightarrow \\frac{1}{x^{\\frac{1}{4}}} = \\frac{3}{4}\\] \\[\\Longleftrightarrow x^{\\frac{1}{4}} = \\frac{4}{3}\\] \\[\\Longleftrightarrow \\sqrt[4]{x} = \\frac{4}{3}\\] \\[\\Longleftrightarrow x = \\left(\\frac{4}{3}\\right)^4=\\frac{256}{81}\\] Then find $y$ as in solution 3, and we get $529."
] |
2010-I-4
| 2,010
| 4
|
Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 515
|
I
|
[
"This can be solved quickly and easily with generating functions. Let $x^n$ represent flipping $n$ heads. The generating functions for these coins are $(1+x)$,$(1+x)$,and $(3+4x)$ in order. The product is $3+10x+11x^2+4x^3$. ($ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, and therefore $2$ tails, here.) The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is $(4 + 11 + 10 + 3)^2 = 28^2 = 784$ and the sum of the squares of each coefficient (the sum of the number of ways that each coefficient can be chosen by the two people) is $4^2 + 11^2 + 10^2 + 3^2=246$. The probability is then $\\frac{4^2 + 11^2 + 10^2 + 3^2}{28^2} = \\frac{246}{784} = \\frac{123}{392}$. (Notice the relationship between the addends of the numerator here and the cases in the following solution.) $123 + 392 = 515",
"We perform casework based upon the number of heads that are flipped. Case 1: No heads. The only possibility is TTT (the third coin being the unfair coin). The probability for this to happen to Jackie is $\\frac {1}{2} \\cdot \\frac {1}{2} \\cdot \\frac {3}{7} = \\frac {3}{28}$ Thus the probability for this to happen to both players is $\\left(\\frac {3}{28}\\right)^2 = \\frac {9}{784}$ Case 2: One head. We can have either HTT, THT, or TTH. The first two happen to Jackie with the same $\\frac {3}{28}$ chance, but the third happens $\\frac {4}{28}$ of the time, since the unfair coin is heads instead of tails. With 3 possibilities for Jackie and 3 for Phil, there are a total of 9 ways for them both to have 1 head. Multiplying and adding up all 9 ways, we have a \\[\\frac {4(3 \\cdot 3) + 4(3 \\cdot 4) + 1(4 \\cdot 4)}{28^{2}} = \\frac {100}{784}\\] overall chance for this case. Case 3: Two heads. With HHT $\\frac {4}{28}$, HTH $\\frac {4}{28}$, and THH $\\frac {3}{28}$ possible, we proceed as in Case 2, obtaining \\[\\frac {1(3 \\cdot 3) + 4(3 \\cdot 4) + 4(4 \\cdot 4)}{28^{2}} = \\frac {121}{784}.\\] Case 4: Three heads. Similar to Case 1, we can only have HHH, which has $\\frac {4}{28}$ chance. Then in this case we get $\\frac {16}{784}$ Finally, we take the sum: $\\frac {9 + 100 + 121 + 16}{784} = \\frac {246}{784} = \\frac {123}{392}$, so our answer is $123 + 392 = 515."
] |
2010-I-5
| 2,010
| 5
|
Positive integers $a$ , $b$ , $c$ , and $d$ satisfy $a > b > c > d$ , $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ .
| 501
|
I
|
[
"Using the difference of squares, $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \\ge a + b + c + d = 2010$, where equality must hold so $b = a - 1$ and $d = c - 1$. Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $501",
"Since $a+b$ must be greater than $1005$, it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $501 possible values."
] |
2010-I-6
| 2,010
| 6
|
Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ .
| 406
|
I
|
[
"Solution 1 [asy] import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype(\"6 2\")+linewidth(0.7)); draw(graph(R,min,max),linetype(\"6 2\")+linewidth(0.7)); dot((1,1)); label(\"$P(x)$\",(max,P(max)),E,fontsize(10)); label(\"$Q(x)$\",(max,Q(max)),E,fontsize(10)); label(\"$R(x)$\",(max,R(max)),E,fontsize(10)); /* axes */ Label f; f.p=fontsize(8); xaxis(-2, 3, Ticks(f, 5, 1)); yaxis(-1, 5, Ticks(f, 6, 1)); [/asy] Let $Q(x) = x^2 - 2x + 2$, $R(x) = 2x^2 - 4x + 3$. Completing the square, we have $Q(x) = (x-1)^2 + 1$, and $R(x) = 2(x-1)^2 + 1$, so it follows that $P(x) \\ge Q(x) \\ge 1$ for all $x$ (by the Trivial Inequality). Also, $1 = Q(1) \\le P(1) \\le R(1) = 1$, so $P(1) = 1$, and $P$ obtains its minimum at the point $(1,1)$. Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$; substituting $P(11) = 181$ yields $c = \\frac 95$. Finally, $P(16) = \\frac 95 \\cdot (16 - 1)^2 + 1 = 406.",
"[asy] import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype(\"6 2\")+linewidth(0.7)); draw(graph(R,min,max),linetype(\"6 2\")+linewidth(0.7)); dot((1,1)); label(\"$P(x)$\",(max,P(max)),E,fontsize(10)); label(\"$Q(x)$\",(max,Q(max)),E,fontsize(10)); label(\"$R(x)$\",(max,R(max)),E,fontsize(10)); /* axes */ Label f; f.p=fontsize(8); xaxis(-2, 3, Ticks(f, 5, 1)); yaxis(-1, 5, Ticks(f, 6, 1)); [/asy] Let $Q(x) = x^2 - 2x + 2$, $R(x) = 2x^2 - 4x + 3$. Completing the square, we have $Q(x) = (x-1)^2 + 1$, and $R(x) = 2(x-1)^2 + 1$, so it follows that $P(x) \\ge Q(x) \\ge 1$ for all $x$ (by the Trivial Inequality). Also, $1 = Q(1) \\le P(1) \\le R(1) = 1$, so $P(1) = 1$, and $P$ obtains its minimum at the point $(1,1)$. Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$; substituting $P(11) = 181$ yields $c = \\frac 95$. Finally, $P(16) = \\frac 95 \\cdot (16 - 1)^2 + 1 = 406.",
"It can be seen that the function $P(x)$ must be in the form $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$. This is because the derivative of $P(x)$ is $2ax - 2a$, and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\\frac{-b}{2a}$). Substituting $(1,1)$ and $(11, 181)$ we obtain two equations: $P(11) = 99a + c = 181$, and $P(1) = -a + c = 1$. Solving, we get $a = \\frac{9}{5}$ and $c = \\frac{14}{5}$, so $P(x) = \\frac{9}{5}x^2 - \\frac{18}{5}x + \\frac {14}{5}$. Therefore, $P(16) = 406.",
"Let $y = x^2 - 2x + 2$; note that $2y - 1 = 2x^2 - 4x + 3$. Setting $y = 2y - 1$, we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$; this is true iff $x = 1$, so $P(1) = 1$. Let $Q(x) = P(x) - x$; clearly $Q(1) = 0$, so we can write $Q(x) = (x - 1)Q'(x)$, where $Q'(x)$ is some linear function. Plug $Q(x)$ into the given inequality: $x^2 - 3x + 2 \\le Q(x) \\le 2x^2 - 5x + 3$ $(x - 1)(x - 2) \\le (x - 1)Q'(x) \\le (x - 1)(2x - 3)$, and thus $x - 2 \\le Q'(x) \\le 2x - 3$ For all $x > 1$; note that the inequality signs are flipped if $x < 1$, and that the division is invalid for $x = 1$. However, $\\lim_{x \\to 1} x - 2 = \\lim_{x \\to 1} 2x - 3 = -1$, and thus by the sandwich theorem $\\lim_{x \\to 1} Q'(x) = -1$; by the definition of a continuous function, $Q'(1) = -1$. Also, $Q(11) = 170$, so $Q'(11) = 170/(11-1) = 17$; plugging in and solving, $Q'(x) = (9/5)(x - 1) - 1$. Thus $Q(16) = 390$, and so $P(16) = 406.",
"Let $Q(x) = P(x) - (x^2-2x+2)$, then $0\\le Q(x) \\le (x-1)^2$ (note this is derived from the given inequality chain). Therefore, $0\\le Q(x+1) \\le x^2 \\Rightarrow Q(x+1) = Ax^2$ for some real value A. $Q(11) = 10^2A \\Rightarrow P(11)-(11^2-22+2)=100A \\Rightarrow 80=100A \\Rightarrow A=\\frac{4}{5}$. $Q(16)=15^2A=180 \\Rightarrow P(16)-(16^2-32+2) = 180 \\Rightarrow P(16)=180+226= 406.",
"Let $P(x) = ax^2 + bx + c$. Plugging in $x = 1$ to the expressions on both sides of the inequality, we see that $a + b + c = 1$. We see from the problem statement that $121a + 11b + c = 181$. Since we know the vertex of $P(x)$ lies at $x = 1$, by symmetry we get $81a -9b + c = 181$ as well. Since we now have three equations, we can solve this trivial system and get our answer of $406.",
"Similar to Solution 5, let $P(x) = ax^2 + bx + c$. Note that $(1,1)$ is a vertex of the polynomial. Additionally, this means that $b = -2a$ (since $\\frac{-b}{2a}$ is the minimum $x$ point). Thus, we have $P(x) = ax^2 - 2ax + c$. Therefore $a - 2a + c = 1$. Moreover, $99a + c = 181$. And so our polynomial is $\\frac{9}{5}x^2 - \\frac{18}{5}x + \\frac{14}{5}$. Plug in $x = 16$ to get $406.",
"Very similar to Solution 6, start by noticing that $P(x)$ is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that $P(x)$ would also have to intersect that point (it must be between the two graphs). Setting $x^2 - 2x + 2 = 2x^2 - 4x + 3$, we find that $x = 1$. Note that both of these graphs have the same vertex (at $x = 1$), and so $P(x)$ must also have the same vertex $(1, 1)$. Setting $P(x) = ax^2 - 2ax + a + 1$ (this is where we have a vertex at $(1, 1)$), we plug in $11$ and find that $a = 1.8$. Evaluating $1.8x^2 - 3.6x + 2.8$ when $x = 16$ (our intended goal), we find that $P(16) = 406."
] |
2010-I-7
| 2,010
| 7
|
Define an ordered triple $(A, B, C)$ of sets to be $\textit{minimally intersecting}$ if $|A \cap B| = |B \cap C| = |C \cap A| = 1$ and $A \cap B \cap C = \emptyset$ . For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $\{1,2,3,4,5,6,7\}$ . Find the remainder when $N$ is divided by $1000$ . Note : $|S|$ represents the number of elements in the set $S$ .
| 760
|
I
|
[
"Let each pair of two sets have one element in common. Label the common elements as $x$, $y$, $z$. Set $A$ will have elements $x$ and $y$, set $B$ will have $y$ and $z$, and set $C$ will have $x$ and $z$. There are $7 \\cdot 6 \\cdot 5 = 210$ ways to choose values of $x$, $y$ and $z$. There are $4$ unpicked numbers, and each number can either go in the first set, second set, third set, or none of them. Since we have $4$ choices for each of $4$ numbers, that gives us $4^4 = 256$. Finally, $256 \\cdot 210 = 53760$, so the answer is $760."
] |
2010-I-8
| 2,010
| 8
|
For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$ , where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$ .
| 132
|
I
|
[
"The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\\pm5,0), (0,\\pm5), (\\pm3,\\pm4), (\\pm4,\\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\\left(\\frac12,\\frac12\\right)$. The distance from $\\left(\\frac12,\\frac12\\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$, is $\\sqrt {\\frac {11}2^2 + \\frac12^2} = \\sqrt {\\frac {122}4}.$ The distance from $\\left(\\frac12,\\frac12\\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$, is $\\sqrt {\\frac92^2 + \\frac72^2} = \\sqrt {\\frac {130}4}.$ The latter is the larger, and is $\\frac {\\sqrt {130}}2$, giving an answer of $130 + 2 = 132. [asy]import graph; size(10.22cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.68,xmax=6.54,ymin=-5.52,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); filldraw((-3,4)--(-2,4)--(-2,5)--(-3,5)--cycle,evevff,blue); filldraw((3,4)--(4,4)--(4,5)--(3,5)--cycle,evevff,blue); filldraw((4,3)--(5,3)--(5,4)--(4,4)--cycle,evevff,blue); filldraw((5,0)--(6,0)--(6,1)--(5,1)--cycle,evevff,blue); filldraw((4,-3)--(5,-3)--(5,-2)--(4,-2)--cycle,evevff,blue); filldraw((3,-3)--(3,-4)--(4,-4)--(4,-3)--cycle,evevff,blue); filldraw((0,-5)--(1,-5)--(1,-4)--(0,-4)--cycle,evevff,blue); filldraw((-3,-4)--(-2,-4)--(-2,-3)--(-3,-3)--cycle,evevff,blue); filldraw((-4,-3)--(-3,-3)--(-3,-2)--(-4,-2)--cycle,evevff,blue); filldraw((-4,3)--(-3,3)--(-3,4)--(-4,4)--cycle,evevff,blue); filldraw((-5,0)--(-4,0)--(-4,1)--(-5,1)--cycle,evevff,blue); filldraw((0,6)--(0,5)--(1,5)--(1,6)--cycle,evevff,blue); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); draw(circle((0,0),5),linewidth(1.6)); draw(circle((0.5,0.5),5.7),linetype(\"2 2\")); draw((-3,4)--(-2,4),zzttqq); draw((-2,4)--(-2,5),zzttqq); draw((-2,5)--(-3,5),zzttqq); draw((-3,5)--(-3,4),zzttqq); draw((3,4)--(4,4),zzttqq); draw((4,4)--(4,5),zzttqq); draw((4,5)--(3,5),zzttqq); draw((3,5)--(3,4),zzttqq); draw((4,3)--(5,3),zzttqq); draw((5,3)--(5,4),zzttqq); draw((5,4)--(4,4),zzttqq); draw((4,4)--(4,3),zzttqq); draw((5,0)--(6,0),zzttqq); draw((6,0)--(6,1),zzttqq); draw((6,1)--(5,1),zzttqq); draw((5,1)--(5,0),zzttqq); draw((4,-3)--(5,-3),zzttqq); draw((5,-3)--(5,-2),zzttqq); draw((5,-2)--(4,-2),zzttqq); draw((4,-2)--(4,-3),zzttqq); draw((3,-3)--(3,-4),zzttqq); draw((3,-4)--(4,-4),zzttqq); draw((4,-4)--(4,-3),zzttqq); draw((4,-3)--(3,-3),zzttqq); draw((0,-5)--(1,-5),zzttqq); draw((1,-5)--(1,-4),zzttqq); draw((1,-4)--(0,-4),zzttqq); draw((0,-4)--(0,-5),zzttqq); draw((-3,-4)--(-2,-4),zzttqq); draw((-2,-4)--(-2,-3),zzttqq); draw((-2,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-4),zzttqq); draw((-4,-3)--(-3,-3),zzttqq); draw((-3,-3)--(-3,-2),zzttqq); draw((-3,-2)--(-4,-2),zzttqq); draw((-4,-2)--(-4,-3),zzttqq); draw((-4,3)--(-3,3),zzttqq); draw((-3,3)--(-3,4),zzttqq); draw((-3,4)--(-4,4),zzttqq); draw((-4,4)--(-4,3),zzttqq); draw((-5,0)--(-4,0),zzttqq); draw((-4,0)--(-4,1),zzttqq); draw((-4,1)--(-5,1),zzttqq); draw((-5,1)--(-5,0),zzttqq); draw((0,6)--(0,5),zzttqq); draw((0,5)--(1,5),zzttqq); draw((1,5)--(1,6),zzttqq); draw((1,6)--(0,6),zzttqq); dot((0,5),ds); dot((3,4),ds); dot((4,3),ds); dot((5,0),ds); dot((4,-3),ds); dot((3,-4),ds); dot((0,-5),ds); dot((-3,-4),ds); dot((-4,-3),ds); dot((-4,3),ds); dot((-3,4),ds); dot((-2,4),ds); dot((-2,5),ds); dot((-3,5),ds); dot((4,4),ds); dot((4,5),ds); dot((3,5),ds); dot((5,3),ds); dot((5,4),ds); dot((4,4),ds); dot((6,0),ds); dot((6,1),ds); dot((5,1),ds); dot((5,-3),ds); dot((5,-2),ds); dot((4,-2),ds); dot((3,-3),ds); dot((4,-4),ds); dot((4,-3),ds); dot((1,-5),ds); dot((1,-4),ds); dot((0,-4),ds); dot((-2,-4),ds); dot((-2,-3),ds); dot((-3,-3),ds); dot((-3,-2),ds); dot((-4,-2),ds); dot((-3,3),ds); dot((-3,4),ds); dot((-4,4),ds); dot((-5,0),ds); dot((-4,0),ds); dot((-4,1),ds); dot((-5,1),ds); dot((0,6),ds); dot((1,5),ds); dot((1,6),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]",
"When observing the equation $\\lfloor x \\rfloor ^2 + \\lfloor y \\rfloor ^2 = 25$, it is easy to see that it is a circle graph. So, we can draw a coordinate plane and find some points. In quadrant $1$, $x < 6$ and $y < 6$. Note that $\\lfloor 5.999...\\rfloor = 5$, but if we add more $9's$ after the $5$, it will get infinitely close to $6$, so we can use $6$ as a bounding line. Also, with the same logic, when $x = 6$, $y = 1$ (the equal sign represents as $x$ approaches..., not equal to...) So, in quadrant one, we have points $(6,1)$ and $(1,6)$. Moving to quadrant $2$, we must note that $\\lfloor -4.999...\\rfloor = -5$, so the circle will not be centered at $(0,0)$. In quadrant 2, $y$ is still positive, so we can have $y = 1$. When $y = 1$, $x = -5$, so we have our next point $(-5,1)$. With this method, other points can be found in quadrants $3$ and $4$. Additionally, $3 ^2 + 4 ^2 = 5 ^2$, and with the same approaching limit, we know that quadrant $1$ also has lattice points $(4,5)$ and $(5,4)$. We need a point that passes through the circle's center (we don't need to find the center). If we focus on $(5,4)$, the \"opposite\" point is $(-4,-3)$ located in quadrant 3. Using the distance formula, we find that the distance between the two points is $\\sqrt {130}$. Since the line connecting those two points passes through the circle's center, it is the diameter. So, the radius can be found by dividing $\\sqrt {130}$ by $2$ to get $\\frac {\\sqrt {130}}2$, and $m+n=130 + 2 = 132. ~hwan ~edited by ALANARCHERMAN"
] |
2010-I-9
| 2,010
| 9
|
Let $(a,b,c)$ be a real solution of the system of equations $x^3 - xyz = 2$ , $y^3 - xyz = 6$ , $z^3 - xyz = 20$ . The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 158
|
I
|
[
"Solution 1 Add the three equations to get $a^3 + b^3 + c^3 = 28 + 3abc$. Now, let $abc = p$. $a = \\sqrt [3]{p + 2}$, $b = \\sqrt [3]{p + 6}$ and $c = \\sqrt [3]{p + 20}$, so $p = abc = (\\sqrt [3]{p + 2})(\\sqrt [3]{p + 6})(\\sqrt [3]{p + 20})$. Now cube both sides; the $p^3$ terms cancel out. Solve the remaining quadratic to get $p = - 4, - \\frac {15}{7}$. To maximize $a^3 + b^3 + c^3$ choose $p = - \\frac {15}{7}$ and so the sum is $28 - \\frac {45}{7} = \\frac {196 - 45}{7}$ giving $151 + 7 = 158.",
"Add the three equations to get $a^3 + b^3 + c^3 = 28 + 3abc$. Now, let $abc = p$. $a = \\sqrt [3]{p + 2}$, $b = \\sqrt [3]{p + 6}$ and $c = \\sqrt [3]{p + 20}$, so $p = abc = (\\sqrt [3]{p + 2})(\\sqrt [3]{p + 6})(\\sqrt [3]{p + 20})$. Now cube both sides; the $p^3$ terms cancel out. Solve the remaining quadratic to get $p = - 4, - \\frac {15}{7}$. To maximize $a^3 + b^3 + c^3$ choose $p = - \\frac {15}{7}$ and so the sum is $28 - \\frac {45}{7} = \\frac {196 - 45}{7}$ giving $151 + 7 = 158.",
"This is almost the same as solution 1. Note $a^3 + b^3 + c^3 = 28 + 3abc$. Next, let $k = a^3$. Note that $b = \\sqrt [3]{k + 4}$ and $c = \\sqrt [3]{k + 18}$, so we have $28 + 3\\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22$. Move 28 over, divide both sides by 3, then cube to get $k^3-6k^2+12k-8 = k^3+22k^2+18k$. The $k^3$ terms cancel out, so solve the quadratic to get $k = -2, -\\frac{1}{7}$. We maximize $abc$ by choosing $k = -\\frac{1}{7}$, which gives us $a^3+b^3+c^3 = 3k + 22 = \\frac{151}{7}$. Thus, our answer is $151+7=158.",
"We have that $x^3 = 2 + xyz$, $y^3 = 6 + xyz$, and $z^3 = 20 + xyz$. Multiplying the three equations, and letting $m = xyz$, we have that $m^3 = (2+m)(6+m)(20+m)$, and reducing, that $7m^2 + 43m + 60 = 0$, which has solutions $m = -\\frac{15}{7}, -4$. Adding the three equations and testing both solutions, we find the answer of $\\frac{151}{7}$, so the desired quantity is $151 + 7 = 158."
] |
2010-I-10
| 2,010
| 10
|
Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ .
| 202
|
I
|
[
"If we choose $a_3$ and $a_1$ such that $(10^3)(a_3) + (10)(a_1) \\leq 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3 = 0$ or $a_3 = 1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3 = 2$ then $a_1 = 0$ or $a_1 = 1$. Thus $N = 100 + 100 + 2 = 202.",
"Note that $a_2\\cdot 10^2 + a_0$ is the base $100$ representation of any number from $0$ to $9999$, and similarly $10(a_3\\cdot 10^2 + a_1)$ is ten times the base $100$ representation of any number from $0$ to $9999$. Thus, the number of solutions is just the number of solutions to $2010 = 10a+b$ where $0\\le a, b\\le 9999$, which is equal to $202.",
"Note that $a_0 \\equiv 2010\\ (\\textrm{mod}\\ 10)$ and $a_1 \\equiv 2010 - a_0\\ (\\textrm{mod}\\ 100)$. It's easy to see that exactly 10 values in $0 \\leq a_0 \\leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$. Thus, there are $10 \\times 10 = 100$ possible choices for $a_0$ and $a_1$. We next note that if $a_0$ and $a_1$ are chosen, then a valid value of $a_3$ determines $a_2$, so we dive into some simple casework: If $2010 - 10a_1 - a_0 \\geq 2000$, there are 3 valid choices for $a_3$. There are only 2 possible cases where $2010 - 10a_1 - a_0 \\geq 2000$, namely $(a_1, a_0) = (1,0), (10,0)$. Thus, there are $3 \\times 2 = 6$ possible representations in this case. If $2010 - 10a_1 - a_0 < 1000$, $a_3$ can only equal 0. However, this case cannot occur, as $10a_1+a_0\\leq 990+99 = 1089$. Thus, $2010-10a_1-a_0 \\geq 921$. However, $2010-10a_1-a_0 = 1000a_3 + 100a_2 \\equiv 0\\ (\\textrm{mod}\\ 100)$. Thus, we have $2010-10a_1-a_0 \\geq 1000$ always. If $1000 \\leq 2010 - 10a_1 - a_0 < 2000$, then there are 2 valid choices for $a_3$. Since there are 100 possible choices for $a_0$ and $a_1$, and we have already checked the other cases, it follows that $100 - 2 - 0 = 98$ choices of $a_0$ and $a_1$ fall under this case. Thus, there are $2 \\times 98 = 196$ possible representations in this case. Our answer is thus $6 + 0 + 196 = 202.",
"We immediately see that $a_3$ can only be $0$, $1$ or $2$. We also note that the maximum possible value for $10a_1 + a_0$ is $990 + 99 = 1089$. We then split into cases: Case 1: $a_3 = 0$. We try to find possible values of $a_2$. We plug in $a_3 = 0$ and $10a_1 + a_0 = 1089$ to our initial equation, which gives us $2010 = 0 + 100a_2 + 1089$. Thus $a_2 \\geq 10$. We also see that $a_2 \\leq 20$. We now take these values of $a_2$ and find the number of pairs $(a_1, a_0)$ that work. If $a_2 = 10$, $10a_1 + a_0 = 1010$. We see that there are $8$ possible pairs in this case. Using the same logic, there are $10$ ways for $a_2 = 11, 12 \\ldots 19$. For $a_2 = 20$, we get the equation $10a_1 + a_0 = 10$, for 2 ways. Thus, for $a_3 = 0$, there are $8 + 10 \\cdot 9 + 2 = 100$ ways. Case 2: $a_3 = 1$. This case is almost identical to the one above, except $0 \\leq a_2 \\leq 10$. We also get 100 ways. Case 3: $a_3 = 2$. If $a_3 = 2$, our initial equation becomes $100a_2 + 10a_1 + a_0 = 10$. It is obvious that $a_2 = 0$, and we are left with $10a_1 + a_0 = 10$. We saw above that there are $2$ ways. Totaling everything, we get that there are $100 + 100 + 2 = 202 ways.",
"We will represent the problem using generating functions. Consider the generating function \\[f(x) = (1+x^{1000}+x^{2000}+\\cdots+x^{99000})(1+x^{100}+x^{200}+\\cdots+x^{9900})(1+x^{10}+x^{20}+\\cdots+x^{990})(1+x+x^2+\\cdots+x^{99})\\] where the first factor represents $a_3$, the second factor $a_2$, and so forth. We want to find the coefficient of $x^{2010}$ in the expansion of $f(x)$. Now rewriting each factor using the geometric series yields \\[f(x) = \\frac{\\cancel{x^{100}-1}}{x-1} \\cdot \\frac{\\cancel{x^{1000}-1}}{x^{10}-1} \\cdot \\frac{x^{10000}-1}{\\cancel{x^{100}-1}} \\cdot \\frac{x^{100000}-1}{\\cancel{x^{1000}-1}}=\\frac{x^{10000}-1}{x-1} \\cdot \\frac{x^{100000}-1}{x^{10}-1} = (1+x+x^2+\\cdots + x^{9999})(1+x^{10}+x^{20}+\\cdots+x^{99990})\\] The coefficient of $x^{2010}$ in this is simply $202, as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor. ~rzlng",
"First note that $a_3$ has to be a single-digit number($0$, $1$, or $2$ to be exact), and that $a_1$ has to be a two-digit multiple of ten. Then, $a_3$, $a_2$, $a_1$ and $a_0$ can be represented as follows: \\begin{align*} a_3 = a \\\\ a_2 = 10b+c \\\\ a_1= 10d+e \\\\ a_0 = 10f \\end{align*} , where $a$, $b$, $c$, $d$, $e$, and $f$ are all(not necessarily nonzero) digits. Now, we can write our given equation as follows: \\begin{align*} 2010 = 1000(a+b) + 100(c+d) + 10(e+f) \\\\ 201 = 100(a+b) + 10(c+d) + (e+f) \\\\ 201 = (100a + 10c + e) + (100b + d + f) \\end{align*} Now, each integer between $0$ and $201$ inclusive can be represented in exactly one way as $100a + 10c + e$, and this corresponds with one unique $100b + d + f$, so it remains to count the number of integers between $0$ and $201$ inclusive. This is easily counted to be $202.",
"Just note that this corresponds to $0 \\leq 10^3 \\cdot a_3 + 10^2 \\cdot a_2 + 10^1 \\cdot a_1 \\leq 2010$, because we can use $a_0$ to fill in the remaining gap. Then, dividing by $10$, we have $0 \\leq \\overline{a_3 a_2 a_1} \\leq 201$, of which there are $202 solutions."
] |
2010-I-11
| 2,010
| 11
|
Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$ . When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$ , the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$ , where $m$ , $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$ .
| 365
|
I
|
[
"[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype(\"4 4\")); label(\"$y \\ge x/3 + 5$\",(max,P(max)),E,fontsize(10)); label(\"$y \\le 10 - |x-8|$\",(max,Q(max)),E,fontsize(10)); label(\"$\\mathcal{R}$\",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1)); [/asy] The inequalities are equivalent to $y \\ge x/3 + 5, y \\le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \\Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \\left(\\frac 92, \\frac {13}2\\right), \\left(\\frac{39}{4}, \\frac{33}{4}\\right)$. The region $\\mathcal{R}$ is a triangle, as shown above. When revolved about the line $y = x/3+5$, the resulting solid is the union of two right cones that share the same base and axis. [asy]size(200); import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3); pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25)); triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A)); real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2; triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 ); draw(C--A--B--G--cycle,linetype(\"4 4\")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype(\"4 4\")); draw(J--C--K); draw(L--B--M); dot(B);dot(C);dot(F); label(\"$h_1$\",(B+F)/2,SE,fontsize(10)); label(\"$h_2$\",(C+F)/2,S,fontsize(10)); label(\"$r$\",(A+F)/2,E,fontsize(10)); [/asy] Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$), and let $r$ denote their common radius. The volume of a cone is given by $\\frac 13 Bh$; since both cones share the same base, then the desired volume is $\\frac 13 \\cdot \\pi r^2 \\cdot (h_1 + h_2)$. The distance from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\\left|\\frac{(8) - 3(10) + 15}{\\sqrt{1^2 + (-3)^2}}\\right| = \\frac{7}{\\sqrt{10}}$. The distance between $\\left(\\frac 92, \\frac {13}2\\right)$ and $\\left(\\frac{39}{4}, \\frac{33}{4}\\right)$ is given by $h_1 + h_2 = \\sqrt{\\left(\\frac{18}{4} - \\frac{39}{4}\\right)^2 + \\left(\\frac{26}{4} - \\frac{33}{4}\\right)^2} = \\frac{7\\sqrt{10}}{4}$. Thus, the answer is $\\frac{343\\sqrt{10}\\pi}{120} = \\frac{343\\pi}{12\\sqrt{10}} \\Longrightarrow 343 + 12 + 10 = 365. (Note to MAA: Is it a coincidence that this is the number of days in a non-leap year?)"
] |
2010-I-12
| 2,010
| 12
|
Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$ . Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$ , $b$ , and $c$ (not necessarily distinct) such that $ab = c$ . Note : a partition of $S$ is a pair of sets $A$ , $B$ such that $A \cap B = \emptyset$ , $A \cup B = S$ .
| 243
|
I
|
[
"We claim that $243$ is the minimal value of $m$. Let the two partitioned sets be $A$ and $B$; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality, we place $3$ in $A$. Then $9$ must be placed in $B$, so $81$ must be placed in $A$, and $27$ must be placed in $B$. Then $243$ cannot be placed in any set, so we know $m$ is less than or equal to $243$. For $m \\le 242$, we can partition $S$ into $S \\cap \\{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\\}$ and $S \\cap \\{9, 10, 11 ... 80\\}$, and in neither set are there values where $ab=c$ (since $8 < (3\\text{ to }8)^2 < 81$ and $81^2>242$ and $(9\\text{ to }80)^2 > 80$). Thus $m = 243. FYI, this. is a bad solution",
"Consider $\\{3,4,12\\}$. We could have any two of the three be together in the same set, and the third in the other set. Thus, we have $\\{3,4\\}, \\{3,12\\}, \\{4,12\\}$. We will try to 'place' numbers in either set such that we never have $a\\cdot b = c$, until we reach a point where we MUST have $a\\cdot b =c$. We begin with $\\{3,12\\}$. Notice that $a,b,c$ do not have to be distinct, meaning we could have $3\\cdot 3=9$. Thus $9$ must be with $4$. Notice that no matter in which set $36$ is placed, we will be forced to have $a\\cdot b =c$, since $3*12=36$ and $4*9=36$. We could have $\\{4,12\\}$. Similarly, $16$ must be with $3$, and no matter to which set $48$ is placed into, we will be forced to have $a \\cdot b =c$. Now we have $\\{3,4\\}$. $9$ must be with $12$. Then $81$ must be with $\\{3,4\\}$. Since $27$ can't be placed in the same set as $\\{3,4,81\\}$, $27$ must go with $\\{9,12\\}$. But then no matter where $243$ is placed we will have $a\\cdot b =c$. Thus, $243. ~skibbysiggy"
] |
2010-I-13
| 2,010
| 13
|
Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ , $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ , $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ .
| 69
|
I
|
[
"Diagram [asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((126,0)--(63,109.1192),zzttqq); draw((63,109.1192)--(0,0),zzttqq); draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21)); draw((0,-154.31785)--(252,-154.31785)); draw((252,-154.31785)--(252,0)); draw((0,0)--(84,0)); draw((84,0)--(252,0)); draw((63,109.1192)--(63,0)); draw((84,0)--(84,-154.31785)); draw(arc((126,0),126,0,180)); /* points and labels */ dot((0,0)); label(\"$A$\",(-16.43287,-9.3374),NE/2); dot((252,0)); label(\"$B$\",(255.242,5.00321),NE/2); dot((0,-154.31785)); label(\"$D$\",(3.48464,-149.55669),NE/2); dot((252,-154.31785)); label(\"$C$\",(255.242,-149.55669),NE/2); dot((126,0)); label(\"$O$\",(129.36332,5.00321),NE/2); dot((63,109.1192)); label(\"$N$\",(44.91307,108.57427),NE/2); label(\"$126$\",(28.18236,40.85473),NE/2); dot((84,0)); label(\"$U$\",(87.13819,5.00321),NE/2); dot((113.69848,-154.31785)); label(\"$T$\",(116.61611,-149.55669),NE/2); dot((63,0)); label(\"$N'$\",(66.42398,5.00321),NE/2); label(\"$84$\",(41.72627,-12.5242),NE/2); label(\"$168$\",(167.60494,-12.5242),NE/2); dot((84,-154.31785)); label(\"$T'$\",(87.13819,-149.55669),NE/2); dot((252,0)); label(\"$I$\",(255.242,5.00321),NE/2); clip((-71.4052,-225.24323)--(-71.4052,171.51361)--(504.60925,171.51361)--(504.60925,-225.24323)--cycle); [/asy] Solution 1 The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$. Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$. Then $AB = 6$ so $OA = OB = 3$. Since $ON$ is a radius of the semicircle, $ON = 3$. Thus $OAN$ is an equilateral triangle. Let $X$, $Y$, and $Z$ be the areas of triangle $OUN$, sector $ONB$, and trapezoid $UBCT$ respectively. $X = \\frac {1}{2}(UO)(NO)\\sin{O} = \\frac {1}{2}(1)(3)\\sin{60^\\circ} = \\frac {3}{4}\\sqrt {3}$ $Y = \\frac {1}{3}\\pi(3)^2 = 3\\pi$ To find $Z$ we have to find the length of $TC$. Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$. Notice that $UNN'$ and $TUT'$ are similar. Thus: $\\frac {TT'}{UT'} = \\frac {UN'}{NN'} \\implies \\frac {TT'}{h} = \\frac {1/2}{3\\sqrt {3}/2} \\implies TT' = \\frac {\\sqrt {3}}{9}h$. Then $TC = T'C - T'T = UB - TT' = 4 - \\frac {\\sqrt {3}}{9}h$. So: $Z = \\frac {1}{2}(BU + TC)(CB) = \\frac {1}{2}\\left(8 - \\frac {\\sqrt {3}}{9}h\\right)h = 4h - \\frac {\\sqrt {3}}{18}h^2$ Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$. Then $L = X + Y + Z = \\frac {3}{4}\\sqrt {3} + 3\\pi + 4h - \\frac {\\sqrt {3}}{18}h^2$ Obviously, the $L$ is greater than the area on the other side of line $l$. This other area is equal to the total area minus $L$. Thus: $\\frac {2}{1} = \\frac {L}{6h + \\frac {9}{2}{\\pi} - L} \\implies 12h + 9\\pi = 3L$. Now just solve for $h$. \\begin{align*} 12h + 9\\pi & = \\frac {9}{4}\\sqrt {3} + 9\\pi + 12h - \\frac {\\sqrt {3}}{6}h^2 \\\\ 0 & = \\frac {9}{4}\\sqrt {3} - \\frac {\\sqrt {3}}{6}h^2 \\\\ h^2 & = \\frac {9}{4}(6) \\\\ h & = \\frac {3}{2}\\sqrt {6} \\end{align*} Don't forget to un-rescale at the end to get $AD = \\frac {3}{2}\\sqrt {6} \\cdot 42 = 63\\sqrt {6}$. Finally, the answer is $63 + 6 = 069. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$. Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$. Then $AB = 6$ so $OA = OB = 3$. Since $ON$ is a radius of the semicircle, $ON = 3$. Thus $OAN$ is an equilateral triangle. Let $X$, $Y$, and $Z$ be the areas of triangle $OUN$, sector $ONB$, and trapezoid $UBCT$ respectively. $X = \\frac {1}{2}(UO)(NO)\\sin{O} = \\frac {1}{2}(1)(3)\\sin{60^\\circ} = \\frac {3}{4}\\sqrt {3}$ $Y = \\frac {1}{3}\\pi(3)^2 = 3\\pi$ To find $Z$ we have to find the length of $TC$. Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$. Notice that $UNN'$ and $TUT'$ are similar. Thus: $\\frac {TT'}{UT'} = \\frac {UN'}{NN'} \\implies \\frac {TT'}{h} = \\frac {1/2}{3\\sqrt {3}/2} \\implies TT' = \\frac {\\sqrt {3}}{9}h$. Then $TC = T'C - T'T = UB - TT' = 4 - \\frac {\\sqrt {3}}{9}h$. So: $Z = \\frac {1}{2}(BU + TC)(CB) = \\frac {1}{2}\\left(8 - \\frac {\\sqrt {3}}{9}h\\right)h = 4h - \\frac {\\sqrt {3}}{18}h^2$ Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$. Then $L = X + Y + Z = \\frac {3}{4}\\sqrt {3} + 3\\pi + 4h - \\frac {\\sqrt {3}}{18}h^2$ Obviously, the $L$ is greater than the area on the other side of line $l$. This other area is equal to the total area minus $L$. Thus: $\\frac {2}{1} = \\frac {L}{6h + \\frac {9}{2}{\\pi} - L} \\implies 12h + 9\\pi = 3L$. Now just solve for $h$. \\begin{align*} 12h + 9\\pi & = \\frac {9}{4}\\sqrt {3} + 9\\pi + 12h - \\frac {\\sqrt {3}}{6}h^2 \\\\ 0 & = \\frac {9}{4}\\sqrt {3} - \\frac {\\sqrt {3}}{6}h^2 \\\\ h^2 & = \\frac {9}{4}(6) \\\\ h & = \\frac {3}{2}\\sqrt {6} \\end{align*} Don't forget to un-rescale at the end to get $AD = \\frac {3}{2}\\sqrt {6} \\cdot 42 = 63\\sqrt {6}$. Finally, the answer is $63 + 6 = 069. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"Let $O$ be the center of the semicircle. It follows that $AU + UO = AN = NO = 126$, so triangle $ANO$ is equilateral. Let $Y$ be the foot of the altitude from $N$, such that $NY = 63\\sqrt{3}$ and $NU = 21$. Finally, denote $DT = a$, and $AD = x$. Extend $U$ to point $Z$ so that $Z$ is on $CD$ and $UZ$ is perpendicular to $CD$. It then follows that $ZT = a-84$. Since $NYU$ and $UZT$ are similar, $\\frac {x}{a-84} = \\frac {63\\sqrt{3}}{21} = 3\\sqrt{3}$ Given that line $NT$ divides $R$ into a ratio of $1:2$, we can also say that $(x)(\\frac{84+a}{2}) + \\frac {126^2\\pi}{6} - (63)(21)(\\sqrt{3}) = (\\frac{1}{3})(252x + \\frac{126^2\\pi}{2})$ where the first term is the area of trapezoid $AUTD$, the second and third terms denote the areas of $\\frac{1}{6}$ a full circle, and the area of $NUO$, respectively, and the fourth term on the right side of the equation is equal to $R$. Cancelling out the $\\frac{126^2\\pi}{6}$ on both sides, we obtain $(x)(\\frac{84+a}{2}) - \\frac{252x}{3} = (63)(21)(\\sqrt{3})$ By adding and collecting like terms, $\\frac{3ax - 252x}{6} = (63)(21)(\\sqrt{3})$ $\\frac{(3x)(a-84)}{6} = (63)(21)(\\sqrt{3})$. Since $a - 84 = \\frac{x}{3\\sqrt{3}}$, $\\frac {(3x)(\\frac{x}{3\\sqrt{3}})}{6} = (63)(21)(\\sqrt{3})$ $\\frac {x^2}{\\sqrt{3}} = (63)(126)(\\sqrt{3})$ $x^2 = (63)(126)(3) = (2)(3^5)(7^2)$ $x = AD = (7)(3^2)(\\sqrt{6}) = 63\\sqrt{6}$, so the answer is $069. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"Note that the total area of $\\mathcal{R}$ is $252DA + \\frac {126^2 \\pi}{2}$ and thus one of the regions has area $84DA + \\frac {126^2 \\pi}{6}$ As in the above solutions we discover that $\\angle AON = 60^\\circ$, thus sector $ANO$ of the semicircle has $\\frac{1}{3}$ of the semicircle's area. Similarly, dropping the $N'T'$ perpendicular we observe that $[AN'T'D] = 84DA$, which is $\\frac{1}{3}$ of the total rectangle. Denoting the region to the left of $\\overline {NT}$ as $\\alpha$ and to the right as $\\beta$, it becomes clear that if $[\\triangle UT'T] = [\\triangle NUO]$ then the regions will have the desired ratio. Using the 30-60-90 triangle, the slope of $NT$, is ${-3\\sqrt{3}}$, and thus $[\\triangle UT'T] = \\frac {DA^2}{6\\sqrt{3}}$. $[NUO]$ is most easily found by $\\frac{absin(c)}{2}$: $[\\triangle NUO] = \\frac {126*42 * \\frac {\\sqrt{3}}{2}}{2}$ Equating, $\\frac {126*42 * \\frac {\\sqrt{3}}{2}}{2} = \\frac {DA^2}{6\\sqrt{3}}$ Solving, $63 * 21 * 3 * 6 = DA^2$ $DA = 63 \\sqrt{6} \\longrightarrow 069. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"Like above solutions, note that $ANO$ is equilateral with side length $126,$ where $O$ is the midpoint of $AB.$ Then, if we let $DA=a$ and set origin at $D=(0,0),$ we get $N=(63,a+63\\sqrt{3}), U=(84,a).$ Line $NU$ is then $y-a=\\sqrt{27}(x-84),$ so it intersects $CA,$ the $x$-axis, at $x=(a/\\sqrt{27}+84),$ giving us point $T.$ Now the area of region $R$ is $252a+\\pi(126)^2 / 2,$ so one third of that is $84a+\\pi(126)^2 / 6.$ The area of the smaller piece of $R$ is $[AUTD] + [ANU] + [\\text{lune} AN] = \\frac{1}{2} \\cdot a(84+\\frac{a}{\\sqrt{27}}+84)+\\frac{1}{2} \\cdot 84 \\cdot 63 \\sqrt{3}+ \\frac{\\pi (126)^2}{6}-\\frac{1}{2} \\cdot 126 \\cdot 63 \\sqrt{2}$ $=\\frac{a^2}{2\\sqrt{27}}+84a-21\\cdot 63\\sqrt{2} + \\frac{\\pi(126)^2}{6}.$ Setting this equal to $84a+\\pi(126)^2 / 6$ and canceling the $84a + \\pi(126)^2$ yields $\\frac{a^2}{2\\sqrt{27}}=21 \\cdot 63 \\sqrt{3},$ so $a = 63 \\sqrt{6}$ and the anser is $069. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"Like above solutions, note that $ANO$ is equilateral with side length $126,$ where $O$ is the midpoint of $AB.$ Then, if we let $DA=a$ and set origin at $D=(0,0),$ we get $N=(63,a+63\\sqrt{3}), U=(84,a).$ Line $NU$ is then $y-a=\\sqrt{27}(x-84),$ so it intersects $CA,$ the $x$-axis, at $x=(a/\\sqrt{27}+84),$ giving us point $T.$ Now the area of region $R$ is $252a+\\pi(126)^2 / 2,$ so one third of that is $84a+\\pi(126)^2 / 6.$ The area of the smaller piece of $R$ is $[AUTD] + [ANU] + [\\text{lune} AN] = \\frac{1}{2} \\cdot a(84+\\frac{a}{\\sqrt{27}}+84)+\\frac{1}{2} \\cdot 84 \\cdot 63 \\sqrt{3}+ \\frac{\\pi (126)^2}{6}-\\frac{1}{2} \\cdot 126 \\cdot 63 \\sqrt{2}$ $=\\frac{a^2}{2\\sqrt{27}}+84a-21\\cdot 63\\sqrt{2} + \\frac{\\pi(126)^2}{6}.$ Setting this equal to $84a+\\pi(126)^2 / 6$ and canceling the $84a + \\pi(126)^2$ yields $\\frac{a^2}{2\\sqrt{27}}=21 \\cdot 63 \\sqrt{3},$ so $a = 63 \\sqrt{6}$ and the anser is $069. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"Once we establish that $\\Delta ANO$ is equilateral, we have \\[[{\\rm Sector } BON] = 2[{\\rm Sector } AON], [BCT'U]=2[ADT'U]\\] \\[\\Rightarrow [\\overset{\\large\\frown}{NB} CT'UO]=2[\\overset{\\large\\frown}{NA} DT'UO]\\] On the other hand, \\[[\\overset{\\large\\frown}{NB} CT]=2[\\overset{\\large\\frown}{NA} DT]\\] Therefore, $[UT'T]=[NUO]$. Now, $UO=42, NU=21 \\Rightarrow [UT'T]=[NUO]=2[NN'U]$. Also $\\Delta UT'T \\sim \\Delta NN'U$. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops",
"Once we establish that $\\Delta ANO$ is equilateral, we have \\[[{\\rm Sector } BON] = 2[{\\rm Sector } AON], [BCT'U]=2[ADT'U]\\] \\[\\Rightarrow [\\overset{\\large\\frown}{NB} CT'UO]=2[\\overset{\\large\\frown}{NA} DT'UO]\\] On the other hand, \\[[\\overset{\\large\\frown}{NB} CT]=2[\\overset{\\large\\frown}{NA} DT]\\] Therefore, $[UT'T]=[NUO]$. Now, $UO=42, NU=21 \\Rightarrow [UT'T]=[NUO]=2[NN'U]$. Also $\\Delta UT'T \\sim \\Delta NN'U$. Therefore, \\[DA=UT'=\\sqrt{2} NN'=\\sqrt{2} \\left(\\frac{\\sqrt{3}}{2}\\cdot 126\\right)=63\\sqrt{6}\\longrightarrow 069\\] ~asops"
] |
2010-I-14
| 2,010
| 14
|
For each positive integer $n,$ let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$ . Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .
| 109
|
I
|
[
"Observe that $f$ is strictly increasing in $n$. We realize that we need $100$ terms to add up to around $300$, so we need some sequence of $2$s, $3$s, and then $4$s. It follows that $n \\approx 100$ (alternatively, use binary search to get to this, with $n\\le 1000$). Manually checking shows that $f(109) = 300$ and $f(110) > 300$. Thus, our answer is $109.",
"Because we want the value for which $f(n)=300$, the average value of the 100 terms of the sequence should be around $3$. For the value of $\\lfloor \\log_{10} (kn) \\rfloor$ to be $3$, $1000 \\le kn < 10000$. We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let $k=50$, so $50n=\\frac{10000+1000}{2}=\\frac{11000}{2}=5500$, and $n = 110$. $f(110) = 301$, so we want to lower $n$. Testing $109$ yields $300$, so our answer is still $109.",
"For any $n$ where the sum is close to $300$, all the terms in the sum must be equal to $2$, $3$ or $4$. Let $M$ be the number of terms less than or equal to $3$ and $N$ be the number of terms equal to $2$ (also counted in $M$). With this definition of $M$ and $N$ the total will be $400 - M - N \\le 300$, from which $M + N \\ge 100$. Now $M+1$ is the smallest integer $k$ for which $\\log_{10}(kn) \\ge 4$ or $kn \\ge 10000$, thus \\[M = \\left\\lfloor\\frac{9999}{n}\\right\\rfloor.\\] Similarly, \\[N = \\left\\lfloor\\frac{999}{n}\\right\\rfloor = \\left\\lfloor\\frac{M}{10}\\right\\rfloor.\\] Therefore, \\[M + \\left\\lfloor \\frac{M}{10} \\right\\rfloor \\ge 100 \\implies M \\ge \\left\\lceil\\frac{1000}{11}\\right\\rceil = 91 \\implies n \\le \\left\\lfloor\\frac{9999}{91}\\right\\rfloor = 109.\\] Since we want the largest possible $n$, the answer is $109.",
"Since we're working with base-$10$ logarithms, we can start by testing out $n$'s that are powers of $10$. For $n = 1$, the terms in the sum are $\\lfloor \\log_{10} (1)\\rfloor, \\lfloor \\log_{10} (2)\\rfloor, \\lfloor \\log_{10} (3) \\rfloor , . . . , \\lfloor \\log_{10} (100) \\rfloor$. For numbers $1$-$9$, $\\lfloor \\log_{10} (kn) \\rfloor = 0$. Then we have $90$ numbers, namely $10$-$99$, for which $\\lfloor \\log_{10} (kn) \\rfloor = 1$. The last number we have is $100$, which gives us $\\lfloor \\log_{10} (kn) \\rfloor = 2$. This sum gives us only $90 + 2 = 92$, which is much too low. However, applying the same counting technique for $n = 10$, our sum comes out to be $9 + 180 + 3 = 192$, since there are $9$ terms for which $\\lfloor \\log_{10} (kn) \\rfloor = 1$, $90$ terms for which $\\lfloor \\log_{10} (kn) \\rfloor = 2$, and one term for which $\\lfloor \\log_{10} (kn) \\rfloor = 3$. So we go up one more power of $10$ and get $18 + 270 + 4 = 292$, which is very close to what we are looking for. Now we only have to bump up the value of $n$ a bit and check our sum. Each increase in $n$ by $1$ actually increases the value of our sum by $1$ as well (except for $n = 101$), because whenever a $4$ is added to the sum, a $3$ is taken away. It doesn't take long to check and see that the value of $n$ we're looking for is $109. ~ anellipticcurveoverq",
"Since we're working with base-$10$ logarithms, we can start by testing out $n$'s that are powers of $10$. For $n = 1$, the terms in the sum are $\\lfloor \\log_{10} (1)\\rfloor, \\lfloor \\log_{10} (2)\\rfloor, \\lfloor \\log_{10} (3) \\rfloor , . . . , \\lfloor \\log_{10} (100) \\rfloor$. For numbers $1$-$9$, $\\lfloor \\log_{10} (kn) \\rfloor = 0$. Then we have $90$ numbers, namely $10$-$99$, for which $\\lfloor \\log_{10} (kn) \\rfloor = 1$. The last number we have is $100$, which gives us $\\lfloor \\log_{10} (kn) \\rfloor = 2$. This sum gives us only $90 + 2 = 92$, which is much too low. However, applying the same counting technique for $n = 10$, our sum comes out to be $9 + 180 + 3 = 192$, since there are $9$ terms for which $\\lfloor \\log_{10} (kn) \\rfloor = 1$, $90$ terms for which $\\lfloor \\log_{10} (kn) \\rfloor = 2$, and one term for which $\\lfloor \\log_{10} (kn) \\rfloor = 3$. So we go up one more power of $10$ and get $18 + 270 + 4 = 292$, which is very close to what we are looking for. Now we only have to bump up the value of $n$ a bit and check our sum. Each increase in $n$ by $1$ actually increases the value of our sum by $1$ as well (except for $n = 101$), because whenever a $4$ is added to the sum, a $3$ is taken away. It doesn't take long to check and see that the value of $n$ we're looking for is $109. ~ anellipticcurveoverq"
] |
2010-I-15
| 2,010
| 15
|
In $\triangle{ABC}$ with $AB = 12$ , $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .
| 45
|
I
|
[
"[asy] /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200); /* segments and figures */ draw((0,0)--(15,0)); draw((15,0)--(6.66667,9.97775)); draw((6.66667,9.97775)--(0,0)); draw((7.33333,0)--(6.66667,9.97775)); draw(circle((4.66667,2.49444),2.49444)); draw(circle((9.66667,2.49444),2.49444)); draw((4.66667,0)--(4.66667,2.49444)); draw((9.66667,2.49444)--(9.66667,0)); /* points and labels */ label(\"r\",(10.19662,1.92704),SE); label(\"r\",(5.02391,1.8773),SE); dot((0,0)); label(\"$A$\",(-1.04408,-0.60958),NE); dot((15,0)); label(\"$C$\",(15.41907,-0.46037),NE); dot((6.66667,9.97775)); label(\"$B$\",(6.66525,10.23322),NE); label(\"$15$\",(6.01866,-1.15669),NE); label(\"$13$\",(11.44006,5.50815),NE); label(\"$12$\",(2.28834,5.75684),NE); dot((7.33333,0)); label(\"$M$\",(7.56053,-1.000),NE); label(\"$H_1$\",(3.97942,-1.200),NE); label(\"$H_2$\",(9.54741,-1.200),NE); dot((4.66667,2.49444)); label(\"$I_1$\",(3.97942,2.92179),NE); dot((9.66667,2.49444)); label(\"$I_2$\",(9.54741,2.92179),NE); clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle); [/asy] Solution 1 Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\\frac {[ABM]}{[CBM]} = \\frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\\frac {[ABM]}{[CBM]} = \\frac {p(ABM)}{p(CBM)} = \\frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \\frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$. By Stewart's Theorem, we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$, we have $432 = \\frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$ Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $6x$ is an integer because we can divide the polynomial by $2$. The only such $x$ in the above-stated range is $\\frac {22}3$. Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\\frac {22}3$. Then $CM = \\frac {23}3$, and our desired ratio $\\frac {AM}{CM} = \\frac {22}{23}$, giving us an answer of $045. ~blueprimes",
"Let $AM = x$, then $CM = 15 - x$. Also let $BM = d$ Clearly, $\\frac {[ABM]}{[CBM]} = \\frac {x}{15 - x}$. We can also express each area by the rs formula. Then $\\frac {[ABM]}{[CBM]} = \\frac {p(ABM)}{p(CBM)} = \\frac {12 + d + x}{28 + d - x}$. Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \\frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$. By Stewart's Theorem, we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$, we have $432 = \\frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$ Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $6x$ is an integer because we can divide the polynomial by $2$. The only such $x$ in the above-stated range is $\\frac {22}3$. Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\\frac {22}3$. Then $CM = \\frac {23}3$, and our desired ratio $\\frac {AM}{CM} = \\frac {22}{23}$, giving us an answer of $045. ~blueprimes",
"Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$. Let the incenters of $\\triangle ABM$ and $\\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$. From $r = \\sqrt {(s - a)(s - b)(s - c)}/s$, we find that \\begin{align*}r^2 & = \\frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\\\ & = \\frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \\end{align*} Let the incircle of $\\triangle ABM$ meet $AM$ at $P$ and the incircle of $\\triangle BCM$ meet $CM$ at $Q$. Then note that $I_1 P Q I_2$ is a rectangle. Also, $\\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\\angle AMB$ and $\\angle CMB$ respectively and $\\angle AMB + \\angle CMB = 180^\\circ$. By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC)/2 = - x + y + 1$. Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$, so by similar triangles we find that $r^2 = MP \\cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields \\begin{align*} 2y^2 - 30 = 2xy + 5x - 7y \\\\ 2y^2 - 70 = - 2xy - 5x + 7y, \\end{align*} and adding these we have \\[4y^2 - 100 = 0\\implies y = 5\\implies x = 11/3 \\\\ \\implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \\implies 045. ~blueprimes",
"Let the incircle of $ABM$ hit $AM$, $AB$, $BM$ at $X_{1},Y_{1},Z_{1}$, and let the incircle of $CBM$ hit $MC$, $BC$, $BM$ at $X_{2},Y_{2},Z_{2}$. Draw the incircle of $ABC$, and let it be tangent to $AC$ at $X$. Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$, and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$. By standard computations, $AX=\\dfrac{AB+AC-BC}{2}=7$ and $CX=\\dfrac{BC+AC-AB}{2}=8$. Now, let $AX_{1}=7x$ and $CX_{2}=8x$. We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$. Then, $BZ_{1}=12-7x$. We also have $CY_{2}=CX_{2}=8x$, so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$. Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$. Also,$X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$. Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range). Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$. To finish, we will compute area ratios. $\\dfrac{[ABM]}{[CBM]}=\\dfrac{AM}{MC}=\\dfrac{8-x}{7+x}$. Also, since their inradii are equal, we get $\\dfrac{[ABM]}{[CBM]}=\\dfrac{40-16x}{40-14x}$. Equating and cross multiplying yields the quadratic $3x^{2}-8x+4=0$, so $x=2/3,2$. However, observe that $AX_{1}+CX_{2}=15x<15$, so we take $x=2/3$. Our ratio is therefore $\\dfrac{8-2/3}{7+2/3}=\\dfrac{22}{23}$, giving the answer $045. ~blueprimes",
"Suppose the incircle of $ABM$ touches $AM$ at $X$, and the incircle of $CBM$ touches $CM$ at $Y$. Then \\[r = AX \\tan(A/2) = CY \\tan(C/2)\\] We have $\\cos A = \\frac{12^2+15^2-13^2}{2\\cdot 12\\cdot 15} = \\frac{200}{30\\cdot 12}=\\frac{5}{9}$, $\\tan(A/2) = \\sqrt{\\frac{1-\\cos A}{1+\\cos A}} = \\sqrt{\\frac{9-5}{9+5}} = \\frac{2}{\\sqrt{14}}$ $\\cos C = \\frac{13^2+15^2-12^2}{2\\cdot 13\\cdot 15} = \\frac{250}{30\\cdot 13} = \\frac{25}{39}$, $\\tan(C/2) = \\sqrt{\\frac{39-25}{39+25}}=\\frac{\\sqrt{14}}{8}$, Therefore $AX/CY = \\tan(C/2)/\\tan(A/2) = \\frac{14}{2\\cdot 8}= \\frac{7}{8}.$ And since $AX=\\frac{1}{2}(12+AM-BM)$, $CY = \\frac{1}{2}(13+CM-BM)$, \\[\\frac{12+AM-BM}{13+CM-BM} = \\frac{7}{8}\\] \\[96+8AM-8BM = 91 +7CM-7BM\\] \\[BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \\dots\\dots (1)\\] Now, $\\frac{AM}{CM} = \\frac{[ABM]}{[CBM]} = \\frac{\\frac{1}{2}(12+AM+BM)r}{\\frac{1}{2}(13+CM+BM)r}=\\frac{12+AM+BM}{13+CM+BM}= \\frac{12+BM}{13+BM} = \\frac{17+15(AM-7)}{18+15(AM-7)}$ \\[\\frac{AM}{15} = \\frac{17+15(AM-7)}{35+30(AM-7)} = \\frac{15AM-88}{30AM-175}\\] \\[6AM^2 - 35AM = 45AM-264\\] \\[3AM^2 -40AM+132=0\\] \\[(3AM-22)(AM-6)=0\\] So $AM=22/3$ or $6$. But from (1) we know that $5+15(AM-7)>0$, or $AM>7-1/3>6$, so $AM=22/3$, $CM=15-22/3=23/3$, $AM/CM=22/23$. Solution 5 Let the common inradius equal r, $BM = x$, $AM = y$, $MC = z$ From the prespective of $\\triangle{ABM}$ and $\\triangle{BMC}$ we get: $S_{ABM} = rs = r \\cdot (\\frac{12+x+y}{2})$ $S_{BMC} = rs = r \\cdot (\\frac{13+x+z}{2})$ Add two triangles up, we get $\\triangle{ABC}$ : $S_{ABC} = S_{ABM} + S_{BMC} = r \\cdot \\frac{25+2x+y+z}{2}$ Since $y + z = 15$, we get: $r = \\frac{S_{ABC}}{20 + x}$ By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$, we can get: $r \\cdot cot(\\frac{\\angle A}{2}) =r \\cdot A H_1 = r \\cdot \\frac{12+y-x}{2}$ and $r \\cdot cot(\\frac{\\angle C}{2}) = r \\cdot H_2 C = r \\cdot \\frac{13+z-x}{2}$ Adding these up, we get: $r \\cdot (cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) = \\frac{25+y+z-2x}{2} = \\frac{40-2x}{2} = 20-x$ $r = \\frac{20-x}{cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})}$ Now, we have 2 values equal to r, we can set them equal to each other: $\\frac{S_{ABC}}{20 + x} = \\frac{20-x}{cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})}$ If we let R denote the incircle of ABC, note: AC = $(cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) \\cdot R = 15$ and $S_{ABC} = \\frac{12+13+15}{2} \\cdot R = 20 \\cdot R$. By cross multiplying the equation above, we get: $400 - x^2 = (cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) \\cdot S_{ABC} = (cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) \\cdot R \\cdot 20 = 15 \\cdot 20 = 300$ We can find out x: $x = 10$. Now, we can find ratio of y and z: $\\frac{AM}{CM} = \\frac{y}{z} = \\frac{S_{ABM}}{S_{BCM}} = \\frac{r \\cdot \\frac{22+y}{2} }{r \\cdot \\frac{23+z}{2}} = \\frac{22+y}{23+z} = \\frac{22}{23}$ The answer is $045. ~blueprimes",
"Let the common inradius equal r, $BM = x$, $AM = y$, $MC = z$ From the prespective of $\\triangle{ABM}$ and $\\triangle{BMC}$ we get: $S_{ABM} = rs = r \\cdot (\\frac{12+x+y}{2})$ $S_{BMC} = rs = r \\cdot (\\frac{13+x+z}{2})$ Add two triangles up, we get $\\triangle{ABC}$ : $S_{ABC} = S_{ABM} + S_{BMC} = r \\cdot \\frac{25+2x+y+z}{2}$ Since $y + z = 15$, we get: $r = \\frac{S_{ABC}}{20 + x}$ By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$, we can get: $r \\cdot cot(\\frac{\\angle A}{2}) =r \\cdot A H_1 = r \\cdot \\frac{12+y-x}{2}$ and $r \\cdot cot(\\frac{\\angle C}{2}) = r \\cdot H_2 C = r \\cdot \\frac{13+z-x}{2}$ Adding these up, we get: $r \\cdot (cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) = \\frac{25+y+z-2x}{2} = \\frac{40-2x}{2} = 20-x$ $r = \\frac{20-x}{cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})}$ Now, we have 2 values equal to r, we can set them equal to each other: $\\frac{S_{ABC}}{20 + x} = \\frac{20-x}{cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})}$ If we let R denote the incircle of ABC, note: AC = $(cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) \\cdot R = 15$ and $S_{ABC} = \\frac{12+13+15}{2} \\cdot R = 20 \\cdot R$. By cross multiplying the equation above, we get: $400 - x^2 = (cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) \\cdot S_{ABC} = (cot(\\frac{\\angle A}{2})+cot(\\frac{\\angle C}{2})) \\cdot R \\cdot 20 = 15 \\cdot 20 = 300$ We can find out x: $x = 10$. Now, we can find ratio of y and z: $\\frac{AM}{CM} = \\frac{y}{z} = \\frac{S_{ABM}}{S_{BCM}} = \\frac{r \\cdot \\frac{22+y}{2} }{r \\cdot \\frac{23+z}{2}} = \\frac{22+y}{23+z} = \\frac{22}{23}$ The answer is $045. ~blueprimes",
"Let $CM=x, AM=rx, BM=d$. $x+rx=15\\Rightarrow x=\\frac{15}{1+r}$. Similar to Solution 1, we have \\[r=\\frac{[AMB]}{[CMB]}=\\frac{12+rx+d}{13+x+d} \\Rightarrow d=\\frac{13r-12}{1-r}\\] as well as \\[12^2\\cdot x + 13^2 rx=15x\\cdot rx+15d^2 (\\text{via Stewart's Theorem})\\] \\[\\frac{(12^2 + 13^2r) \\cdot 15}{1+r} - \\frac{15r\\cdot 15^2}{(1+r)^2}=\\frac{15(13r-12)^2}{(1-r)^2}\\] \\[\\frac{169r^2+88r+144}{(1+r)^2}=\\frac{(13r-12)^2}{(1-r)^2} =\\frac{169r^2-312r+144}{(1-r)^2} =\\frac{400r}{4r}=100\\] (here we used the fact that if $\\frac{a}{b} = \\frac{c}{d} = k,$ then $\\frac{a-c}{b-d}=k$ as well.) Notice $\\frac{12}{13} < r < 1$, so $\\frac{13r-12}{1-r} = 10$ and $r = \\frac{22}{23}. ~blueprimes",
"Let $CM=x, AM=rx, BM=d$. $x+rx=15\\Rightarrow x=\\frac{15}{1+r}$. Similar to Solution 1, we have \\[r=\\frac{[AMB]}{[CMB]}=\\frac{12+rx+d}{13+x+d} \\Rightarrow d=\\frac{13r-12}{1-r}\\] as well as \\[12^2\\cdot x + 13^2 rx=15x\\cdot rx+15d^2 (\\text{via Stewart's Theorem})\\] \\[\\frac{(12^2 + 13^2r) \\cdot 15}{1+r} - \\frac{15r\\cdot 15^2}{(1+r)^2}=\\frac{15(13r-12)^2}{(1-r)^2}\\] \\[\\frac{169r^2+88r+144}{(1+r)^2}=\\frac{(13r-12)^2}{(1-r)^2} =\\frac{169r^2-312r+144}{(1-r)^2} =\\frac{400r}{4r}=100\\] (here we used the fact that if $\\frac{a}{b} = \\frac{c}{d} = k,$ then $\\frac{a-c}{b-d}=k$ as well.) Notice $\\frac{12}{13} < r < 1$, so $\\frac{13r-12}{1-r} = 10$ and $r = \\frac{22}{23}. ~blueprimes",
"Let $BM = d, AM = x, CM = 15 - x$. Observe that we have the equation by the incircle formula: \\[\\frac{[ABM]}{12 + AM + MB} = \\frac{[CBM]}{13 + CM + MB} \\implies \\frac{AM}{CM} = \\frac{12 + MB}{13 + MB} \\implies \\frac{x}{15 - x} = \\frac{12 + d}{13 + d}.\\] Now let $X$ be the point of tangency between the incircle of $\\triangle ABC$ and $AC$. Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\\triangle ABM$ and $\\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$. By homothety we have \\[\\frac{AP}{7} = \\frac{CQ}{8} \\implies 8(AP) = 7(CQ) \\implies 8(12 + x - d) = 7(13 + 15 - x - d) \\implies d = 15x - 100.\\] Substituting into the first equation derived earlier it is left to solve \\[\\frac{x}{15 - x} = \\frac{15x - 88}{15x - 87} \\implies 3x^2 - 40x + 132 \\implies (x - 6)(3x - 22) = 0.\\] Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \\frac{22}{3}$ so $\\frac{AM}{CM} = \\frac{\\frac{22}{3}}{15 - \\frac{22}{3}} = \\frac{22}{23}.$ The requested sum is $22 + 23 = 45. ~blueprimes",
"Let $BM = d, AM = x, CM = 15 - x$. Observe that we have the equation by the incircle formula: \\[\\frac{[ABM]}{12 + AM + MB} = \\frac{[CBM]}{13 + CM + MB} \\implies \\frac{AM}{CM} = \\frac{12 + MB}{13 + MB} \\implies \\frac{x}{15 - x} = \\frac{12 + d}{13 + d}.\\] Now let $X$ be the point of tangency between the incircle of $\\triangle ABC$ and $AC$. Additionally, let $P$ and $Q$ be the points of tangency between the incircles of $\\triangle ABM$ and $\\triangle CBM$ with $AC$ respectively. Some easy calculation yields $AX = 7, CX = 8$. By homothety we have \\[\\frac{AP}{7} = \\frac{CQ}{8} \\implies 8(AP) = 7(CQ) \\implies 8(12 + x - d) = 7(13 + 15 - x - d) \\implies d = 15x - 100.\\] Substituting into the first equation derived earlier it is left to solve \\[\\frac{x}{15 - x} = \\frac{15x - 88}{15x - 87} \\implies 3x^2 - 40x + 132 \\implies (x - 6)(3x - 22) = 0.\\] Now $x = 6$ yields $d = -10$ which is invalid, hence $x = \\frac{22}{3}$ so $\\frac{AM}{CM} = \\frac{\\frac{22}{3}}{15 - \\frac{22}{3}} = \\frac{22}{23}.$ The requested sum is $22 + 23 = 45. ~blueprimes"
] |
2010-II-1
| 2,010
| 1
|
Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000$ .
| 640
|
II
|
[
"If an integer is divisible by $36$, it must also be divisible by $9$ since $9$ is a factor of $36$. It is a well-known fact that, if $N$ is divisible by $9$, the sum of the digits of $N$ is a multiple of $9$. Hence, if $N$ contains all the even digits, the sum of the digits would be $0 + 2 + 4 + 6 + 8 = 20$, which is not divisible by $9$ and thus $36$. The next logical try would be $8640$, which happens to be divisible by $36$. Thus, $N = 8640 \\equiv 640."
] |
2010-II-2
| 2,010
| 2
|
A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 281
|
II
|
[
"Any point outside the square with side length $\\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\\frac{3}{5}$ that has the same center and orientation as the unit square has $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$. [asy] unitsize(1mm); defaultpen(linewidth(.8pt)); draw((0,0)--(0,30)--(30,30)--(30,0)--cycle); draw((6,6)--(6,24)--(24,24)--(24,6)--cycle); draw((10,10)--(10,20)--(20,20)--(20,10)--cycle); fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray); fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white); [/asy] Since the area of the unit square is $1$, the probability of a point $P$ with $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares. $\\left(\\frac{3}{5}\\right)^2-\\left(\\frac{1}{3}\\right)^2=\\frac{9}{25}-\\frac{1}{9}=\\frac{56}{225}$ Thus, the answer is $56 + 225 = 281",
"First, let's figure out $d(P) \\geq \\frac{1}{3}$ which is\\[\\left(\\frac{3}{5}\\right)^2=\\frac{9}{25}.\\]Then, $d(P) \\geq \\frac{1}{5}$ is a square inside $d(P) \\geq \\frac{1}{3}$, so\\[\\left(\\frac{1}{3}\\right)^2=\\frac{1}{9}.\\]Therefore, the probability that $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$ is\\[\\frac{9}{25}-\\frac{1}{9}=\\frac{56}{225}\\]So, the answer is $56+225=281",
"First, lets assume that point $P$ is closest to a side $S$ of the square. If it is $\\frac{1}{5}$ far from $S$, then it should be at least $\\frac{1}{5}$ from both the adjacent sides of $S$ in the square. This leaves a segment of $1 - 2 \\cdot \\frac{1}{5} = \\frac{3}{5}$. If the distance from $P$ to $S$ is $\\frac{1}{3}$, then notice the length of the side-ways segment for $P$ is $1 - 2 \\cdot \\frac{1}{3} = \\frac{1}{3}$. Notice that as the distance from $P$ to $S$ increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides $\\frac{3}{5}$ and $\\frac{1}{3}$ with height $\\frac{1}{3} - \\frac{1}{5} = \\frac{2}{15}$. This trapezoid has area (or probability for one side) $\\frac{1}{2} \\cdot \\left(\\frac{1}{3}+\\frac{3}{5}\\right)\\cdot \\frac{2}{15} = \\frac{14}{225}$. Since the square has $4$ sides, we multiply by $4$. Hence, the probability is $\\frac{56}{225}$. The answer is $281. ~Saucepan_man02"
] |
2010-II-3
| 2,010
| 3
|
Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$ . Find the greatest positive integer $n$ such that $2^n$ divides $K$ .
| 150
|
II
|
[
"In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\\cdots(19^1)$ (or alternatively, $19! \\cdot 18! \\cdots 1!$.) When we count the number of factors of $2$, we have 4 groups, factors that are divisible by $2$ at least once, twice, three times and four times. Numbers that are divisible by $2$ at least once: $2, 4, \\cdots, 18$ Exponent corresponding to each one of them $18, 16, \\cdots 2$ Sum $=2+4+\\cdots+18=\\frac{(20)(9)}{2}=90$ Numbers that are divisible by $2$ at least twice: $4, 8, \\cdots, 16$ Exponent corresponding to each one of them $16, 12, \\cdots 4$ Sum $=4+8+\\cdots+16=\\frac{(20)(4)}{2}=40$ Numbers that are divisible by $2$ at least three times: $8,16$ Exponent corresponding to each one of them $12, 4$ Sum $=12+4=16$ Number that are divisible by $2$ at least four times: $16$ Exponent corresponding to each one of them $4$ Sum $=4$ Summing these give an answer of $150."
] |
2010-II-4
| 2,010
| 4
|
Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 52
|
II
|
[
"Solution 1 There are $12 \\cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart. If we number the gates $1$ through $12$, then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$, $6$, $7$, $8$ have eight. Therefore, the number of valid gate assignments is \\[2\\cdot(4+5+6+7)+4\\cdot8 = 2 \\cdot 22 + 4 \\cdot 8 = 76\\] so the probability is $\\frac{76}{132} = \\frac{19}{33}$. The answer is $19 + 33 = 052.",
"There are $12 \\cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart. If we number the gates $1$ through $12$, then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$, $6$, $7$, $8$ have eight. Therefore, the number of valid gate assignments is \\[2\\cdot(4+5+6+7)+4\\cdot8 = 2 \\cdot 22 + 4 \\cdot 8 = 76\\] so the probability is $\\frac{76}{132} = \\frac{19}{33}$. The answer is $19 + 33 = 052.",
"As before, derive that there are $132$ possibilities for Dave's original and replacement gates. Now suppose that Dave has to walk $100k$ feet to get to his new gate. This means that Dave's old and new gates must be $k$ gates apart. (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.) There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$. As a result, the total number of valid gate arrangements is \\[2\\cdot 11 + 2\\cdot 10 + 2\\cdot 9 + 2\\cdot 8 = 76\\] and so the requested probability is $\\tfrac{19}{33}$ for a final answer of $052."
] |
2010-II-5
| 2,010
| 5
|
Positive numbers $x$ , $y$ , and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$ . Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ .
| 75
|
II
|
[
"Using the properties of logarithms, $\\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\\log_{10}x)(\\log_{10}y) + (\\log_{10}x)(\\log_{10}z) + (\\log_{10}y)(\\log_{10}z)= 468$ by using the fact that $\\log_{10}ab = \\log_{10}a + \\log_{10}b$. Through further simplification, we find that $\\log_{10}x+\\log_{10}y+\\log_{10}z = 81$. It can be seen that there is enough information to use the formula $\\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, as we have both $\\ a+b+c$ and $\\ 2ab+2ac+2bc$, and we want to find $\\sqrt {a^2 + b^2 + c^2}$. After plugging in the values into the equation, we find that $\\ (\\log_{10}x)^2 + (\\log_{10}y)^2 + (\\log_{10}z)^2$ is equal to $\\ 6561 - 936 = 5625$. However, we want to find $\\sqrt {(\\log_{10}x)^2 + (\\log_{10}y)^2 + (\\log_{10}z)^2}$, so we take the square root of $\\ 5625$, or $075.",
"Let $a=\\log_{10}x$, $b=\\log_{10}y,$ and $c=\\log_{10}z$. We have $a+b+c=81$ and $a(b+c)+bc=ab+ac+bc=468$. Since these two equations look a lot like Vieta's for a cubic, create the polynomial $x^3-81x^2+468x=0$ (leave the constant term as $0$ to make things easy). Dividing by $x$ yields $x^2-81x+468=0$. Now we use the quadratic formula: $x=\\frac{81\\pm\\sqrt{81^2-4\\cdot468}}{2}$ $x=\\frac{81\\pm\\sqrt{4689}}{2}$ $x=\\frac{81+3\\sqrt{521}}{2}$, $x=\\frac{81-3\\sqrt{521}}{2}$ Since the question asks for $\\sqrt{a^2+b^2+c^2}$ (remember one of the values was the solution $x=0$ that we divided out in the beginning), we find: $\\sqrt{\\left(\\frac{81+3\\sqrt{521}}{2}\\right)^2+\\left(\\frac{81-3\\sqrt{521}}{2}\\right)^2}$ $=\\sqrt{2\\cdot\\frac{81^2+(3\\sqrt{521})^2}{4}}$ $=\\sqrt{\\frac{11250}{2}}$ $=075 ~bad_at_mathcounts"
] |
2010-II-6
| 2,010
| 6
|
Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients.
| 8
|
II
|
[
"You can factor the polynomial into two quadratic factors or a linear and a cubic factor. For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that \\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\\] Therefore, again setting coefficients equal, $a + c = 0\\Longrightarrow a=-c$, $b + d + ac = 0\\Longrightarrow b+d=a^2$ , $ad + bc = - n$, and so $bd = 63$. Since $b+d=a^2$, the only possible values for $(b,d)$ are $(1,63)$ and $(7,9)$. From this we find that the possible values for $n$ are $\\pm 8 \\cdot 62$ and $\\pm 4 \\cdot 2$. For the case of one linear and one cubic factor, doing a similar expansion and matching of the coefficients gives the smallest $n$ in that case to be $48$. Therefore, the answer is $4 \\cdot 2 = 008.",
"Let $x^4-nx+63=(x^2+ax+b)(x^2+cx+d)$. From this, we get that $bd=63\\implies d=\\frac{63}{b}$ and $a+c=0\\implies c=-a$. Plugging this back into the equation, we get $x^4-nx+63=(x^2+ax+b)\\left(x^2-ax+\\frac{63}{b}\\right)$. Expanding gives us $x^4-nx+63=x^4+\\left(-a^2+b+\\frac{63}{b}\\right)x^2+\\left(\\frac{63a}{b}-ab\\right)x+63$. Therefore $-a^2+b+\\frac{63}{b}=0$. Simplifying gets us $b(a^2-b)=63$. Since $a$ and $b$ must be integers, we can use guess and check for values of $b$ because $b$ must be a factor of $63$. Note that $b$ cannot be negative because $a$ would be imaginary. After guessing and checking, we find that the possible values of $(a,b)$ are $(\\pm 8, 1), (\\pm 4, 7), (\\pm 4, 9),$ and $(\\pm 8, 63)$. We have that $n=ab-\\frac{63a}{b}$. Plugging in our values for $a$ and $b$, we get that the smallest positive integer value $n$ can be is $008. -Heavytoothpaste"
] |
2010-II-7
| 2,010
| 7
|
Let $P(z)=z^3+az^2+bz+c$ , where $a$ , $b$ , and $c$ are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ , $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ .
| 136
|
II
|
[
"Set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$. Since $a,b,c\\in{R}$, the imaginary part of $a,b,c$ must be $0$. Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$, and therefore: $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$. Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: $x(x+6i)(2x-4-6i)$. The imaginary part is $6x^2-24x$, which is 0, and therefore $x=4$, since $x=0$ doesn't work. So now, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$, and therefore: $a=-12, b=84, c=-208$. Finally, we have $|a+b+c|=|-12+84-208|=136.",
"Set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$. Since $a,b,c\\in{R}$, the imaginary part of $a,b,c$ must be $0$. Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$, and therefore: $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$. Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: $x(x+6i)(2x-4-6i)$. The imaginary part is $6x^2-24x$, which is 0, and therefore $x=4$, since $x=0$ doesn't work. So now, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$, and therefore: $a=-12, b=84, c=-208$. Finally, we have $|a+b+c|=|-12+84-208|=136.",
"Same as solution 1 except that when you get to $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$, you don't need to find the imaginary part of $c$. We know that $x_1$ is a real number, which means that $x_2$ and $x_3$ are complex conjugates. Therefore, $x=2x-4$.",
"Note that at least one of $w+3i$, $w+9i$, or $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one. Let $w=x+yi$, where $x$ and $y$ are reals. Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$, so $w=x-3i$. Now note that since $w+3i$ is real, $w+9i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+9i}=2w-4$, so $\\overline{x+6i}=2(x-3i)-4$, implying that $x=4$, so $w=4-3i$. Case 2: $w+9i$ is real. This means that $x+yi+9i$ is real, so again setting imaginary part to zero we get $y=-9$, so $w=x-9i$. Now by the same logic as above $w+3i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+3i}=2w-4$, so $\\overline{x-6i}=2(x-9i)-4$, so $x+6i=2x-4-18i$, which has no solution as $x$ is real. Case 3: $2w-4$ is real. Going through the same steps, we get $y=0$, so $w=x$. Now $w+3i$ and $w+6i$ are complex conjugates, but $w=x$, which means that $\\overline{x+3i}=x+6i$, so $x-3i=x+6i$, which has no solutions. Thus case 1 is the only one that works, so $w=4-3i$ and our polynomial is $(z-(4))(z-(4+6i))(z-(4-6i))$. Note that instead of expanding this, we can save time by realizing that the answer format is $|a+b+c|$, so we can plug in $z=1$ to our polynomial to get the sum of coefficients, which will give us $a+b+c+1$. Plugging in $z=1$ into our polynomial, we get $(-3)(-3-6i)(-3+6i)$ which evaluates to $-135$. Since this is $a+b+c+1$, we subtract 1 from this to get $a+b+c=-136$, so $|a+b+c|=136. ~chrisdiamond10",
"Note that at least one of $w+3i$, $w+9i$, or $2w-4$ is real by complex conjugate roots. We now separate into casework based on which one. Let $w=x+yi$, where $x$ and $y$ are reals. Case 1: $w+3i$ is real. This implies that $x+yi+3i$ is real, so by setting the imaginary part equal to zero we get $y=-3$, so $w=x-3i$. Now note that since $w+3i$ is real, $w+9i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+9i}=2w-4$, so $\\overline{x+6i}=2(x-3i)-4$, implying that $x=4$, so $w=4-3i$. Case 2: $w+9i$ is real. This means that $x+yi+9i$ is real, so again setting imaginary part to zero we get $y=-9$, so $w=x-9i$. Now by the same logic as above $w+3i$ and $2w-4$ are complex conjugates. Thus $\\overline{w+3i}=2w-4$, so $\\overline{x-6i}=2(x-9i)-4$, so $x+6i=2x-4-18i$, which has no solution as $x$ is real. Case 3: $2w-4$ is real. Going through the same steps, we get $y=0$, so $w=x$. Now $w+3i$ and $w+6i$ are complex conjugates, but $w=x$, which means that $\\overline{x+3i}=x+6i$, so $x-3i=x+6i$, which has no solutions. Thus case 1 is the only one that works, so $w=4-3i$ and our polynomial is $(z-(4))(z-(4+6i))(z-(4-6i))$. Note that instead of expanding this, we can save time by realizing that the answer format is $|a+b+c|$, so we can plug in $z=1$ to our polynomial to get the sum of coefficients, which will give us $a+b+c+1$. Plugging in $z=1$ into our polynomial, we get $(-3)(-3-6i)(-3+6i)$ which evaluates to $-135$. Since this is $a+b+c+1$, we subtract 1 from this to get $a+b+c=-136$, so $|a+b+c|=136. ~chrisdiamond10",
"By vieta's we know the sum of the roots must be $-a$, a real number. That means $4w+12i-4$ is a real number, meaning $w$ has an imaaginary component of $-3i$. Now we write $w = x-3i$. Then, $w+3i$ is the real root, meaning the other two are complex conjugates. We have $\\overline{x+6i} = 2x-4-6i$, and solving, we get $x=4$. Then, $f(x) = (x-4)(x-4-6i)(x-4+6i) = (x-4)(x^2-8x+52)$. We get $|a+b+c| = |-12+84-208| = 136. -skibbysiggy ~hashbrown2009"
] |
2010-II-9
| 2,010
| 9
|
Let $ABCDEF$ be a regular hexagon. Let $G$ , $H$ , $I$ , $J$ , $K$ , and $L$ be the midpoints of sides $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , and $AF$ , respectively. The segments $\overline{AH}$ , $\overline{BI}$ , $\overline{CJ}$ , $\overline{DK}$ , $\overline{EL}$ , and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 11
|
II
|
[
"[asy] defaultpen(0.8pt+fontsize(12pt)); pair A,B,C,D,E,F; pair G,H,I,J,K,L; A=dir(0); B=dir(60); C=dir(120); D=dir(180); E=dir(240); F=dir(300); draw(A--B--C--D--E--F--cycle,blue); G=(A+B)/2; H=(B+C)/2; I=(C+D)/2; J=(D+E)/2; K=(E+F)/2; L=(F+A)/2; int i; for (i=0; i<6; i+=1) { draw(rotate(60*i)*(A--H),dotted); } pair M,N,O,P,Q,R; M=extension(A,H,B,I); N=extension(B,I,C,J); O=extension(C,J,D,K); P=extension(D,K,E,L); Q=extension(E,L,F,G); R=extension(F,G,A,H); draw(M--N--O--P--Q--R--cycle,red); label('$A$',A,(1,0)); label('$B$',B,NE); label('$C$',C,NW); label('$D$',D, W); label('$E$',E,SW); label('$F$',F,SE); label('$G$',G,NE); label('$H$',H, (0,1)); label('$I$',I,NW); label('$J$',J,SW); label('$K$',K, S); label('$L$',L,SE); label('$M$',M); label('$N$',N); label('$O$',(0,0),NE); dot((0,0)); [/asy] Let $M$ be the intersection of $\\overline{AH}$ and $\\overline{BI}$. Let $N$ be the intersection of $\\overline{BI}$ and $\\overline{CJ}$. Let $O$ be the center. Solution 1 Without loss of generality, let $BC=2.$ Note that $\\angle BMH$ is the vertical angle to an angle of the regular hexagon, so it has a measure of $120^\\circ$. Because $\\triangle ABH$ and $\\triangle BCI$ are rotational images of one another, we get that $\\angle{MBH}=\\angle{HAB}$ and hence $\\triangle ABH \\sim \\triangle BMH \\sim \\triangle BCI$. Using a similar argument, $NI=MH$, and \\[MN=BI-NI-BM=BI-(BM+MH).\\] Applying the Law of cosines on $\\triangle BCI$, $BI=\\sqrt{2^2+1^2-2(2)(1)(\\cos(120^\\circ))}=\\sqrt{7}$ \\begin{align*}\\frac{BC+CI}{BI}&=\\frac{3}{\\sqrt{7}}=\\frac{BM+MH}{BH} \\\\ BM+MH&=\\frac{3BH}{\\sqrt{7}}=\\frac{3}{\\sqrt{7}} \\\\ MN&=BI-(BM+MH)=\\sqrt{7}-\\frac{3}{\\sqrt{7}}=\\frac{4}{\\sqrt{7}} \\\\ \\frac{\\text{Area of smaller hexagon}}{\\text{Area of bigger hexagon}}&=\\left(\\frac{MN}{BC}\\right)^2=\\left(\\frac{2}{\\sqrt{7}}\\right)^2=\\frac{4}{7}\\end{align*} Thus, the answer is $4 + 7 = 011. Diagram (by dragoon) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC82L2ZiNDljZmZiNjUzYWE4NGRmNmIwYTljMWQxZDU2ZDc1ZmNiMDQ3LmpwZWc=&rn=RDQ3ODA2RjUtMzlDNi00QzQ3LUE2OTYtMjlCQkE4NThDNkRBLmpwZWc=",
"Without loss of generality, let $BC=2.$ Note that $\\angle BMH$ is the vertical angle to an angle of the regular hexagon, so it has a measure of $120^\\circ$. Because $\\triangle ABH$ and $\\triangle BCI$ are rotational images of one another, we get that $\\angle{MBH}=\\angle{HAB}$ and hence $\\triangle ABH \\sim \\triangle BMH \\sim \\triangle BCI$. Using a similar argument, $NI=MH$, and \\[MN=BI-NI-BM=BI-(BM+MH).\\] Applying the Law of cosines on $\\triangle BCI$, $BI=\\sqrt{2^2+1^2-2(2)(1)(\\cos(120^\\circ))}=\\sqrt{7}$ \\begin{align*}\\frac{BC+CI}{BI}&=\\frac{3}{\\sqrt{7}}=\\frac{BM+MH}{BH} \\\\ BM+MH&=\\frac{3BH}{\\sqrt{7}}=\\frac{3}{\\sqrt{7}} \\\\ MN&=BI-(BM+MH)=\\sqrt{7}-\\frac{3}{\\sqrt{7}}=\\frac{4}{\\sqrt{7}} \\\\ \\frac{\\text{Area of smaller hexagon}}{\\text{Area of bigger hexagon}}&=\\left(\\frac{MN}{BC}\\right)^2=\\left(\\frac{2}{\\sqrt{7}}\\right)^2=\\frac{4}{7}\\end{align*} Thus, the answer is $4 + 7 = 011. Diagram (by dragoon) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC82L2ZiNDljZmZiNjUzYWE4NGRmNmIwYTljMWQxZDU2ZDc1ZmNiMDQ3LmpwZWc=&rn=RDQ3ODA2RjUtMzlDNi00QzQ3LUE2OTYtMjlCQkE4NThDNkRBLmpwZWc=",
"We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$, then $B$ is at $(\\cos(60^\\circ),\\sin(60^\\circ))=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)$, $C$ is at $(\\cos(120^\\circ),\\sin(120^\\circ))=\\left(-\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)$, $D$ is at $(\\cos(180^\\circ),\\sin(180^\\circ))=(-1,0)$, \\begin{align*}&H=\\frac{B+C}{2}=\\left(0,\\frac{\\sqrt{3}}{2}\\right) \\\\ &I=\\frac{C+D}{2}=\\left(-\\frac{3}{4},\\frac{\\sqrt{3}}{4}\\right)\\end{align*} Line $AH$ has the slope of $-\\frac{\\sqrt{3}}{2}$ and the equation of $y=-\\frac{\\sqrt{3}}{2}(x-1)$ Line $BI$ has the slope of $\\frac{\\sqrt{3}}{5}$ and the equation $y-\\frac{3}{2}=\\frac{\\sqrt{3}}{5}\\left(x-\\frac{1}{2}\\right)$ Let's solve the system of equation to find $M$ \\begin{align*}-\\frac{\\sqrt{3}}{2}(x-1)-\\frac{3}{2}&=\\frac{\\sqrt{3}}{5}\\left(x-\\frac{1}{2}\\right) \\\\ -5\\sqrt{3}x&=2\\sqrt{3}x-\\sqrt{3} \\\\ x&=\\frac{1}{7} \\\\ y&=-\\frac{\\sqrt{3}}{2}(x-1)=\\frac{3\\sqrt{3}}{7}\\end{align*} Finally, \\begin{align*}&\\sqrt{x^2+y^2}=OM=\\frac{1}{7}\\sqrt{1^2+(3\\sqrt{3})^2}=\\frac{1}{7}\\sqrt{28}=\\frac{2}{\\sqrt{7}} \\\\ &\\frac{\\text{Area of smaller hexagon}}{\\text{Area of bigger hexagon}}=\\left(\\frac{OM}{OA}\\right)^2=\\left(\\frac{2}{\\sqrt{7}}\\right)^2=\\frac{4}{7}\\end{align*} Thus, the answer is $011. Diagram (by dragoon) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC82L2ZiNDljZmZiNjUzYWE4NGRmNmIwYTljMWQxZDU2ZDc1ZmNiMDQ3LmpwZWc=&rn=RDQ3ODA2RjUtMzlDNi00QzQ3LUE2OTYtMjlCQkE4NThDNkRBLmpwZWc=",
"We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$, then $B$ is at $(\\cos(60^\\circ),\\sin(60^\\circ))=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)$, $C$ is at $(\\cos(120^\\circ),\\sin(120^\\circ))=\\left(-\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)$, $D$ is at $(\\cos(180^\\circ),\\sin(180^\\circ))=(-1,0)$, \\begin{align*}&H=\\frac{B+C}{2}=\\left(0,\\frac{\\sqrt{3}}{2}\\right) \\\\ &I=\\frac{C+D}{2}=\\left(-\\frac{3}{4},\\frac{\\sqrt{3}}{4}\\right)\\end{align*} Line $AH$ has the slope of $-\\frac{\\sqrt{3}}{2}$ and the equation of $y=-\\frac{\\sqrt{3}}{2}(x-1)$ Line $BI$ has the slope of $\\frac{\\sqrt{3}}{5}$ and the equation $y-\\frac{3}{2}=\\frac{\\sqrt{3}}{5}\\left(x-\\frac{1}{2}\\right)$ Let's solve the system of equation to find $M$ \\begin{align*}-\\frac{\\sqrt{3}}{2}(x-1)-\\frac{3}{2}&=\\frac{\\sqrt{3}}{5}\\left(x-\\frac{1}{2}\\right) \\\\ -5\\sqrt{3}x&=2\\sqrt{3}x-\\sqrt{3} \\\\ x&=\\frac{1}{7} \\\\ y&=-\\frac{\\sqrt{3}}{2}(x-1)=\\frac{3\\sqrt{3}}{7}\\end{align*} Finally, \\begin{align*}&\\sqrt{x^2+y^2}=OM=\\frac{1}{7}\\sqrt{1^2+(3\\sqrt{3})^2}=\\frac{1}{7}\\sqrt{28}=\\frac{2}{\\sqrt{7}} \\\\ &\\frac{\\text{Area of smaller hexagon}}{\\text{Area of bigger hexagon}}=\\left(\\frac{OM}{OA}\\right)^2=\\left(\\frac{2}{\\sqrt{7}}\\right)^2=\\frac{4}{7}\\end{align*} Thus, the answer is $011. Diagram (by dragoon) https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC82L2ZiNDljZmZiNjUzYWE4NGRmNmIwYTljMWQxZDU2ZDc1ZmNiMDQ3LmpwZWc=&rn=RDQ3ODA2RjUtMzlDNi00QzQ3LUE2OTYtMjlCQkE4NThDNkRBLmpwZWc=",
"Use the diagram. Now notice that all of the \"overlapping triangles\" are congruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1."
] |
2010-II-10
| 2,010
| 10
|
Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ .
| 163
|
II
|
[
"Solution 1 Let $f(x) = a(x-r)(x-s)$. Then $ars=2010=2\\cdot3\\cdot5\\cdot67$. First consider the case where $r$ and $s$ (and thus $a$) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$, $r$, and $s$. However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$, we have $r=s$. The other $80$ cases are double counting, so there are $40$. We must now consider the various cases of signs. For the $40$ cases where $|r|\\neq |s|$, there are a total of four possibilities, For the case $|r|=|s|=1$, there are only three possibilities, $(r,s) = (1,1); (1,-1); (-1,-1)$ as $(-1,1)$ is not distinguishable from the second of those three. You may ask: How can one of ${r, s}$ be positive and the other negative? $a$ will be negative as a result. That way, it's still $+2010$ that gets multiplied. Thus the grand total is $4\\cdot40 + 3 = 163.",
"Let $f(x) = a(x-r)(x-s)$. Then $ars=2010=2\\cdot3\\cdot5\\cdot67$. First consider the case where $r$ and $s$ (and thus $a$) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$, $r$, and $s$. However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$, we have $r=s$. The other $80$ cases are double counting, so there are $40$. We must now consider the various cases of signs. For the $40$ cases where $|r|\\neq |s|$, there are a total of four possibilities, For the case $|r|=|s|=1$, there are only three possibilities, $(r,s) = (1,1); (1,-1); (-1,-1)$ as $(-1,1)$ is not distinguishable from the second of those three. You may ask: How can one of ${r, s}$ be positive and the other negative? $a$ will be negative as a result. That way, it's still $+2010$ that gets multiplied. Thus the grand total is $4\\cdot40 + 3 = 163.",
"We use Burnside's Lemma. The set being acted upon is the set of integer triples $(a,r,s)$ such that $ars=2010$. Because $r$ and $s$ are indistinguishable, the permutation group consists of the identity and the permutation that switches $r$ and $s$. In cycle notation, the group consists of $(a)(r)(s)$ and $(a)(r \\: s)$. There are $4 \\cdot 3^4$ fixed points of the first permutation (after distributing the primes among $a$, $r$, $s$ and then considering their signs. We have 4 ways since we can keep them all positive, first 2 negative, first and third negative, or last two negative) and $2$ fixed points of the second permutation ($r=s=\\pm 1$). By Burnside's Lemma, there are $\\frac{1}{2} (4 \\cdot 3^4+2)= 163.",
"The equation can be written in the form of $k(x-a)(x-b)$, where $|k|$ is a divisor of $2010$. Factor $2010=2\\cdot 3\\cdot 5\\cdot 67$. We divide into few cases to study. Firstly, if $|k|=1$, the equation can be $-(x-a)(x+b)$ or $(x-a)(x-b)$ or $(x+a)(x+b)$, there are $2^4+2^4=32$ ways If $|k|\\in \\{2,3,5,67\\}$. Take $|k|=2$ as an example, follow the procedure above, there are $2^3+2^3=16$ ways, and there are $\\binom {4}{1}=4$ ways to choose $|k|$, so there are $16\\cdot 4=64$ ways If $|k|$ is the product of two factors of $2010$, there are $8$ ways for each. Thus there are $\\binom {4}{2}\\cdot 8=48$ ways If $|k|$ is the product of three factors of $2010$, there are $\\binom {4}{3}\\cdot 4=16$ ways In the end, $|k|=2010$, only $2010(x-1)(x-1); 2010(x+1)(x+1); -2010(x-1)(x+1)$ work, there are $3$ ways Thus, $32+64+48+16+3=163 ~bluesoul"
] |
2010-II-12
| 2,010
| 12
|
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter.
| 676
|
II
|
[
"Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$. Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$. By Heron's Formula, we have \\[\\sqrt{s(8x)(8x)(s-16x)}=\\sqrt{s(7x)(7x)(s-14x)}\\] \\[8\\sqrt{s-16x}=7\\sqrt{s-14x}\\] \\[64s-1024x=49s-686x\\] \\[15s=338x\\] Since $15$ and $338$ are coprime, to minimize, we must have $s=338$ and $x=15$. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by $2$, which gives us a final answer of $676.",
"Let the first triangle have sides $16n,a,a$, so the second has sides $14n,a+n,a+n$. The height of the first triangle is $\\frac{7}{8}$ the height of the second triangle. Therefore, we have \\[a^2-64n^2=\\frac{49}{64}((a+n)^2-49n^2).\\] Multiplying this, we get \\[64a^2-4096n^2=49a^2+98an-2352n^2,\\] which simplifies to \\[15a^2-98an-1744n^2=0.\\] Solving this for $a$, we get $a=n\\cdot\\frac{218}{15}$, so $n=15$ and $a=218$ and the perimeter is $15\\cdot16+218+218=676. ~john0512"
] |
2010-II-13
| 2,010
| 13
|
The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each pick a card from the deck randomly and without replacement. The two people with lower numbered cards form a team, and the two people with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$ , and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 263
|
II
|
[
"Once the two cards are drawn, there are $\\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$, which occurs in $\\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$, which occurs in $\\dbinom{43-a}{2}$ ways. Thus, \\[p(a)=\\frac{\\dbinom{43-a}{2}+\\dbinom{a-1}{2}}{1225}.\\] Simplifying, we get $p(a)=\\frac{(43-a)(42-a)+(a-1)(a-2)}{2\\cdot1225}$, so we need $(43-a)(42-a)+(a-1)(a-2)\\ge (1225)$. If $a=22+b$, then \\begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\\ge (1225) \\\\ b^2\\ge \\frac{385}{2} &= 192.5 >13^2 \\end{align*} So $b> 13$ or $b< -13$, and $a=22+b<9$ or $a>35$, so $a=8$ or $a=36$. Thus, $p(8) = \\frac{616}{1225} = \\frac{88}{175}$, and the answer is $88+175 = 263.",
"Given that Alex and Dylan hold the cards $a$ and $a+9$, we need to calculate the probability that they end up on the same team. This happens in two scenarios: 1. Both on the Lower Team: This occurs if the other two cards drawn are both greater than $a+9$. 2. Both on the Higher Team: This occurs if the other two cards drawn are both less than $a$. The total number of ways to choose the other two cards from the remaining 50 cards is $\\binom{50}{2} = 1225$. The number of favorable outcomes is the sum of: The number of ways to choose 2 cards greater than $a+9$: $\\binom{43-a}{2}$ The number of ways to choose 2 cards less than $a$: $\\binom{a-1}{2}$ Thus, the probability $p(a)$ is: \\[p(a) = \\frac{\\binom{a-1}{2} + \\binom{43-a}{2}}{\\binom{50}{2}} = \\frac{\\frac{(a-1)(a-2)}{2} + \\frac{(43-a)(42-a)}{2}}{1225}.\\] Finding the Minimum $a$ for $p(a) \\geq \\frac{1}{2}$ Solving the inequality: \\[\\frac{\\binom{a-1}{2} + \\binom{43-a}{2}}{1225} \\geq \\frac{1}{2}\\] leads to: \\[2(a^2 - 44a + 291) \\geq 0.\\] This quadratic inequality is satisfied when $a \\leq 8$ or $a \\geq 36$. The minimum value of $p(a)$ that satisfies $p(a) \\geq \\frac{1}{2}$ occurs at $a = 8$ (or symmetrically at $a = 36$), giving: \\[p(8) = \\frac{616}{1225} = \\frac{88}{175}\\] where $88$ and $175$ are coprime. Final Answer \\[88 + 175 = 263\\] Answer: $263$ ~Athmyx"
] |
2010-II-14
| 2,010
| 14
|
Triangle $ABC$ with right angle at $C$ , $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ , $q$ , $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ .
| 7
|
II
|
[
"Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\\angle{DOA} = 2\\angle ACP = \\angle{APC} = \\angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$. Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean Theorem, $OE = \\sqrt {2^2 - \\frac {3^2}{2^2}} = \\sqrt {\\frac {7}{4}}$. Now note that $EP = \\frac {1}{2}$. By the Pythagorean Theorem, $OP = \\sqrt {\\frac {7}{4} + \\frac {1^2}{2^2}} = \\sqrt {2}$. Hence it now follows that, \\[\\frac {AP}{BP} = \\frac {AO + OP}{BO - OP} = \\frac {2 + \\sqrt {2}}{2 - \\sqrt {2}} = 3 + 2\\sqrt {2}\\] This gives that the answer is $007\", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); [/asy]",
"Let $AC=b$, $BC=a$ by convention. Also, Let $AP=x$ and $BP=y$. Finally, let $\\angle ACP=\\theta$ and $\\angle APC=2\\theta$. We are then looking for $\\frac{AP}{BP}=\\frac{x}{y}$ Now, by arc interceptions and angle chasing we find that $\\triangle BPD \\sim \\triangle CPA$, and that therefore $BD=yb.$ Then, since $\\angle ABD=\\theta$ (it intercepts the same arc as $\\angle ACD$) and $ADB$ is right, $\\cos\\theta=\\frac{DB}{AB}=\\frac{by}{4}$. Using law of sines on $APC$, we additionally find that $\\frac{b}{\\sin 2\\theta}=\\frac{x}{\\sin\\theta}.$ Simplification by the double angle formula $\\sin 2\\theta=2\\sin \\theta\\cos\\theta$ yields $\\cos \\theta=\\frac{b}{2x}$. We equate these expressions for $\\cos\\theta$ to find that $xy=2$. Since $x+y=AB=4$, we have enough information to solve for $x$ and $y$. We obtain $x,y=2 \\pm \\sqrt{2}$ Since we know $x>y$, we use $\\frac{x}{y}=\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}=3+2\\sqrt{2}$",
"Let $\\angle{ACP}$ be equal to $x$. Then by Law of Sines, $PB = -\\frac{\\cos{x}}{\\cos{3x}}$ and $AP = \\frac{\\sin{x}}{\\sin{3x}}$. We then obtain $\\cos{3x} = 4\\cos^3{x} - 3\\cos{x}$ and $\\sin{3x} = 3\\sin{x} - 4\\sin^3{x}$. Solving, we determine that $\\sin^2{x} = \\frac{4 \\pm \\sqrt{2}}{8}$. Plugging this in gives that $\\frac{AP}{PB} = \\frac{\\sqrt{2}+1}{\\sqrt{2}-1} = 3 + 2\\sqrt{2}$. The answer is $007 to finish. ~happypi31415)",
"Let $\\alpha=\\angle{ACP}$, $\\beta=\\angle{ABC}$, and $x=BP$. By Law of Sines, $\\frac{1}{\\sin(\\beta)}=\\frac{x}{\\sin(90-\\alpha)}\\implies \\sin(\\beta)=\\frac{\\cos(\\alpha)}{x}$ (1), and $\\frac{4-x}{\\sin(\\alpha)}=\\frac{4\\sin(\\beta)}{\\sin(2\\alpha)} \\implies 4-x=\\frac{2\\sin(\\beta)}{\\cos(\\alpha)}$. (2) Then, substituting (1) into (2), we get $4-x=\\frac{2}{x} \\implies x^2-4x+2=0 \\implies x=2-\\sqrt{2} \\implies \\frac{4-x}{x}=\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}=3+2\\sqrt{2}$ The answer is $007. ~Rowechen",
"Let $\\alpha=\\angle{ACP}$, $\\beta=\\angle{ABC}$, and $x=BP$. By Law of Sines, $\\frac{1}{\\sin(\\beta)}=\\frac{x}{\\sin(90-\\alpha)}\\implies \\sin(\\beta)=\\frac{\\cos(\\alpha)}{x}$ (1), and $\\frac{4-x}{\\sin(\\alpha)}=\\frac{4\\sin(\\beta)}{\\sin(2\\alpha)} \\implies 4-x=\\frac{2\\sin(\\beta)}{\\cos(\\alpha)}$. (2) Then, substituting (1) into (2), we get $4-x=\\frac{2}{x} \\implies x^2-4x+2=0 \\implies x=2-\\sqrt{2} \\implies \\frac{4-x}{x}=\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}=3+2\\sqrt{2}$ The answer is $007. ~Rowechen",
"Let $\\angle{ACP}=x$. Then, $\\angle{APC}=2x$ and $\\angle{A}=180-3x$. Let the foot of the angle bisector of $\\angle{APC}$ on side $AC$ be $D$. Then, $CD=DP$ and $\\triangle{DAP}\\sim{\\triangle{APC}}$ due to the angles of these triangles. Let $CD=a$. By the Angle Bisector Theorem, $\\frac{1}{a}=\\frac{AP}{AD}$, so $AD=a\\cdot{AP}$. Moreover, since $CD=DP=a$, by similar triangle ratios, $\\frac{AP}{a+a\\cdot{AP}}=a$. Therefore, $AP = \\frac{a^2}{1-a^2}$. Construct the perpendicular from $D$ to $AP$ and denote it as $F$. Denote the midpoint of $CP$ as $M$. Since $PD$ is an angle bisector, $PF$ is congruent to $PM$, so $PF=\\frac{1}{2}$. Also, $\\triangle{DFA}\\sim{\\triangle{BCA}}$. Thus, $\\frac{FA}{AC}=\\frac{AD}{AB}\\Longrightarrow\\frac{\\frac{a^2}{1-a^2}-\\frac{1}{2}}{a+\\frac{a^3}{1-a^2}}=\\frac{\\frac{a^3}{1-a^3}}{4}$. After some major cancellation, we have $7a^4-8a^2+2=0$, which is a quadratic in $a^2$. Thus, $a^2 = \\frac{4\\pm\\sqrt{2}}{7}$. Taking the negative root implies $AP<BP$, contradiction. Thus, we take the positive root to find that $AP=2+\\sqrt{2}$. Thus, $BP=2-\\sqrt{2}$, and our desired ratio is $\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}\\implies{3+2\\sqrt{2}}$. The answer is $007.",
"Let $O$ be the circumcenter of $\\triangle ABC$. Since $\\triangle ABC$ is a right triangle, $O$ will be on $\\overline{AB}$ and $\\overline{AO} \\cong \\overline{OB} \\cong \\overline{OC} = 2$. Let $\\overline{OP} = x$. Next, let's do some angle chasing. Label $\\angle ACP = \\theta^{\\circ}$, and $\\angle APC = 2\\theta^{\\circ}$. Thus, $\\angle PAC = (180-3\\theta)^{\\circ}$, and by isosceles triangles, $\\angle ACO = (180-3\\theta)^{\\circ}$. Then, by angle subtraction, $\\angle OCP = (\\theta - (180-3\\theta))^{\\circ} = (4\\theta - 180)^{\\circ}$. Using the Law of Sines: \\[\\frac{x}{\\sin (4\\theta-180)^{\\circ}}=\\frac{2}{\\sin (2\\theta)^{\\circ}}\\]Using trigonometric identies, $\\sin (4\\theta-180)^{\\circ}=-\\sin (4\\theta)^{\\circ}=-2\\sin (2\\theta)^{\\circ}\\cos (2\\theta)^{\\circ}$. Plugging this back into the Law of Sines formula gives us: \\[\\frac{x}{-2\\sin (2\\theta)^{\\circ}\\cos (2\\theta)^{\\circ}}=\\frac{2}{\\sin (2\\theta)^{\\circ}}\\] \\[-4\\sin (2\\theta)^{\\circ}\\cos (2\\theta)^{\\circ}=x\\sin (2\\theta)^{\\circ}\\] \\[-4\\cos (2\\theta)^{\\circ}=x\\] \\[\\cos(2\\theta)^{\\circ}=\\frac{-x}4\\] Next, using the Law of Cosines: \\[2^2=1^2+x^2-2\\cdot 1\\cdot x\\cdot \\cos (2\\theta)^{\\circ}\\] Substituting $\\cos(2\\theta)^{\\circ}=\\frac{-x}4$ gives us: \\[2^2=1^2+x^2-2\\cdot 1\\cdot x\\cdot \\frac{-x}4\\] \\[4=1+x^2+\\frac{x^2}{2}\\] Solving for x gives $x=\\sqrt 2$ Finally: $\\frac{\\overline{AP}}{\\overline{BP}}=\\frac{\\overline{AO}+\\overline{OP}}{\\overline{BO}-\\overline{OP}}=\\frac{2+\\sqrt 2}{2-\\sqrt 2}=3+2\\sqrt2$, which gives us an answer of $3+2+2=007. ~adyj"
] |
2010-II-15
| 2,010
| 15
|
In triangle $ABC$ , $AC=13$ , $BC=14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$ .
| 218
|
II
|
[
"Define the function $f:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}$ by \\[f(X)=\\text{Pow}_{(AMN)}(X)-\\text{Pow}_{(ADE)}(X)\\] for points $X$ in the plane. Then $f$ is linear, so $\\frac{BQ}{CQ}=\\frac{f(B)-f(Q)}{f(Q)-f(C)}$. But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$, $(ADE)$ thus \\[\\frac{BQ}{CQ}=-\\frac{f(B)}{f(C)}=-\\frac{BN\\cdot BA-BE\\cdot BA}{CM\\cdot CA-CD\\cdot CA}\\] Let $AC=b$, $BC=a$ and $AB=c$. Note that $BN=\\tfrac{c}{2}$ and $CM=\\tfrac{b}{2}$ because they are midpoints, while $BE=\\frac{ac}{a+b}$ and $CD=\\frac{ab}{a+c}$ by Angle Bisector Theorem. Thus we can rewrite this expression as \\begin{align*}&-\\frac{\\tfrac{c^{2}}{2}-\\tfrac{ac^{2}}{a+b}}{\\tfrac{b^{2}}{2}-\\tfrac{ab^{2}}{a+c}} \\\\ =&-\\left(\\frac{c^{2}}{b^{2}}\\right)\\left(\\frac{\\tfrac{1}{2}-\\tfrac{a}{a+b}}{\\tfrac{1}{2}-\\tfrac{a}{a+c}}\\right) \\\\ =&\\left(\\frac{c^{2}}{b^{2}}\\right)\\left(\\frac{a+c}{a+b}\\right) \\\\ =&\\left(\\frac{225}{169}\\right)\\left(\\frac{29}{27}\\right) \\\\ =&~\\frac{725}{507}\\end{align*} so $m-n=218.",
"Define the function $f:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}$ by \\[f(X)=\\text{Pow}_{(AMN)}(X)-\\text{Pow}_{(ADE)}(X)\\] for points $X$ in the plane. Then $f$ is linear, so $\\frac{BQ}{CQ}=\\frac{f(B)-f(Q)}{f(Q)-f(C)}$. But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$, $(ADE)$ thus \\[\\frac{BQ}{CQ}=-\\frac{f(B)}{f(C)}=-\\frac{BN\\cdot BA-BE\\cdot BA}{CM\\cdot CA-CD\\cdot CA}\\] Let $AC=b$, $BC=a$ and $AB=c$. Note that $BN=\\tfrac{c}{2}$ and $CM=\\tfrac{b}{2}$ because they are midpoints, while $BE=\\frac{ac}{a+b}$ and $CD=\\frac{ab}{a+c}$ by Angle Bisector Theorem. Thus we can rewrite this expression as \\begin{align*}&-\\frac{\\tfrac{c^{2}}{2}-\\tfrac{ac^{2}}{a+b}}{\\tfrac{b^{2}}{2}-\\tfrac{ab^{2}}{a+c}} \\\\ =&-\\left(\\frac{c^{2}}{b^{2}}\\right)\\left(\\frac{\\tfrac{1}{2}-\\tfrac{a}{a+b}}{\\tfrac{1}{2}-\\tfrac{a}{a+c}}\\right) \\\\ =&\\left(\\frac{c^{2}}{b^{2}}\\right)\\left(\\frac{a+c}{a+b}\\right) \\\\ =&\\left(\\frac{225}{169}\\right)\\left(\\frac{29}{27}\\right) \\\\ =&~\\frac{725}{507}\\end{align*} so $m-n=218.",
"Let $Y = MN \\cap AQ$. $\\frac {BQ}{QC} = \\frac {NY}{MY}$ since $\\triangle AMN \\sim \\triangle ACB$. Since quadrilateral $AMPN$ is cyclic, $\\triangle MYA \\sim \\triangle PYN$ and $\\triangle MYP \\sim \\triangle AYN$, yielding $\\frac {YM}{YA} = \\frac {MP}{AN}$ and $\\frac {YA}{YN} = \\frac {AM}{PN}$. Multiplying these together yields *$\\frac {YN}{YM} = \\left(\\frac {AN}{AM}\\right) \\left(\\frac {PN}{PM}\\right)$. $\\frac {AN}{AM} = \\frac {\\frac {AB}{2}}{\\frac {AC}{2}} = \\frac {15}{13}$. Now we claim that $\\triangle PMD \\sim \\triangle PNE$. To prove this, we can use cyclic quadrilaterals. From $AMPN$, $\\angle PNY \\cong \\angle PAM$ and $\\angle ANM \\cong \\angle APM$. So, $m\\angle PNA = m\\angle PNY + m\\angle ANM = m\\angle PAM + m\\angle APM = 180-m\\angle PMA$ and $\\angle PNA \\cong \\angle PMD$. From $ADPE$, $\\angle PDE \\cong \\angle PAE$ and $\\angle EDA \\cong \\angle EPA$. Thus, $m\\angle MDP = m\\angle PDE + m\\angle EDA = m\\angle PAE + m\\angle EPA = 180-m\\angle PEA$ and $\\angle PDM \\cong \\angle PEN$. Thus, from AA similarity, $\\triangle PMD \\sim \\triangle PNE$. Therefore, $\\frac {PN}{PM} = \\frac {NE}{MD}$, which can easily be computed by the angle bisector theorem to be $\\frac {145}{117}$. It follows that *$\\frac {BQ}{CQ} = \\frac {15}{13} \\cdot \\frac {145}{117} = \\frac {725}{507}$, giving us an answer of $725 - 507 = 218, for any cevian AD of a triangle ABC. For the sine ratios use Law of Sines on triangles APM and APN, \\[\\frac{PM}{\\sin \\angle PAM}=\\frac{AP}{\\sin \\angle AMP}=\\frac{AP}{\\sin \\angle ANP}=\\frac{PN}{\\sin \\angle PAN}\\]. The information needed to use the Ratio Lemma can be found from the similar triangle section above. Source: [1] by Zhero",
"This problem can be solved with barycentric coordinates. Let triangle $ABC$ be the reference triangle with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Thus, $N=(1:1:0)$ and $M=(1:0:1)$. Using the Angle Bisector Theorem, we can deduce that $D=(14:0:15)$ and $E = (14:13:0)$. Plugging the coordinates for triangles $ANM$ and $AED$ into the circle formula, we deduce that the equation for triangle $ANM$ is $-a^2yz-b^2zx-c^2xy+(\\frac{c^2}{2}y+\\frac{b^2}{2}z)(x+y+z)=0$ and the equation for triangle $AED$ is $-a^2yz-b^2zx-c^2xy+(\\frac{14c^2}{27}y+\\frac{14b^2}{29}z)(x+y+z)=0$. Solving the system of equations, we get that $\\frac{c^2y}{54}=\\frac{b^2z}{58}$. This equation determines the radical axis of circles $ANM$ and $AED$, on which points $P$ and $Q$ lie. Thus, solving for $\\frac{z}{y}$ gets the desired ratio of lengths, and $\\frac{z}{y}=\\frac{58c^2}{54b^2}$ and plugging in the lengths $b=13$ and $c=15$ gets $\\frac{725}{507}$. From this we get the desired answer of $725-507=218. -wertguk",
"Observe that $P$ is the center of spiral symmetry of segments $DM$ and $EN$. Using the angle bisector theorem, we compute $DM = \\frac{13}{58}$ and $EN = \\frac{5}{18}$. Hence, it follows that the side-length ratio of triangles $DMP$ to $ENP$ is $\\frac{117}{145}$ (note that the two triangles are similar by the spiral symmetry). This implies that the ratio of the height from $P$ to $AC$ to the height from $P$ to $AB$ is $\\frac{117}{145}$, so we compute the area ratio $\\frac{APC}{APB} = \\frac{507}{725}$. From the above, we see that the barycentric coordinates of $P$ are of the form $(x : 507 : 725)$. Hence, it follows that the point $Q$ has the coordinates $(0 : 507 : 725)$, so $\\frac{BQ}{CQ} = \\frac{725}{507}$ and our answer is $218. ~hgomamogh",
"Observe that $P$ is the center of spiral symmetry of segments $DM$ and $EN$. Using the angle bisector theorem, we compute $DM = \\frac{13}{58}$ and $EN = \\frac{5}{18}$. Hence, it follows that the side-length ratio of triangles $DMP$ to $ENP$ is $\\frac{117}{145}$ (note that the two triangles are similar by the spiral symmetry). This implies that the ratio of the height from $P$ to $AC$ to the height from $P$ to $AB$ is $\\frac{117}{145}$, so we compute the area ratio $\\frac{APC}{APB} = \\frac{507}{725}$. From the above, we see that the barycentric coordinates of $P$ are of the form $(x : 507 : 725)$. Hence, it follows that the point $Q$ has the coordinates $(0 : 507 : 725)$, so $\\frac{BQ}{CQ} = \\frac{725}{507}$ and our answer is $218. ~hgomamogh"
] |
2011-I-1
| 2,011
| 1
|
Jar $A$ contains four liters of a solution that is $45\%$ acid. Jar $B$ contains five liters of a solution that is $48\%$ acid. Jar $C$ contains one liter of a solution that is $k\%$ acid. From jar $C$ , $\frac{m}{n}$ liters of the solution is added to jar $A$ , and the remainder of the solution in jar $C$ is added to jar B. At the end both jar $A$ and jar $B$ contain solutions that are $50\%$ acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ .
| 85
|
I
|
[
"Jar A contains $\\frac{11}{5}$ liters of water, and $\\frac{9}{5}$ liters of acid; jar B contains $\\frac{13}{5}$ liters of water and $\\frac{12}{5}$ liters of acid. The gap between the amount of water and acid in the first jar, $\\frac{2}{5}$, is double that of the gap in the second jar, $\\frac{1}{5}$. Therefore, we must add twice as much of jar C into the jar $A$ over jar $B$. So, we must add $\\frac{2}{3}$ of jar C into jar $A$, so $m = 2, n=3$. Since jar C contains $1$ liter of solution, we are adding $\\frac{2}{3}$ of a liter of solution to jar $A$. In order to close the gap between water and acid, there must be $\\frac{2}{5}$ more liters of acid than liters of water in these $\\frac{2}{3}$ liters of solution. So, in the $\\frac{2}{3}$ liters of solution, there are $\\frac{2}{15}$ liters of water, and $\\frac{8}{15}$ liters of acid. So, 80% of the $\\frac{2}{3}$ sample is acid, so overall, in jar C, 80% of the sample is acid. Therefore, our answer is $80 + 2 + 3 = 85. ~ ihatemath123",
"There are $\\frac{45}{100}(4)=\\frac{9}{5}$ L of acid in Jar A. There are $\\frac{48}{100}(5)=\\frac{12}{5}$ L of acid in Jar B. And there are $\\frac{k}{100}$ L of acid in Jar C. After transferring the solutions from jar C, there will be $4+\\frac{m}{n}$ L of solution in Jar A and $\\frac{9}{5}+\\frac{k}{100}\\cdot\\frac{m}{n}$ L of acid in Jar A. $6-\\frac{m}{n}$ L of solution in Jar B and $\\frac{12}{5}+\\frac{k}{100}\\cdot \\left(1-\\frac{m}{n}\\right)=\\frac{12}{5}+\\frac{k}{100}-\\frac{mk}{100n}$ of acid in Jar B. Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution. \\[\\frac{18}{5}+\\frac{km}{50n}=4+\\frac{m}{n}\\] \\[\\frac{24}{5}-\\frac{km}{50n}+\\frac{k}{50}=6-\\frac{m}{n}\\] Add the equations to get \\[\\frac{42}{5}+\\frac{k}{50}=10\\] Solving gives $k=80$. If we substitute back in the original equation we get $\\frac{m}{n}=\\frac{2}{3}$ so $3m=2n$. Since $m$ and $n$ are relatively prime, $m=2$ and $n=3$. Thus $k+m+n=80+2+3=085.",
"One might cleverly change the content of both Jars. Since the end result of both Jars are $50\\%$ acid, we can turn Jar A into a 1 gallon liquid with $50\\%-4(5\\%) = 30\\%$ acid and Jar B into 1 gallon liquid with $50\\%-5(2\\%) =40\\%$ acid. Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so $\\dfrac{2}{3}$ of Jar C will be pour into Jar A. Thus, $m=2$ and $n=3$. $\\dfrac{30\\% + \\frac{2}{3} \\cdot k\\%}{\\frac{5}{3}} = 50\\%$ Solving for $k$ yields $k=80$ So the answer is $80+2+3 = 085",
"One may first combine all three jars in to a single container. That container will have $10$ liters of liquid, and it should be $50\\%$ acidic. Thus there must be $5$ liters of acid. Jar A contained $45\\% \\cdot 4L$, or $1.8L$ of acid, and jar B $48\\% \\cdot 5L$ or $2.4L$. Solving for the amount of acid in jar C, $k = (5 - 2.4 - 1.8) = .8$, or $80\\%$. Once one knowss that the jar C is $80\\%$ acid, use solution 1 to figure out m and n for $k+m+n=80+2+3=085."
] |
2011-I-2
| 2,011
| 2
|
In rectangle $ABCD$ , $AB = 12$ and $BC = 10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ , $DF = 8$ , $\overline{BE} \parallel \overline{DF}$ , $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , where $m$ , $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$ .
| 36
|
I
|
[
"Let us call the point where $\\overline{EF}$ intersects $\\overline{AD}$ point $G$, and the point where $\\overline{EF}$ intersects $\\overline{BC}$ point $H$. Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\\frac{BH}{GD} = \\frac{9}{8}$. Since $BC=10$, $BH+GD=BH+HC=BC=10$. ($HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system: $\\frac{BH}{GD}=\\frac{9}8$ $BH+GD=10$ Solving this system (easiest by substitution), we get that: $BH=\\frac{90}{17}$ $GD=\\frac{80}{17}$ Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles: $\\sqrt{9^2-\\left(\\frac{90}{17}\\right)^2}$ and $\\sqrt{8^2-\\left(\\frac{80}{17}\\right)^2}$ Notice that adding these two sides would give us twelve plus the overlap $EF$. This means that: $EF= \\sqrt{9^2-\\left(\\frac{90}{17}\\right)^2}+\\sqrt{8^2-\\left(\\frac{80}{17}\\right)^2}-12=3\\sqrt{21}-12$ Since $21$ isn't divisible by any perfect square, our answer is: $3+21+12=36",
"Extend lines $BE$ and $CD$ to meet at point $G$. Draw the altitude $GH$ from point $G$ to line $BA$ extended. $GE=DF=8,$ $GB=17$ In right $\\bigtriangleup GHB$, $GH=10$, $GB=17$, thus by Pythagoras Theorem we have: $HB=\\sqrt{17^2-10^2}=3\\sqrt{21}$ $HA=EF=3\\sqrt{21}-12$ Thus our answer is: $3+21+12=36",
"We notice that since $\\overline{BE}||\\overline{DF}$, $\\overline{EF}$ is the diagonal of a rectangle. Now, let us extend the width lines to intersect with $\\overline{BE}$ and $\\overline{DF}$, respectively, to form this rectangle. Let us call the length of $\\overline{EF}$ $d$, the perpendicular distance between $\\overline{EF}$ and $\\overline{AD}$ $k$, and the perpendicular distance between $\\overline{EF}$ and $\\overline{CD}$ $x$. We now can begin the similar triangles. When drawing the diagram (where $E$ is closer to $\\overline{AD}$ than $F$), we put the similar right triangles in the same position so that we can begin solving for our variables and finding ratios. Since all of these triangles are similar, we find that $\\frac{dx}{8}=\\frac{10d-dx}{9}$, which solves for $x=\\frac{80}{17}$. Completing the same thing for $k$, we see $\\frac{12d-dk}{9}=\\frac{dk+d^2}{8}$, which solves for $k=\\frac{96-9d}{17}$. We are now ready to find both the length and the width of the rectangle. We find these to be $\\frac{10d}{17}$ and $\\frac{12d+d^2}{17}$. Now let us use the Pythagorean to solve for $d$. We square the length and the width and multiply both sides by 289 (the denominator of the LHS) to reach the equation $(12+d^2)^2+100d^2=289d^2$. Expanding and simplifying, we find that $d^4+24d^3=45d^2$. Dividing both sides by $d^2$ and moving everything to the LHS, we see that $d^2+24d-45=0$. Applying the quadratic formula, $d=\\frac{-24\\pm \\sqrt{756}}{2}=-12\\pm \\sqrt{189}$. Since $d>0$, $d=\\sqrt{189}-12=3\\sqrt{21}-12$. Finally, $m+n+p=3+21+12=036. -Gideontz",
"We notice that since $\\overline{BE}||\\overline{DF}$, $\\overline{EF}$ is the diagonal of a rectangle. Now, let us extend the width lines to intersect with $\\overline{BE}$ and $\\overline{DF}$, respectively, to form this rectangle. Let us call the length of $\\overline{EF}$ $d$, the perpendicular distance between $\\overline{EF}$ and $\\overline{AD}$ $k$, and the perpendicular distance between $\\overline{EF}$ and $\\overline{CD}$ $x$. We now can begin the similar triangles. When drawing the diagram (where $E$ is closer to $\\overline{AD}$ than $F$), we put the similar right triangles in the same position so that we can begin solving for our variables and finding ratios. Since all of these triangles are similar, we find that $\\frac{dx}{8}=\\frac{10d-dx}{9}$, which solves for $x=\\frac{80}{17}$. Completing the same thing for $k$, we see $\\frac{12d-dk}{9}=\\frac{dk+d^2}{8}$, which solves for $k=\\frac{96-9d}{17}$. We are now ready to find both the length and the width of the rectangle. We find these to be $\\frac{10d}{17}$ and $\\frac{12d+d^2}{17}$. Now let us use the Pythagorean to solve for $d$. We square the length and the width and multiply both sides by 289 (the denominator of the LHS) to reach the equation $(12+d^2)^2+100d^2=289d^2$. Expanding and simplifying, we find that $d^4+24d^3=45d^2$. Dividing both sides by $d^2$ and moving everything to the LHS, we see that $d^2+24d-45=0$. Applying the quadratic formula, $d=\\frac{-24\\pm \\sqrt{756}}{2}=-12\\pm \\sqrt{189}$. Since $d>0$, $d=\\sqrt{189}-12=3\\sqrt{21}-12$. Finally, $m+n+p=3+21+12=036. -Gideontz"
] |
2011-I-3
| 2,011
| 3
|
Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A = (24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B = (5,6)$ . The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha + \beta$ .
| 31
|
I
|
[
"Given that $L$ has slope $\\frac{5}{12}$ and contains the point $A=(24,-1)$, we may write the point-slope equation for $L$ as $y+1=\\frac{5}{12}(x-24)$. Since $M$ is perpendicular to $L$ and contains the point $B=(5,6)$, we have that the slope of $M$ is $-\\frac{12}{5}$, and consequently that the point-slope equation for $M$ is $y-6=-\\frac{12}{5}(x-5)$. Converting both equations to the form $0=Ax+By+C$, we have that $L$ has the equation $0=5x-12y-132$ and that $M$ has the equation $0=12x+5y-90$. Applying the point-to-line distance formula, $\\frac{|Ax+By+C|}{\\sqrt{A^2+B^2}}$, to point $P$ and lines $L$ and $M$, we find that the distance from $P$ to $L$ and $M$ are $\\frac{526}{13}$ and $\\frac{123}{13}$, respectively. Since $A$ and $B$ lie on the positive axes of the shifted coordinate plane, we may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the $x$-coordinate of $P$ is negative, and is therefore $-\\frac{123}{13}$; similarly, the $y$-coordinate of $P$ is positive and is therefore $\\frac{526}{13}$. Thus, we have that $\\alpha=-\\frac{123}{13}$ and that $\\beta=\\frac{526}{13}$. It follows that $\\alpha+\\beta=-\\frac{123}{13}+\\frac{526}{13}=\\frac{403}{13}=031.",
"The equations for the axes are $\\frac{5}{12} (x-24) = y+1$ and $-\\frac{12}{5}(x-5) = y - 6$. We can solve the system to find that they intersect at the point $\\left( \\frac{1740}{169},\\frac{-1134}{169} \\right)$ The unit basis vectors of our new axes are $\\begin{pmatrix} 12/13 \\\\ 5/13 \\end{pmatrix}$ and $\\begin{pmatrix} -5/13 \\\\ 12/13 \\end{pmatrix}$ for the $x$ and $y$ axes respectively (taking into account which direction is positive). Then, we solve the following system for $\\alpha$ and $\\beta$ : \\[\\alpha \\begin{pmatrix} 12/13 \\\\ 5/13 \\end{pmatrix} + \\beta \\begin{pmatrix} -5/13 \\\\ 12/13 \\end{pmatrix} + \\begin{pmatrix} 1740/169 \\\\ -1134/169 \\end{pmatrix} = \\begin{pmatrix} -146 \\\\ 27 \\end{pmatrix}\\] Painful bashing gives $\\alpha = -\\frac{123}{13}$ and $\\beta = \\frac{526}{13}$. Adding gives $\\alpha + \\beta = \\frac{403}{13} = 031 separately. ~jd9",
"The equations for the axes are $\\frac{5}{12} (x-24) = y+1$ and $-\\frac{12}{5}(x-5) = y - 6$. We can solve the system to find that they intersect at the point $\\left( \\frac{1740}{169},\\frac{-1134}{169} \\right)$ The unit basis vectors of our new axes are $\\begin{pmatrix} 12/13 \\\\ 5/13 \\end{pmatrix}$ and $\\begin{pmatrix} -5/13 \\\\ 12/13 \\end{pmatrix}$ for the $x$ and $y$ axes respectively (taking into account which direction is positive). Then, we solve the following system for $\\alpha$ and $\\beta$ : \\[\\alpha \\begin{pmatrix} 12/13 \\\\ 5/13 \\end{pmatrix} + \\beta \\begin{pmatrix} -5/13 \\\\ 12/13 \\end{pmatrix} + \\begin{pmatrix} 1740/169 \\\\ -1134/169 \\end{pmatrix} = \\begin{pmatrix} -146 \\\\ 27 \\end{pmatrix}\\] Painful bashing gives $\\alpha = -\\frac{123}{13}$ and $\\beta = \\frac{526}{13}$. Adding gives $\\alpha + \\beta = \\frac{403}{13} = 031 separately. ~jd9"
] |
2011-I-4
| 2,011
| 4
|
In triangle $ABC$ , $AB = 125$ , $AC = 117$ , and $BC = 120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$ .
| 56
|
I
|
[
"Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively. [asy] defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot(\"$A$\",A,dir(200),p); dot(\"$B$\",B,right,p); dot(\"$C$\",C,up,p); dot(\"$L$\",L,2*dir(70),p); dot(\"$N$\",N,2*dir(-90),p); dot(\"$M$\",M,2*dir(-90),p); dot(\"$P$\",extension(A,B,C,M),2*down,p); dot(\"$Q$\",extension(A,B,C,N),2*down,p); label(\"$125$\",A--B,down,fontsize(10)); label(\"$117$\",A--C,2*dir(130),fontsize(10)); label(\"$120$\",B--C,1.5*dir(30),fontsize(10)); [/asy] Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$, $\\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$, and $M$ is the midpoint of ${CP}$. For the same reason, $AQ=AC=117$, and $N$ is the midpoint of ${CQ}$. Hence $MN=\\tfrac 12 PQ$. Since \\[PQ=BP+AQ-AB=120+117-125=112,\\] so $MN=056.",
"Let $I$ be the incenter of $ABC$. Since $I$ lies on $BM$ and $AN$, $IM \\perp MC$ and $IN \\perp NC$, so $\\angle IMC + \\angle INC = 180^\\circ$. This means that $CMIN$ is a cyclic quadrilateral. By the Law of Sines, $\\frac{MN}{\\sin \\angle MIN} = \\frac{2R}{\\sin \\angle CMI} = 2R = CI$, where $R$ is the radius of the circumcircle of $CMIN$. Since $\\sin \\angle MIN = \\sin \\angle BIA = \\sin (90^\\circ + \\tfrac 12 \\angle BCA) = \\cos \\tfrac 12 \\angle BCA = \\cos \\angle BCI$, we have that $MN = CI \\cdot \\sin \\angle MIN = CI \\cdot \\cos \\angle BCI$. Letting $H$ be the point of contact of the incircle of $ABC$ with side $BC$, we have $MN = CI \\cdot \\cos \\angle BCI = CI \\cdot \\frac{CH}{CI} = CH$. Thus, $MN = s - AB = \\frac{117+120-125}{2}=056.",
"Project $I$ onto $AC$ and $BC$ as $D$ and $E$. $ID$ and $IE$ are both in-radii of $\\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$. Since $IC$ is the hypotenuse for the 4 triangles ($\\triangle INC, \\triangle IMC, \\triangle IDC,$ and $\\triangle IEC$), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\\omega$ which is also the circumcircle of $\\triangle CMN$ and $\\triangle CDE$. To find $MN$, we can use the Law of Cosines on $\\angle MON \\implies MN^2 = 2R^2(1 - \\cos{2\\angle MCN})$ where $O$ is the center of $\\omega$. Now, the circumradius $R$ can be found with Pythagorean Theorem with $\\triangle CDI$ or $\\triangle CEI$: $r^2 + 56^2 = (2R)^2$. To find $r$, we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \\sqrt{181 \\cdot 61 \\cdot 56 \\cdot 64} \\implies r = \\sqrt{\\frac{61 \\cdot 56 \\cdot 64}{181}} \\implies 2R^2 = \\frac{393120}{181}$. To find $\\angle MCN$, we can find $\\angle MIN$ since $\\angle MCN = 180 - \\angle MIN$. $\\angle MIN = \\angle MIC + \\angle NIC = 180 - \\angle BIC + 180 - \\angle AIC = 180 - (180 - \\frac{\\angle A + \\angle C}{2}) + 180 - (180 - \\frac{\\angle B + \\angle C}{2}) = \\frac{\\angle A + \\angle B + \\angle C}{2} + \\frac{\\angle C}{2}$. Thus, $\\angle MCN = 180 - \\frac{\\angle A + \\angle B + \\angle C}{2} - \\frac{\\angle C}{2}$ and since $\\angle A + \\angle B + \\angle C = 180$, we have $\\angle A + \\angle B + \\angle C - \\frac{\\angle A + \\angle B + \\angle C}{2} - \\frac{\\angle C}{2} = \\frac{\\angle A + \\angle B}{2}$. Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \\cos{\\angle A + \\angle B}) = 2R^2(1 + \\cos{\\angle C})$. To find $\\cos{\\angle C}$, we use LoC on $\\triangle ABC \\implies \\cos{\\angle C} = \\frac{120^2 + 117^2 - 125^2}{2 \\cdot 117 \\cdot 120} = \\frac{41 \\cdot 19}{117 \\cdot 15}$. Our formula now becomes $MN^2 = \\frac{393120}{181} + \\frac{2534}{15 \\cdot 117}$. After simplifying, we get $MN^2 = 3136 \\implies MN = 056. --lucasxia01",
"Project $I$ onto $AC$ and $BC$ as $D$ and $E$. $ID$ and $IE$ are both in-radii of $\\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$. Since $IC$ is the hypotenuse for the 4 triangles ($\\triangle INC, \\triangle IMC, \\triangle IDC,$ and $\\triangle IEC$), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\\omega$ which is also the circumcircle of $\\triangle CMN$ and $\\triangle CDE$. To find $MN$, we can use the Law of Cosines on $\\angle MON \\implies MN^2 = 2R^2(1 - \\cos{2\\angle MCN})$ where $O$ is the center of $\\omega$. Now, the circumradius $R$ can be found with Pythagorean Theorem with $\\triangle CDI$ or $\\triangle CEI$: $r^2 + 56^2 = (2R)^2$. To find $r$, we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \\sqrt{181 \\cdot 61 \\cdot 56 \\cdot 64} \\implies r = \\sqrt{\\frac{61 \\cdot 56 \\cdot 64}{181}} \\implies 2R^2 = \\frac{393120}{181}$. To find $\\angle MCN$, we can find $\\angle MIN$ since $\\angle MCN = 180 - \\angle MIN$. $\\angle MIN = \\angle MIC + \\angle NIC = 180 - \\angle BIC + 180 - \\angle AIC = 180 - (180 - \\frac{\\angle A + \\angle C}{2}) + 180 - (180 - \\frac{\\angle B + \\angle C}{2}) = \\frac{\\angle A + \\angle B + \\angle C}{2} + \\frac{\\angle C}{2}$. Thus, $\\angle MCN = 180 - \\frac{\\angle A + \\angle B + \\angle C}{2} - \\frac{\\angle C}{2}$ and since $\\angle A + \\angle B + \\angle C = 180$, we have $\\angle A + \\angle B + \\angle C - \\frac{\\angle A + \\angle B + \\angle C}{2} - \\frac{\\angle C}{2} = \\frac{\\angle A + \\angle B}{2}$. Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \\cos{\\angle A + \\angle B}) = 2R^2(1 + \\cos{\\angle C})$. To find $\\cos{\\angle C}$, we use LoC on $\\triangle ABC \\implies \\cos{\\angle C} = \\frac{120^2 + 117^2 - 125^2}{2 \\cdot 117 \\cdot 120} = \\frac{41 \\cdot 19}{117 \\cdot 15}$. Our formula now becomes $MN^2 = \\frac{393120}{181} + \\frac{2534}{15 \\cdot 117}$. After simplifying, we get $MN^2 = 3136 \\implies MN = 056. --lucasxia01",
"Because $\\angle CMI = \\angle CNI = 90$, $CMIN$ is cyclic. Applying Ptolemy's theorem on CMIN: $CN \\cdot MI+CM \\cdot IN=CI \\cdot MN$ $CI^2(\\cos \\angle ICN \\sin \\angle ICM + \\cos \\angle ICM \\sin \\angle ICN) = CI \\cdot MN$ $MN = CI \\sin \\angle MCN$ by sine angle addition formula. $\\angle MCN = 180 - \\angle MIN = 90 - \\angle BCI$. Let $H$ be where the incircle touches $BC$, then $CI \\cos \\angle BCI = CH = \\frac{a+b-c}{2}$. $a=120, b=117, c=125$, for a final answer of $056. Note: This is similar to Solution 2 after the first four lines",
"Applying Ptolemy's Theorem on the cyclic quadrilateral $MINC$, we find that $MI\\cdot CN + IN\\cdot MC = MN\\cdot IC$. $\\angle CIN=\\frac{\\alpha+\\gamma}{2}$ and $\\angle MIC=\\frac{\\beta+\\gamma}{2}$ by the Exterior Angle Theorem, so from properties of sine and cosine, we can find that $MI=IC\\cdot\\cos\\left(\\frac{\\beta+\\gamma}{2}\\right),$ $MC=IC\\cdot\\sin\\left(\\frac{\\beta+\\gamma}{2}\\right),$ $IN=IC\\cdot\\cos\\left(\\frac{\\alpha+\\gamma}{2}\\right),$ $CN=IC\\cdot\\sin\\left(\\frac{\\alpha+\\gamma}{2}\\right).$ Plugging in the values and simplifying results in $MN = IC\\cdot\\sin\\left(\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)$ by the angle-addition identity $\\sin(A+B)=\\sin(A)\\cos(B)+\\sin(B)\\cos(A)$. Before we continue, we would like to simplify the value in the sine function. We see that $\\frac{\\alpha+\\beta+2\\gamma}{2}=\\frac{\\gamma}{2}+\\frac{\\alpha+\\beta+\\gamma}{2}=\\frac{\\gamma}{2}+90$. Using the fact that $\\cos(A)=\\sin(90-A)$ results in $\\sin\\left(\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)=\\sin\\left(90+\\frac{\\gamma}{2}\\right)=\\sin\\left(90-(-\\frac{\\gamma}{2})\\right)=\\cos\\left(-\\frac{\\gamma}{2}\\right)=\\cos\\left(\\frac{\\gamma}{2}\\right).$ How do we simplify $IC$? Well, we can perform the Law of Sines on triangle $AIC$. This results in: $\\frac{AC}{\\sin(\\angle AIC)}=\\frac{IC}{\\sin\\left(\\frac{\\alpha}{2}\\right)}$ The value of $\\angle AIC$ is $\\frac{\\alpha+2\\beta+\\gamma}{2}$ by the Exterior Angle Theorem on $\\triangle ABI$, so the value of $\\sin(\\angle AIC)$ is equivalent to the value of $\\cos\\left(\\frac{\\beta}{2}\\right)$ by a similar argument as above. Then rearranging yields $IC = b\\cdot\\frac{\\sin\\left(\\frac{\\alpha}{2}\\right)}{\\cos\\left(\\frac{\\beta}{2}\\right)}$. Going back to the previous formula $MN = IC\\cdot\\sin\\left(\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)$ and substituting values yields: $MN = b\\cdot\\frac{\\sin\\left(\\frac{\\alpha}{2}\\right)\\cos\\left(\\frac{\\gamma}{2}\\right)}{\\cos\\left(\\frac{\\beta}{2}\\right)}$. Finally, using the formulae $\\sin\\left(\\frac{\\alpha}{2}\\right)=\\sqrt{\\frac{(s-b)(s-c)}{bc}}$ and $\\cos\\left(\\frac{\\alpha}{2}\\right)=\\sqrt{\\frac{s(s-a)}{bc}}$ (where $s$ is half the perimeter of the triangle), we reach our final value: $MN = b\\cdot\\frac{\\sqrt{\\frac{(s-b)(s-c)}{bc}}\\cdot\\sqrt{\\frac{s(s-c)}{ab}}}{\\sqrt{\\frac{s(s-b)}{ac}}}$ $=\\frac{\\sqrt{s(s-b)(s-c)^2}}{\\sqrt{s(s-b)}}$ $=\\sqrt{(s-c)^2}$ $=s-c$ $=181-125$ $=056",
"Applying Ptolemy's Theorem on the cyclic quadrilateral $MINC$, we find that $MI\\cdot CN + IN\\cdot MC = MN\\cdot IC$. $\\angle CIN=\\frac{\\alpha+\\gamma}{2}$ and $\\angle MIC=\\frac{\\beta+\\gamma}{2}$ by the Exterior Angle Theorem, so from properties of sine and cosine, we can find that $MI=IC\\cdot\\cos\\left(\\frac{\\beta+\\gamma}{2}\\right),$ $MC=IC\\cdot\\sin\\left(\\frac{\\beta+\\gamma}{2}\\right),$ $IN=IC\\cdot\\cos\\left(\\frac{\\alpha+\\gamma}{2}\\right),$ $CN=IC\\cdot\\sin\\left(\\frac{\\alpha+\\gamma}{2}\\right).$ Plugging in the values and simplifying results in $MN = IC\\cdot\\sin\\left(\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)$ by the angle-addition identity $\\sin(A+B)=\\sin(A)\\cos(B)+\\sin(B)\\cos(A)$. Before we continue, we would like to simplify the value in the sine function. We see that $\\frac{\\alpha+\\beta+2\\gamma}{2}=\\frac{\\gamma}{2}+\\frac{\\alpha+\\beta+\\gamma}{2}=\\frac{\\gamma}{2}+90$. Using the fact that $\\cos(A)=\\sin(90-A)$ results in $\\sin\\left(\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)=\\sin\\left(90+\\frac{\\gamma}{2}\\right)=\\sin\\left(90-(-\\frac{\\gamma}{2})\\right)=\\cos\\left(-\\frac{\\gamma}{2}\\right)=\\cos\\left(\\frac{\\gamma}{2}\\right).$ How do we simplify $IC$? Well, we can perform the Law of Sines on triangle $AIC$. This results in: $\\frac{AC}{\\sin(\\angle AIC)}=\\frac{IC}{\\sin\\left(\\frac{\\alpha}{2}\\right)}$ The value of $\\angle AIC$ is $\\frac{\\alpha+2\\beta+\\gamma}{2}$ by the Exterior Angle Theorem on $\\triangle ABI$, so the value of $\\sin(\\angle AIC)$ is equivalent to the value of $\\cos\\left(\\frac{\\beta}{2}\\right)$ by a similar argument as above. Then rearranging yields $IC = b\\cdot\\frac{\\sin\\left(\\frac{\\alpha}{2}\\right)}{\\cos\\left(\\frac{\\beta}{2}\\right)}$. Going back to the previous formula $MN = IC\\cdot\\sin\\left(\\frac{\\alpha+\\beta+2\\gamma}{2}\\right)$ and substituting values yields: $MN = b\\cdot\\frac{\\sin\\left(\\frac{\\alpha}{2}\\right)\\cos\\left(\\frac{\\gamma}{2}\\right)}{\\cos\\left(\\frac{\\beta}{2}\\right)}$. Finally, using the formulae $\\sin\\left(\\frac{\\alpha}{2}\\right)=\\sqrt{\\frac{(s-b)(s-c)}{bc}}$ and $\\cos\\left(\\frac{\\alpha}{2}\\right)=\\sqrt{\\frac{s(s-a)}{bc}}$ (where $s$ is half the perimeter of the triangle), we reach our final value: $MN = b\\cdot\\frac{\\sqrt{\\frac{(s-b)(s-c)}{bc}}\\cdot\\sqrt{\\frac{s(s-c)}{ab}}}{\\sqrt{\\frac{s(s-b)}{ac}}}$ $=\\frac{\\sqrt{s(s-b)(s-c)^2}}{\\sqrt{s(s-b)}}$ $=\\sqrt{(s-c)^2}$ $=s-c$ $=181-125$ $=056"
] |
2011-I-5
| 2,011
| 5
|
The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements.
| 144
|
I
|
[
"First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$. It is simplest to do this by looking at each of the digits $\\bmod{3}$. We see that the numbers $1, 4,$ and $7$ are congruent to $1 \\pmod{3}$, that the numbers $2, 5,$ and $8$ are congruent to $2 \\pmod{3}$, and that the numbers $3, 6,$ and $9$ are congruent to $0 \\pmod{3}$. In order for a sum of three of these numbers to be a multiple of three, the mod $3$ sum must be congruent to $0$. Quick inspection reveals that the only possible combinations are $0+0+0, 1+1+1, 2+2+2,$ and $0+1+2$. However, every set of three consecutive vertices must sum to a multiple of three, so using any of $0+0+0, 1+1+1$, or $2+2+2$ would cause an adjacent sum to include exactly $2$ digits with the same $\\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\\bmod{3}$ values. We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \\pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \\pmod{3}$, or the reverse can be true. We set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit. The other two $0 \\pmod{3}$ numbers can be arranged in $2!=2$ ways. The three $1 \\pmod{3}$ and three $2 \\pmod{3}$ can both be arranged in $3!=6$ ways. Therefore, the desired result is $2(2 \\times 6 \\times 6)=144.",
"Notice that there are three triplets of congruent integers $\\mod 3$: $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=144 total arrangements."
] |
2011-I-6
| 2,011
| 6
|
Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .
| 11
|
I
|
[
"If the vertex is at $\\left(\\frac{1}{4}, -\\frac{9}{8}\\right)$, the equation of the parabola can be expressed in the form \\[y=a\\left(x-\\frac{1}{4}\\right)^2-\\frac{9}{8}.\\] Expanding, we find that \\[y=a\\left(x^2-\\frac{x}{2}+\\frac{1}{16}\\right)-\\frac{9}{8},\\] and \\[y=ax^2-\\frac{ax}{2}+\\frac{a}{16}-\\frac{9}{8}.\\] From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$, where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$, $-\\frac{a}{2}=b$, and $\\frac{a}{16}-\\frac{9}{8}=c$. Adding up all of these gives us \\[\\frac{9a-18}{16}=a+b+c.\\] We know that $a+b+c$ is an integer, so $9a-18$ must be divisible by $16$. Let $9a=z$. If ${z-18}\\equiv {0} \\pmod{16}$, then ${z}\\equiv {2} \\pmod{16}$. Therefore, if $9a=2$, $a=\\frac{2}{9}$. Adding up gives us $2+9=011",
"Complete the square. Since $a>0$, the parabola must be facing upwards. $a+b+c=\\text{integer}$ means that $f(1)$ must be an integer. The function can be recasted into $a\\left(x-\\frac{1}{4}\\right)^2-\\frac{9}{8}$ because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than $-\\frac{9}{8}$ is $-1$. So the $y$-coordinate must change by $\\frac{1}{8}$ and the $x$-coordinate must change by $1-\\frac{1}{4}=\\frac{3}{4}$. Thus, $a\\left(\\frac{3}{4}\\right)^2=\\frac{1}{8}\\implies \\frac{9a}{16}=\\frac{1}{8}\\implies a=\\frac{2}{9}$. So $2+9=011.",
"To do this, we can use the formula for the minimum (or maximum) value of the $x$ coordinate at a vertex of a parabola, $-\\frac{b}{2a}$ and equate this to $\\frac{1}{4}$. Solving, we get $-\\frac{a}{2}=b$. Enter $x=\\frac{1}{4}$ to get $-\\frac{9}{8}=\\frac{a}{16}+\\frac{b}{4}+c=-\\frac{a}{16}+c$ so $c=\\frac{a-18}{16}$. This means that $\\frac{9a-18}{16}\\in \\mathbb{Z}$ so the minimum of $a>0$ is when the fraction equals -1, so $a=\\frac{2}{9}$. Therefore, $p+q=2+9=011. -Gideontz",
"Write this as $a\\left( x- \\frac 14 \\right)^2 - \\frac 98$. Since $a+b+c$ is equal to the value of this expression when you plug $x=1$ in, we just need $\\frac{9a}{16}- \\frac 98$ to be an integer. Since $a>0$, we also have $\\frac{9a}{16}>0$ which means $\\frac{9a}{16}- \\frac 98 > -\\frac{9}{8}$. The least possible value of $a$ is when this is equal to $-1$, or $a=\\frac 29$, which gives answer $11$. -bobthegod78, krwang, Simplest14",
"Take the derivative to get that the vertex is at $2ax+b=0$ and note that this implies $\\frac{1}{2} \\cdot a = -b$ and proceed with any of the solutions above. ~Dhillonr25 $\\textbf{Appendix to Solution 5 (You don't remember differential calculus)}$ Note that the quadratic formula for finding roots of parabolas is $x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$, so if you average the two x-values of the roots the $\\pm\\sqrt{b^2-4ac}$ part will cancel out and leave you with $x=-\\frac{b}{2a}$. Then proceed as above. ~WhatdoHumanitariansEat",
"Take the derivative to get that the vertex is at $2ax+b=0$ and note that this implies $\\frac{1}{2} \\cdot a = -b$ and proceed with any of the solutions above. ~Dhillonr25 $\\textbf{Appendix to Solution 5 (You don't remember differential calculus)}$ Note that the quadratic formula for finding roots of parabolas is $x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$, so if you average the two x-values of the roots the $\\pm\\sqrt{b^2-4ac}$ part will cancel out and leave you with $x=-\\frac{b}{2a}$. Then proceed as above. ~WhatdoHumanitariansEat"
] |
2011-I-7
| 2,011
| 7
|
Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ , $x_1$ , $\dots$ , $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\]
| 16
|
I
|
[
"$m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}$. Now, divide by $m^{x_0}$ to get $1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}$. Notice that since we can choose all nonnegative $x_0,...,x_{2011}$, we can make $x_n-x_0$ whatever we desire. WLOG, let $x_0\\geq...\\geq x_{2011}$ and let $a_n=x_n-x_0$. Notice that, also, $m^{a_{2011}}$ doesn't matter if we are able to make $m^{a_1} +m^{a_2} + .... + m^{a_{2010}}$ equal to $1-\\left(\\frac{1}{m}\\right)^x$ for any power of $x$. Consider $m=2$. We can achieve a sum of $1-\\left(\\frac{1}{2}\\right)^x$ by doing $\\frac{1}{2}+\\frac{1}{4}+...$ (the \"simplest\" sequence). If we don't have $\\frac{1}{2}$, to compensate, we need $2\\cdot \\frac{1}{4}$'s. Now, let's try to generalize. The \"simplest\" sequence is having $\\frac{1}{m}$ $m-1$ times, $\\frac{1}{m^2}$ $m-1$ times, $\\ldots$. To make other sequences, we can split $m-1$ $\\frac{1}{m^i}$s into $m(m-1)$ $\\text{ }\\frac{1}{m^{i+1}}$s since $(m-1)\\cdot\\frac{1}{m^{i}} = m(m-1)\\cdot\\frac{1}{m^{i+1}}$. Since we want $2010$ terms, we have $\\sum$ $(m-1)\\cdot m^x=2010$. However, since we can set $x$ to be anything we want (including 0), all we care about is that $(m-1)\\text{ }|\\text{ } 2010$ which happens $016 times.",
"Let \\[P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}\\]. The problem then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_{2011}$ are nonnegative integers. We plug in $m = 1$ and see that \\[P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010\\] Now, we can say that \\[P(m) = (m-1)Q(m) - 2010\\] for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$, $(m-1)Q(m) = 2010$. Thus, if $P(m) = 0$, then $m-1 | 2010$ . Now, we need to show that \\[m^{x_{0}}=\\sum_{k = 1}^{2011}m^{x_{k}}\\forall m-1|2010\\] We try with the first few $m$ that satisfy this. For $m = 2$, we see we can satisfy this if $x_0 = 2010$, $x_1 = 2009$, $x_2 = 2008$, $\\cdots$ , $x_{2008} = 2$, $x_{2009} = 1$, $x_{2010} = 0$, $x_{2011} = 0$, because \\[2^{2009} + 2^{2008} + \\cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \\cdots + 2^1 + 2^1 = \\cdots\\] (based on the idea $2^n + 2^n = 2^{n+1}$, leading to a chain of substitutions of this kind) \\[= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}\\]. Thus $2$ is a possible value of $m$. For other values, for example $m = 3$, we can use the same strategy, with \\[x_{2011} = x_{2010} = x_{2009} = 0\\], \\[x_{2008} = x_{2007} = 1\\], \\[x_{2006} = x_{2005} = 2\\], \\[\\cdots\\], \\[x_2 = x_1 = 1004\\] and \\[x_0 = 1005\\], because \\[3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\\cdots+3^{1004} +3^{1004}\\] \\[=\\cdots = 3^{1004} +3^{1004}+3^{1004} = 3^{1005}\\] It's clearly seen we can use the same strategy for all $m-1 |2010$. We count all positive $m$ satisfying $m-1 |2010$, and see there are $016",
"One notices that $m-1 \\mid 2010$ if and only if there exist non-negative integers $x_0,x_1,\\ldots,x_{2011}$ such that $m^{x_0} = \\sum_{k=1}^{2011}m^{x_k}$. To prove the forward case, we proceed by directly finding $x_0,x_1,\\ldots,x_{2011}$. Suppose $m$ is an integer such that $m^{x_0} = \\sum_{k=1}^{2011}m^{x_k}$. We will count how many $x_k = 0$, how many $x_k = 1$, etc. Suppose the number of $x_k = 0$ is non-zero. Then, there must be at least $m$ such $x_k$ since $m$ divides all the remaining terms, so $m$ must also divide the sum of all the $m^0$ terms. Thus, if we let $x_k = 0$ for $k = 1,2,\\ldots,m$, we have, \\[m^{x_0} = m + \\sum_{k=m+1}^{2011}m^{x_k}.\\] Well clearly, $m^{x_0}$ is greater than $m$, so $m^2 \\mid m^{x_0}$. $m^2$ will also divide every term, $m^{x_k}$, where $x_k \\geq 2$. So, all the terms, $m^{x_k}$, where $x_k < 2$ must sum to a multiple of $m^2$. If there are exactly $m$ terms where $x_k = 0$, then we must have at least $m-1$ terms where $x_k = 1$. Suppose there are exactly $m-1$ such terms and $x_k = 1$ for $k = m+1,m+2,2m-1$. Now, we have, \\[m^{x_0} = m^2 + \\sum_{k=2m}^{2011}m^{x_k}.\\] One can repeat this process for successive powers of $m$ until the number of terms reaches 2011. Since there are $m + j(m-1)$ terms after the $j$th power, we will only hit exactly 2011 terms if $m-1$ is a factor of 2010. To see this, \\[m+j(m-1) = 2011 \\Rightarrow m-1+j(m-1) = 2010 \\Rightarrow (m-1)(j+1) = 2010.\\] Thus, when $j = 2010/(m-1) - 1$ (which is an integer since $m-1 \\mid 2010$ by assumption, there are exactly 2011 terms. To see that these terms sum to a power of $m$, we realize that the sum is a geometric series: \\[1 + (m-1) + (m-1)m+(m-1)m^2 + \\cdots + (m-1)m^j = 1+(m-1)\\frac{m^{j+1}-1}{m-1} = m^{j+1}.\\] Thus, we have found a solution for the case $m-1 \\mid 2010$. Now, for the reverse case, we use the formula \\[x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\\cdots+1).\\] Suppose $m^{x_0} = \\sum_{k=1}^{2011}m^{x^k}$ has a solution. Subtract 2011 from both sides to get \\[m^{x_0}-1-2010 = \\sum_{k=1}^{2011}(m^{x^k}-1).\\] Now apply the formula to get \\[(m-1)a_0-2010 = \\sum_{k=1}^{2011}[(m-1)a_k],\\] where $a_k$ are some integers. Rearranging this equation, we find \\[(m-1)A = 2010,\\] where $A = a_0 - \\sum_{k=1}^{2011}a_k$. Thus, if $m$ is a solution, then $m-1 \\mid 2010$. So, there is one positive integer solution corresponding to each factor of 2010. Since $2010 = 2\\cdot 3\\cdot 5\\cdot 67$, the number of solutions is $2^4 = 016.",
"The problem is basically asking how many integers $m$ have a power that can be expressed as the sum of 2011 other powers of $m$ (not necessarily distinct). Notice that $2+2+4+8+16=32$, $3+3+3+9+9+27+27+81+81=243$, and $4+4+4+4+16+16+16+64+64+64+256+256+256=1024$. Thus, we can safely assume that the equation $2011 = (m-1)x + m$ must have an integer solution $x$. To find the number of $m$-values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant $m$ to make the equation equal a friendlier number, $2010$, instead of the ugly prime number $2011$: $2010 = (m-1)x+(m-1)$. Factor the equation and we get $2010 = (m-1)(x+1)$. The number of values of $m-1$ that allow $x+1$ to be an integer is quite obviously the number of factors of $2010$. Factoring $2010$, we obtain $2010 = 2 \\times 3 \\times 5 \\times 67$, so the number of positive integers $m$ that satisfy the required condition is $2^4 = 016. -fidgetboss_4000",
"The problem is basically asking how many integers $m$ have a power that can be expressed as the sum of 2011 other powers of $m$ (not necessarily distinct). Notice that $2+2+4+8+16=32$, $3+3+3+9+9+27+27+81+81=243$, and $4+4+4+4+16+16+16+64+64+64+256+256+256=1024$. Thus, we can safely assume that the equation $2011 = (m-1)x + m$ must have an integer solution $x$. To find the number of $m$-values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant $m$ to make the equation equal a friendlier number, $2010$, instead of the ugly prime number $2011$: $2010 = (m-1)x+(m-1)$. Factor the equation and we get $2010 = (m-1)(x+1)$. The number of values of $m-1$ that allow $x+1$ to be an integer is quite obviously the number of factors of $2010$. Factoring $2010$, we obtain $2010 = 2 \\times 3 \\times 5 \\times 67$, so the number of positive integers $m$ that satisfy the required condition is $2^4 = 016. -fidgetboss_4000",
"First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of $m$ we can multiply to get to non-negative powers. Now we see that our problem is just a matter of m-chopping blocks. What is meant by $m$-chopping is taking an existing block of say $m^{k}$ and turning it into $m$ blocks of $m^{k-1}$. This process increases the total number of blocks by $m-1$ per chop.The problem wants us to find the number of positive integers $m$ where some number of chops will turn $1$ block into $2011$ such blocks, thus increasing the total amount by $2010= 2 \\cdot 3 \\cdot 5 \\cdot 67$. Thus $m-1 | 2010$, and a cursory check on extreme cases will confirm that there are indeed $016s.",
"The problem is basically saying that we want to find $2011$ powers of $m$ that add together to equal another power of $m$. If we had a power of $m$, $m^n$, then to get to $m^{n+1}$ or $m \\cdot (m^n)$, we have to add $m^n$, $m-1$times. Then when we are at $m^{n+1}$, to get to $m^{n+2}$, it is similar. So we have to have $m^{\\text{some number}} = m^n + (m-1){m^n} + (m-1)(m^{n+1})\\dots$. This expression has $1 + (m-1) + (m-1) + \\dots = 1 + k(m-1)$ terms, and the number of terms must be equal to $2011$ (as stated in the problem). Then $1+k(m-1) = 2011$, and we get the equation $k(m-1) = 2010 = 2\\cdot3\\cdot5\\cdot67$. $2010$ has $16$ factors, and setting $m-1$ equal to each of these factors makes $016.",
"The problem is basically saying that we want to find $2011$ powers of $m$ that add together to equal another power of $m$. If we had a power of $m$, $m^n$, then to get to $m^{n+1}$ or $m \\cdot (m^n)$, we have to add $m^n$, $m-1$times. Then when we are at $m^{n+1}$, to get to $m^{n+2}$, it is similar. So we have to have $m^{\\text{some number}} = m^n + (m-1){m^n} + (m-1)(m^{n+1})\\dots$. This expression has $1 + (m-1) + (m-1) + \\dots = 1 + k(m-1)$ terms, and the number of terms must be equal to $2011$ (as stated in the problem). Then $1+k(m-1) = 2011$, and we get the equation $k(m-1) = 2010 = 2\\cdot3\\cdot5\\cdot67$. $2010$ has $16$ factors, and setting $m-1$ equal to each of these factors makes $016.",
"Take$\\pmod {m-1}$ of both sides \\[m^{x_0} \\equiv 1 \\pmod {m-1}\\] \\[\\sum_{k = 1}^{2011} m^{x_k} \\equiv 2011 \\pmod {m-1}\\] Setting both of these to be congruent to eachother, we get: \\[1 \\equiv 2011 \\pmod {m-1}\\] Subtract $1$ from both sides to get: \\[0 \\equiv 2010 \\pmod {m-1}\\] $m-1$ must be a factor of $2010$. $2010$ has $16$ factors, so there are $16.",
"Take$\\pmod {m-1}$ of both sides \\[m^{x_0} \\equiv 1 \\pmod {m-1}\\] \\[\\sum_{k = 1}^{2011} m^{x_k} \\equiv 2011 \\pmod {m-1}\\] Setting both of these to be congruent to eachother, we get: \\[1 \\equiv 2011 \\pmod {m-1}\\] Subtract $1$ from both sides to get: \\[0 \\equiv 2010 \\pmod {m-1}\\] $m-1$ must be a factor of $2010$. $2010$ has $16$ factors, so there are $16.",
"Let $f(a)$ be a polynomial in terms of $a$, listed below, which has roots at the values $m$ that satisfy the problem condition: \\[f(a) = a^{x_0} - \\sum_{k=1}^{2011} a^{x_k}\\] It is evident to see that the polynomial has integer coefficients, which will be used later in the solution. $f(1) = 1 - 2011 = -2010$. We use the fact that polynomials $P(x)$ with integer coefficients satisfy the condition that $(a-b)|f(a)-f(b)$ for some integers $a$ and $b$ to conclude that \\[(m-1)|f(m)-f(1)\\] Since $m$ is a solution of the polynomial, we have that $f(m) = 0$. Therefore, we have that \\[(m-1)|2010\\] Therefore, we arrive at the condition that $m-1$ must be a factor of $2010$. Since $2010=2\\cdot 3\\cdot 5\\cdot 67$, $2010$ has $16$ factors, so there are $016. -Vivdax",
"Let $f(a)$ be a polynomial in terms of $a$, listed below, which has roots at the values $m$ that satisfy the problem condition: \\[f(a) = a^{x_0} - \\sum_{k=1}^{2011} a^{x_k}\\] It is evident to see that the polynomial has integer coefficients, which will be used later in the solution. $f(1) = 1 - 2011 = -2010$. We use the fact that polynomials $P(x)$ with integer coefficients satisfy the condition that $(a-b)|f(a)-f(b)$ for some integers $a$ and $b$ to conclude that \\[(m-1)|f(m)-f(1)\\] Since $m$ is a solution of the polynomial, we have that $f(m) = 0$. Therefore, we have that \\[(m-1)|2010\\] Therefore, we arrive at the condition that $m-1$ must be a factor of $2010$. Since $2010=2\\cdot 3\\cdot 5\\cdot 67$, $2010$ has $16$ factors, so there are $016. -Vivdax"
] |
2011-I-9
| 2,011
| 9
|
Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$ . Find $24 \cot^2 x$ .
| 192
|
I
|
[
"We can rewrite the given expression as \\[\\sqrt{24^3\\sin^3 x}=24\\cos x\\] Square both sides and divide by $24^2$ to get \\[24\\sin ^3 x=\\cos ^2 x\\] Rewrite $\\cos ^2 x$ as $1-\\sin ^2 x$ \\[24\\sin ^3 x=1-\\sin ^2 x\\] \\[24\\sin ^3 x+\\sin ^2 x - 1=0\\] Testing values using the rational root theorem gives $\\sin x=\\frac{1}{3}$ as a root, $\\sin^{-1} \\frac{1}{3}$ does fall in the first quadrant so it satisfies the interval. There are now two ways to finish this problem. First way: Since $\\sin x=\\frac{1}{3}$, we have \\[\\sin ^2 x=\\frac{1}{9}\\] Using the Pythagorean Identity gives us $\\cos ^2 x=\\frac{8}{9}$. Then we use the definition of $\\cot ^2 x$ to compute our final answer. $24\\cot ^2 x=24\\frac{\\cos ^2 x}{\\sin ^2 x}=24\\left(\\frac{\\frac{8}{9}}{\\frac{1}{9}}\\right)=24(8)=192.",
"Like Solution 1, we can rewrite the given expression as \\[24\\sin^3x=\\cos^2x\\] Divide both sides by $\\sin^3x$. \\[24 = \\cot^2x\\csc x\\] Square both sides. \\[576 = \\cot^4x\\csc^2x\\] Substitute the identity $\\csc^2x = \\cot^2x + 1$. \\[576 = \\cot^4x(\\cot^2x + 1)\\] Let $a = \\cot^2x$. Then \\[576 = a^3 + a^2\\]. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = 192."
] |
2011-I-10
| 2,011
| 10
|
The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ .
| 503
|
I
|
[
"Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater). Break up the problem into two cases: an even number of sides $2n$, or an odd number of sides $2n-1$. For polygons with $2n$ sides, the circumdiameter has endpoints on $2$ vertices. There are $n-1$ points on one side of a diameter, plus $1$ of the endpoints of the diameter for a total of $n$ points. For polygons with $2n - 1$ points, the circumdiameter has $1$ endpoint on a vertex and $1$ endpoint on the midpoint of the opposite side. There are also $n - 1$ points on one side of the diameter, plus the vertex for a total of $n$ points on one side of the diameter. Case 1: $2n$-sided polygon. There are clearly $\\binom{2n}{3}$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously $2n$ choices for this point. From there, the other two points must be within the $n-1$ points remaining on the same side of the diameter. So our desired probability is $\\frac{2n\\binom{n-1}{2}}{\\binom{2n}{3}}$ $=\\frac{n(n-1)(n-2)}{\\frac{2n(2n-1)(2n-2)}{6}}$ $=\\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}$ $=\\frac{3(n-2)}{2(2n-1)}$ so $\\frac{93}{125}=\\frac{3(n-2)}{2(2n-1)}$ $186(2n-1)=375(n-2)$. $372n-186=375n-750$ $3n=564$ $n=188$ and so the polygon has $376$ sides. Case 2: $2n-1$-sided polygon. Similarly, $\\binom{2n-1}{3}$ total triangles. Again choose the leftmost point, with $2n-1$ choices. For the other two points, there are again $\\binom{n-1}{2}$ possibilities. The probability is $\\frac{(2n-1)\\binom{n-1}{2}}{\\binom{2n-1}{3}}$ $=\\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}$ $=\\frac{3(n-2)}{2(2n-3)}$ so $\\frac{93}{125}=\\frac{3(n-2)}{2(2n-3)}$ $186(2n-3)=375(n-2)$ $375n-750=372n-558$ $3n=192$ $n=64$ and our polygon has $127$ sides. Adding, $127+376=503",
"We use casework on the locations of the vertices, if we choose the locations of vertices $v_a, v_b, v_c$ on the n-gon (where the vertices of the n-gon are $v_0, v_1, v_2, ... v_{n-1},$ in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that $a<b<c$. By symmetry, we can assume WLOG that the location of vertex A is vertex $v_0$. Now, vertex B can be any of $v_1, v_2, ... v_{n-2}$. We start in on casework. Case 1: vertex B is at one of the locations $v_{n-2}, v_{n-3}, ... v_{\\lfloor n/2 \\rfloor +1}$. (The ceiling function is necessary for the cases in which n is odd.) Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle. There are $\\lceil n/2 \\rceil - 2$ choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n). If vertex B is placed at $v_m$, there are $n - m - 1$ possible places for vertex C. Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is $\\frac{(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil) - 1}{2}$. Case 2: vertex B is at one of the locations not covered in the first case. Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in $v_0$, then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at $v_0$, then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at $v_0$, and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case. Therefore, there are $\\frac{3(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{2}$ total obtuse triangles obtainable. The total number of triangles obtainable is $1+2+3+...+(n-2) = \\frac{(n-2)(n-1)}{2}$. The ratio of obtuse triangles obtainable to all triangles obtainable is therefore $\\frac{\\frac{3(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{2}}{\\frac{(n-2)(n-1)}{2}} = \\frac{3(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{(n-2)(n-1)} = \\frac{93}{125}$. So, $\\frac{(n- \\lceil n/2 \\rceil - 2)(n - \\lceil n/2 \\rceil - 1)}{(n-2)(n-1)} = \\frac{31}{125}$. Now, we have that $(n-2)(n-1)$ is divisible by $125 = 5^3$. It is now much easier to perform trial-and-error on possible values of n, because we see that $n \\equiv 1,2 \\pmod{125}$. We find that $n = 127$ and $n = 376$ both work, so the final answer is $127 + 376 = 503."
] |
2011-I-11
| 2,011
| 11
|
Let $R$ be the set of all possible remainders when a number of the form $2^n$ , $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$ . Find the remainder when $S$ is divided by 1000.
| 7
|
I
|
[
"Note that $x \\equiv y \\pmod{1000} \\Leftrightarrow x \\equiv y \\pmod{125}$ and $x \\equiv y \\pmod{8}$. So we must find the first two integers $i$ and $j$ such that $2^i \\equiv 2^j \\pmod{125}$ and $2^i \\equiv 2^j \\pmod{8}$ and $i \\neq j$. Note that $i$ and $j$ will be greater than 2 since remainders of $1, 2, 4$ will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that $2^{100}\\equiv 1\\pmod{125}$ (see Euler's theorem) and $2^0,2^1,2^2,\\ldots,2^{99}$ are all distinct modulo 125 (proof below). Thus, $i = 103$ and $j =3$ are the first two integers such that $2^i \\equiv 2^j \\pmod{1000}$. All that is left is to find $S$ in mod $1000$. After some computation: \\[S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \\equiv 8 - 1 \\mod 1000 = 007, giving us the needed contradiction.",
"Notice that our sum of remainders looks like \\[S = 1 + 2 + 4 + 2^3 + 2^4 + \\cdots.\\] We want to find the remainder of $S$ upon division by $1000.$ Since $1000$ decomposes into primes as $1000 = 2^3 \\cdot 5^3$, we can check the remainders of $S$ modulo $2^3$ and modulo $5^3$ separately. Checking $S$ modulo $2^3$ is easy, so lets start by computing the remainder of $S$ upon division by $5^3.$ To do this, let's figure out when our sequence finally repeats. Notice that since the remainder when dividing any term of $S$ (after the third term) by $1000$ will be a multiple of $2^3$, when this summation finally repeats, the first term to repeat will be not be $1$ since $2^3 \\nmid 1.$ Instead, the first term to repeat will be $2^3$, and then the sequence will continue once again $2^4, 2^5, \\cdots.$ Now, to compute $S$ modulo $125$, we want to find the least positive integer $d$ such that $2^d \\equiv 1 \\pmod {125}$ since then $d$ will just be the number of terms of $S$ (after the third term!) before the sequence repeats. In other words, our sequence will be of the form $S = 1 + 2 + 4 + \\left(2^3 + 2^4 + \\cdots + 2^{2 + d}\\right)$ and then we will have $2^{d + 3} \\equiv 2^3 \\pmod {125}$, and the sequence will repeat from there. Here, $d$ simply represents the order of $2$ modulo $125$, denoted by $d = \\text{ord}_{125}(2).$ To begin with, we'll use a well-known property of the order to get a bound on $d.$ Since $\\gcd(2, 125) = 1$ and $\\phi(125) = 100$, we know by Euler's Theorem that $2^{100} \\equiv 1 \\pmod {125}.$ However, we do not know that $100$ is the least $d$ satisfying $2^d \\equiv 1 \\pmod {125}.$ Nonetheless, it is a well known property of the order that $\\text{ord}_{125}(2) = d | \\phi(125) = 100.$ Therefore, we can conclude that $d$ must be a positive divisor of $100.$ Now, this still leaves a lot of possibilities, so let's consider a smaller modulus for the moment, say $\\mod 5.$ Clearly, we must have that $2^d \\equiv 1 \\pmod 5.$ Since $2^4 \\equiv 1 \\pmod 5$ and powers of two will then cycle every four terms, we know that $2^d \\equiv 1 \\pmod 5 \\iff 4 | d.$ Combining this relation with $d | 100$, it follows that $d \\in \\{4, 20, 100\\}.$ Now, it is trivial to verify that $d \\ne 4.$ In addition, we know that \\[2^{20} = \\left(2^{10}\\right)^2 = \\left(1024\\right)^2 \\equiv 24^2 = 576 \\not\\equiv 1 \\pmod {125}.\\] Therefore, we conclude that $d \\ne 20.$ Hence, we must have $d = 100.$ (Notice that we could have guessed this by Euler's, but we couldn't have been certain without investigating the order more thoroughly). Now, since we have found $d = 100$, we know that \\[S = 1 + 2 + 4 + 2^3 + 2^4 + \\cdots + 2^{102}.\\] There are two good ways to finish from here: The first way is to use a trick involving powers of $2.$ Notice that \\[S = 1 + 2 + 4 + ... + 2^{102} = 2^{103} - 1.\\] Certainly \\[S = 2^{103} - 1 \\equiv -1 \\equiv 7 \\pmod {8}.\\] In addition, since we already computed $\\text{ord}_{125}(2) = d = 100$, we know that \\[S = 2^{103} - 1 = 2^{100} \\cdot 2^3 - 1 \\equiv 2^3 - 1 \\equiv 7 \\pmod {125}.\\] Therefore, since $S \\equiv 7 \\pmod{8}$ and $S \\equiv 7 \\pmod{125}$, we conclude that $S \\equiv 007 Therefore, \\[S = R + T \\equiv R \\equiv 007 \\pmod{1000}.\\]",
"Obviously, $2^i$ will have to repeat at some point, and our goal is just to find when it repeats. Suppose $2^a$ is the first time the powers of 2 repeat mod 1000, and that it is the same as $2^b$ where $b < a.$ We have \\[2^a \\equiv 2^b \\mod 1000 \\rightarrow 2^a - 2^b \\equiv 0 \\mod 1000\\] We can factor out a $2^b$ to get \\[2^b\\left(2^{a-b} - 1\\right) \\equiv 0 \\mod 1000\\] Now, let's apply CRT. Obviously, if it is 0 mod 1000, this means directly that it is 0 mod 8 and 0 mod 125. Since $2^{a-b} - 1$ has to be odd, this necessarily means $b \\ge 3$ for $8 \\div 2^b.$ This means that 125 has to divide $2^{a-b} - 1.$ We wish to find the minimum $a - b$ such that this is true, or similarly, we just wish to find $\\text{ord}_{125}(2).$ Note that by Euler's Totient Theorem, that $2^{100} \\equiv 1 \\mod 125.$ After checking, and seeing that $2^{50} \\equiv -1 \\mod 125$, this directly means that $\\text{ord}_{125}(2) = 100$ or $a - b = 100.$ Since $b \\ge 3$, the smallest $(a, b)$ pair is $(103, 3).$ What this really means is that the sequence of remainders will start repeating at $2^{103}$ or namely that $\\{2^0, 2^1, ..., 2^{102}\\}$ are all distinct residues mod 1000. Adding these yields \\[2^{103} - 1 \\equiv 8 - 1 \\equiv 7 \\mod 1000\\]",
"If we write out the remainders when powers of $2$ are divided by $1000$, we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down. The answer extraction of the problem is equivalent to asking for $1+2+4 + \\dots + 2^n \\pmod{1000}$, where $2^{n+1}$ is the first number whose remainder when divided by $1000$ is a repeat. This expression is equivalent to $2^{n+1}-1 \\pmod{1000}$. So our answer will just be the first repeated remainder minus $1$. (Everything up to this point has been perfectly rigorous; now, we will destroy this.) Note that $1$, $2$ and $4$ cannot be the first repeated remainders, due to the $2$, $4$ and $8$ divisibility rules, respectively (i.e. $2^0$, $2^1$ and $2^2$ are the only powers of $2$ that leave a remainder of $1$, $2$ and $4$, respectively, when divided by $1000$.) There is no reason why $8$ could not be the first repeated remainder, so it probably is. Thus, our answer is $8-1 = 7 reappears at all in the sequence, it must be the first repeated remainder.) ~ihatemath123",
"If we write out the remainders when powers of $2$ are divided by $1000$, we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down. The answer extraction of the problem is equivalent to asking for $1+2+4 + \\dots + 2^n \\pmod{1000}$, where $2^{n+1}$ is the first number whose remainder when divided by $1000$ is a repeat. This expression is equivalent to $2^{n+1}-1 \\pmod{1000}$. So our answer will just be the first repeated remainder minus $1$. (Everything up to this point has been perfectly rigorous; now, we will destroy this.) Note that $1$, $2$ and $4$ cannot be the first repeated remainders, due to the $2$, $4$ and $8$ divisibility rules, respectively (i.e. $2^0$, $2^1$ and $2^2$ are the only powers of $2$ that leave a remainder of $1$, $2$ and $4$, respectively, when divided by $1000$.) There is no reason why $8$ could not be the first repeated remainder, so it probably is. Thus, our answer is $8-1 = 7 reappears at all in the sequence, it must be the first repeated remainder.) ~ihatemath123",
"We can simply list out the last $3$ digits of all the powers of $2$ until we find a pattern. Note: This solution is very tedious and should not be used unless there is enough remaining time and you can't think of any other way $001, 002, 004, 008, 016, 032, 064, 128, 256, 512, 024, 048, 096, 192, 384, 768, 536, 072, 144, 288, 576, 152, 304, 608,$ $216, 432, 864, 728, 456, 912, 824, 648, 296, 592, 184, 368, 736, 472, 944, 888, 776, 552, 104, 208, 416, 832, 664, 328,$ $656, 312, 624, 248, 496, 992 \\cdots$ We can see that after $496$ comes $992$ which can be written as $-8 \\pmod {1000}$. That means that all the terms after $8$ will repeat but negated modulo $1000$. Note that $n \\pmod m + -n \\pmod m = 0 \\pmod m$. As we are looking for the sum of all of the possible values mod $1000$, the answer is just $1 + 2 + 4 = 007 ~EvanZ",
"We can simply list out the last $3$ digits of all the powers of $2$ until we find a pattern. Note: This solution is very tedious and should not be used unless there is enough remaining time and you can't think of any other way $001, 002, 004, 008, 016, 032, 064, 128, 256, 512, 024, 048, 096, 192, 384, 768, 536, 072, 144, 288, 576, 152, 304, 608,$ $216, 432, 864, 728, 456, 912, 824, 648, 296, 592, 184, 368, 736, 472, 944, 888, 776, 552, 104, 208, 416, 832, 664, 328,$ $656, 312, 624, 248, 496, 992 \\cdots$ We can see that after $496$ comes $992$ which can be written as $-8 \\pmod {1000}$. That means that all the terms after $8$ will repeat but negated modulo $1000$. Note that $n \\pmod m + -n \\pmod m = 0 \\pmod m$. As we are looking for the sum of all of the possible values mod $1000$, the answer is just $1 + 2 + 4 = 007 ~EvanZ",
"Note that the problem essentially asks for the sum of all $x$ in the residue class of $1000$ that come from a power of $2$. We can split up this residue class into finding a solution for $x$ in the modular system \\[x \\equiv 2^n \\bmod 8\\] \\[x \\equiv 2^n \\bmod 125\\] Clearly, if $n \\geq 3$, we get the first equation to be \\[x \\equiv 0 \\bmod 8\\] If $n \\leq 2$, we get our residues to be $1$,$2$, and $4$ which add to $7$. Now, we just need to worry about the $125$ mod equation. If we take a look at the powers of $2$ modulo $5$, we see that we get every residue except for $0$. If we consider it modulo $25$, we see that we get every residue except for $0,5,10,15,20$ (ie. all multiples of $5$). This is similar to hensel's lemma but it isn't and from here we assume the pattern will continue in that the residues will only be exempt of the multiples of $5$ in modulo $125$. Therefore, we can add all multiples of $8$ from $0$ to $1000$ and subtract the multiples of $40$ to get $0 \\bmod 1000$. Therefore, our final answer is $7. ~Vedoral",
"Note that the problem essentially asks for the sum of all $x$ in the residue class of $1000$ that come from a power of $2$. We can split up this residue class into finding a solution for $x$ in the modular system \\[x \\equiv 2^n \\bmod 8\\] \\[x \\equiv 2^n \\bmod 125\\] Clearly, if $n \\geq 3$, we get the first equation to be \\[x \\equiv 0 \\bmod 8\\] If $n \\leq 2$, we get our residues to be $1$,$2$, and $4$ which add to $7$. Now, we just need to worry about the $125$ mod equation. If we take a look at the powers of $2$ modulo $5$, we see that we get every residue except for $0$. If we consider it modulo $25$, we see that we get every residue except for $0,5,10,15,20$ (ie. all multiples of $5$). This is similar to hensel's lemma but it isn't and from here we assume the pattern will continue in that the residues will only be exempt of the multiples of $5$ in modulo $125$. Therefore, we can add all multiples of $8$ from $0$ to $1000$ and subtract the multiples of $40$ to get $0 \\bmod 1000$. Therefore, our final answer is $7. ~Vedoral"
] |
2011-I-12
| 2,011
| 12
|
Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent.
| 594
|
I
|
[
"Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men: _(2)_(2)_(2)_ _(3)_(3)_ _(2)_(4)_ _(4)_(2)_ _(6)_ For the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the $n$ women. Since there are $n+1$ possible places to insert the dividers, and we need to choose any three of these locations, we have $\\dbinom{n+1}{3}$ ways. The second, third, and fourth cases are like the first, only that we need to insert two dividers among the $n+1$ possible locations. Each gives us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$ ways. The last case gives us $\\dbinom{n+1}{1}=n+1$ ways. Therefore, the total number of possible ways where there are no isolated men is \\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\] The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or \\[2\\dbinom{n+1}{2}+(n+1).\\] Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that \\[\\dfrac{2\\dbinom{n+1}{2}+(n+1)}{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\] After simplification, we arrive at \\[\\dfrac{6(n+1)}{n^2+8n+6}\\le\\dfrac{1}{100}.\\] Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\\ge594$. Clearly $n>592$, or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=594 satisfies the inequality."
] |
2011-I-13
| 2,011
| 13
|
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r - \sqrt{s}}{t}$ , where $r$ , $s$ , and $t$ are positive integers. Find $r + s + t$ .
| 330
|
I
|
[
"Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \\[\\frac{AX+BY+CZ+D}{\\sqrt{A^2+B^2+C^2}},\\] so the (directed) distance from any point $(x,y,z)$ to the plane is $ax+by+cz+d$. So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$, and by rearranging and summing, \\[(10-d)^2+(11-d)^2+(12-d)^2= 100\\cdot(a^2+b^2+c^2)=100.\\] Solving the equation is easier if we substitute $11-d=y$, to get $3y^2+2=100$, or $y=\\sqrt {98/3}$. The distance from the origin to the plane is simply $d$, which is equal to $11-\\sqrt{98/3} =(33-\\sqrt{294})/3$, so $33+294+3=330.",
"Let the vertices with distance $10,11,12$ be $B,C,D$, respectively. An equilateral triangle $\\triangle BCD$ is formed with side length $10\\sqrt{2}$. We care only about the $z$ coordinate: $B=10,C=11,D=12$. It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\\text{centroid}=(10+11+12)/3=11$. Designate the midpoint of $BD$ as $M$. Notice that median $CM$ is parallel to the plane because the $\\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$, and the median contains $C$ and the $\\text{centroid}$. We seek the angle $\\theta$ of the line:$(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\\triangle BCD$, $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $\\textit{in slope}$ to $BD$. Since $BD$ makes a $2-14-10\\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\\circ$. Therefore, $\\sin\\theta=\\frac{14}{10\\sqrt{2}}=\\frac{7\\sqrt{2}}{10}$. It is also known that the centroid of $\\triangle BCD$ is a third of the way between vertex $A$ and $H$, the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\\sqrt{3}$. So the distance from the $\\text{centroid}$ to $A$ is $10/\\sqrt{3}$. So, the $\\Delta z$ from $A$ to the centroid is $\\frac{10}{\\sqrt{3}}\\sin\\theta=\\frac{10}{\\sqrt{3}}\\left(\\frac{7\\sqrt{2}}{10}\\right)=\\frac{7\\sqrt{6}}{3}$. Thus the distance from $A$ to the plane is $11-\\frac{7\\sqrt{6}}{3}=\\frac{33-7\\sqrt{6}}{3}=\\frac{33-\\sqrt{294}}{3}$, and $33+294+3=330.",
""
] |
2011-I-14
| 2,011
| 14
|
Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ , $M_3$ , $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ , $\overline{A_3 A_4}$ , $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ , $R_3 \perp R_5$ , $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ , $R_3$ and $R_5$ , $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ , $B_3$ , $B_5$ , $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ .
| 37
|
I
|
[
"We use coordinates. Let the octagon have side length $2$ and center $(0, 0)$. Then all of its vertices have the form $(\\pm 1, \\pm\\left(1+\\sqrt{2}\\right))$ or $(\\pm\\left(1+\\sqrt{2}\\right), \\pm 1)$. By symmetry, $B_{1}B_{3}B_{5}B_{7}$ is a square. Thus lines $\\overleftrightarrow{B_{1}B_{3}}$ and $\\overleftrightarrow{B_{5}B_{7}}$ are parallel, and its side length is the distance between these two lines. However, this is given to be the side length of the octagon, or $2$. Suppose the common slope of the lines is $m$ and let $m=\\tan\\theta$. Then, we want to find \\[\\cos 2\\left(90-\\theta\\right)=2\\cos^{2}\\left(90-\\theta\\right)-1=2\\sin^{2}\\theta-1.\\] It can easily be seen that the equations of the lines are \\begin{align*} B_{1}B_{3}: y-mx+m\\left(1+\\sqrt{2}\\right)=0 \\\\ B_{5}B_{7}: y-mx-m\\left(1+\\sqrt{2}\\right)=0.\\end{align*} By the distance between parallel lines formula, a corollary of the point to line distance formula, the distance between these two lines is \\[\\frac{|c_{2}-c_{1}|}{\\sqrt{a^{2}+b^{2}}}=\\frac{2m\\left(1+\\sqrt{2}\\right)}{\\sqrt{m^{2}+1}}.\\] Since we want this to equal $2$, we have \\begin{align*}\\frac{2m\\left(1+\\sqrt{2}\\right)}{\\sqrt{m^{2}+1}}&=2 \\\\ 4m^{2}\\left(3+2\\sqrt{2}\\right)&=4m^{2}+4 \\\\ \\left(12+8\\sqrt{2}\\right)m^{2}&=4m^{2}+4 \\\\ \\left(8+8\\sqrt{2}\\right)m^{2}&=4 \\\\ m^{2}&=\\frac{4}{8+8\\sqrt{2}} \\\\ \\Rightarrow m^{2}=\\tan^{2}\\theta=\\frac{\\sin^{2}\\theta}{\\cos^{2}\\theta}&=\\frac{1}{2+2\\sqrt{2}}.\\end{align*} Since $\\sin^{2}\\theta+\\cos^{2}\\theta=1,$ we have $\\sin^{2}\\theta=\\frac{1}{3+2\\sqrt{2}}$. Thus \\[2\\sin^{2}\\theta-1=\\frac{2}{3+2\\sqrt{2}}-1=\\frac{-1-2\\sqrt{2}}{3+2\\sqrt{2}}=\\frac{\\left(-1-2\\sqrt{2}\\right)\\left(3-2\\sqrt{2}\\right)}{\\left(3+2\\sqrt{2}\\right)\\left(3-2\\sqrt{2}\\right)}=\\frac{5-4\\sqrt{2}}{1}=5-\\sqrt{32}.\\] The answer is $037.",
"Let $\\theta=\\angle M_1 M_3 B_1$. Thus we have that $\\cos 2 \\angle A_3 M_3 B_1=\\cos \\left(2\\theta + \\frac{\\pi}{2} \\right)=-\\sin2\\theta$. Since $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ is a regular octagon and $B_1 B_3 = A_1 A_2$, let $k=A_1 A_2 = A_2 A_3 = B_1 B_3$. Extend $\\overline{A_1 A_2}$ and $\\overline{A_3 A_4}$ until they intersect. Denote their intersection as $I_1$. Through similar triangles & the $45-45-90$ triangles formed, we find that $M_1 M_3=\\frac{k}{2}(2+\\sqrt2)$. We also have that $\\triangle M_7 B_7 M_1 =\\triangle M_1 B_1 M_3$ through ASA congruence ($\\angle B_7 M_7 M_1 =\\angle B_1 M_1 M_3$, $M_7 M_1 = M_1 M_3$, $\\angle B_7 M_1 M_7 =\\angle B_1 M_3 M_1$). Therefore, we may let $n=M_1 B_7 = M_3 B_1$. Thus, we have that $\\sin\\theta=\\frac{n-k}{\\frac{k}{2}(2+\\sqrt2)}$ and that $\\cos\\theta=\\frac{n}{\\frac{k}{2}(2+\\sqrt2)}$. Therefore $\\cos\\theta-\\sin\\theta=\\frac{k}{\\frac{k}{2}(2+\\sqrt2)}=\\frac{2}{2+\\sqrt2}=2-\\sqrt2$. Squaring gives that $\\sin^2\\theta - 2\\sin\\theta\\cos\\theta + \\cos^2\\theta = 6-4\\sqrt2$ and consequently that $-2\\sin\\theta\\cos\\theta = 5-4\\sqrt2 = -\\sin2\\theta$ through the identities $\\sin^2\\theta + \\cos^2\\theta = 1$ and $\\sin2\\theta = 2\\sin\\theta\\cos\\theta$. Thus we have that $\\cos 2 \\angle A_3 M_3 B_1=5-4\\sqrt2=5-\\sqrt{32}$. Therefore $m+n=5+32=037.",
"Let $A_1A_2 = 2$. Then $B_1$ and $B_3$ are the projections of $M_1$ and $M_5$ onto the line $B_1B_3$, so $2=B_1B_3=-M_1M_5\\cos x$, where $x = \\angle A_3M_3B_1$. Then since $M_1M_5 = 2+2\\sqrt{2}, \\cos x = \\dfrac{-2}{2+2\\sqrt{2}}= 1-\\sqrt{2}$, $\\cos 2x = 2\\cos^2 x -1 = 5 - 4\\sqrt{2} = 5-\\sqrt{32}$, and $m+n=037.",
"Notice that $R_3$ and $R_7$ are parallel ($B_1B_3B_5B_7$ is a square by symmetry and since the rays are perpendicular) and $B_1B_3=B_3B_5=s=$ the distance between the parallel rays. If the regular hexagon as a side length of $s$, then $M_3M_7$ has a length of $s+s\\sqrt{2}$. Let $X$ be on $R_3$ such that $M_7X$ is perpendicular to $M_3X$, and $\\phi=\\angle M_7M_3X$. The distance between $R_3$ and $R_7$ is $s=M_7X$, so $\\sin\\phi=\\frac{s}{s+s\\sqrt{2}}=\\frac{1}{1+\\sqrt{2}}$. Since we are considering a regular hexagon, $M_3$ is directly opposite to $M_7$ and $\\angle A_3M_3B_1=90 ^\\circ +\\phi$. All that's left is to calculate $\\cos 2\\angle A_3M_3B_1=\\cos^2(90^\\circ+\\phi)-\\sin^2(90^\\circ+\\phi)=\\sin^2\\phi-\\cos^2\\phi$. By drawing a right triangle or using the Pythagorean identity, $\\cos^2\\phi=\\frac{2+2\\sqrt2}{3+2\\sqrt2}$ and $\\cos 2\\angle A_3M_3B_1=\\frac{-1-2\\sqrt2}{3+2\\sqrt2}=5-4\\sqrt2=5-\\sqrt{32}$, so $m+n=037.",
"Assume that $A_1A_2=1.$ Denote the center $O$, and the midpoint of $B_1$ and $B_3$ as $B_2$. Then we have that\\[\\cos\\angle A_3M_3B_1=\\cos(\\angle A_3M_3O+\\angle OM_3B_1)=-\\sin(\\angle OM_3B_1)=-\\frac{OB_2}{OM_3}=-\\frac{1/2}{1/2+\\sqrt2/2}=-\\frac{1}{\\sqrt2+1}=1-\\sqrt2.\\]Thus, by the cosine double-angle theorem,\\[\\cos2\\angle A_3M_3B_1=2(1-\\sqrt2)^2-1=5-\\sqrt{32},\\]so $m+n=037."
] |
2011-I-15
| 2,011
| 15
|
For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ , $b$ , and $c$ . Find $|a| + |b| + |c|$ .
| 98
|
I
|
[
"From Vieta's formulas, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$. Thus $a = -(b+c)$. All three of $a$, $b$, and $c$ are non-zero: say, if $a=0$, then $b=-c=\\pm\\sqrt{2011}$ (which is not an integer). $\\textsc{wlog}$, let $|a| \\ge |b| \\ge |c|$. If $a > 0$, then $b,c < 0$ and if $a < 0$, then $b,c > 0,$ from the fact that $a+b+c=0$. We have \\[-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc\\] Thus $a^2 = 2011 + bc$. We know that $b$, $c$ have the same sign, so product $bc$ is always positive. So $|a| \\ge 45 = \\lceil \\sqrt{2011} \\rceil$. Also, if we fix $a$, $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence, \\[2011 = a^2 - bc > \\tfrac{3}{4}a^2 \\qquad \\Longrightarrow \\qquad a ^2 < \\tfrac{4}{3}\\cdot 2011 = 2681+\\tfrac{1}{3}\\] So $|a| \\le 51$. Thus we have bounded $a$ as $45\\le |a| \\le 51$, i.e. $45\\le |b+c| \\le 51$ since $a=-(b+c)$. Let's analyze $bc=(b+c)^2-2011$. Here is a table: $|a|$$bc=a^2-2011$ $45$$14$ $46$$105$ $47$$198$ $48$$293$ $49$$390$ We can tell we don't need to bother with $45$, $105 = (3)(5)(7)$, So $46$ won't work. $198/47 > 4$, $198$ is not divisible by $5$, $198/6 = 33$, which is too small to get $47$. $293/48 > 6$, $293$ is not divisible by $7$ or $8$ or $9$, we can clearly tell that $10$ is too much. Hence, $|a| = 49$, $a^2 -2011 = 390$. $b = 39$, $c = 10$. Answer: $098",
"Starting off like the previous solution, we know that $a + b + c = 0$, and $ab + bc + ac = -2011$. Therefore, $c = -b-a$. Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$. Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$. Therefore, a sum ($a+b$) squared minus a product ($ab$) gives $2011$.. We can guess and check different $a+b$’s starting with $45$ since $44^2 < 2011$. $45^2 = 2025$ therefore $ab = 2025-2011 = 14$. Since no factors of $14$ can sum to $45$ ($1+14$ being the largest sum), a + b cannot equal $45$. $46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$. $5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either. We can continue to do this until we reach $49$. $49^2 = 2401$ making $ab = 390 = 2 * 3 * 5* 13$. $3 * 13 + 2* 5 = 49$, so one root is $10$ and another is $39$. The roots sum to zero, so the last root must be $-49$. $|-49|+10+39 = 098.",
"Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$. Now, due to the first equation, let us say that $a+b=-c$, meaning that $a,b>0$ and $c<0$. Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$, respectively. Since $c$ is less than 0, it equals $-a-b$. Therefore, $|c|=|-a-b|=a+b$, meaning $|a|+|b|+|c|=2(a+b)$. We now apply Newton's sums to get that $a^2+b^2+ab=2011$,or $(a+b)^2-ab=2011$. Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$, so $2(a+b)=098.",
"We have $(x-a)\\cdot (x-b)\\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc$ As a result, we have $a+b+c=0$ $ab+bc+ac=-2011$ $abc=-m$ So, $a=-b-c$ As a result, $ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011$ Solve $b=\\frac {-c+\\sqrt{c^2-4(c^2-2011)}}{2}$ and $\\Delta =8044-3c^2=k^2$, where $k$ is an integer Cause $89<\\sqrt{8044}<90$ So, after we tried for $2$ times, we get $k=88$ and $c=10$ then $b=39$, $a=-b-c=-49$ As a result, $|a|+|b|+|c|=10+39+49=098",
"First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\\implies b^2+bc+c^2=2011$. Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\\equiv \\{1,2\\} \\pmod{3}$, and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$, testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$, $c=39, -49$. Choosing $c$ positive we get $a=-49$, so $|a|+|b|+|c|=10+49+39=098 ~firebolt360",
"First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\\implies b^2+bc+c^2=2011$. Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\\equiv \\{1,2\\} \\pmod{3}$, and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$, testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$, $c=39, -49$. Choosing $c$ positive we get $a=-49$, so $|a|+|b|+|c|=10+49+39=098 ~firebolt360",
"Note that $-c=b+a$, so $c^2=a^2+2ab+b^2$, or $-c^2+ab=-a^2-ab-b^2$. Also, $ab+bc+ca=-2011$, so $(a+b)c+ab=-c^2+ab=-2011$. Substituting $-c^2+ab=-a^2-ab-b^2$, we can obtain $a^2+ab+b^2=2011$, or $\\frac{a^3-b^3}{a-b}=2011$. If it is not known that $2011$ is prime, it may be proved in $5$ minutes or so by checking all primes up to $43$. If $2011$ divided either of $a, b$, then in order for $a^3-b^3$ to contain an extra copy of $2011$, both $a, b$ would need to be divisible by $2021$. But then $c$ would also be divisible by $2011$, and the sum $ab+bc+ca$ would clearly be divisible by $2011^2$. By LTE, $v_{2011}(a^3-b^3)=v_{2011}(a-b)$ if $a-b$ is divisible by $2011$ and neither $a,b$ are divisible by $2011$. Thus, the only possibility remaining is if $a-b$ did not divide $2011$. Let $a=k+b$. Then, we have $(b+k)^3-b^3=2011k$. Rearranging gives $3b(b+k)=2011-k^2$. As in the above solutions, we may eliminate certain values of $k$ by using mods. Then, we may test values until we obtain $k=29$, and $a=10$. Thus, $b=39$, $c=-49$, and our answer is $49+39+10=098$."
] |
2011-II-1
| 2,011
| 1
|
Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$ .
| 37
|
II
|
[
"Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $\\frac{1}{2} - 2x =\\frac{2}{9} (1-x)$, or $9 - 36x = 4 - 4x$, or $32x = 5$ or $x = 5/32$. Therefore, $m + n = 5 + 32 = 037.",
"WLOG, Gary purchased \\( n \\) liters and consumed \\( m \\) liters. After this, he purchased \\( \\frac{n}{2} \\) liters, and consumed \\( 2m \\) liters. He originally wasted \\( n-m \\) liters, but now he wasted \\( \\frac{n}{2} - 2m \\). \\[ \\frac{n}{2} - 2m = \\frac{4}{18} \\cdot (n-m) \\] \\[ 9n - 36m = 4n - 4m \\implies 5n = 32m \\implies \\frac{m}{n} = \\frac{5}{32}. \\] Thus, the answer is $37 ~idk123456"
] |
2011-II-2
| 2,011
| 2
|
On square $ABCD$ , point $E$ lies on side $AD$ and point $F$ lies on side $BC$ , so that $BE=EF=FD=30$ . Find the area of the square $ABCD$ .
| 810
|
II
|
[
"Drawing the square and examining the given lengths, [asy] size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label(\"$A$\", A, plain.SW); pair B = (3, 0); dot(B); label(\"$B$\", B, plain.SE); pair C = (3, 3); dot(C); label(\"$C$\", C, plain.NE); pair D = (0, 3); dot(D); label(\"$D$\", D, plain.NW); pair E = (0, 1); dot(E); label(\"$E$\", E, plain.W); pair F = (3, 2); dot(F); label(\"$F$\", F, plain.E); label(\"$\\frac x3$\", E--A); label(\"$\\frac x3$\", F--C); label(\"$x$\", A--B); label(\"$x$\", C--D); label(\"$\\frac {2x}3$\", B--F); label(\"$\\frac {2x}3$\", D--E); label(\"$30$\", B--E); label(\"$30$\", F--E); label(\"$30$\", F--D); draw(B--C--D--F--E--B--A--D); [/asy] you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Solving for $x$, we get $x=9\\sqrt{10}$, and $x^2=810.$ Area of the square is $810."
] |
2011-II-3
| 2,011
| 3
|
The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.
| 143
|
II
|
[
"Solution 1 The average angle in an 18-gon is $160^\\circ$. In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to $160^\\circ$. Thus for some positive (the sequence is increasing and thus non-constant) integer $d$, the middle two terms are $(160-d)^\\circ$ and $(160+d)^\\circ$. Since the step is $2d$ the last term of the sequence is $(160 + 17d)^\\circ$, which must be less than $180^\\circ$, since the polygon is convex. This gives $17d < 20$, so the only suitable positive integer $d$ is 1. The first term is then $(160-17)^\\circ = 143 ~Arcticturn",
"The average angle in an 18-gon is $160^\\circ$. In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to $160^\\circ$. Thus for some positive (the sequence is increasing and thus non-constant) integer $d$, the middle two terms are $(160-d)^\\circ$ and $(160+d)^\\circ$. Since the step is $2d$ the last term of the sequence is $(160 + 17d)^\\circ$, which must be less than $180^\\circ$, since the polygon is convex. This gives $17d < 20$, so the only suitable positive integer $d$ is 1. The first term is then $(160-17)^\\circ = 143 ~Arcticturn",
"Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\\circ}$. Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\\frac{360}{18}=20^\\circ$. We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is $20$. Since there are even number of exterior angles, the middle two must be $19^\\circ$ and $21^\\circ$, and the difference between terms must be $2$. Check to make sure the smallest exterior angle is greater than $0$: $19-2(8)=19-16=3^\\circ$. It is, so the greatest exterior angle is $21+2(8)=21+16=37^\\circ$ and the smallest interior angle is $180-37=143 ~Arcticturn",
"The sum of the angles in a 18-gon is $(18-2) \\cdot 180^\\circ = 2880 ^\\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18a+153d = 2880 ^\\circ,$ or $2a+17d = 320^\\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=143 ~Arcticturn",
"Each individual angle in a $18$-gon is $\\frac {(18-2) \\cdot 180^\\circ}{18} = 160^\\circ$. Since no angle in a convex polygon can be larger than $180^\\circ$, the smallest angle possible is in the set $159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177$. Our smallest possible angle is $143 ~Arcticturn"
] |
2011-II-4
| 2,011
| 4
|
In triangle $ABC$ , $AB=20$ and $AC=11$ . The angle bisector of angle $A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of intersection of $AC$ and the line $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 51
|
II
|
[
"[asy] pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP(\"A\",D(A))--MP(\"B\",D(B),N)--MP(\"C\",D(C))--cycle); D(A--MP(\"D\",D(D),NE)--MP(\"D'\",D(D2))); D(B--MP(\"P\",D(P))); D(MP(\"M\",M,NW)); MP(\"20\",(B+D)/2,ENE); MP(\"11\",(C+D)/2,ENE); [/asy] Let $D'$ be on $\\overline{AC}$ such that $BP \\parallel DD'$. It follows that $\\triangle BPC \\sim \\triangle DD'C$, so \\[\\frac{PC}{D'C} = 1 + \\frac{BD}{DC} = 1 + \\frac{AB}{AC} = \\frac{31}{11}\\] by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that $AP = PD'$. Thus, \\[\\frac{CP}{PA} = \\frac{1}{\\frac{PD'}{PC}} = \\frac{1}{1 - \\frac{D'C}{PC}} = \\frac{31}{20},\\] and $m+n = 51.",
"Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$, so we assign $m(B) = 11, m(C) = 20, m(D) = 31$. Since $AM = MD$, then $m(A) = 31$, and $\\frac{CP}{PA} = \\frac{m(A) }{ m(C)} = \\frac{31}{20}$, so $m+n = 51.",
"Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$, so we assign $m(B) = 11, m(C) = 20, m(D) = 31$. Since $AM = MD$, then $m(A) = 31$, and $\\frac{CP}{PA} = \\frac{m(A) }{ m(C)} = \\frac{31}{20}$, so $m+n = 51.",
"By Menelaus' Theorem on $\\triangle ACD$ with transversal $PB$, \\[1 = \\frac{CP}{PA} \\cdot \\frac{AM}{MD} \\cdot \\frac{DB}{CB} = \\frac{CP}{PA} \\cdot \\left(\\frac{1}{1+\\frac{AC}{AB}}\\right) \\quad \\Longrightarrow \\quad \\frac{CP}{PA} = \\frac{31}{20}.\\] So $m+n = 051.",
"We will use barycentric coordinates. Let $A = (1, 0, 0)$, $B = (0, 1, 0)$, $C = (0, 0, 1)$. By the Angle Bisector Theorem, $D = [0:11:20] = \\left(0, \\frac{11}{31}, \\frac{20}{31}\\right)$. Since $M$ is the midpoint of $AD$, $M = \\frac{A + D}{2} = \\left(\\frac{1}{2}, \\frac{11}{62}, \\frac{10}{31}\\right)$. Therefore, the equation for line BM is $20x = 31z$. Let $P = (x, 0, 1-x)$. Using the equation for $BM$, we get \\[20x = 31(1-x)\\] \\[x = \\frac{31}{51}\\] Therefore, $\\frac{CP}{PA} = \\frac{1-x}{x} = \\frac{31}{20}$ so the answer is $051.",
"Let $DC=x$. Then by the Angle Bisector Theorem, $BD=\\frac{20}{11}x$. By the Ratio Lemma, we have that $\\frac{PC}{AP}=\\frac{\\frac{31}{11}x\\sin\\angle PBC}{20\\sin\\angle ABP}.$ Notice that $[\\triangle BAM]=[\\triangle BMD]$ since their bases have the same length and they share a height. By the sin area formula, we have that \\[\\frac{1}{2}\\cdot20\\cdot BM\\cdot \\sin\\angle ABP=\\frac{1}{2}\\cdot \\frac{20}{11}x\\cdot BM\\cdot\\sin\\angle PBC.\\] Simplifying, we get that $\\frac{\\sin\\angle PBC}{\\sin\\angle ABP}=\\frac{11}{x}.$ Plugging this into what we got from the Ratio Lemma, we have that $\\frac{PC}{AP}=\\frac{31}{20}\\implies051.",
"First, we will find $\\frac{MP}{BP}$. By Menelaus on $\\triangle BDM$ and the line $AC$, we have \\[\\frac{BC}{CD}\\cdot\\frac{DA}{AM}\\cdot\\frac{MP}{PB}=1\\implies \\frac{62MP}{11BP}=1\\implies \\frac{MP}{BP}=\\frac{11}{62}.\\] This implies that $\\frac{MB}{BP}=1-\\frac{MP}{BP}=\\frac{51}{62}$. Then, by Menelaus on $\\triangle AMP$ and line $BC$, we have \\[\\frac{AD}{DM}\\cdot\\frac{MB}{BP}\\cdot\\frac{PC}{CA}=1\\implies \\frac{PC}{CA}=\\frac{31}{51}.\\] Therefore, $\\frac{PC}{AP}=\\frac{31}{51-31}=\\frac{31}{20}.$ The answer is $051. -brainiacmaniac31",
"First, we will find $\\frac{MP}{BP}$. By Menelaus on $\\triangle BDM$ and the line $AC$, we have \\[\\frac{BC}{CD}\\cdot\\frac{DA}{AM}\\cdot\\frac{MP}{PB}=1\\implies \\frac{62MP}{11BP}=1\\implies \\frac{MP}{BP}=\\frac{11}{62}.\\] This implies that $\\frac{MB}{BP}=1-\\frac{MP}{BP}=\\frac{51}{62}$. Then, by Menelaus on $\\triangle AMP$ and line $BC$, we have \\[\\frac{AD}{DM}\\cdot\\frac{MB}{BP}\\cdot\\frac{PC}{CA}=1\\implies \\frac{PC}{CA}=\\frac{31}{51}.\\] Therefore, $\\frac{PC}{AP}=\\frac{31}{51-31}=\\frac{31}{20}.$ The answer is $051. -brainiacmaniac31",
"[email protected], vvsss",
"[email protected], vvsss",
"Assume $ABC$ is a right triangle at $A$. Line $AD = x$ and $BC = \\tfrac{-11}{20}x + 11$. These two lines intersect at $D$ which have coordinates $(\\frac{220}{31},\\frac{220}{31})$ and thus $M$ has coordinates $(\\frac{110}{31},\\frac{110}{31})$. Thus, the line $BM = \\tfrac{11}{51} \\cdot (20-x)$. When $x = 0$, $P$ has $y$ coordinate equal to $\\frac{11\\cdot20}{51} \\frac{AP + CP}{AP} = 1 + \\frac{CP}{AP}$ = $\\tfrac{51}{20} = 1 + \\frac{CP}{AP},$ which equals ${\\tfrac{31}{20}},$ giving an answer of $51",
"Assume $ABC$ is a right triangle at $A$. Line $AD = x$ and $BC = \\tfrac{-11}{20}x + 11$. These two lines intersect at $D$ which have coordinates $(\\frac{220}{31},\\frac{220}{31})$ and thus $M$ has coordinates $(\\frac{110}{31},\\frac{110}{31})$. Thus, the line $BM = \\tfrac{11}{51} \\cdot (20-x)$. When $x = 0$, $P$ has $y$ coordinate equal to $\\frac{11\\cdot20}{51} \\frac{AP + CP}{AP} = 1 + \\frac{CP}{AP}$ = $\\tfrac{51}{20} = 1 + \\frac{CP}{AP},$ which equals ${\\tfrac{31}{20}},$ giving an answer of $51",
"We start by using Menelaus' theorem on $\\triangle ABD$ and $EC$. So, we see that $\\frac{BC}{DC}\\cdot\\frac{DM}{AM}\\cdot\\frac{AE}{EB}=1$. By Angle Bisector theorem, $\\frac{BC}{DC}=\\frac{31}{11}$, and therefore after plugging in our values we get $\\frac{AE}{EB}=\\frac{11}{31}$. Then, by Ceva's on the whole figure, we have $\\frac{CP}{PA}\\cdot\\frac{AE}{EB}\\cdot\\frac{BD}{DC}=1$. Plugging in our values, we get $\\frac{CP}{PA}=\\frac{31}{20}$, giving an answer of $51. ~ESAOPS",
"We start by using Menelaus' theorem on $\\triangle ABD$ and $EC$. So, we see that $\\frac{BC}{DC}\\cdot\\frac{DM}{AM}\\cdot\\frac{AE}{EB}=1$. By Angle Bisector theorem, $\\frac{BC}{DC}=\\frac{31}{11}$, and therefore after plugging in our values we get $\\frac{AE}{EB}=\\frac{11}{31}$. Then, by Ceva's on the whole figure, we have $\\frac{CP}{PA}\\cdot\\frac{AE}{EB}\\cdot\\frac{BD}{DC}=1$. Plugging in our values, we get $\\frac{CP}{PA}=\\frac{31}{20}$, giving an answer of $51. ~ESAOPS"
] |
2011-II-5
| 2,011
| 5
|
The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms.
| 542
|
II
|
[
"Since the sum of the first $2011$ terms is $200$, and the sum of the first $4022$ terms is $380$, the sum of the second $2011$ terms is $180$. This is decreasing from the first 2011, so the common ratio is less than one. Because it is a geometric sequence and the sum of the first 2011 terms is $200$, second $2011$ is $180$, the ratio of the second $2011$ terms to the first $2011$ terms is $\\frac{9}{10}$. Following the same pattern, the sum of the third $2011$ terms is $\\frac{9}{10}*180 = 162$. Thus, $200+180+162=542$, so the sum of the first $6033$ terms is $542.",
"Solution by e_power_pi_times_i The sum of the first $2011$ terms can be written as $\\dfrac{a_1(1-k^{2011})}{1-k}$, and the first $4022$ terms can be written as $\\dfrac{a_1(1-k^{4022})}{1-k}$. Dividing these equations, we get $\\dfrac{1-k^{2011}}{1-k^{4022}} = \\dfrac{10}{19}$. Noticing that $k^{4022}$ is just the square of $k^{2011}$, we substitute $x = k^{2011}$, so $\\dfrac{1}{x+1} = \\dfrac{10}{19}$. That means that $k^{2011} = \\dfrac{9}{10}$. Since the sum of the first $6033$ terms can be written as $\\dfrac{a_1(1-k^{6033})}{1-k}$, dividing gives $\\dfrac{1-k^{2011}}{1-k^{6033}}$. Since $k^{6033} = \\dfrac{729}{1000}$, plugging all the values in gives $542.",
"The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$. The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$. Notice that the latter sum of terms can be expressed as $(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$. We also know that the latter sum of terms can be obtained by subtracting 200 from 380, which then means that $r^{2011} = 9/10$. The terms from 4023 to 6033 can be expressed as $(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$, which is equivalent to $((9/10)^2)(200) = 162$. Adding 380 and 162 gives the answer of $542."
] |
2011-II-6
| 2,011
| 6
|
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there?
| 80
|
II
|
[
"Rearranging the inequality we get $d-c > b-a$. Let $e = 11$, then $(a, b-a, c-b, d-c, e-d)$ is a partition of 11 into 5 positive integers or equivalently: $(a-1, b-a-1, c-b-1, d-c-1, e-d-1)$ is a partition of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partition 6 into 5 non-negative parts is $\\binom{6+4}4 = \\binom{10}4 = 210$. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry, there are as many partitions where the fourth is less than the second. So, if $N$ is the number of partitions where the second element is equal to the fourth, our answer is $(210-N)/2$. We find $N$ as a sum of 4 cases: two parts equal to zero, $\\binom82 = 28$ ways, two parts equal to one, $\\binom62 = 15$ ways, two parts equal to two, $\\binom42 = 6$ ways, two parts equal to three, $\\binom22 = 1$ way. Therefore, $N = 28 + 15 + 6 + 1 = 50$ and our answer is $(210 - 50)/2 = 080",
"Let us consider our quadruple $(a,b,c,d)$ as the following image $xaxbcxxdxx$. The location of the letter $a$, $b$, $c$, $d$ represents its value and $x$ is a place holder. Clearly the quadruple is interesting if there are more place holders between $c$ and $d$ than there are between $a$ and $b$. $0$ holders between $a$ and $b$ means we consider $a$ and $b$ as one unit $ab$ and $c$ as $cx$ yielding $\\binom83 = 56$ ways; $1$ holder between $a$ and $b$ means we consider $a$ and $b$ as one unit $axb$ and $c$ as $cxx$ yielding $\\binom 63 = 20$ ways; $2$ holders between $a$ and $b$ means we consider $a$ and $b$ as one unit $axxb$ and $c$ as $cxxx$ yielding $\\binom43 = 4$ ways. Since there cannot be $3$ holders between $a$ and $b$ so our total is $56+20+4=080.",
"We first start out when the value of $a=1$. Doing casework, we discover that $d=5,6,7,8,9,10$. We quickly find a pattern. Now, doing this for the rest of the values of $a$ and $d$, we see that the answer is simply: $(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=080",
"We first start out when the value of $a=1$. Doing casework, we discover that $d=5,6,7,8,9,10$. We quickly find a pattern. Now, doing this for the rest of the values of $a$ and $d$, we see that the answer is simply: $(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=080",
"Notice that if $a+d>b+c$, then $(11-a)+(11-d)<(11-b)+(11-c)$, so there is a bijection between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$. Quick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50. To count this, consider our numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. Notice that if, for example, $a+d=b+c=8$, that the average of $a,d$ and $b,c$ must both be $4$. In this way, there is a symmetry for this case, centered at $4$. If instead, say, $a+d=b+c=7$, an odd number, then there is symmetry with $(a,d);(b,c)$ about $3.5$. Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at $2.5,3,8,8.5$, there is $1$ case for each and so on, until centered at $5.5$, there are $10$ possible cases. Adding these all, we have $2(1+3+6)+10=50$. Thus the answer is $\\frac{\\binom{10}{4}-50}{2} = 080",
"Notice that if $a+d>b+c$, then $(11-a)+(11-d)<(11-b)+(11-c)$, so there is a bijection between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$. Quick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50. To count this, consider our numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. Notice that if, for example, $a+d=b+c=8$, that the average of $a,d$ and $b,c$ must both be $4$. In this way, there is a symmetry for this case, centered at $4$. If instead, say, $a+d=b+c=7$, an odd number, then there is symmetry with $(a,d);(b,c)$ about $3.5$. Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at $2.5,3,8,8.5$, there is $1$ case for each and so on, until centered at $5.5$, there are $10$ possible cases. Adding these all, we have $2(1+3+6)+10=50$. Thus the answer is $\\frac{\\binom{10}{4}-50}{2} = 080",
"Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is $(10*9*8*7/24)=210$ Next, intuitively, the number of quadruples where $a+d>b+c$ is equal to the number of quadruples where $a+d<b+c$. So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when $a-d$ ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where $a+d=b+c$. So the answer will be $\\frac{210-50}{2} = 080",
"Let $b = a+x$ and $c = a+x+y$ and $d = a+x+y+z$ for positive integers $x,y,z.$ In order to satisfy the other condition we need $z > x$ so we let $z = x+k.$ Now the only other condition we need to satisfy so $a+2x+y+k \\le 10.$ This condition can be transformed into $a+2x+y+k+b = 11$ for positive $a,x,y,k,b.$ Now we use generating functions to finish. We find the generating function of the whole expression is $(x + x^2 + \\cdots)^4 \\cdot (x^2+x^4 + \\cdots)$ and we are looking for the $x^{11}$ coefficient. This simplifies to finding the $x^5$ coefficient of $(1+x+\\cdots)^4 \\cdot (1+x^2+\\cdots) =\\frac{1}{(1-x)^4} \\cdot\\frac{1}{1+x}.$ Now this expression simplifies to \\[\\left(\\binom{4}{4}+\\binom{5}{4} + \\cdots +\\binom{4+k}{4}x^k\\right)(1-x+x^2-x^3 \\cdots).\\] The $x^5$ coefficient ends up to be $\\binom{9}{4} -\\binom{8}{4} +\\binom{7}{4} -\\binom{6}{4} +\\binom{5}{4} -\\binom{4}{4} = 126 - 70 + 35 - 15 + 5 - 1 = 080",
"First, let $a=1$ and $d=10$. If $b=2$, then $c$ can be from $3$ to $8$. If $b=3$, then $4$ to $7$. If $b=4$, then $c$ is between $5$ and $6$. We find a pattern that whenever $b$ increases by $1$, when $a$ and $d$ are stationary, then the possible values of $c$ decrease by 2, unless it gets to zero or negative, in which case that case ends. Counting up, we have $6+4+2=12$ different possibilities when $a=1$ and $d=10$. For $a=1$ and $d=9$, $b=2$, then $c$ can be from $3$ to $7$. If $b=3$, then $c$ can be from $4$ to $6$, and so on. Notice that the possible values for each case of $b$ gets one less than if $d$ were one greater, unless that number is zero, in which it stays zero. We then use this pattern to find all the values: $12+9+6+4+2+1+9+6+4+2+1+6+4+2+1+4+2+1+2+1+1 \\Rightarrow \\\\ 12\\cdot1+9\\cdot2+6\\cdot3+4\\cdot4+2\\cdot5+1\\cdot6= \\\\ 12+18+18+16+10+6=\\textbf{080}, is deleted after each row.",
"First, let $a=1$ and $d=10$. If $b=2$, then $c$ can be from $3$ to $8$. If $b=3$, then $4$ to $7$. If $b=4$, then $c$ is between $5$ and $6$. We find a pattern that whenever $b$ increases by $1$, when $a$ and $d$ are stationary, then the possible values of $c$ decrease by 2, unless it gets to zero or negative, in which case that case ends. Counting up, we have $6+4+2=12$ different possibilities when $a=1$ and $d=10$. For $a=1$ and $d=9$, $b=2$, then $c$ can be from $3$ to $7$. If $b=3$, then $c$ can be from $4$ to $6$, and so on. Notice that the possible values for each case of $b$ gets one less than if $d$ were one greater, unless that number is zero, in which it stays zero. We then use this pattern to find all the values: $12+9+6+4+2+1+9+6+4+2+1+6+4+2+1+4+2+1+2+1+1 \\Rightarrow \\\\ 12\\cdot1+9\\cdot2+6\\cdot3+4\\cdot4+2\\cdot5+1\\cdot6= \\\\ 12+18+18+16+10+6=\\textbf{080}, is deleted after each row.",
"Rearranging the equation obtains $b-a<d-c$. Let $a-0=e_0$, $b-a=e_3+e_1$, $c-b=e_2$, $d-c=e_3$, $11-d=e_4$. Add up all of these newly defined equations to obtain $e_0+e_1+e_2+2e_3+e_4=11$. Note that since all $e_n$ were defined to be $e_n\\ge1$, to form our stars and bars argument we can let $d_n+1=e_n$ for all $n$. Then we obtain $d_0+d_1+d_2+2d_3+d_4=5$ where $d_n$ is nonnegative. Now, we can move the $2d_3$ term to the other side and perform casework. If $2d_3=0$: 5 objects for 4 variables -> $\\binom{8}{3}$ If $2d_3=2$: 3 objects for 4 variables -> $\\binom{6}{3}$ If $2d_3=4$: 1 object for 4 variables -> $\\binom{4}{3}$ Adding all of these cases up, we get $56+20+4=080 as our requested answer. ~sigma"
] |
2011-II-7
| 2,011
| 7
|
Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let $m$ be the maximum number of red marbles for which such an arrangement is possible, and let $N$ be the number of ways he can arrange the $m+5$ marbles to satisfy the requirement. Find the remainder when $N$ is divided by $1000$ .
| 3
|
II
|
[
"We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 \"not the same colors\" and 0 \"same colors.\" Now, for every red marble we add, we will add one \"same color\" pair and keep all 10 \"not the same color\" pairs. It follows that we can add 10 more red marbles for a total of $m = 16$. We can place those ten marbles in any of 6 \"boxes\": To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as $\\binom{n+k}{k}$ where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each \"box\" must contain at least one, are not counted here) and 5 \"bars,\" the green marbles. So the answer is $\\binom{15}{5} = 3003$, take the remainder when divided by 1000 to get the answer: $003. -jackshi2006",
"Begin with the above solution's reasoning to find that the ideal sequence is RGRGRGRGRGR + 10 Rs. To count the number of arrangements where the Gs are separate, group a G and an R together (GR) which will be rearranged/counted as one. However this also introduces the case of having GRGR.... which results in too few number of marbles whose right-hand neighbor is different, so we fix an R in the beginning of our sequence which also removes it from our counting calculation. Finally we are looking at the arrangement of 5 (GR)s + 10 Rs, all indistinguishable, which is $\\frac{15!}{5!10!}=3003\\Longrightarrow003. ~clarkculus",
"Let a collection of consecutive letters be a \"blob.\" Let there be n blobs. The marbles such that the marble to their right has a different color are the ones that are the last marble of a \"blob.\" The only exception to this is the very last marble in this list. There are \\( n-1 \\) of these. The marbles such that the marble to their right has the same color are any marbles which are not at the end of a blob. The only exception to this is the last marble, which does not go in any category. The problem states that the number of marbles like this is equal to the number of marbles with a different color. Therefore, there are \\( (n-1) + (n-1) + 1 \\) marbles in total (the last \\( +1 \\) is for the last marble at the end). We know there are \\( m+5 \\) marbles, so \\( m+5 = 2n-1 \\). Solving for \\( n \\), we get \\( n = \\frac{m}{2} + 3 \\). Notice that \\( n \\leq 11 \\) because the \"best-case scenario\" is that each \\( G \\) forms its own blob, and there is a red blob in between any two \\( G \\)'s and a red blob on both ends. Here is how it should look: <red blob> G <red blob> G <red blob> G <red blob> G <red blob> G <red blob>. In this case, there are \\( 11 \\) blobs. Thus, \\( \\frac{m}{2} + 3 \\leq 11 \\implies m \\leq 16 \\). This is now a stars and bars problem. For \\( m = 16 \\), we need to create the optimal solution: \\( 11 \\) blobs. Since every <red blob> must have at least one red marble, we can just give each blob one red at the start, leaving us with \\( 16-6 = 10 \\) red marbles left to separate into \\( 6 \\) groups. By stars and bars, this is \\( \\binom{15}{5} = 3003 \\). Thus, the answer is \\( 003 \\).",
"Let a collection of consecutive letters be a \"blob.\" Let the blobs have sizes s_1, s_2, ... s_n given the existence of n blobs. The marbles such that the marble to their right has a different color are the ones that are the last marble of a \"blob.\" The only exception to this is the very last marble in this list. There are \\( n-1 \\) of these. The marbles such that the marble to their right has the same color is every marble in a blob but the last, which is \\[\\sum_{i=1}^{n} (s_i - 1) = \\left(\\sum_{i=1}^{n} s_i\\right) - n = m + 5 - n.\\] Equating our two expression $n-1 = m+5-n \\rightarrow 2n=m+6.$ Solving for \\( n \\), we get \\( n = \\frac{m}{2} + 3 \\). Notice that \\( n \\leq 11 \\) because the \"best-case scenario\" is that each \\( G \\) forms its own blob, and there is a red blob in between any two \\( G \\)'s and a red blob on both ends. Here is how it should look: <red blob> G <red blob> G <red blob> G <red blob> G <red blob> G <red blob>. In this case, there are \\( 11 \\) blobs. Thus, \\( \\frac{m}{2} + 3 \\leq 11 \\implies m \\leq 16 \\). This is now a stars and bars problem. For \\( m = 16 \\), we need to create the optimal solution: \\( 11 \\) blobs. Since every <red blob> must have at least one red marble, we can just give each blob one red at the start, leaving us with \\( 16-6 = 10 \\) red marbles left to separate into \\( 6 \\) groups. By stars and bars, this is \\( \\binom{15}{5} = 3003 \\). Thus, the answer is \\( 003 \\).",
"Let a collection of consecutive letters be a \"blob.\" Let the blobs have sizes s_1, s_2, ... s_n given the existence of n blobs. The marbles such that the marble to their right has a different color are the ones that are the last marble of a \"blob.\" The only exception to this is the very last marble in this list. There are \\( n-1 \\) of these. The marbles such that the marble to their right has the same color is every marble in a blob but the last, which is \\[\\sum_{i=1}^{n} (s_i - 1) = \\left(\\sum_{i=1}^{n} s_i\\right) - n = m + 5 - n.\\] Equating our two expression $n-1 = m+5-n \\rightarrow 2n=m+6.$ Solving for \\( n \\), we get \\( n = \\frac{m}{2} + 3 \\). Notice that \\( n \\leq 11 \\) because the \"best-case scenario\" is that each \\( G \\) forms its own blob, and there is a red blob in between any two \\( G \\)'s and a red blob on both ends. Here is how it should look: <red blob> G <red blob> G <red blob> G <red blob> G <red blob> G <red blob>. In this case, there are \\( 11 \\) blobs. Thus, \\( \\frac{m}{2} + 3 \\leq 11 \\implies m \\leq 16 \\). This is now a stars and bars problem. For \\( m = 16 \\), we need to create the optimal solution: \\( 11 \\) blobs. Since every <red blob> must have at least one red marble, we can just give each blob one red at the start, leaving us with \\( 16-6 = 10 \\) red marbles left to separate into \\( 6 \\) groups. By stars and bars, this is \\( \\binom{15}{5} = 3003 \\). Thus, the answer is \\( 003 \\)."
] |
2011-II-8
| 2,011
| 8
|
Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ .
| 784
|
II
|
[
"The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$. If we write $z=a+bi$, then the real part of $z$ is $a$ and the real part of $iz$ is $-b$. The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$, and the red dots represent those roots $z$ for which the real part of $iz$ is greater than the real part of $z$. Now, the sum of the real parts of the blue dots is easily seen to be $8+16\\cos\\frac{\\pi}{6}=8+8\\sqrt{3}$ and the negative of the sum of the imaginary parts of the red dots is easily seen to also be $8+8\\sqrt{3}$. Hence our desired sum is $16+16\\sqrt{3}=16+\\sqrt{768}$, giving the answer $784 times two ~CrazyVideoGamez",
"As a small note, we could factor the equation as $z^{12} = 2^{36} \\implies (\\frac{z}{8})^{12} = 1$ in which these are just the 12th roots of unity for $\\frac{z}{8}$ which might be easier to work with. Then we just add up the optimal real parts as in solution 1 and then multiply by 8 at the end. Solution $1.\\overline{6}$ Note that $\\sin(x) = \\sin(x + \\pi/2 - \\pi/2) = -\\cos(x + \\pi/2)$. When summing the real part of $8i(\\cos(x) + i\\sin(x))$ is $-8\\sin(x) = 8\\cos(x + \\pi/2)$. So, translate all the red dots by $90^{\\circ}$ and sum the cosines of angles of the blue and (rotated) red dots. Then, it's easy to see that points 2, 3, and 4 cancel out with 5, 6, and 7, so you're left with $8 \\cdot (\\cos \\pi/6 + \\cos -\\pi/6 + \\cos 0)$ times two ~CrazyVideoGamez",
"Note that $\\sin(x) = \\sin(x + \\pi/2 - \\pi/2) = -\\cos(x + \\pi/2)$. When summing the real part of $8i(\\cos(x) + i\\sin(x))$ is $-8\\sin(x) = 8\\cos(x + \\pi/2)$. So, translate all the red dots by $90^{\\circ}$ and sum the cosines of angles of the blue and (rotated) red dots. Then, it's easy to see that points 2, 3, and 4 cancel out with 5, 6, and 7, so you're left with $8 \\cdot (\\cos \\pi/6 + \\cos -\\pi/6 + \\cos 0)$ times two ~CrazyVideoGamez",
"The equation $z^{12}-2^{36}=0$ can be factored as follows: \\[(z^6-2^{18})(z^6+2^{18})=0\\] \\[(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\\cdot2^6)({(z^2-2^6)}^2+z^2\\cdot2^6)=0\\] \\[(z^2-2^6)(z^2+2^6)(z^2+2^6-z\\cdot2^3)(z^2+2^6+z\\cdot2^3) (z^2-2^6-iz\\cdot2^3)(z^2-2^6+i z\\cdot2^3)=0\\] Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by $-1$ and since we have 6 factors, that’s 1 root per factor. We just need to solve for $z$ in each factor and pick whether or not to multiply by $i$ and $-1$ for each one depending on the one that yields the highest real value. After that process, we get $8+8+2((4\\sqrt{3}+4)+(4\\sqrt{3}-4))$ Adding the values up yields $16+16\\sqrt{3}$, or $16+\\sqrt{768}$, and $16+768=784. -Solution by Someonenumber011.",
"Clearly, the roots are: $2^3*(\\cos{\\frac{k\\pi}{12}}+i\\sin{\\frac{k\\pi}{12}}), k\\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$. Now, realize for $z=a+bi$, $\\operatorname{Re}(iz)=-b$. $\\operatorname{Re}(z)=a$. $\\operatorname{Re}(z)<\\operatorname{Re}(iz)$ is true when $a<-b$. This means: When $a>0$, $b<-a<0$. When $a<0$, $0<b<-a$. For the 12 roots of the polynomial in the original equation, $8\\cos{k\\pi/12}=\\operatorname{Re}(z), k\\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$. $-8\\sin{k\\pi/12}=\\operatorname{Im}(iz), k\\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$. So $8\\cos{k\\pi/12}<-8\\sin{k\\pi/12}$. $-\\cos{k\\pi/12}>\\sin{k\\pi/12}$. This can be easily true for roots that are in the 3rd quadrant in the complex plane. This cannot be true for roots in the 1st quadrant because that would yield a negative number bigger than a positive one. Consider the roots in the 2nd and 4th quadrants. Calculate the roots, choose, and then add the ones up. You will get $784. ~hastapasta",
"Use De Moivre's Theorem to brute force all the roots out. Then choose the greater value of $\\operatorname{Re}(z), \\operatorname{Re}(iz)$. After adding everything up, you get $784. ~hastapasta",
"Use De Moivre's Theorem to brute force all the roots out. Then choose the greater value of $\\operatorname{Re}(z), \\operatorname{Re}(iz)$. After adding everything up, you get $784. ~hastapasta"
] |
2011-II-9
| 2,011
| 9
|
Let $x_1$ , $x_2$ , $\dots$ , $x_6$ be nonnegative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 1$ , and $x_1x_3x_5 + x_2x_4x_6 \ge {\frac{1}{540}}$ . Let $p$ and $q$ be relatively prime positive integers such that $\frac{p}{q}$ is the maximum possible value of $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2$ . Find $p + q$ .
| 559
|
II
|
[
"Solution 1 Note that neither the constraint nor the expression we need to maximize involves products $x_i x_j$ with $i \\equiv j \\pmod 3$. Factoring out say $x_1$ and $x_4$ we see that the constraint is $x_1(x_3x_5) + x_4(x_2x_6) \\ge {\\frac1{540}}$, while the expression we want to maximize is $x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)$. Adding the left side of the constraint to the expression, we get: $(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most $\\frac1{27}$. Since we have added at least $\\frac{1}{540}$ the desired maximum is at most $\\frac1{27} - \\frac1{540} =\\frac{19}{540}$. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to $\\frac1{540}$ with $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =\\frac13$—for example, by choosing $x_1$ and $x_2$ small enough—so our answer is $540 + 19 = 559 you receive complex answers (which contradict the problem statement), but the final answer is correct.",
"Note that neither the constraint nor the expression we need to maximize involves products $x_i x_j$ with $i \\equiv j \\pmod 3$. Factoring out say $x_1$ and $x_4$ we see that the constraint is $x_1(x_3x_5) + x_4(x_2x_6) \\ge {\\frac1{540}}$, while the expression we want to maximize is $x_1(x_2x_3 + x_5x_6 + x_6x_2) + x_4(x_2x_3 + x_5x_6 + x_3x_5)$. Adding the left side of the constraint to the expression, we get: $(x_1 + x_4)(x_2x_3 + x_5x_6 + x_6x_2 + x_3x_5) = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$. This new expression is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most $\\frac1{27}$. Since we have added at least $\\frac{1}{540}$ the desired maximum is at most $\\frac1{27} - \\frac1{540} =\\frac{19}{540}$. It is easy to see that this upper bound can in fact be achieved by ensuring that the constraint expression is equal to $\\frac1{540}$ with $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 =\\frac13$—for example, by choosing $x_1$ and $x_2$ small enough—so our answer is $540 + 19 = 559 you receive complex answers (which contradict the problem statement), but the final answer is correct.",
"There's a symmetry between $x_1, x_3, x_5$ and $x_2,x_4,x_6$. Therefore, a good guess is that $a = x_1 = x_3 = x_5$ and $b = x_2 = x_4 = x_6$, at which point we know that $a+b = 1/3$, $a^3+b^3 \\geq 1/540$, and we are trying to maximize $3a^2b+3ab^2$. Then, \\[3a^2b+3ab^2 = (a+b)^3-a^3-b^3 \\leq \\frac{1}{27} - \\frac{1}{540} = \\frac{19}{540} you receive complex answers (which contradict the problem statement), but the final answer is correct.",
"There's a symmetry between $x_1, x_3, x_5$ and $x_2,x_4,x_6$. Therefore, a good guess is that $a = x_1 = x_3 = x_5$ and $b = x_2 = x_4 = x_6$, at which point we know that $a+b = 1/3$, $a^3+b^3 \\geq 1/540$, and we are trying to maximize $3a^2b+3ab^2$. Then, \\[3a^2b+3ab^2 = (a+b)^3-a^3-b^3 \\leq \\frac{1}{27} - \\frac{1}{540} = \\frac{19}{540} you receive complex answers (which contradict the problem statement), but the final answer is correct."
] |
2011-II-10
| 2,011
| 10
|
A circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000.
| 57
|
II
|
[
"Let $E$ and $F$ be the midpoints of $\\overline{AB}$ and $\\overline{CD}$, respectively, such that $\\overline{BE}$ intersects $\\overline{CF}$. Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$. $B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$. The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so $\\triangle OEB$ and $\\triangle OFC$ are right triangles (with $\\angle OEB$ and $\\angle OFC$ being the right angles). By the Pythagorean Theorem, $OE = \\sqrt{25^2 - 15^2} = 20$, and $OF = \\sqrt{25^2 - 7^2} = 24$. Let $x$, $a$, and $b$ be lengths $OP$, $EP$, and $FP$, respectively. OEP and OFP are also right triangles, so $x^2 = a^2 + 20^2 \\to a^2 = x^2 - 400$, and $x^2 = b^2 + 24^2 \\to b^2 = x^2 - 576$ We are given that $EF$ has length 12, so, using the Law of Cosines with $\\triangle EPF$: $12^2 = a^2 + b^2 - 2ab \\cos (\\angle EPF) = a^2 + b^2 - 2ab \\cos (\\angle EPO + \\angle FPO)$ Substituting for $a$ and $b$, and applying the Cosine of Sum formula: $144 = (x^2 - 400) + (x^2 - 576) - 2 \\sqrt{x^2 - 400} \\sqrt{x^2 - 576} \\left( \\cos \\angle EPO \\cos \\angle FPO - \\sin \\angle EPO \\sin \\angle FPO \\right)$ $\\angle EPO$ and $\\angle FPO$ are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines: $144 = 2x^2 - 976 - 2 \\sqrt{(x^2 - 400)(x^2 - 576)} \\left(\\frac{\\sqrt{x^2 - 400}}{x} \\frac{\\sqrt{x^2 - 576}}{x} - \\frac{20}{x} \\frac{24}{x} \\right)$ Combine terms and multiply both sides by $x^2$: $144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960 \\sqrt{(x^2 - 400)(x^2 - 576)}$ Combine terms again, and divide both sides by 64: $13 x^2 = 7200 - 15 \\sqrt{x^4 - 976 x^2 + 230400}$ Square both sides: $169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51,840,000$ This reduces to $x^2 = \\frac{4050}{7} = (OP)^2$; $4050 + 7 \\equiv 057.",
"We begin as in the first solution. Once we see that $\\triangle EOF$ has side lengths $12$, $20$, and $24$, we can compute its area with Heron's formula: \\[K = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{28\\cdot 16\\cdot 8\\cdot 4} = 32\\sqrt{14}.\\] Thus, the circumradius of triangle $\\triangle EOF$ is $R = \\frac{abc}{4K} = \\frac{45}{\\sqrt{14}}$. Looking at $EPFO$, we see that $\\angle OEP = \\angle OFP = 90^\\circ$, which makes it a cyclic quadrilateral. This means $\\triangle EOF$'s circumcircle and $EPFO$'s inscribed circle are the same. Since $EPFO$ is cyclic with diameter $OP$, we have $OP = 2R = \\frac{90}{\\sqrt{14}}$, so $OP^2 = \\frac{4050}{7}$ and the answer is $057.",
"We begin as the first solution have $OE=20$ and $OF=24$. Because $\\angle PEO+\\angle PFO=180^{\\circ}$, Quadrilateral $EPFO$ is inscribed in a Circle. Assume point $I$ is the center of this circle. $\\because \\angle OEP=90^{\\circ}$ $\\therefore$ point $I$ is on $OP$ Link $EI$ and $FI$, Made line $IK\\bot EF$, then $\\angle EIK=\\angle EOF$ On the other hand, $\\cos\\angle EOF=\\frac{EO^2+OF^2-EF^2}{2\\cdot EO\\cdot OF}=\\frac{13}{15}=\\cos\\angle EIK$ $\\sin\\angle EOF=\\sin\\angle EIK=\\sqrt{1-\\frac{13^2}{15^2}}=\\frac{2\\sqrt{14}}{15}$ As a result, $IE=IO=\\frac{45}{\\sqrt 14}$ Therefore, $OP^2=4\\cdot \\frac{45^2}{14}=\\frac{4050}{7}.$ As a result, $m+n=4057\\equiv 057",
"Let $OP=x$. Proceed as the first solution in finding that quadrilateral $EPFO$ has side lengths $OE=20$, $OF=24$, $EP=\\sqrt{x^2-20^2}$, and $PF=\\sqrt{x^2-24^2}$, and diagonals $OP=x$ and $EF=12$. We note that quadrilateral $EPFO$ is cyclic and use Ptolemy's theorem to solve for $x$: \\[20\\cdot \\sqrt{x^2-24^2} + 12\\cdot x = 24\\cdot \\sqrt{x^2-20^2}\\] Solving, we have $x^2=\\frac{4050}{7}$ so the answer is $057. -Solution by blueberrieejam ~bluesoul changes the equation to a right equation, the previous equation isn't solvable",
"Let $M$ be the midpoint of $AB$ and $N$ of $CD$. As $\\angle OMP = \\angle ONP$, quadrilateral $OMPN$ is cyclic with diameter $OP$. By Cyclic quadrilaterals note that $\\angle MPO = \\angle MNO$. The area of $\\triangle MNP$ can be computed by Herons as \\[[MNO] = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{28\\cdot 16\\cdot 8\\cdot 4} = 32\\sqrt{14}.\\] The area is also $\\frac{1}{2}ON \\cdot MN \\sin{\\angle MNO}$. Therefore, \\begin{align*} \\sin{\\angle MNO} &= \\frac{2[MNO]}{ON \\cdot MN} \\\\ &= \\frac{2}{9}\\sqrt{14} \\\\ \\sin{\\angle MNO} &= \\frac{OM}{OP} \\\\ &= \\frac{2}{9}\\sqrt{14} \\\\ OP &= \\frac{90\\sqrt{14}}{14} \\\\ OP^2 &= \\frac{4050}{7} \\implies 057. \\end{align*} ~ Aaryabhatta1",
"Let $M$ be the midpoint of $AB$ and $N$ of $CD$. As $\\angle OMP = \\angle ONP$, quadrilateral $OMPN$ is cyclic with diameter $OP$. By Cyclic quadrilaterals note that $\\angle MPO = \\angle MNO$. The area of $\\triangle MNP$ can be computed by Herons as \\[[MNO] = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{28\\cdot 16\\cdot 8\\cdot 4} = 32\\sqrt{14}.\\] The area is also $\\frac{1}{2}ON \\cdot MN \\sin{\\angle MNO}$. Therefore, \\begin{align*} \\sin{\\angle MNO} &= \\frac{2[MNO]}{ON \\cdot MN} \\\\ &= \\frac{2}{9}\\sqrt{14} \\\\ \\sin{\\angle MNO} &= \\frac{OM}{OP} \\\\ &= \\frac{2}{9}\\sqrt{14} \\\\ OP &= \\frac{90\\sqrt{14}}{14} \\\\ OP^2 &= \\frac{4050}{7} \\implies 057. \\end{align*} ~ Aaryabhatta1",
"Define $M$ and $N$ as the midpoints of $AB$ and $CD$, respectively. Because $\\angle OMP = \\angle ONP = 90^{\\circ}$, we have that $ONPM$ is a cyclic quadrilateral. Hence, $\\angle PNM = \\angle POM.$ Then, let these two angles be denoted as $\\alpha$. Now, assume WLOG that $PD = x < 7$ and $PB = y < 15$ (We can do this because one of $PD$ or $PC$ must be less than 7, and similarly for $PB$ and $PA$). Then, by Power of a Point on P with respect to the circle with center $O$, we have that \\[(14-x)x = (30-y)y\\] \\[(7-x)^{2}+176=(15-y)^{2}.\\] Then, let $z = (7-x)^{2}$. From Law of Cosines on $\\triangle NMP$, we have that \\[\\textrm{cos } \\angle MNP = \\frac{NP^{2}+MN^{2}-MP^{2}}{2 \\cdot NP \\cdot MN}\\] \\[\\textrm{cos } \\alpha = \\frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \\cdot (7-x)}.\\] Plugging in $z$ in gives \\[\\textrm{cos } \\alpha = \\frac{-32}{24 \\cdot \\sqrt{z}}\\] \\[\\textrm{cos } \\alpha = \\frac{-4}{3\\sqrt{z}}\\] \\[\\textrm{cos }^{2} \\alpha = \\frac{16}{9z}.\\] Hence, \\[\\textrm{tan }^{2} \\alpha = \\frac{\\frac{9z-16}{9z}}{\\frac{16}{9z}} = \\frac{9z-16}{16}.\\] Then, we also know that \\[\\textrm{tan } \\alpha = \\textrm{tan } \\angle MOP = \\frac{MP}{OM} = \\frac{14-y}{20}.\\] Squaring this, we get \\[\\textrm{tan }^{2} \\alpha = \\frac{z+176}{400}.\\] Equating our expressions for $z$, we get $\\frac{z+176}{400} = \\frac{9z-16}{16}.$ Solving gives us that $z = \\frac{18}{7}$. Since $\\angle ONP = 90^{\\circ}$, from the Pythagorean Theorem, $OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \\frac{4050}{7}$, and thus the answer is $4050+7 = 4057$, which when divided by a thousand leaves a remainder of $57 -Mr.Sharkman Note: my solution was very long and tedious. It was definitely was the least elegant solution. The only thing I like about it is it contains no quadratic equations (unless you count LoC).",
"Let $E$ and $F$ be the midpoints of $\\overline{AB}$ and $\\overline{CD}$, respectively, such that $\\overline{BE}$ intersects $\\overline{CF}$. Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$. $B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$. Since $\\overline{OE}\\perp \\overline{AB}$ and $\\overline{OF}\\perp \\overline{CD}$, $OE = \\sqrt{OB^2-BE^2}=20$ and $OF = \\sqrt{OC^2-OF^2}=24$ With law of cosines, $\\cos \\angle EOF = \\frac{OE^2+OF^2-EF^2}{2\\cdot OE\\cdot OF} = \\frac{13}{15}$ Since $EF < OF$, $\\angle EOF$ is acute angle. $\\sin \\angle EOF = \\sqrt{1-\\cos^2 \\angle EOF} = \\frac{\\sqrt{56}}{15}$ and $\\tan \\angle EOF = \\frac{\\sqrt{56}}{13}$ Let $\\overline{OF}$ line be $x$ axis. Line $\\overline{DC}$ equation is $x = OF$. Since line $\\overline{AB}$ passes point $E$ and perpendicular to $\\overline{OD}$, its equation is $y - E_y = -\\frac{1}{\\tan \\angle EOF} (x - E_x)$ where $E_x = OE\\cos{\\angle EOF}$ , $E_y = OE\\sin{\\angle EOF}$ Since $P$ is the intersection of $\\overline{AB}$ and $\\overline{CD}$, $P_x = OF = 24$ $P_y = E_y -\\frac{1}{\\tan \\angle EOF} (OF - E_x) = - \\frac{3\\sqrt{14}}{7}$ (Negative means point $P$ is between point $F$ and $C$) $OP^2 = P_x^2 + P_y^2 = \\frac{4050}{7}$ and the answer is $057 is the diagonal or side of quadrilateral. Equation for wrong case cannot be solved. For example,",
"We work in the complex plane where \\( O = 0 \\). Let \\( M_1, M_2 \\), be midpoints of chords \\( AB \\), and \\( CD \\) respectively. \\( AB=30 \\rightarrow AM=15 \\). Since \\( AM=15 \\), and \\( OA=25 \\), \\( OM_1 = 20 \\) by the Pythagorean theorem. WLOG, Let \\( M_1 = 20 + 0i \\). Think about the locus of all points \\( M_2 \\). By similar logic as above it is any \\( M_2 \\), such that \\( OM_2=24 \\). Note that \\( M_1M_2 = 12 \\). Therefore by LoC on \\( \\triangle OM_1M_2 \\), if \\( \\angle O = \\theta \\), \\( \\cos\\theta = \\frac{13}{15} \\rightarrow \\sin\\theta = \\frac{2\\sqrt{14}}{15} \\). Note that a rotation by \\( \\theta \\) and a scaling of \\( \\frac{24}{20} = \\frac{6}{5} \\) transforms \\( AB \\) to \\( DC \\). This transformation = \\( \\text{cis}\\theta \\cdot \\frac{6}{5} = \\frac{2}{5} \\cdot \\frac{13 + 2\\sqrt{14}i}{5} \\). Applying the transformation we on \\( A,B \\) respectively we find: (We are back in Cartesian coordinates now) \\[ D = \\frac{2}{5} \\cdot (52 - 6\\sqrt{14}, 39 + 8\\sqrt{14}), \\] \\[ C = \\frac{2}{5} \\cdot (52 + 6\\sqrt{14}, -39 + 8\\sqrt{14}). \\] To make bashing easier first ignore the \\( \\frac{2}{5} \\). e.g. let \\( D = 0.4D' \\) and \\( C = 0.4C' \\) and first find the equation of the line passing through \\( C' \\) and \\( D' \\). After bashing a bit, we get \\[ y = \\frac{-13}{2\\sqrt{14}} \\cdot x + \\frac{225}{7}\\sqrt{14}. \\] Now to account for the \\( \\frac{2}{5} \\) thing we say the actual line is this: \\[ y = \\frac{-13}{2\\sqrt{14}} \\cdot x + \\frac{2}{5} \\cdot \\frac{225}{7} \\sqrt{14}, \\] \\[ y = \\frac{-13}{2\\sqrt{14}} \\cdot x + \\frac{90}{7} \\sqrt{14}. \\] \\( P \\) is the intersection of \\( CD \\) and \\( AB \\). We have the equation of \\( CD \\), and \\( AB \\) is simply \\( x = 20 \\), so Letting \\( x = 20 \\) we find \\[ y = \\frac{25\\sqrt{14}}{7}. \\] \\[ OP^2 = 20^2 + \\left( \\frac{25\\sqrt{14}}{7} \\right)^2 = 5^2 \\cdot \\left( 4^2 + \\left( \\frac{5\\sqrt{2}}{\\sqrt{7}} \\right)^2 \\right) = 25 \\cdot \\frac{810}{7} = \\frac{4050}{7}. \\] Thus, the answer is \\( 057 \\).",
"We work in the complex plane where \\( O = 0 \\). Let \\( M_1, M_2 \\), be midpoints of chords \\( AB \\), and \\( CD \\) respectively. \\( AB=30 \\rightarrow AM=15 \\). Since \\( AM=15 \\), and \\( OA=25 \\), \\( OM_1 = 20 \\) by the Pythagorean theorem. WLOG, Let \\( M_1 = 20 + 0i \\). Think about the locus of all points \\( M_2 \\). By similar logic as above it is any \\( M_2 \\), such that \\( OM_2=24 \\). Note that \\( M_1M_2 = 12 \\). Therefore by LoC on \\( \\triangle OM_1M_2 \\), if \\( \\angle O = \\theta \\), \\( \\cos\\theta = \\frac{13}{15} \\rightarrow \\sin\\theta = \\frac{2\\sqrt{14}}{15} \\). Note that a rotation by \\( \\theta \\) and a scaling of \\( \\frac{24}{20} = \\frac{6}{5} \\) transforms \\( AB \\) to \\( DC \\). This transformation = \\( \\text{cis}\\theta \\cdot \\frac{6}{5} = \\frac{2}{5} \\cdot \\frac{13 + 2\\sqrt{14}i}{5} \\). Applying the transformation we on \\( A,B \\) respectively we find: (We are back in Cartesian coordinates now) \\[ D = \\frac{2}{5} \\cdot (52 - 6\\sqrt{14}, 39 + 8\\sqrt{14}), \\] \\[ C = \\frac{2}{5} \\cdot (52 + 6\\sqrt{14}, -39 + 8\\sqrt{14}). \\] To make bashing easier first ignore the \\( \\frac{2}{5} \\). e.g. let \\( D = 0.4D' \\) and \\( C = 0.4C' \\) and first find the equation of the line passing through \\( C' \\) and \\( D' \\). After bashing a bit, we get \\[ y = \\frac{-13}{2\\sqrt{14}} \\cdot x + \\frac{225}{7}\\sqrt{14}. \\] Now to account for the \\( \\frac{2}{5} \\) thing we say the actual line is this: \\[ y = \\frac{-13}{2\\sqrt{14}} \\cdot x + \\frac{2}{5} \\cdot \\frac{225}{7} \\sqrt{14}, \\] \\[ y = \\frac{-13}{2\\sqrt{14}} \\cdot x + \\frac{90}{7} \\sqrt{14}. \\] \\( P \\) is the intersection of \\( CD \\) and \\( AB \\). We have the equation of \\( CD \\), and \\( AB \\) is simply \\( x = 20 \\), so Letting \\( x = 20 \\) we find \\[ y = \\frac{25\\sqrt{14}}{7}. \\] \\[ OP^2 = 20^2 + \\left( \\frac{25\\sqrt{14}}{7} \\right)^2 = 5^2 \\cdot \\left( 4^2 + \\left( \\frac{5\\sqrt{2}}{\\sqrt{7}} \\right)^2 \\right) = 25 \\cdot \\frac{810}{7} = \\frac{4050}{7}. \\] Thus, the answer is \\( 057 \\)."
] |
2011-II-11
| 2,011
| 11
|
Let $M_n$ be the $n \times n$ matrix with entries as follows: for $1 \le i \le n$ , $m_{i,i} = 10$ ; for $1 \le i \le n - 1$ , $m_{i+1,i} = m_{i,i+1} = 3$ ; all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n$ . Then $\sum_{n=1}^{\infty} \frac{1}{8D_n+1}$ can be represented as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ . Note: The determinant of the $1 \times 1$ matrix $[a]$ is $a$ , and the determinant of the $2 \times 2$ matrix $\left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] = ad - bc$ ; for $n \ge 2$ , the determinant of an $n \times n$ matrix with first row or first column $a_1$ $a_2$ $a_3$ $\dots$ $a_n$ is equal to $a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n$ , where $C_i$ is the determinant of the $(n - 1) \times (n - 1)$ matrix formed by eliminating the row and column containing $a_i$ .
| 73
|
II
|
[
"\\[D_{1}=\\begin{vmatrix} 10 \\end{vmatrix} = 10, \\quad D_{2}=\\begin{vmatrix} 10 & 3 \\\\ 3 & 10 \\\\ \\end{vmatrix} =(10)(10) - (3)(3) = 91, \\quad D_{3}=\\begin{vmatrix} 10 & 3 & 0 \\\\ 3 & 10 & 3 \\\\ 0 & 3 & 10 \\\\ \\end{vmatrix}.\\] Using the expansionary/recursive definition of determinants (also stated in the problem): $D_{3}=\\left| {\\begin{array}{ccc} 10 & 3 & 0 \\\\ 3 & 10 & 3 \\\\ 0 & 3 & 10 \\\\ \\end{array} } \\right|=10\\left| {\\begin{array}{cc} 10 & 3 \\\\ 3 & 10 \\\\ \\end{array} } \\right| - 3\\left| {\\begin{array}{cc} 3 & 3 \\\\ 0 & 10 \\\\ \\end{array} } \\right| + 0\\left| {\\begin{array}{cc} 3 & 10 \\\\ 0 & 3 \\\\ \\end{array} } \\right| = 10D_{2} - 9D_{1} = 820$ This pattern repeats because the first element in the first row of $M_{n}$ is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to $10D_{n-1}$. The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix $X_{n}$. $X_{n}$ has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields $3D_{n-2} + 0(\\text{other things})=3D_{n-2}$. Thus, the $3 \\det(X_{n})$ expression turns into $9D_{n-2}$. Thus, the equation $D_{n}=10D_{n-1}-9D_{n-2}$ holds for all n > 2. This equation can be rewritten as $D_{n}=10(D_{n-1}-D_{n-2}) + D_{n-2}$. This version of the equation involves the difference of successive terms of a recursive sequence. Calculating $D_{0}$ backwards from the recursive formula and $D_{4}$ from the formula yields $D_{0}=1, D_{4}=7381$. Examining the differences between successive terms, a pattern emerges. $D_{0}=1=9^{0}$, $D_{1}-D_{0}=10-1=9=9^{1}$, $D_{2}-D_{1}=91-10=81=9^{2}$, $D_{3}-D_{2}=820-91=729=9^{3}$, $D_{4}-D_{3}=7381-820=6561=9^{4}$. Thus, $D_{n}=D_{0} + 9^{1}+9^{2}+ . . . +9^{n}=\\sum_{i=0}^{n}9^{i}=\\frac{(1)(9^{n+1}-1)}{9-1}=\\frac{9^{n+1}-1}{8}$. Thus, the desired sum is $\\sum_{n=1}^{\\infty}\\frac{1}{8\\frac{9^{n+1}-1}{8}+1}=\\sum_{n=1}^{\\infty}\\frac{1}{9^{n+1}-1+1} = \\sum_{n=1}^{\\infty}\\frac{1}{9^{n+1}}$ This is an infinite geometric series with first term $\\frac{1}{81}$ and common ratio $\\frac{1}{9}$. Thus, the sum is $\\frac{\\frac{1}{81}}{1-\\frac{1}{9}}=\\frac{\\frac{1}{81}}{\\frac{8}{9}}=\\frac{9}{(81)(8)}=\\frac{1}{(9)(8)}=\\frac{1}{72}$. Thus, $p + q = 1 + 72 = 073.",
"Manually, we can find \\( D_1 = 10 \\), \\( D_2 = 91 \\), \\( D_3 = 820 \\). \\[ \\frac{1}{8D_1+1} = \\frac{1}{81}, \\quad \\frac{1}{8D_2+1} = \\frac{1}{729}, \\quad \\frac{1}{8D_3+1} = \\frac{1}{6561}. \\] We notice that \\( 81 = 9^2 \\), \\( 729 = 9^3 \\), and \\( 6561 = 9^4 \\), and believe that this cannot be a coincidence. Assume \\( \\frac{1}{8D_n+1} = \\frac{1}{9^{n+1}} \\). Then this problem asks us to find the sum of a geometric series with first term \\( \\frac{1}{81} \\) and common ratio \\( \\frac{1}{9} \\). \\[ \\frac{\\frac{1}{81}}{1 - \\frac{1}{9}} = \\frac{\\frac{1}{81}}{\\frac{8}{9}} = \\frac{1}{72}. \\] Thus, the answer is \\(73\\).",
"Manually, we can find \\( D_1 = 10 \\), \\( D_2 = 91 \\), \\( D_3 = 820 \\). \\[ \\frac{1}{8D_1+1} = \\frac{1}{81}, \\quad \\frac{1}{8D_2+1} = \\frac{1}{729}, \\quad \\frac{1}{8D_3+1} = \\frac{1}{6561}. \\] We notice that \\( 81 = 9^2 \\), \\( 729 = 9^3 \\), and \\( 6561 = 9^4 \\), and believe that this cannot be a coincidence. Assume \\( \\frac{1}{8D_n+1} = \\frac{1}{9^{n+1}} \\). Then this problem asks us to find the sum of a geometric series with first term \\( \\frac{1}{81} \\) and common ratio \\( \\frac{1}{9} \\). \\[ \\frac{\\frac{1}{81}}{1 - \\frac{1}{9}} = \\frac{\\frac{1}{81}}{\\frac{8}{9}} = \\frac{1}{72}. \\] Thus, the answer is \\(73\\).",
"From Solution 1, \\( D_n = 10D_{n-1} - 9D_{n-2} \\). From Solution 1, we also know \\( D_2 = 91 \\). We can manually compute \\( D_1 = 10 \\). Iterating backwards, \\(D_0 = 1 \\). The characteristic polynomial of this recurrence is \\[x^2 - 10x + 9,\\] which has roots \\( 1 \\) and \\( 9 \\). Therefore, the explicit formula for \\( D_n \\) is: \\[D_n = \\alpha_1 + 9^n \\alpha_2.\\] We can solve for \\( \\alpha_1 \\) and \\( \\alpha_2 \\) by setting \\( n=0 \\) and \\( n=1 \\) respectively. This gives the following equations: \\[\\alpha_1 + \\alpha_2 = 1,\\] \\[\\alpha_1 + 9\\alpha_2 = 10.\\] Solving these equations, we find \\[\\alpha_1 = -\\frac{1}{8}, \\quad \\alpha_2 = \\frac{9}{8}.\\] Substituting these back into the explicit formula, we find \\[D_n = \\frac{9^{n+1} - 1}{8}.\\] Notice that \\[\\frac{1}{8D_{n+1} + 1} = 9^{-(n+1)}.\\] The summation asks us to find the sum of a geometric series with ratio \\( \\frac{1}{9} \\) and first term \\( \\frac{1}{81} \\). The sum of this series is \\[\\frac{\\frac{1}{81}}{1 - \\frac{1}{9}} = \\frac{\\frac{1}{81}}{\\frac{8}{9}} = \\frac{1}{72}.\\] Thus, the answer is $73",
"From Solution 1, \\( D_n = 10D_{n-1} - 9D_{n-2} \\). From Solution 1, we also know \\( D_2 = 91 \\). We can manually compute \\( D_1 = 10 \\). Iterating backwards, \\(D_0 = 1 \\). The characteristic polynomial of this recurrence is \\[x^2 - 10x + 9,\\] which has roots \\( 1 \\) and \\( 9 \\). Therefore, the explicit formula for \\( D_n \\) is: \\[D_n = \\alpha_1 + 9^n \\alpha_2.\\] We can solve for \\( \\alpha_1 \\) and \\( \\alpha_2 \\) by setting \\( n=0 \\) and \\( n=1 \\) respectively. This gives the following equations: \\[\\alpha_1 + \\alpha_2 = 1,\\] \\[\\alpha_1 + 9\\alpha_2 = 10.\\] Solving these equations, we find \\[\\alpha_1 = -\\frac{1}{8}, \\quad \\alpha_2 = \\frac{9}{8}.\\] Substituting these back into the explicit formula, we find \\[D_n = \\frac{9^{n+1} - 1}{8}.\\] Notice that \\[\\frac{1}{8D_{n+1} + 1} = 9^{-(n+1)}.\\] The summation asks us to find the sum of a geometric series with ratio \\( \\frac{1}{9} \\) and first term \\( \\frac{1}{81} \\). The sum of this series is \\[\\frac{\\frac{1}{81}}{1 - \\frac{1}{9}} = \\frac{\\frac{1}{81}}{\\frac{8}{9}} = \\frac{1}{72}.\\] Thus, the answer is $73"
] |
2011-II-12
| 2,011
| 12
|
Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 97
|
II
|
[
"Use complementary probability and Principle of Inclusion-Exclusion. If we consider the delegates from each country to be indistinguishable and number the chairs, we have \\[\\frac{9!}{(3!)^3} = \\frac{9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4}{6\\cdot6} = 6\\cdot8\\cdot7\\cdot5 = 30\\cdot56\\] total ways to seat the candidates. Of these, there are $3 \\times 9 \\times \\frac{6!}{(3!)^2}$ ways to have the candidates of at least some one country sit together. This comes to \\[\\frac{27\\cdot6\\cdot5\\cdot4}6 = 27\\cdot 20.\\] Among these there are $3 \\times 9 \\times 4$ ways for candidates from two countries to each sit together. This comes to $27\\cdot 4.$ Finally, there are $9 \\times 2 = 18.$ ways for the candidates from all the countries to sit in three blocks (9 clockwise arrangements, and 9 counter-clockwise arrangements). So, by PIE, the total count of unwanted arrangements is $27\\cdot 20 - 27\\cdot 4 + 18 = 16\\cdot27 + 18 = 18\\cdot25.$ So the fraction \\[\\frac mn = \\frac{30\\cdot 56 - 18\\cdot 25}{30\\cdot 56} = \\frac{56 - 15}{56} = \\frac{41}{56}.\\] Thus $m + n = 56 + 41 = 097",
"We use complementary counting. We can order the $9$ people around a circle in $\\frac{9!}{9} = 8!$ ways. Now we count when there is at least one delegate surrounded by people from only his/her country. Let the countries be $A,B,C$. Suppose that the group $XXX$ (for some country $X$) appears. To account for circular over counting we fix this group at the top. There are $6!$ ways to arrange the rest of the delegates and $3!$ ways to arrange inside the group. Since there are three countries this group can belong to, the total is $6!*3!*3$. But notice that when the group $XXX$ AND $YYY$ both appear is over counted. So, fix one group at the top. for the last country, we can insert $0,1,2,3$ people in between the two groups. Since we can choose two countries in $\\binom{3}{2} = 3$ ways, the total is $3!*3!*3!*4*3$ ways. Unfortunately, there is over count in the over count. If each country has their delegates together, this case must be added back. Think of these three groups as a whole, and there are $3!/3 = 2$ ways of arranging. In each group there is $3!$ ways of arrangements. So, $3!*3!*3!*2$ is the total for this case. We now find the complementary probability total. This is \\[\\frac{6!*3!*3 - 3!*3!*3!*4*3 + 3!*3!*3!*2}{8!} = \\frac{15}{56}\\] so the actual probability is $1-\\frac{15}{56} = \\frac{41}{56}$ for an answer of $097. ~Leonard_my_dude~",
"$1680= \\binom{9}{3} \\binom{6}{3}$ is the total. $540 = 3* 9* \\binom{6}{3}$ is the case where one country has three people in a row. (Three is for selection of country and nine is for rotation of the seats.) $108 = 3* 9* 4$ is the case where two countries has three people in a row. (Three and nine are for the same reasons above and four is for putting three people in a row for the remaining six seats.) $18$ is the case where three countries has three people in a row. $\\frac{1680-540+108-18}{1680}=\\frac{41}{56}$ so the answer is $097 By maxamc",
"$1680= \\binom{9}{3} \\binom{6}{3}$ is the total. $540 = 3* 9* \\binom{6}{3}$ is the case where one country has three people in a row. (Three is for selection of country and nine is for rotation of the seats.) $108 = 3* 9* 4$ is the case where two countries has three people in a row. (Three and nine are for the same reasons above and four is for putting three people in a row for the remaining six seats.) $18$ is the case where three countries has three people in a row. $\\frac{1680-540+108-18}{1680}=\\frac{41}{56}$ so the answer is $097 By maxamc"
] |
2011-II-13
| 2,011
| 13
|
Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ , respectively. Given that $AB = 12$ and $\angle O_1PO_2 = 120 ^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ .
| 96
|
II
|
[
"Denote the midpoint of $\\overline{DC}$ be $E$ and the midpoint of $\\overline{AB}$ be $F$. Because they are the circumcenters, both Os lie on the perpendicular bisectors of $AB$ and $CD$ and these bisectors go through $E$ and $F$. It is given that $\\angle O_{1}PO_{2}=120^{\\circ}$. Because $O_{1}P$ and $O_{1}B$ are radii of the same circle, the have the same length. This is also true of $O_{2}P$ and $O_{2}D$. Because $m\\angle CAB=m\\angle ACD=45^{\\circ}$, $m\\stackrel{\\frown}{PD}=m\\stackrel{\\frown}{PB}=2(45^{\\circ})=90^{\\circ}$. Thus, $O_{1}PB$ and $O_{2}PD$ are isosceles right triangles. Using the given information above and symmetry, $m\\angle DPB = 120^{\\circ}$. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees. Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles $O_{1}BF$ and $O_{2}DE$ have measures of 30 degrees. Thus, both triangles $O_{1}BF$ and $O_{2}DE$ are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, $DO_{2}=BO_{1}=4\\sqrt{3}$. Because of 45-45-90 right triangles, $PB=PD=4\\sqrt{6}$. Now, letting $x = AP$ and using Law of Cosines on $\\triangle ABP$, we have \\[96=144+x^{2}-24x\\frac{\\sqrt{2}}{2}\\] \\[0=x^{2}-12x\\sqrt{2}+48\\] Using the quadratic formula, we arrive at \\[x = \\sqrt{72} \\pm \\sqrt{24}\\] Taking the positive root, $AP=\\sqrt{72}+ \\sqrt{24}$ and the answer is thus 096$",
"This takes a slightly different route than Solution 1. Solution 1 proves that $\\angle{DPB}=120^{\\circ}$ and that $\\overline{BP} = \\overline{DP}$. Construct diagonal $\\overline{BD}$ and using the two statements above it quickly becomes clear that $\\angle{BDP} = \\angle{DBP} = 30^{\\circ}$ by isosceles triangle base angles. Let the midpoint of diagonal $\\overline{AC}$ be $M$, and since the diagonals are perpendicular, both triangle $DMP$ and triangle $BMP$ are 30-60-90 right triangles. Since $\\overline{AB} = 12$, $\\overline{AC} = \\overline{BD} = 12\\sqrt{2}$ and $\\overline{BM} = \\overline{DM} = 6\\sqrt{2}$. 30-60-90 triangles' sides are in the ratio $1 : \\sqrt{3} : 2$, so $\\overline{MP} = \\frac{6\\sqrt{2}}{\\sqrt{3}} = 2\\sqrt {6}$. $\\overline{AP} = \\overline{MP} + \\overline{BM} = 6\\sqrt{2} + 2\\sqrt{6} = \\sqrt{72} + \\sqrt{24}$. Hence, $72 + 24 = 096$.",
"Use vectors. In an $xy$ plane, let $(-s,0)$ be $A$, $(0,s)$ be $B$, $(s,0)$ be $C$, $(0,-s)$ be $D$, and $(p,0)$ be P, where $s=|AB|/\\sqrt{2}=6\\sqrt{2}$. It remains to find $p$. The line $y=-x$ is the perpendicular bisector of $AB$ and $CD$, so $O_1$ and $O_2$ lies on the line. Now compute the perpendicular bisector of $AP$. The center has coordinate $(\\frac{p-s}{2},0)$, and the segment is part of the $x$-axis, so the perpendicular bisector has equation $x=\\frac{p-s}{2}$. Since $O_1$ is the circumcenter of triangle $ABP$, it lies on the perpendicular bisector of both $AB$ and $AP$, so \\[O_1=(\\frac{p-s}{2},-\\frac{p-s}{2})\\] Similarly, \\[O_2=(\\frac{p+s}{2},-\\frac{p+s}{2})\\] The relation $\\angle O_1PO_2=120^\\circ$ can now be written using dot product as \\[\\vec{PO_1}\\cdot\\vec{PO_2}=|\\vec{PO_1}|\\cdot|\\vec{PO_2}|\\cos 120^\\circ=-\\frac{1}{2}|\\vec{PO_1}|\\cdot|\\vec{PO_2}|\\] Computation of both sides yields \\[\\frac{p^2-s^2}{p^2+s^2}=-\\frac{1}{2}\\] Solve for $p$ gives $p=s/\\sqrt{3}=2\\sqrt{6}$, so $AP=s+p=6\\sqrt{2}+2\\sqrt{6}=\\sqrt{72}+\\sqrt{24}$. The answer is 72+24$\\Rightarrow096",
"Translate $\\triangle{ABP}$ so that the image of $AB$ coincides $DC$. Let the image of $P$ be $P’$. $\\angle{DPC}=\\angle{CPB}$ by symmetry, and $\\angle{APB}=\\angle{DP’C}$ because translation preserves angles. Thus $\\angle{DP’C}+\\angle{CPD}=\\angle{CPB}+\\angle{APB}=180^\\circ$. Therefore, quadrilateral $CPDP’$ is cyclic. Thus the image of $O_1$ coincides with $O_2$. $O_1P$ is parallel to $O_2P’$ so $\\angle{P’O_2P}=\\angle{O_1PO_2}=120^\\circ$, so $\\angle{PDP’}=60^\\circ$ and $\\angle{PDC}=15^\\circ$, thus $\\angle{ADP}=75^{\\circ}$. Let $M$ be the foot of the perpendicular from $D$ to $AC$. Then $\\triangle{AMD}$ is a 45-45-90 triangle and $\\triangle{DMP}$ is a 30-60-90 triangle. Thus $AM=6\\sqrt{2}$ and $MP=\\frac{6\\sqrt{2}}{\\sqrt{3}}$. This gives us $AP=AM+MP=\\sqrt{72}+\\sqrt{24}$, and the answer is $72+24=096",
"Reflect $O_1$ across $AP$ to $O_1'$. By symmetry $O_1’$ is the circumcenter of $\\triangle{ADP}$ $\\angle{DO_1’P}$ = $2*\\angle{DAP} = 90^\\circ$, so $\\angle{O_1’PD}=45^\\circ$ similarly $\\angle{DO_2P}$ = $2*\\angle{DCP} = 90^\\circ$, so $\\angle{O_2PD}=45^\\circ$ Therefore $\\angle{O_1’PO_2}=90^\\circ$, so that $\\angle{O_1’PO_1} =120^\\circ - 90^\\circ = 30^\\circ$ By symmetry, $\\angle{O_1'PA} = \\angle{APO_1} = 0.5*\\angle{O_1’PO_1} = 15^\\circ$ Therefore, since $O_1’$ is the circumcenter of $\\triangle{ADP}$, $\\angle{ADP}$ = $0.5*(180^\\circ - 2*\\angle{O_1'PA}) = 75^\\circ$ Therefore $\\angle{APD} = 180^\\circ - 45^\\circ - 75^\\circ = 60^\\circ$ Using sine rule in $\\triangle{ADP}$, $AP = (12 * \\sin 75^\\circ) / \\sin 60^\\circ =\\sqrt{72}+\\sqrt{24}$, and the answer is $72+24=096 By Kris17",
"Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set $A=(0, 12)$, $B=(12,12)$, $C=(12, 0)$, $D=(0, 0)$. Let this $P=(a, 12-a)$ for some $a$. We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that $O_{1}$ is the intersection of $x=6$ and, knowing the midpoint of $AP$ to be $(\\frac{a}{2}, \\frac{12-a}{2})$ and thus the equation to be $y=x+(12-a)$, we get $(6, 18-a)$. Likewise for $O_{2}$ it's $(6, 6-a)$. Now what do we see? $O_{1}P=O_{2}P$ (just look at the coordinates)! So both of those distances are $4\\sqrt{3}$. Solving for $a$ we get it to be $6+2\\sqrt{3}$, since $AP>CP$. Multiply by $\\sqrt{2}$ because we are looking for $AP$ to get the answer of $096.",
"Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set $A=(0, 12)$, $B=(12,12)$, $C=(12, 0)$, $D=(0, 0)$. Let this $P=(a, 12-a)$ for some $a$. We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that $O_{1}$ is the intersection of $x=6$ and, knowing the midpoint of $AP$ to be $(\\frac{a}{2}, \\frac{12-a}{2})$ and thus the equation to be $y=x+(12-a)$, we get $(6, 18-a)$. Likewise for $O_{2}$ it's $(6, 6-a)$. Now what do we see? $O_{1}P=O_{2}P$ (just look at the coordinates)! So both of those distances are $4\\sqrt{3}$. Solving for $a$ we get it to be $6+2\\sqrt{3}$, since $AP>CP$. Multiply by $\\sqrt{2}$ because we are looking for $AP$ to get the answer of $096.",
"Let $\\angle APD = \\theta$. Then $\\angle ADP = 180^{\\circ}-45^{\\circ}-\\theta=135^{\\circ}-\\theta \\implies \\angle PDC=\\theta-45^{\\circ}$. Realize that because $O_1$ is a circumcenter, $\\angle DO_1P=\\angle DCP=45^{\\circ} \\implies \\angle DPO_1=\\frac{180^{\\circ}-\\angle DO_1P}{2}=45^{\\circ}$. Then $\\angle O_2PA=75^{\\circ}-\\theta \\implies \\angle AOP = 180^{\\circ}-2\\angle O_2PA = 2\\theta+30^{\\circ} \\implies \\angle ABP=\\theta + 15^{\\circ} \\implies \\angle PBC=75-\\theta$. Now, because $P$ lies on diagonal $AC$, $\\triangle PDC \\cong \\triangle PBC \\implies \\angle PDC = \\angle PBC \\implies \\theta-45^{\\circ}=75^{\\circ}-\\theta \\implies \\theta = 60^{\\circ}$. To finish, we look at $\\triangle ADP$. Drop a perpendicular from $D$ to $AP$ at $E$. Then $\\triangle ADE$ is a $45-45-90$ and $\\triangle PDE$ is a $30-60-90$. Therefore, $DE=EA=6\\sqrt{2}, EP=2\\sqrt{6}$, so $AP=AE+EP=6\\sqrt{2}+2\\sqrt{6}=\\sqrt{72}+\\sqrt{24} \\implies 096 ~msc",
"Let $\\angle APD = \\theta$. Then $\\angle ADP = 180^{\\circ}-45^{\\circ}-\\theta=135^{\\circ}-\\theta \\implies \\angle PDC=\\theta-45^{\\circ}$. Realize that because $O_1$ is a circumcenter, $\\angle DO_1P=\\angle DCP=45^{\\circ} \\implies \\angle DPO_1=\\frac{180^{\\circ}-\\angle DO_1P}{2}=45^{\\circ}$. Then $\\angle O_2PA=75^{\\circ}-\\theta \\implies \\angle AOP = 180^{\\circ}-2\\angle O_2PA = 2\\theta+30^{\\circ} \\implies \\angle ABP=\\theta + 15^{\\circ} \\implies \\angle PBC=75-\\theta$. Now, because $P$ lies on diagonal $AC$, $\\triangle PDC \\cong \\triangle PBC \\implies \\angle PDC = \\angle PBC \\implies \\theta-45^{\\circ}=75^{\\circ}-\\theta \\implies \\theta = 60^{\\circ}$. To finish, we look at $\\triangle ADP$. Drop a perpendicular from $D$ to $AP$ at $E$. Then $\\triangle ADE$ is a $45-45-90$ and $\\triangle PDE$ is a $30-60-90$. Therefore, $DE=EA=6\\sqrt{2}, EP=2\\sqrt{6}$, so $AP=AE+EP=6\\sqrt{2}+2\\sqrt{6}=\\sqrt{72}+\\sqrt{24} \\implies 096 ~msc",
"Both $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$. Claim: $O_1O_2=12$ and $O_1P=O_2P$. Proof. Translate $O_1$ and $P$ $12$ units down, and let their images be $O_1'$ and $P'$, respectively. Note that $\\triangle ABP\\cong\\triangle DCP'$. Additionally, \\[\\angle CP'D = \\angle BPA = 180^{\\circ} - \\angle BPC = 180^{\\circ} - \\angle CPD,\\] so $CPDP'$ is cyclic. This means $O_1'$ and $O_2$ coincide, so $O_1O_2=12$. This also means the circumradii of both triangles are equal, so $O_1P=O_2P$. $\\blacksquare$. Let the perpendicular from $P$ intersect $O_1O_2$ at $X$ and $AD$ at $Y$. Since $\\triangle O_1XP$ is 30-60-90, $XP=\\frac{6}{\\sqrt{3}} = 2\\sqrt3$. Since $YX=6$, $PY=6+2\\sqrt3$, so $AP=6\\sqrt2+2\\sqrt6 = \\sqrt{72}+\\sqrt{24} \\implies96. ~rayfish",
"Both $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$. Claim: $O_1O_2=12$ and $O_1P=O_2P$. Proof. Translate $O_1$ and $P$ $12$ units down, and let their images be $O_1'$ and $P'$, respectively. Note that $\\triangle ABP\\cong\\triangle DCP'$. Additionally, \\[\\angle CP'D = \\angle BPA = 180^{\\circ} - \\angle BPC = 180^{\\circ} - \\angle CPD,\\] so $CPDP'$ is cyclic. This means $O_1'$ and $O_2$ coincide, so $O_1O_2=12$. This also means the circumradii of both triangles are equal, so $O_1P=O_2P$. $\\blacksquare$. Let the perpendicular from $P$ intersect $O_1O_2$ at $X$ and $AD$ at $Y$. Since $\\triangle O_1XP$ is 30-60-90, $XP=\\frac{6}{\\sqrt{3}} = 2\\sqrt3$. Since $YX=6$, $PY=6+2\\sqrt3$, so $AP=6\\sqrt2+2\\sqrt6 = \\sqrt{72}+\\sqrt{24} \\implies96. ~rayfish",
"\\[BP = DP, \\angle PAB =\\angle PCD = 45^\\circ \\implies O_{1}P =O_{2}P\\] by the Law of Sines. \\[AB= CD, O_{1}A = O_{2}D = O_{1}B = O_{2}C \\implies \\triangle AO_{1}B = \\triangle DO_{2}C.\\] They have translational symmetry. The translation vector is \\[\\vec {AD} \\implies O_{1}O_{2} = AD.\\] \\[\\triangle PO_{1}O_{2} = \\triangle O_{1}AB \\implies \\angle AO_{1}B = 120^\\circ \\implies\\] \\[\\angle APB = 60^\\circ \\implies \\angle ABP = 180^\\circ – 60^\\circ – 45^\\circ = 75^\\circ.\\] By the Law of Sines \\[AP = \\frac {AB \\cdot \\sin 75^\\circ}{\\sin 60^\\circ} = \\frac {AB \\cdot \\sin (30^\\circ + 45^\\circ)} {\\sin 60^\\circ}\\] \\[AP = AB \\cdot (\\sin 45^\\circ + \\cos 45^\\circ \\cdot \\tan 30^\\circ),\\] \\[AP = \\frac {AB}{\\sqrt{2}} (1 + \\frac {1}{\\sqrt{3}}) = 6 \\sqrt {2} + 2 \\sqrt {6} \\implies \\textbf{096}.\\] [email protected], vvsss"
] |
2011-II-14
| 2,011
| 14
|
There are $N$ permutations $(a_1, a_2, \dots, a_{30})$ of $1, 2, \dots, 30$ such that for $m \in \{2,3,5\}$ , $m$ divides $a_{n+m} - a_n$ for all integers $n$ with $1 \le n < n+m \le 30$ . Find the remainder when $N$ is divided by 1000.
| 440
|
II
|
[
"Solution 1 Be wary of \"position\" versus \"number\" in the solution! Each POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$. The $n$th position is defined by this ordered triple where $i$ is $n \\mod 2$, $j$ is $n \\mod 3$, and $k$ is $n \\mod 5$. There are 2 choices for $i$, 3 for $j$, and 5 for $k$, yielding $2 \\cdot 3 \\cdot 5=30$ possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if $i$ is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If $j$ is the same, then the two numbers must be equivalent $\\mod 3$, and if $k$ is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have $1 \\mod 2$! It's that the POSITION which NUMBER 1 occupies has $1 \\mod 2$! The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement. The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed. Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of $1\\cdot 1 \\cdot 3=3$ choices for the placement of 3. As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of $1 \\cdot 1 \\cdot 2=2$ possible placements for 4. 5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5. All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed. Thus, $N = 30 \\cdot 8 \\cdot 3 \\cdot 2=30 \\cdot 48=1440$. Thus, the remainder when $N$ is divided by $1000$ is $440 ~~Mathcounts Griiinder",
"Be wary of \"position\" versus \"number\" in the solution! Each POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$. The $n$th position is defined by this ordered triple where $i$ is $n \\mod 2$, $j$ is $n \\mod 3$, and $k$ is $n \\mod 5$. There are 2 choices for $i$, 3 for $j$, and 5 for $k$, yielding $2 \\cdot 3 \\cdot 5=30$ possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if $i$ is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If $j$ is the same, then the two numbers must be equivalent $\\mod 3$, and if $k$ is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have $1 \\mod 2$! It's that the POSITION which NUMBER 1 occupies has $1 \\mod 2$! The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement. The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed. Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of $1\\cdot 1 \\cdot 3=3$ choices for the placement of 3. As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of $1 \\cdot 1 \\cdot 2=2$ possible placements for 4. 5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5. All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed. Thus, $N = 30 \\cdot 8 \\cdot 3 \\cdot 2=30 \\cdot 48=1440$. Thus, the remainder when $N$ is divided by $1000$ is $440 ~~Mathcounts Griiinder",
"We observe that the condition on the permutations means that two numbers with indices congruent $\\mod m$ are themselves congruent $\\mod m$ for $m \\in \\{ 2,3,5\\}.$ Furthermore, suppose that $a_n \\equiv k \\mod m.$ Then, there are $30/m$ indices congruent to $n \\mod m,$ and $30/m$ numbers congruent to $k \\mod m,$ because 2, 3, and 5 are all factors of 30. Therefore, since every index congruent to $n$ must contain a number congruent to $k,$ and no number can appear twice in the permutation, only the indices congruent to $n$ contain numbers congruent to $k.$ In other words, $a_i \\equiv a_j \\mod m \\iff i \\equiv j \\mod m.$ But it is not necessary that $\\textcolor{red}{(a_i\\equiv i)\\cup (a_j\\equiv j)\\mod m}$. In fact, if that were the case, there would only be one way to assign the indices, since $2,3,5$ are relatively prime to each other and $30=\\text{lcm}(2,3,5)$: $\\{a_1,a_2,\\dots a_{30}\\}\\in\\{1,2,\\dots 30\\}\\text{ }respectively$. This tells us that in a valid permutation, the congruence classes $\\mod m$ are simply swapped around, and if the set $S$ is a congruence class $\\mod m$ for $m =$ 2, 3, or 5, the set $\\{ a_i \\vert i \\in S \\}$ is still a congruence class $\\mod m.$ Clearly, each valid permutation of the numbers 1 through 30 corresponds to exactly one permutation of the congruence classes modulo 2, 3, and 5. Also, if we choose some permutations of the congruence classes modulo 2, 3, and 5, they correspond to exactly one valid permutation of the numbers 1 through 30. This can be shown as follows: First of all, the choice of permutations of the congruence classes gives us every number in the permutation modulo 2, 3, and 5, so by the Chinese Remainder Theorem, it gives us every number $\\mod 2\\cdot 3\\cdot 5 = 30.$ Because the numbers must be between 1 and 30 inclusive, it thus uniquely determines what number goes in each index. Furthermore, two different indices cannot contain the same number. We will prove this by contradiction, calling the indices $a_i$ and $a_j$ for $i \\neq j.$ If $a_i=a_j,$ then they must have the same residues modulo 2, 3, and 5, and so $i \\equiv j$ modulo 2, 3, and 5. Again using the Chinese Remainder Theorem, we conclude that $i \\equiv j \\mod 30,$ so because $i$ and $j$ are both between 1 and 30 inclusive, $i = j,$ giving us a contradiction. Therefore, every choice of permutations of the congruence classes modulo 2, 3, and 5 corresponds to exactly one valid permutation of the numbers 1 through 30. In other words, each set of assignment from $a_j\\rightarrow j\\mod (2,3,5)$ determines a unique string of $30$ numbers. For example: $\\left[(0,1)\\rightarrow (1,0)\\right]\\cap\\left[(0,1,2)\\rightarrow (0,2,1)\\right]\\cap\\left[(0,1,2,3,4)\\rightarrow (4,2,3,0,1)\\right]$: \\[\\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \\hline 2&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0\\\\ \\hline 3&0&2&1&0&2&1&0&2&1&0&2&1&0&2&1&0&2&1&0&2&1&0&2&1&0&2&1&0&2&1\\\\ \\hline 5&4&2&3&0&1&4&2&3&0&1&4&2&3&0&1&4&2&3&0&1&4&2&3&0&1&4&2&3&0&1\\\\ \\hline\\hline 30&9&2&13&0&11&4&27&8&25&6&29&22&3&20&1&24&17&28&15&26&19&12&23&10&21&14&7&18&5&16\\\\ \\hline \\end{array}\\] We have now established a bijection between valid permutations of the numbers 1 through 30 and permutations of the congruence classes modulo 2, 3, and 5, so $N$ is equal to the number of permutations of congruence classes. There are always $m$ congruence classes $\\mod m,$ so $N = 2! \\cdot 3! \\cdot 5! = 2 \\cdot 6 \\cdot 120 = 1440 \\equiv 440 \\mod 1000.$ Solution 3 (2-sec solve) Note that $30=2\\cdot 3\\cdot 5$. Since $\\gcd(2, 3, 5)=1$, by CRT, for each value $k=0\\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\\equiv a\\pmod{2}$, $k\\equiv b\\pmod{3}$, and $k\\equiv c\\pmod{5}$. Therefore, we can independently assign the residues modulo $2, 3, 5$, so $N=2!\\cdot 3!\\cdot 5!=1440$, and the answer is $440 ~~Mathcounts Griiinder",
"Note that $30=2\\cdot 3\\cdot 5$. Since $\\gcd(2, 3, 5)=1$, by CRT, for each value $k=0\\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\\equiv a\\pmod{2}$, $k\\equiv b\\pmod{3}$, and $k\\equiv c\\pmod{5}$. Therefore, we can independently assign the residues modulo $2, 3, 5$, so $N=2!\\cdot 3!\\cdot 5!=1440$, and the answer is $440 ~~Mathcounts Griiinder",
"Note that $30=2\\cdot 3\\cdot 5$. Since $\\gcd(2, 3, 5)=1$, by CRT, for each value $k=0\\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\\equiv a\\pmod{2}$, $k\\equiv b\\pmod{3}$, and $k\\equiv c\\pmod{5}$. Therefore, we can independently assign the residues modulo $2, 3, 5$, so $N=2!\\cdot 3!\\cdot 5!=1440$, and the answer is $440 ~~Mathcounts Griiinder",
"First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\\equiv 0 \\pmod 2$ or $\\equiv 1 \\pmod 2$. Another way to say this is each element in the sequence $a_1,a_3,a_5,a_7...$ would be the same mod 2, and similarly for the other sequence. There are 2! = 2 ways to assign the mods to the sequences. Now when $m$ is equal to 3, the sequences are $T_1 = a_1, a_4,a_7..$, $T_2 = a_2,a_5,a_8,a_11...$, and $T_3 = a_3,a_6,a_9...$. Again, for each sequence, all of its elements are congruent either $0$,$1$, or $2$ mod $3$. There are $3! = 6$ ways to assign the mods to the sequences. Finally do the same thing for $m = 5$. There are $5!$ ways. In total there are $2 * 6*120 = 1440$ and the answer is $440 ~~Mathcounts Griiinder",
"First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\\equiv 0 \\pmod 2$ or $\\equiv 1 \\pmod 2$. Another way to say this is each element in the sequence $a_1,a_3,a_5,a_7...$ would be the same mod 2, and similarly for the other sequence. There are 2! = 2 ways to assign the mods to the sequences. Now when $m$ is equal to 3, the sequences are $T_1 = a_1, a_4,a_7..$, $T_2 = a_2,a_5,a_8,a_11...$, and $T_3 = a_3,a_6,a_9...$. Again, for each sequence, all of its elements are congruent either $0$,$1$, or $2$ mod $3$. There are $3! = 6$ ways to assign the mods to the sequences. Finally do the same thing for $m = 5$. There are $5!$ ways. In total there are $2 * 6*120 = 1440$ and the answer is $440 ~~Mathcounts Griiinder",
"Note, for the purposes of this problem, it is easier to have our residue classes mod \\( m \\) to be \\( \\{1, 2, \\dots, m\\} \\) instead of \\( \\{0, 1, 2, \\dots, m-1\\} \\). We will start by finding a general insight for \\( m \\) given that \\( m \\mid 30 \\). \\[m \\mid (a_{i + m} - a_i) \\implies a_{i + m} \\equiv a_i \\pmod{m}\\] Notice that for every \\( p \\) between \\( 1 \\) and \\( m \\) inclusive, we have \\[a_p \\equiv a_{p+m} \\equiv a_{p+2m} \\equiv a_{p+3m} ... \\pmod{m}.\\] Notice that \\( \\{p, p+m, p+2m, \\dots, p+km\\} \\) all fall under the residue class \\( p \\pmod{m} \\). Let this list of numbers with indices that are \\( p \\pmod{m} \\) all be congruent to \\( \\sigma_m(p) \\pmod{m} \\). Notice that \\( \\sigma_m \\) maps numbers from \\( \\{1, 2, \\dots, m\\} \\) to \\( \\{1, 2, \\dots, m\\} \\). $\\textbf{Claim: \\( \\sigma_m \\) is bijective.}$ Since \\( \\sigma_m \\) is a valid function mapping two sets of the same size, we simply prove injectivity. Assume for the sake of contradiction, for \\( p \\neq q \\), let \\( \\sigma_m(p) \\equiv \\sigma_m(q) \\pmod{m} \\), and let them be congruent to some constant \\( x \\pmod{m} \\). Since \\( 1 \\leq p, q \\leq m \\) and \\( p \\neq q \\), this implies \\( p \\) and \\( q \\) are not congruent mod \\( m \\). \\[\\{p, p+m, p+2m, \\dots \\} \\cap \\{q, q+m, q+2m, \\dots\\} = \\emptyset,\\] since these sets are distinct mod \\( m \\). Therefore, \\( \\sigma_m \\) maps \\( \\frac{2 \\cdot 30}{m} \\) distinct values to some \\( x \\pmod{30} \\). However, this implies that there are \\( \\frac{2 \\cdot 30}{m} \\) distinct values between \\( 1 \\) and \\( 30 \\) falling under the same residue class, which is obviously false. Thus, we have proven the claim. Since \\( \\sigma_m \\) is injective and maps between two sets of identical size, it is bijective. Therefore, there are \\( m! \\) ways to create such a mapping called \\( \\sigma_m \\). For this problem, there are \\( 2! \\cdot 3! \\cdot 5! \\) ways to find \\( \\sigma_2 \\), \\( \\sigma_3 \\), and \\( \\sigma_5 \\). (Notice \\( \\sigma_k \\) lies in the symmetric group \\( S_k \\).) $\\textbf{Claim: Finding \\( \\sigma_2 \\), \\( \\sigma_3 \\), and \\( \\sigma_5 \\) completely determines \\( a_i \\).}$ If we know \\( \\sigma_2(i) \\), \\( \\sigma_3(i) \\), and \\( \\sigma_5(i) \\), we know \\( i \\pmod{2} \\), \\( i \\pmod{3} \\), and \\( i \\pmod{5} \\). By the Chinese Remainder Theorem (CRT), we know \\( i \\pmod{30} \\), and since \\( 1 \\leq i \\leq 30 \\), this completely determines \\( i \\). Therefore, \\( 2! \\cdot 3! \\cdot 5! = 2 \\cdot 6 \\cdot 120 = 1440 \\). Therefore, the answer is $440 ~~Mathcounts Griiinder",
"Note, for the purposes of this problem, it is easier to have our residue classes mod \\( m \\) to be \\( \\{1, 2, \\dots, m\\} \\) instead of \\( \\{0, 1, 2, \\dots, m-1\\} \\). We will start by finding a general insight for \\( m \\) given that \\( m \\mid 30 \\). \\[m \\mid (a_{i + m} - a_i) \\implies a_{i + m} \\equiv a_i \\pmod{m}\\] Notice that for every \\( p \\) between \\( 1 \\) and \\( m \\) inclusive, we have \\[a_p \\equiv a_{p+m} \\equiv a_{p+2m} \\equiv a_{p+3m} ... \\pmod{m}.\\] Notice that \\( \\{p, p+m, p+2m, \\dots, p+km\\} \\) all fall under the residue class \\( p \\pmod{m} \\). Let this list of numbers with indices that are \\( p \\pmod{m} \\) all be congruent to \\( \\sigma_m(p) \\pmod{m} \\). Notice that \\( \\sigma_m \\) maps numbers from \\( \\{1, 2, \\dots, m\\} \\) to \\( \\{1, 2, \\dots, m\\} \\). $\\textbf{Claim: \\( \\sigma_m \\) is bijective.}$ Since \\( \\sigma_m \\) is a valid function mapping two sets of the same size, we simply prove injectivity. Assume for the sake of contradiction, for \\( p \\neq q \\), let \\( \\sigma_m(p) \\equiv \\sigma_m(q) \\pmod{m} \\), and let them be congruent to some constant \\( x \\pmod{m} \\). Since \\( 1 \\leq p, q \\leq m \\) and \\( p \\neq q \\), this implies \\( p \\) and \\( q \\) are not congruent mod \\( m \\). \\[\\{p, p+m, p+2m, \\dots \\} \\cap \\{q, q+m, q+2m, \\dots\\} = \\emptyset,\\] since these sets are distinct mod \\( m \\). Therefore, \\( \\sigma_m \\) maps \\( \\frac{2 \\cdot 30}{m} \\) distinct values to some \\( x \\pmod{30} \\). However, this implies that there are \\( \\frac{2 \\cdot 30}{m} \\) distinct values between \\( 1 \\) and \\( 30 \\) falling under the same residue class, which is obviously false. Thus, we have proven the claim. Since \\( \\sigma_m \\) is injective and maps between two sets of identical size, it is bijective. Therefore, there are \\( m! \\) ways to create such a mapping called \\( \\sigma_m \\). For this problem, there are \\( 2! \\cdot 3! \\cdot 5! \\) ways to find \\( \\sigma_2 \\), \\( \\sigma_3 \\), and \\( \\sigma_5 \\). (Notice \\( \\sigma_k \\) lies in the symmetric group \\( S_k \\).) $\\textbf{Claim: Finding \\( \\sigma_2 \\), \\( \\sigma_3 \\), and \\( \\sigma_5 \\) completely determines \\( a_i \\).}$ If we know \\( \\sigma_2(i) \\), \\( \\sigma_3(i) \\), and \\( \\sigma_5(i) \\), we know \\( i \\pmod{2} \\), \\( i \\pmod{3} \\), and \\( i \\pmod{5} \\). By the Chinese Remainder Theorem (CRT), we know \\( i \\pmod{30} \\), and since \\( 1 \\leq i \\leq 30 \\), this completely determines \\( i \\). Therefore, \\( 2! \\cdot 3! \\cdot 5! = 2 \\cdot 6 \\cdot 120 = 1440 \\). Therefore, the answer is $440 ~~Mathcounts Griiinder"
] |
2011-II-15
| 2,011
| 15
|
Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\left\lfloor\sqrt{P(x)}\right\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ , $b$ , $c$ , $d$ , and $e$ are positive integers. Find $a + b + c + d + e$ .
| 850
|
II
|
[
"Table of values of $P(x)$: \\begin{align*} P(5) &= 1 \\\\ P(6) &= 9 \\\\ P(7) &= 19 \\\\ P(8) &= 31 \\\\ P(9) &= 45 \\\\ P(10) &= 61 \\\\ P(11) &= 79 \\\\ P(12) &= 99 \\\\ P(13) &= 121 \\\\ P(14) &= 145 \\\\ P(15) &= 171 \\\\ \\end{align*} In order for $\\lfloor \\sqrt{P(x)} \\rfloor = \\sqrt{P(\\lfloor x \\rfloor)}$ to hold, $\\sqrt{P(\\lfloor x \\rfloor)}$ must be an integer and hence $P(\\lfloor x \\rfloor)$ must be a perfect square. This limits $x$ to $5 \\le x < 6$ or $6 \\le x < 7$ or $13 \\le x < 14$ since, from the table above, those are the only values of $x$ for which $P(\\lfloor x \\rfloor)$ is an perfect square. However, in order for $\\sqrt{P(x)}$ to be rounded down to $P(\\lfloor x \\rfloor)$, $P(x)$ must be less than the next perfect square after $P(\\lfloor x \\rfloor)$ (for the said intervals). Now, we consider the three cases: Case $5 \\le x < 6$: $P(x)$ must be less than the first perfect square after $1$, which is $4$, i.e.: $1 \\le P(x) < 4$ (because $\\lfloor \\sqrt{P(x)} \\rfloor = 1$ implies $1 \\le \\sqrt{P(x)} < 2$) Since $P(x)$ is increasing for $x \\ge 5$, we just need to find the value $v \\ge 5$ where $P(v) = 4$, which will give us the working range $5 \\le x < v$. \\begin{align*} v^2 - 3v - 9 &= 4 \\\\ v &= \\frac{3 + \\sqrt{61}}{2} \\end{align*} So in this case, the only values that will work are $5 \\le x < \\frac{3 + \\sqrt{61}}{2}$. Case $6 \\le x < 7$: $P(x)$ must be less than the first perfect square after $9$, which is $16$. \\begin{align*} v^2 - 3v - 9 &= 16 \\\\ v &= \\frac{3 + \\sqrt{109}}{2} \\end{align*} So in this case, the only values that will work are $6 \\le x < \\frac{3 + \\sqrt{109}}{2}$. Case $13 \\le x < 14$: $P(x)$ must be less than the first perfect square after $121$, which is $144$. \\begin{align*} v^2 - 3v - 9 &= 144 \\\\ v &= \\frac{3 + \\sqrt{621}}{2} \\end{align*} So in this case, the only values that will work are $13 \\le x < \\frac{3 + \\sqrt{621}}{2}$. Now, we find the length of the working intervals and divide it by the length of the total interval, $15 - 5 = 10$: \\begin{align*} \\frac{\\left( \\frac{3 + \\sqrt{61}}{2} - 5 \\right) + \\left( \\frac{3 + \\sqrt{109}}{2} - 6 \\right) + \\left( \\frac{3 + \\sqrt{621}}{2} - 13 \\right)}{10} \\\\ &= \\frac{\\sqrt{61} + \\sqrt{109} + \\sqrt{621} - 39}{20} \\end{align*} Thus, the answer is $61 + 109 + 621 + 39 + 20 = 850.",
"Make the substitution $y=2x-3$, so $P(x)=\\frac{y^2-45}{4}.$ We're looking for solutions to \\[\\bigg\\lfloor{\\sqrt{\\frac{y^2-45}{4}}\\bigg\\rfloor}=\\sqrt{\\frac{\\lfloor{y\\rfloor}^2-45}{4}}\\]with the new bounds $y\\in{[7,27]}$. Since the left side is an integer, it must be that $\\frac{\\lfloor{y\\rfloor}^2-45}{4}$ is a perfect square. For simplicity, write $\\lfloor{y\\rfloor}=a$ and \\[a^2-45=4b^2\\implies{(a-2b)(a+2b)=45}.\\]Since $a-2b<a+2b$, it must be that $(a-2b,a+2b)=(1,45),(3,15),(5,9)$, which gives solutions $(23,11),(9,3),(7,1)$, respectively. But this gives us three cases to check: Case 1: $\\bigg\\lfloor{\\sqrt{\\frac{y^2-45}{4}}\\bigg\\rfloor}=11$. In this case, we have \\[11\\leq{\\sqrt{\\frac{y^2-45}{4}}}<12\\implies{y\\in{[23,\\sqrt{621})}}.\\] Case 2: $\\bigg\\lfloor{\\sqrt{\\frac{y^2-45}{4}}\\bigg\\rfloor}=3$. In this case, we have \\[3\\leq{\\sqrt{\\frac{y^2-45}{4}}}<4\\implies{y\\in{[9,\\sqrt{109})}}.\\] Case 3: $\\bigg\\lfloor{\\sqrt{\\frac{y^2-45}{4}}\\bigg\\rfloor}=1$ In this case, we have \\[1\\leq{\\sqrt{\\frac{y^2-45}{4}}}<2\\implies{y\\in{[7,\\sqrt{61})}}.\\] To finish, the total length of the interval from which we choose $y$ is $27-7=20$. The total length of the success intervals is \\[(\\sqrt{61}-7)+(\\sqrt{109}-9)+(\\sqrt{621}-23)=\\sqrt{61}+\\sqrt{109}+\\sqrt{621}-39,\\]which means the probability is \\[\\frac{\\sqrt{61}+\\sqrt{109}+\\sqrt{621}-39}{20}.\\]The requested sum is $850.",
"It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let $A = \\left\\lfloor\\sqrt{P(x)}\\right\\rfloor$ and $B = \\sqrt{P(\\left\\lfloor x \\right\\rfloor)}$. The graph of $A$ and $B$ will look like this, with $A$ having only integral y-values and $B$ having only integral x-values: As both $A$ and $B$ consist of a bunch of line segments, the probability that $A = B$ is the \"length\" of the overlap between the segments of $A$ and $B$ divided by the total length of the segments of $B$. Looking at the graph, we see that $A$ and $B$ will overlap only when $B$ is an integer. Specifically, each region of overlap will begin when $\\sqrt{P(x)}\\ = k, 5 \\le x \\le 15$ has solutions for integral $k$ in the range of $A$, which consists of the integers $1-13$, and end when $A$ jumps up to its next y-value. Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of $\\frac{3 + \\sqrt{45 + 4k^2}}{2}$ for $k$ in the specified range, meaning $45 + 4k^2$ must be a perfect square. Plugging in all the possible values of $k$, we get $k = 1, 3, 11$, corresponding to start points of $x = 5, 6, 13$. As already stated, the endpoints will occur when $A$ jumps up to the next integer $k+1$ at each of these segments, at which point the x-value will be $\\frac{3 + \\sqrt{45 + 4(k+1)^2}}{2}$. On the graph, the overlapping segments of $A$ and $B$ would be represented by the highlighted green segments below: Taking the difference between this second x-value and the start point for each of our start points $x = 5, 6, 13$ and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of $B$) to give us the probability of overlap between $A$ and $B$. Doing so gives us: $\\frac{\\left( \\frac{3 + \\sqrt{61}}{2} - 5 \\right) + \\left( \\frac{3 + \\sqrt{109}}{2} - 6 \\right) + \\left( \\frac{3 + \\sqrt{621}}{2} - 13 \\right)}{10} = \\frac{\\sqrt{61} + \\sqrt{109} + \\sqrt{621} - 39}{20}$ $\\implies{61 + 109 + 621 + 39 + 20 = 850)",
"It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let $A = \\left\\lfloor\\sqrt{P(x)}\\right\\rfloor$ and $B = \\sqrt{P(\\left\\lfloor x \\right\\rfloor)}$. The graph of $A$ and $B$ will look like this, with $A$ having only integral y-values and $B$ having only integral x-values: As both $A$ and $B$ consist of a bunch of line segments, the probability that $A = B$ is the \"length\" of the overlap between the segments of $A$ and $B$ divided by the total length of the segments of $B$. Looking at the graph, we see that $A$ and $B$ will overlap only when $B$ is an integer. Specifically, each region of overlap will begin when $\\sqrt{P(x)}\\ = k, 5 \\le x \\le 15$ has solutions for integral $k$ in the range of $A$, which consists of the integers $1-13$, and end when $A$ jumps up to its next y-value. Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of $\\frac{3 + \\sqrt{45 + 4k^2}}{2}$ for $k$ in the specified range, meaning $45 + 4k^2$ must be a perfect square. Plugging in all the possible values of $k$, we get $k = 1, 3, 11$, corresponding to start points of $x = 5, 6, 13$. As already stated, the endpoints will occur when $A$ jumps up to the next integer $k+1$ at each of these segments, at which point the x-value will be $\\frac{3 + \\sqrt{45 + 4(k+1)^2}}{2}$. On the graph, the overlapping segments of $A$ and $B$ would be represented by the highlighted green segments below: Taking the difference between this second x-value and the start point for each of our start points $x = 5, 6, 13$ and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of $B$) to give us the probability of overlap between $A$ and $B$. Doing so gives us: $\\frac{\\left( \\frac{3 + \\sqrt{61}}{2} - 5 \\right) + \\left( \\frac{3 + \\sqrt{109}}{2} - 6 \\right) + \\left( \\frac{3 + \\sqrt{621}}{2} - 13 \\right)}{10} = \\frac{\\sqrt{61} + \\sqrt{109} + \\sqrt{621} - 39}{20}$ $\\implies{61 + 109 + 621 + 39 + 20 = 850)",
"Note that all the \"bounds\" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is $\\frac{3*3+\\sqrt{9+4(4+9)}-10+\\sqrt{9+4(16+9)}-12+\\sqrt{9+4(144+9)}}{20} \\implies 850 ~Lcz"
] |
2012-I-1
| 2,012
| 1
|
Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$ .
| 40
|
I
|
[
"A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$, there are two possible values for $a$ and $c$, since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$, and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$. There are thus $2 \\cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \\cdot 10 = 040.",
"A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$. If $b$ is odd, then $a$ and $c$ must both be $2 \\pmod 4$ meaning that $a$ and $c$ are $2$ or $6$. If $b$ is even, then $a$ and $c$ must be $0 \\pmod 4$ meaning that $a$ and $c$ are $4$ or $8$. For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \\cdot 2 \\cdot 2 = 040 numbers.",
"For this number to fit the requirements $bc$ and $ba$ must be divisible by 4. So $bc = 00, 04, 08, 12, 16, ... , 92, 96$ and so must $ba$ for each two digits of $bc$. There are two possibilities for $ba$ if $b$ is odd and three possibilities if $b$ is even. So there are $2^{2} \\cdot 5+3^{2} \\cdot 4 = 65$ possibilities but this overcounts when $a$ or $c = 0$. So when $bc = 00, 20, 40, 60, 80$ and the corresponding $ba$ should be removed, so $65 - 5 \\cdot 3 = 50$. But we are still overcounting when $b$ is even because then $a$ can be 0. So the answer is $50 - 10 \\cdot 1 = 040 ~LuisFonseca123"
] |
2012-I-2
| 2,012
| 2
|
The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle term of the original sequence.
| 195
|
I
|
[
"Solution 1 If the sum of the original sequence is $\\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \\sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \\rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $195. ~ cxsmi",
"If the sum of the original sequence is $\\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \\sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \\rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $195. ~ cxsmi",
"After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$. Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\\dfrac{715}{11} = 65$. Since the first, last, and middle terms are centered around the mean, our final answer is $65 \\cdot 3 = 195. ~ cxsmi",
"Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\\frac{2a_1 + 10d}{2} \\cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$. Then the first term of the corresponding arithmetic sequence will be $60$, the sixth (middle) term will be $65$, and the eleventh (largest) term will be $70$. Thus, our final answer is $60 + 65 + 70 = 195. ~ cxsmi",
"Let the sequence be $a,a+d,a+2d,..., a+(k-1)d$. Since there are $k$ terms, we have an equation $k^2 = 836-715 = 121$. Solving, we get $k$ as $11$. Replacing $k$ in our sequence with $11$, we get $a,a+d,a+2d,..., a+10d$. The sum of this sequence is equal to $715$, or $\\frac{(2a+10d)(11)}{2} = 715$, and by simplifying we get $a+5d=65$. We are asked for the first, middle, and end terms, which are $a$,$a+5d$, and $a+10d$ respectively. Their sum is $3a+15$, or $3 \\cdot (a+5d)$. Our desired answer is $65 \\cdot 3 = 195. ~Irfans123",
"Let the sequence be $a,a+d,a+2d,..., a+(k-1)d$. Since there are $k$ terms, we have an equation $k^2 = 836-715 = 121$. Solving, we get $k$ as $11$. Replacing $k$ in our sequence with $11$, we get $a,a+d,a+2d,..., a+10d$. The sum of this sequence is equal to $715$, or $\\frac{(2a+10d)(11)}{2} = 715$, and by simplifying we get $a+5d=65$. We are asked for the first, middle, and end terms, which are $a$,$a+5d$, and $a+10d$ respectively. Their sum is $3a+15$, or $3 \\cdot (a+5d)$. Our desired answer is $65 \\cdot 3 = 195. ~Irfans123"
] |
2012-I-3
| 2,012
| 3
|
Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.
| 216
|
I
|
[
"Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\\text{BBBCCCFFF}$ respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by $9$ to account for the $9$ different ways in which the person to receive the correct meal could be picked. Note, this implies that the dishes are indistinguishable, though the people aren't. For example, two people who order chicken are separate, though if they receive fish, there is only 1 way to order them. The problem we must solve is to distribute meals $\\text{BBCCCFFF}$ to orders $\\text{BBCCCFFF}$ with no matches. The two people who ordered $B$'s can either both get $C$'s, both get $F$'s, or get one $C$ and one $F.$ We proceed with casework. If the two $B$ people both get $C$'s, then the three $F$ meals left to distribute must all go to the $C$ people. The $F$ people then get $BBC$ in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the $F$ meals to the $C$ people, and there is only 1 way to order this, as all three meals are the same. If the two $B$ people both get $F$'s, the situation is identical to the above and three possibilities arise. If the two $B$ people get $CF$ in some order, then the $C$ people must get $FFB$ and the $F$ people must get $CCB.$ This gives $2 \\cdot 3 \\cdot 3 = 18$ possibilities. Summing across the cases we see there are $24$ possibilities, so the answer is $9 \\cdot 24 = 216.",
"We only need to figure out the number of ways to order the string $BBBCCCFFF$, where exactly one $B$ is in the first three positions, one $C$ is in the $4^{th}$ to $6^{th}$ positions, and one $F$ is in the last three positions. There are $3^3=27$ ways to place the first $3$ meals. Then for the other two people, there are $2$ ways to serve their meals. Thus, there are $(3\\cdot2)^3=216",
"First, choose the person that gets the correct meal. There's obviously $9$ ways to do that. Then, we can casework on who gets what type of meal. WLOG the person who had the correct meal had the chicken meal. The remaining chicken meals can be distributed to either one person who ordered beef and another who ordered fish, or can be distributed to two people who ordered the same type of meal that is not chicken. For the first case, it's apparent that there are $9$ ways of choosing who gets the two chicken meal, and two ways to order the remaining meals. For the second case, there are $6$ ways to choose who gets the two chicken meals and only $1$ way of ordering the remaining meals. Therefore, the answer is $9 \\cdot (9 \\cdot 2 + 6 \\cdot 1) = 216. ~Arcticturn"
] |
2012-I-4
| 2,012
| 4
|
Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$ .
| 279
|
I
|
[
"When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\\frac{1}{6}$ hours per mile, Butch takes $\\frac{1}{4}$ hours per mile, and Sundance takes $\\frac{2}{5}$ hours per mile: \\[\\frac{a}{6} + \\frac{n-a}{4} = \\frac{n-a}{6} + \\frac{2a}{5} \\rightarrow a = \\frac{5}{19}n.\\] The smallest possible integral value of $n$ is $19$, so we plug in $n = 19$ and $a = 5$ and get $t = \\frac{13}{3}$ hours, or $260$ minutes. So our answer is $19 + 260 = 279. Note that this solution is not rigorous because it is not guaranteed that they will switch properly to form this combination."
] |
2012-I-5
| 2,012
| 5
|
Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained.
| 330
|
I
|
[
"When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instances of such numbers. With the $10$ in place, the seven remaining $1$'s can be distributed in any of the remaining $11$ spaces, so the answer is ${11 \\choose 7} = 330."
] |
2012-I-6
| 2,012
| 6
|
The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$
| 71
|
I
|
[
"Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \\neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$nd root of unity, and thus, by De Moivre's theorem, the imaginary part of $z$ will be of the form $\\sin{\\frac{2k\\pi}{142}} = \\sin{\\frac{k\\pi}{71}},$ where $k \\in \\{1, 2, \\ldots, 70\\}.$ Note that $71$ is prime and $k<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71.$ Thus, $n = 071"
] |
2012-I-9
| 2,012
| 9
|
Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$
| 49
|
I
|
[
"Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that \\[2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) = 2.\\] Then \\begin{align*} 2\\log_{x}(2y) = 2 &\\implies x=2y\\\\ 2\\log_{2x}(4z) = 2 &\\implies 2x=4z\\\\ \\log_{2x^4}(8yz) = 2 &\\implies 4x^8 = 8yz \\end{align*} Solving these equations, we quickly see that $4x^8 = (2y)(4z) = x(2x) \\rightarrow x=2^{-1/6}$ and then $y=z=2^{-1/6 - 1} = 2^{-7/6}.$ Finally, our desired value is $2^{-1/6} \\cdot (2^{-7/6})^5 \\cdot 2^{-7/6} = 2^{-43/6}$ and thus $p+q = 43 + 6 = 049.",
"Notice that $2y\\cdot 4z=8yz$, $2\\log_2(2y)=\\log_2\\left(4y^2\\right)$ and $2\\log_2(4z)=\\log_2\\left(16z^2\\right)$. From this, we see that $8yz$ is the geometric mean of $4y^2$ and $16z^2$. So, for constant $C\\ne 0$: \\[\\frac{\\log 4y^2}{\\log x}=\\frac{\\log 8yz}{\\log 2x^4}=\\frac{\\log 16z^2}{\\log 2x}=C\\] Since $\\log 4y^2,\\log 8yz,\\log 16z^2$ are in an arithmetic progression, so are $\\log x,\\log 2x^4,\\log 2x$. Therefore, $2x^4$ is the geometric mean of $x$ and $2x$ \\[2x^4=\\sqrt{x\\cdot 2x}\\implies 4x^8=2x^2\\implies 2x^6=1\\implies x=2^{-1/6}\\] We can plug $x$ in to any of the two equal fractions aforementioned. So, without loss of generality: \\[\\frac{\\log 4y^2}{\\log x}=\\frac{\\log 16z^2}{\\log 2x}\\implies \\log\\left(4y^2\\right)\\log\\left(2x\\right)=\\log\\left(16z^2\\right)\\log\\left(x\\right)\\] \\[\\implies \\log\\left(4y^2\\right)\\cdot \\frac{5}{6}\\log 2=\\log\\left(16z^2\\right)\\cdot \\frac{-1}{6}\\log 2\\] \\[\\implies 5\\log\\left(4y^2\\right)=-\\log\\left(16z^2\\right)\\implies 5\\log\\left(4y^2\\right)+\\log\\left(16z^2\\right)=0\\] \\[\\implies \\left(4y^2\\right)^5\\cdot 16z^2=1\\implies 16384y^{10}z^2=1\\implies y^{10}z^2=\\frac{1}{16384}\\implies y^5z=\\frac{1}{128}\\] Thus $xy^5z=2^{-\\frac{1}{6}-7}=2^{-\\frac{43}{6}}$ and $43+6=049.",
"Since we are given that $xy^5z = 2^{-p/q}$, we may assume that $x, y$, and $z$ are all powers of two. We shall thus let $x = 2^X$, $y = 2^Y$, and $z = 2^Z$. Let $a = \\log_{2^X}(2^{Y+1})$. From this we get the system of equations: \\[\\] $(1)$\\[a = \\log_{2^X}(2^{Y+1}) \\Rightarrow aX = Y + 1\\] $(2)$\\[a = \\log_{2^{X + 1}}(2^{Z + 2}) \\Rightarrow aX + a = Z + 2\\] $(3)$\\[2a = \\log_{2^{4X + 1}}(2^{Y + Z + 3}) \\Rightarrow 8aX + 2a = Y + Z + 3\\] Plugging equation $(1)$ into equation $(2)$ yields $Y + a = Z + 1$. Plugging equation $(1)$ into equation $(3)$ and simplifying yields $7Y + 2a + 6 = Z + 1$, and substituting $Y + a$ for $Z + 1$ and simplifying yields $Y + 1 = \\frac{-a}{6}$. But $Y + 1 = aX$, so $aX = \\frac{-a}{6}$, so $X = \\frac{-1}{6}$. Knowing this, we may substitute $\\frac{-1}{6}$ for $X$ in equations $(1)$ and $(2)$, yielding $\\frac{-a}{6} = Y + 1$ and $\\frac{5a}{6} = Z + 2$. Thus, we have that $-5(Y + 1) = Z + 2 \\rightarrow 5Y + Z = -7$. We are looking for $xy^5z = 2^{X+ 5Y + Z}$. $X = \\frac{-1}{6}$ and $5Y + Z = -7$, so $xy^5z = 2^{-43/6} = \\frac{1}{2^{43/6}}$. The answer is $43+6=049.",
"By the Mediant theorem, we know that \\[\\frac{\\log (4y^2)}{\\log (x)} = \\frac{\\log (16z^2)}{\\log (2x)} = \\frac{2\\log (8yz)}{\\log (2x^2)}.\\] Substituting into the original equation yields us $\\frac{2\\log (8yz)}{\\log (2x^2)} = \\frac{\\log (8yz)}{\\log (2x^4)} \\Rightarrow 2\\log (2x^4) = \\log (2x^2) \\Rightarrow x=2^{-1/6}.$ For some constant $C\\not= 0,$ Let $2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) = C.$ Then, we obtain the system of equations \\[y=2^{-13C/12}\\] \\[z=2^{-19C/12}\\] \\[8yz=2^{C/3}.\\] Multiplying the first two equations and dividing by the third, we find $C=1.$ Thus, \\[xy^5z=2^{-1/6} \\cdot 2^{-65/12} \\cdot 2^{-19/12}=2^{-43/6} \\Rightarrow p+q=049.\\] ~Kscv ~minor edits by makethan",
"By the Mediant theorem, we know that \\[\\frac{\\log (4y^2)}{\\log (x)} = \\frac{\\log (16z^2)}{\\log (2x)} = \\frac{2\\log (8yz)}{\\log (2x^2)}.\\] Substituting into the original equation yields us $\\frac{2\\log (8yz)}{\\log (2x^2)} = \\frac{\\log (8yz)}{\\log (2x^4)} \\Rightarrow 2\\log (2x^4) = \\log (2x^2) \\Rightarrow x=2^{-1/6}.$ For some constant $C\\not= 0,$ Let $2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) = C.$ Then, we obtain the system of equations \\[y=2^{-13C/12}\\] \\[z=2^{-19C/12}\\] \\[8yz=2^{C/3}.\\] Multiplying the first two equations and dividing by the third, we find $C=1.$ Thus, \\[xy^5z=2^{-1/6} \\cdot 2^{-65/12} \\cdot 2^{-19/12}=2^{-43/6} \\Rightarrow p+q=049.\\] ~Kscv ~minor edits by makethan"
] |
2012-I-10
| 2,012
| 10
|
Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ .
| 170
|
I
|
[
"It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \\not\\equiv s-16 \\pmod{5},$ one term must have all the factors of $5$ and thus must be a multiple of $125.$ Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$ So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently if $s = 500n \\pm 16.$ Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \\cdot 5 - 16 = 2484$ and thus the corresponding element in $\\mathcal{T}$ is $\\frac{2484^2 - 256}{1000} = 6170 \\rightarrow 170.",
"Notice that is $16^2=256$, $1016^2$ ends in $256$. In general, if $x^2$ ends in $256$, $(x+1000)^2=x^2+2000x+1000000$ ends in 256 because $1000|2000x$ and $1000|2000000$. It is clear that we want all numbers whose squares end in $256$ that are less than $1000$. Firstly, we know the number has to end in a $4$ or a $6$. Let's look at the ones ending in $6$. Assume that the second digit of the three digit number is $0$. We find that the last $3$ digits of $\\overline{a06}^2$ is in the form $12a \\cdot 100 + 3 \\cdot 10 + 6$. However, the last two digits need to be a $56$. Thus, similarly trying all numbers from $0$ to $10$, we see that only 1 for the middle digit works. Carrying out the multiplication, we see that the last $3$ digits of $\\overline{a16}^2$ is in the form $(12a + 2) \\cdot 100 + 5 \\cdot 10 + 6$. We want $(12a + 2)\\pmod{10}$ to be equal to $2$. Thus, we see that a is $0$ or $5$. Thus, the numbers that work in this case are $016$ and $516$. Next, let's look at the ones ending in $4$. Carrying out a similar technique as above, we see that the last $3$ digits of $\\overline{a84}^2$ is in the form $((8a+10) \\cdot 100+ 5 \\cdot 10 + 6$. We want $(8a + 10)\\pmod{10}$ to be equal to $2$. We see that only $4$ and $9$ work. Thus, we see that only $484$ and $984$ work. We order these numbers to get $16$, $484$, $516$, $984$... We want the $10th$ number in order which is $2484^2 = 6170.",
"The condition implies $x^2\\equiv 256 \\pmod{1000}$. Rearranging and factoring, \\[(x-16)(x+16)\\equiv 0\\pmod {1000}.\\] This can be expressed with the system of congruences \\[\\begin{cases} (x-16)(x+16)\\equiv 0\\pmod{125} \\\\ (x-16)(x+16)\\equiv 0\\pmod{8} \\end{cases}\\] Observe that $x\\equiv {109} \\pmod {125}$ or $x\\equiv{16}\\pmod {125}$. Similarly, it can be seen that $x\\equiv{0}\\pmod{8}$ or $x\\equiv{4}\\pmod{8}$. By CRT, there are four solutions to this modulo $1000$ (one for each case e.g. $x\\equiv{109}$ and $x\\equiv{0}$ or $x\\equiv{125}$ and $x\\equiv{4}$. These solutions are (working modulo $1000$) \\[\\begin{cases} x=16 \\\\ x=484 \\\\ x= 516 \\\\ x=984 \\end{cases}\\] The tenth solution is $x=2484,$ which gives an answer of $170.",
"An element in S can be represented by $y^2 = 1000a + 256$, where $y^2$ is the element in S. Since the right hand side must be even, we let $y = 2y_1$ and substitute to get $y_1^2 = 250a + 64$. However, the right hand side is still even, so we make the substitution $y_1 = 2y_2$ to get $y_2^2 = 125a/2 + 16$. Because both sides must be an integer, we know that $a = 2a_1$ for some integer $a_1$. Our equation then becomes $y_2^2 = 125a_1 + 16$, and we can simplify no further. Rearranging terms, we get $y_2^2 - 16 = 125a_1$, whence difference of squares gives $(y_2 + 4)(y_2 - 4) = 125a_1$. Note that this equation tells us that one of $y_2 + 4$ and $y_2 - 4$ contains a nonnegative multiple of $125$. Hence, listing out the smallest possible values of $y_2$, we have $y_2 = 4, 121, 129, \\cdots, 621$. The tenth term is $621$, whence $y = 4y_2 = 2484$. The desired result can then be calculated to be $170. - Spacesam",
"From the conditions, we can let every element in $\\mathcal{S}$ be written as $y^2=1000x+256$, where $x$ and $y$ are integers. Since there are no restrictions on $y$, we let $y_1$ be equal to $y+16$ ($y-16$ works as well). Then the $256$ cancels out and we're left with \\[y_1^2+32y_1=1000x\\] which can be factored as \\[y_1(y_1+32)=1000x\\] Since the RHS is even, $y_1$ must be even, so we let $y_1=2y_2$, to get \\[y_2(y_2+16)=250x\\] Again, because the RHS is even, the LHS must be even too, so substituting $y_3=\\frac{1}{2}y_2$ we have \\[y_3(y_3+8)=125\\cdot\\frac{1}{2}x\\] Since the LHS is an integer, the RHS must thus be an integer, so substituting $x=2x_1$ we get \\[y_3(y_3+8)=125x_1\\] Then we can do casework on the values of $y_3$, as only one of $y_3$ and $y_3+8$ can be multiples of $125$ Case 1: $125|y_3$ Since we're trying to find the values of $x_1$, we can let $y_4=\\frac{1}{125}y_3$, to get \\[x_1=y_4(125y_4+8)\\] or \\[x=2y_4(125y_4+8)\\] Case 2: $125|y_3+8$ Similar to Case 1, only the equation is \\[x=2y_4(125y_4-8)\\] In whole, the values of $x$ (i.e. the elements in $\\mathcal{T}$) are of the form \\[x=2k(125k\\pm8)\\] where $k$ is any integer. It can easily be seen that if $k<0$, then $x$ is negative, thus $k\\geq0$. Also, note that when $k=0$, there is only one value, because one of the factors is $0$ ($k$). Thus the $10^{th}$ smallest number in the set $\\mathcal{T}$ is when the $\\pm$ sign takes the minus side and $k=5$, giving $6170$, so the answer is $170",
"From the conditions, we can let every element in $\\mathcal{S}$ be written as $y^2=1000x+256$, where $x$ and $y$ are integers. Since there are no restrictions on $y$, we let $y_1$ be equal to $y+16$ ($y-16$ works as well). Then the $256$ cancels out and we're left with \\[y_1^2+32y_1=1000x\\] which can be factored as \\[y_1(y_1+32)=1000x\\] Since the RHS is even, $y_1$ must be even, so we let $y_1=2y_2$, to get \\[y_2(y_2+16)=250x\\] Again, because the RHS is even, the LHS must be even too, so substituting $y_3=\\frac{1}{2}y_2$ we have \\[y_3(y_3+8)=125\\cdot\\frac{1}{2}x\\] Since the LHS is an integer, the RHS must thus be an integer, so substituting $x=2x_1$ we get \\[y_3(y_3+8)=125x_1\\] Then we can do casework on the values of $y_3$, as only one of $y_3$ and $y_3+8$ can be multiples of $125$ Case 1: $125|y_3$ Since we're trying to find the values of $x_1$, we can let $y_4=\\frac{1}{125}y_3$, to get \\[x_1=y_4(125y_4+8)\\] or \\[x=2y_4(125y_4+8)\\] Case 2: $125|y_3+8$ Similar to Case 1, only the equation is \\[x=2y_4(125y_4-8)\\] In whole, the values of $x$ (i.e. the elements in $\\mathcal{T}$) are of the form \\[x=2k(125k\\pm8)\\] where $k$ is any integer. It can easily be seen that if $k<0$, then $x$ is negative, thus $k\\geq0$. Also, note that when $k=0$, there is only one value, because one of the factors is $0$ ($k$). Thus the $10^{th}$ smallest number in the set $\\mathcal{T}$ is when the $\\pm$ sign takes the minus side and $k=5$, giving $6170$, so the answer is $170"
] |
2012-I-11
| 2,012
| 11
|
A frog begins at $P_0 = (0,0)$ and makes a sequence of jumps according to the following rule: from $P_n = (x_n, y_n),$ the frog jumps to $P_{n+1},$ which may be any of the points $(x_n + 7, y_n + 2),$ $(x_n + 2, y_n + 7),$ $(x_n - 5, y_n - 10),$ or $(x_n - 10, y_n - 5).$ There are $M$ points $(x, y)$ with $|x| + |y| \le 100$ that can be reached by a sequence of such jumps. Find the remainder when $M$ is divided by $1000.$
| 373
|
I
|
[
"First of all, it is easy to see by induction that for any $P(x,y)$ in the frog's jump sequence, $x+y$ will be a multiple of $3$ and $x-y$ will be a multiple of $5.$ The base case $(x,y) = (0,0)$ obviously satisfies the constraints and if $x+y = 3n$ and $x-y = 5m,$ any of the four transformations will sustain this property: \\begin{align*} (x+7)+(y+2) = x+y+9 \\rightarrow 3(n+3) &\\text{ and } (x+7)-(y+2) = x-y+5 \\rightarrow 5(m+1)\\\\ (x+2)+(y+7) = x+y+9 \\rightarrow 3(n+3) &\\text{ and } (x+2)-(y+7) = x-y-5 \\rightarrow 5(m-1)\\\\ (x-5)+(y-10) = x+y-15 \\rightarrow 3(n-5) &\\text{ and } (x-5)-(y-10) = x-y+5 \\rightarrow 5(m+1)\\\\ (x-10)+(y-5) = x+y-15 \\rightarrow 3(n-5) &\\text{ and } (x-10)-(y-5) = x-y-5 \\rightarrow 5(m-1).\\\\ \\end{align*} So we know that any point the frog can reach will satisfy $x+y = 3n$ and $x-y = 5m.$ $\\textbf{Lemma:}$ Any point $(x,y)$ such that there exists 2 integers $m$ and $n$ that satisfy $x+y = 3n$ and $x-y = 5m$ is reachable. $\\textbf{Proof:}$ Denote the total amounts of each specific transformation in the frog's jump sequence to be $a,$ $b,$ $c,$ and $d$ respectively. Then $x=7a+2b-5c-10d$, $y=2a+7b-10c-5d$, $x+y = 9(a+b)-15(c+d) = 3n$, and $x-y = 5(a-b)+5(c-d) = 5m$ together must have integral solutions. But $3(a+b)-5(c+d) = n$ implies $(c+d) \\equiv n \\mod 3$ and thus $(a+b) = \\lfloor{n/3}\\rfloor + 2(c+d).$ Similarly, $(a-b)+(c-d) = m$ implies that $(a-b)$ and $(c-d)$ have the same parity. Now in order for an integral solution to exist, there must always be a way to ensure that the pairs $(a+b)$ and $(a-b)$ and $(c+d)$ and $(c-d)$ have identical parities. The parity of $(a+b)$ is completely dependent on $n,$ so the parities of $(a-b)$ and $(c-d)$ must be chosen to match this value. But the parity of $(c+d)$ can then be adjusted by adding or subtracting $3$ until it is identical to the parity of $(c-d)$ as chosen before, so we conclude that it is always possible to find an integer solution for $(a,b,c,d)$ and thus any point that satisfies $x+y = 3n$ and $x-y = 5m$ can be reached by the frog. To count the number of such points in the region $|x| + |y| \\le 100,$ we first note that any such point will lie on the intersection of one line of the form $y=x-5m$ and another line of the form $y=-x+3n.$ The intersection of two such lines will yield the point $\\left(\\frac{3n+5m}{2},\\frac{3n-5m}{2}\\right),$ which will be integral if and only if $m$ and $n$ have the same parity. Now since $|x| + |y| = |x \\pm y|,$ we find that \\begin{align*} |x + y| = |3n| \\le 100 &\\rightarrow -33 \\le n \\le 33\\\\ |x - y| = |5m| \\le 100 &\\rightarrow -20 \\le m \\le 20. \\end{align*} So there are $34$ possible odd values and $33$ possible even values for $n,$ and $20$ possible odd values and $21$ possible even values for $m.$ Every pair of lines described above will yield a valid accessible point for all pairs of $m$ and $n$ with the same parity, and the number of points $M$ is thus $34 \\cdot 20 + 33 \\cdot 21 = 1373 \\rightarrow 373."
] |
2012-I-12
| 2,012
| 12
|
Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$
| 18
|
I
|
[
"We have $\\angle BCE = \\angle ECD = \\angle DCA = \\tfrac 13 \\cdot 90^\\circ = 30^\\circ$. Drop the altitude from $D$ to $CB$ and call the foot $F$. [asy]import cse5;size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,E0,F; C=origin; B=(10,0); A=(0,5); E0=extension(C,dir(30),A,B); D=extension(C,dir(60),A,B); F=foot(D,C,B); draw(A--B--C--A, black+0.8); draw(C--D--F^^C--E0, gray); dot(\"$A$\",A,N); dot(\"$B$\",B,SE); dot(\"$C$\",C,SW); dot(\"$D$\",D,NE); dot(\"$E$\",E0,2*NE); dot(\"$F$\",F,S); label(\"$8$\",D--E0,2*NE); label(\"$15$\",E0--B,2*NE); label(\"$11a$\",C--B,2*S); label(rotate(60)*\"$8a$\",C--D,2*NW); label(rotate(-90)*\"$4\\sqrt{3}a$\",D--F, E); label(\"$4a$\",C--F,2*S); [/asy] Let $CD = 8a$. Using angle bisector theorem on $\\triangle CDB$, we get $CB = 15a$. Now $CDF$ is a $30$-$60$-$90$ triangle, so $CF = 4a$, $FD = 4a\\sqrt{3}$, and $FB = 11a$. Finally, $\\tan{B} = \\tfrac{DF}{FB}=\\tfrac{4\\sqrt{3}a}{11a} = \\tfrac{4\\sqrt{3}}{11}$. Our final answer is $4 + 3 + 11 = 018.",
"Without loss of generality, set $CB = 1$. Then, by the Angle Bisector Theorem on triangle $DCB$, we have $CD = \\frac{8}{15}$. We apply the Law of Cosines to triangle $DCB$ to get $1 + \\frac{64}{225} - \\frac{8}{15} = BD^{2}$, which we can simplify to get $BD = \\frac{13}{15}$. Now, we have $\\cos \\angle B = \\frac{1 + \\frac{169}{225} - \\frac{64}{225}}{\\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$, so $\\cos \\angle B = \\frac{11}{13}$. In addition, $\\sin \\angle B = \\sqrt{1 - \\frac{121}{169}} = \\frac{4\\sqrt{3}}{13}$, so $\\tan \\angle B = \\frac{4\\sqrt{3}}{11}$. Our final answer is $4+3+11 = 018.",
"(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig) Find values for all angles in terms of $\\angle B$. $\\angle CEB = 150-B$, $\\angle CED = 30+B$, $\\angle CDE = 120-B$, $\\angle CDA = 60+B$, and $\\angle A = 90-B$. Use the law of sines on $\\triangle CED$ and $\\triangle CEB$: In $\\triangle CED$, $\\frac{8}{\\sin 30} = \\frac{CE}{\\sin (120-B)}$. This simplifies to $16 = \\frac{CE}{\\sin (120-B)}$. In $\\triangle CEB$, $\\frac{15}{\\sin 30} = \\frac{CE}{\\sin B}$. This simplifies to $30 = \\frac{CE}{\\sin B}$. Solve for $CE$ and equate them so that you get $16\\sin (120-B) = 30\\sin B$. From this, $\\frac{8}{15} = \\frac{\\sin B}{\\sin (120-B)}$. Use a trig identity on the denominator on the right to obtain: $\\frac{8}{15} = \\frac{\\sin B}{\\sin 120 \\cos B - \\cos 120 \\sin B}$ This simplifies to $\\frac{8}{15} = \\frac{\\sin B}{\\frac{\\sqrt{3}\\cos B}{2} + \\frac{\\sin B}{2}} = \\frac{\\sin B}{\\frac{\\sqrt{3} \\cos B + \\sin B}{2}} = \\frac{2\\sin B}{\\sqrt{3}\\cos B + \\sin B}$ This gives $8\\sqrt{3}\\cos B+8\\sin B=30\\sin B$ Dividing by $\\cos B$, we have ${8\\sqrt{3}}= 22\\tan B$ $\\tan{B} = \\frac{8\\sqrt{3}}{22} = \\frac{4\\sqrt{3}}{11}$. Our final answer is $4 + 3 + 11 = 018.",
"(This solution avoids advanced trigonometry) Let $X$ be the foot of the perpendicular from $D$ to $\\overline{BC}$, and let $Y$ be the foot of the perpendicular from $E$ to $\\overline{BC}$. Now let $EY=x$. Clearly, triangles $EYB$ and $DXB$ are similar with $\\frac{BE}{BD}=\\frac{15}{15+8}=\\frac{15}{23}=\\frac{EY}{DX}$, so $DX=\\frac{23}{15}x$. Since triangles $CDX$ and $CEY$ are 30-60-90 right triangles, we can easily find other lengths in terms of $x$. For example, we see that $CY=x\\sqrt{3}$ and $CX=\\frac{\\frac{23}{15}x}{\\sqrt{3}}=\\frac{23\\sqrt{3}}{45}x$. Therefore $XY=CY-CX=x\\sqrt{3}-\\frac{23\\sqrt{3}}{45}x=\\frac{22\\sqrt{3}}{45}x$. Again using the fact that triangles $EYB$ and $DXB$ are similar, we see that $\\frac{BX}{BY}=\\frac{XY+BY}{BY}=\\frac{XY}{BY}+1=\\frac{23}{15}$, so $BY=\\frac{15}{8}XY=\\frac{15}{8}*\\frac{22\\sqrt{3}}{45}=\\frac{11\\sqrt{3}}{2}$. Thus $\\tan \\angle B = \\frac{x}{\\frac{11\\sqrt{3}}{12}x}=\\frac{4\\sqrt{3}}{11}$, and our answer is $4+3+11=018.",
"(Another solution without trigonometry) Extend $CD$ to point $F$ such that $\\overline{AF} \\parallel \\overline{CB}$. It is then clear that $\\triangle AFD$ is similar to $\\triangle BCD$. Let $AC=p$, $BC=q$. Then $\\tan \\angle B = p/q$. With the Angle Bisector Theorem, we get that $CD=\\frac{8}{15}q$. From 30-60-90 $\\triangle CAF$, we get that $AF=\\frac{1}{\\sqrt{3}}p$ and $FD=FC-CD=\\frac{2}{\\sqrt{3}}p-\\frac{8}{15}q$. From $\\triangle AFD \\sim \\triangle BCD$, we have that $\\frac{FD}{CD}=\\frac{FA}{CB}=\\frac{\\frac{2}{\\sqrt{3}}p-\\frac{8}{15}q}{\\frac{8}{15}q}=\\frac{\\frac{1}{\\sqrt{3}}p}{q}$. Simplifying yields $\\left(\\frac{p}{q}\\right)\\left(\\frac{2\\sqrt{3}}{3}*\\frac{15}{8}-\\frac{\\sqrt{3}}{3}\\right)=1$, and $\\tan \\angle B=\\frac{p}{q}=\\frac{4\\sqrt{3}}{11}$, so our answer is $4+3+11=018.",
"Let $CB = 1$, and let the feet of the altitudes from $D$ and $E$ to $\\overline{CB}$ be $D'$ and $E'$, respectively. Also, let $DE = 8k$ and $EB = 15k$. We see that $BD' = 15k\\cos B$ and $BE' = 23k\\cos B$ by right triangles $\\triangle{BDD'}$ and $\\triangle{BEE'}$. From this we have that $D'E' = 8k\\cos B$. With the same triangles we have $DD' = 23k\\sin B$ and $EE' = 15k\\sin B$. From 30-60-90 triangles $\\triangle{CDD'}$ and $\\triangle{CEE'}$, we see that $CD' = \\frac{23k\\sqrt{3}\\sin B}{3}$ and $CE' = 15k\\sqrt{3}\\sin B$, so $D'E' = \\frac{22k\\sqrt{3}\\sin B}{3}$. From our two values of $D'E'$ we get: $8k\\cos B = \\frac{22k\\sqrt{3}\\sin B}{3}$ $\\frac{\\sin B}{\\cos B} = \\frac{8k}{\\frac{22k\\sqrt{3}}{3}} = \\tan B$ $\\tan B = \\frac{8}{\\frac{22\\sqrt{3}}{3}} = \\frac{24}{22\\sqrt{3}} = \\frac{8\\sqrt{3}}{22} = \\frac{4\\sqrt{3}}{11}$ Our answer is then $4+3+11 = 018.",
"WLOG, let $DE=8$ and $BE=15$. First, by the Law of Sines on $\\triangle CEB$, we find that \\[\\frac{\\sin\\angle B}{CE}=\\frac{\\sin\\angle ECB}{BE}=\\frac{1}{30}\\implies \\sin\\angle B=\\frac{CE}{30}.\\] Now, we will find $CE$. Consider the following diagram: We have constructed equilateral triangle $\\triangle BDP$, and its circumcircle. Since $\\angle DCB=\\angle DPB=60^\\circ$, $C$ lies on $(BDP)$ as well. Let $Q$ be the point diametrically opposite $P$ on $(BCD)$, and let $R$ be the foot of $Q$ on $BD$ (this is the midpoint of $BD$). It is easy to compute that $RQ=\\frac{23}{2\\sqrt3}$ and $ER=\\frac{23}{2}-8=\\frac{7}{2}$. Therefore, by the Pythagorean Theorem, $EQ=\\frac{13}{\\sqrt{3}}$. Now, by Power of a Point, we know that $(DE)(BE)=(EQ)(EC)$, which means that \\[120=\\frac{13EC}{\\sqrt{3}}\\implies EC=\\frac{120\\sqrt{3}}{13}.\\] From before, we know that $\\sin\\angle B=\\frac{EC}{30}\\implies \\sin\\angle B=\\frac{4\\sqrt3}{13}$. It's now easy to compute $\\cos\\angle B$ as well using the Pythagorean identity; we find that $\\cos\\angle B=\\frac{11}{13}$, and thus $\\tan\\angle B=\\frac{4\\sqrt3}{11}$ for an answer of $018. -brainiacmaniac31",
"[email protected], vvsss",
"WLOG, let $DE=8$ and $EB=15$. (this will be redefined later) Define points $A'$, $D'$, $E'$, and $B'$ such that $A'$ is on $AC$, $B'$ is on $CB$ and $D'$ and $E'$ are the intersections of $A'B'$ with $CD$ and $CE$, with $CD' = CE' = 1$, respectively. From cross ratios, we have: \\begin {align*} \\frac{(AE)(DB)}{(AB)(DE)} &= \\frac{(A'E')(D'B')}{(A'B')(D'E')} \\\\ \\frac{(AD+8)(23)}{(AD+23)(8)} & = \\frac{(\\cos(15)+\\sin(15))^2}{(2 \\cos(15))(2 \\sin(15))} \\\\ & \\implies AD = 92/11 \\end {align*} For simplicity, scale everything by $11$, so $AD=92$, $DE=88$ and $EB = 165$. From the ratio lemma, we have: \\begin {align*} \\frac{AC}{CB} &= \\frac{AD \\sin{\\angle BCD}}{DB \\sin{\\angle ACD}} \\\\ \\tan B &= \\frac{92 \\cdot \\sqrt{3}/2}{253 \\cdot 1/2} \\\\ \\tan B &= \\frac{4 \\sqrt{3}}{2}\\\\ &\\implies 018. \\end{align*} ~ boxtheanswer"
] |
2012-I-13
| 2,012
| 13
|
Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$
| 41
|
I
|
[
"Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^\\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$ [asy]import cse5; size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair O = origin; pair A,B,C,Op,Bp,Cp; path c3,c4,c5; c3 = CR(O,3); c4 = CR(O,4); c5 = CR(O,5); draw(c3^^c4^^c5, gray+0.25); A = 5*dir(96.25); Op = rotate(60,A)*O; B = OP(CR(Op,4),c3); Bp = IP(CR(Op,4),c3); C = rotate(-60,A)*B; Cp = rotate(-60,A)*Bp; draw(A--B--C--A, black+0.8); draw(A--Bp--Cp--A, royalblue+0.8); draw(CR(Op,4), heavygreen+0.25); dot(\"$A$\",A,N); dot(\"$X$\",O,E); dot(\"$X'$\",Op,E); dot(\"$C$\",B,SE); dot(\"$C'$\",Bp,NE); dot(\"$B$\",C,2*SW); dot(\"$B'$\",Cp,2*S); [/asy] Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$-$4$-$5$ Pythagorean triple. Now note that $\\angle ABC + \\angle ACB = 120^\\circ$ and $\\angle XCA + \\angle XBA = 90^\\circ,$ so $\\angle XCB+\\angle XBC = 30^\\circ$ and $\\angle BXC = 150^\\circ.$ Applying the law of cosines on triangle $BXC$ yields \\[BC^2 = BX^2+CX^2 - 2 \\cdot BX \\cdot CX \\cdot \\cos150^\\circ = 4^2+3^2-24 \\cdot \\frac{-\\sqrt{3}}{2} = 25+12\\sqrt{3}\\] and thus the area of $ABC$ equals \\[\\frac{\\sqrt{3}}{4}\\cdot BC^2 = 25\\frac{\\sqrt{3}}{4}+9.\\] so our final answer is $3+4+25+9 = 041",
"Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties. We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have two cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur proceed as follows. Using the notation from Solution 1, let $\\angle XAC = \\theta$ so that $\\angle BAX=60-\\theta$. Let $AB=BC=AC=x$. The law of cosines on $\\triangle BAX$ and $\\triangle CAX$ yields \\[\\cos(60-\\theta)=\\frac{x^2+9}{10x}\\quad \\text{ and }\\quad \\cos\\theta=\\frac{x^2+16}{10x}.\\] Solving this system will yield the value of $x$. Since $\\cos\\theta=\\frac{x^2+16}{10x}$ we have that \\[\\sin\\theta=\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}.\\] Substituting these into the equation \\[\\frac{x^2+9}{10x}=\\cos(60-\\theta)=\\tfrac{1}{2}(\\cos\\theta+ \\sqrt{3}\\sin\\theta)\\] we obtain \\[\\frac{x^2+9}{10x}=\\frac{1}{2}\\frac{x^2+16}{10x}+\\frac{\\sqrt{3}}{2}\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}.\\] After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain \\[x^4-50x^2+193=0\\] so that by the quadratic formula $x^2=25\\pm12\\sqrt{3}$. Under the hypothesis that $O$ lies in the interior of triangle $ABC$, $x^2$ must be $25+12\\sqrt{3}$. To see this, note that the other value for $x^2$ is roughly $4.2$ so that $x\\approx 2.05$, but since $AX=5$ and $AX\\leq x$ we have a contradiction. We then obtain the area as in Solution 1. Now, suppose $O$ does not lie in the interior of triangle $ABC$. We then obtain convex quadrilateral $OBAC$ with diagonals $CB$ and $OA$ intersecting at $X$. Here $AX=AB=AC=x$. We may let $\\alpha$ denote the measure of angle $CAX$ so that angle $XAB$ measures $60-\\alpha$. Note that the law of cosines on triangles $CXA$ and $BXA$ yield the same equations as in the first case with $\\theta$ replaced with $\\alpha$. Thus we obtain again $x^2=25\\pm12\\sqrt{3}$. If $x^2=25+12\\sqrt{3}$ then $x\\approx 6.8$, but this is impossible since $AX\\leq 5$ but the shortest possible distance from $A$ to $X$ is the height of equilateral triangle $ABC$ which is $\\approx6.8\\sqrt{3}\\approx5.8$; a contradiction. Hence in this case $x^2=25-12\\sqrt{3}$. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).",
"[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot(\"$C$\", C, dir(0)); dot(\"$A$\", A, dir(90));dot(\"$B$\", B, dir(180)); dot(\"$D$\", D, NE); dot(\"$E$\", E, dir(90));dot(\"$F$\", F, dir(270)); dot(\"$M$\", M, NE); dot(\"$N$\", N, dir(270));dot(\"$L$\", L, NW); dot(\"$X$\", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy] Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\\angle DXE = \\angle EXF = \\angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\\sin(60)$. Similarly, $AB=3\\sin(60)$, and $BC=5\\sin(60)$. So $\\triangle ABC$ is a 3-4-5 right triangle with $\\angle BAC=90$. With some more angle chasing we get \\[\\angle MXC+\\angle LXB = \\angle MAC + \\angle LAB = 180 – \\angle BAC = 90\\] \\[\\angle LXM = 360 – (\\angle MXC + \\angle LXB + \\angle BXC) = 360 –(90+120)=150\\] By Law of Cosines, we have \\[LM^2 = 3^2+4^2-2*3*4\\cos(150)=25+12\\sqrt 3\\] And the area follows. \\[[LMN] = \\frac{25}{4}\\sqrt{3} + 9; 041.\\] By Mathdummy",
"Let $ABC$ be the equilateral triangle with $AB=BC=CA=x.$ Assume the coordinates of the vertices are $A(-\\tfrac{x}{2},0), B(\\tfrac{x}{2},0)$ and $C(0,\\tfrac{\\sqrt{3}}{2}x).$ Let $P(a,b)$ be such that $PA=3, PB=4$ and $PC=5.$ Then \\[\\left (a+\\dfrac{x}{2}\\right )^2 +b^2=3^2,\\] \\[\\left (a-\\dfrac{x}{2}\\right )^2 +b^2=4^2,\\] \\[a^2+\\left (b-\\dfrac{\\sqrt{3}}{2}x\\right )^2 =5^2.\\] Subtraction and addition of the first two equations yield $2ax=-7, 2a^2+\\dfrac{1}{2}x^2+2b^2=25.$ The third equation gives $2a^2+2b^2-2\\sqrt{3}bx+\\dfrac{3}{2}x^2=50.$ Then $x^2-2\\sqrt{3}bx=25.$ We can then solve for $a, b$ in terms of $x$ and have a substitution. We have \\[2\\left (\\dfrac{-7}{2x}\\right )^2 + \\dfrac{1}{2}x^2 + 2\\left (\\dfrac{x^2-25}{2\\sqrt{3}x}\\right )^2 =25.\\] Simplify it we have a quadratic equation for $x^2: x^4-50x^2+193=0.$ So $x^2=25\\pm 12\\sqrt{3}.$ The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2. -JZ",
"[asy] import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12)); pair O=origin; pair Op,Bp,Cp; path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5); var theta=55.75; pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E); filldraw(A--O--B--cycle, rgb(255,255,200)); draw(c3, cyan); draw(c4, green); draw(c5, purple); draw(A--F--C--A--E--C, red+0.8); draw(D--B--C--D--O--C, purple+0.6); draw(A--O^^E--B, cyan+0.6); draw(A--B^^O--F, heavygreen+0.6); draw(A--D); dot(\"$A$\",A,N); dot(\"$O$\",O,dir(A-B)); dot(\"$F$\",F,dir(F-E)); dot(\"$C$\",C,SE); dot(\"$B$\",B,dir(-45)); dot(\"$D$\",D,dir(B-C)); dot(\"$E$\",E,dir(E-F)); label(\"$s$\",A--F,dir(A-E),red); label(\"$s$\",F--C,dir(C-B),red); label(\"$s$\",A--C,dir(C-E),red); label(\"$s$\",A--E,dir(E-C),red); label(\"$s$\",C--E,dir(C),red); label(\"$x$\",B--E,down,royalblue); label(\"$x$\",O--A,dir(B-A),royalblue); label(\"$y$\",A--B,dir(F-A),heavygreen); label(\"$y$\",O--F,dir(C),heavygreen); label(\"$z$\",C--B,dir(E-A),purple); label(\"$z$\",O--C,dir(F),purple); label(\"$m$\",A--D,dir(0),blue+fontsize(9)); markscalefactor=0.03; draw(rightanglemark(C,D,B), gray); MA(\"\\varphi\",A,D,O,0.25,blue); [/asy] We have $x=3$, $y=4$, and $z=5$. Because $AD=m$ is the median of $\\triangle AOB$, by Stewart's Theorem we have \\[m^2=\\frac 12 (x^2+y^2)-\\frac 14 \\cdot z^2\\quad \\Rightarrow \\quad m = \\frac 52.\\] Because $CD$ is the altitude of equilateral triangle $OBC$, we have $CD=\\frac{\\sqrt{3}}2\\cdot z$. Then in $\\triangle ADC$, we have $\\angle ADC=\\varphi+90^\\circ$, so $\\cos(\\angle ADC)=-\\sin\\varphi$, and the Law of Cosines gives\\[s^2=m^2+\\frac 34\\cdot z^2 + mz\\sqrt{3}\\sin\\varphi = 25 \\left(1+\\frac {\\sqrt{3}}{2}\\cdot\\sin\\varphi \\right)\\]To calculate $\\sin\\phi$ we apply the Law of Cosines to $\\triangle ADO$ to get\\[mz\\cos\\varphi = m^2+\\frac 14\\cdot z^2 - x^2 \\quad \\Rightarrow \\quad \\cos\\varphi = \\frac 7{25}\\quad \\Rightarrow \\quad \\sin\\varphi = \\frac {24}{25}.\\] Finally, we get $s^2=25+12\\sqrt{3}$ and and thus the area of $\\triangle ABC$ equals\\[\\frac{\\sqrt{3}}{4}\\cdot s^2 = 25\\frac{\\sqrt{3}}{4}+9.\\] so our final answer is $3+4+25+9 = 041 [email protected], vvsss",
"We note that this question is equivalent to finding the area of an equilateral triangle $ABC$ where there exists a point $P$ inside the triangle where $PA = 3$, $PB=4$, $PC=5$. Now, we reflect $P$ across $AB$, $BC$, and $CA$, call these points $D$, $E$, and $F$, respectively. It is obvious that hexagon $[ADBECF] = 2[ABC]$, so we try to find the area of this hexagon. Note that angle $DAF$ is equal to the sum of angles $DAP$ and $PAF$, which are equal to $2\\angle{BAP}$ and $2\\angle{PAC}$, respectively. Because $\\angle{BAP} + \\angle{PAC} = 60$, we have that $\\angle{DAF} = 120$. By similar reasoning, we know $\\angle{DAF} = \\angle{EBD} = \\angle{FCE} = 120$. Now, split this hexagon into 4 triangles, $DAF$, $EBD$, $FCE$, and $DEF$. It is easy to see that $DAF$, $EBD$, $FCE$ are 30-30-120 triangles with areas $\\dfrac{9\\sqrt{3}}{4}$, $4\\sqrt{3}$, and $\\dfrac{25\\sqrt{3}}{4}$, respectively. We note that $DEF$ is a 3-4-5 right triangle, with side lengths $3\\sqrt{3}$, $4\\sqrt{3}$, and $5\\sqrt{3}$, meaning that $DEF$ has area $18$. Summing, we have $[ADBECF] = 18 + \\dfrac{25\\sqrt{3}}{2}$ Thus, $[ABC] = 9+\\dfrac{25\\sqrt{3}}{4}$, $9+25+3+4 = 41 ~blazing5028",
"We note that this question is equivalent to finding the area of an equilateral triangle $ABC$ where there exists a point $P$ inside the triangle where $PA = 3$, $PB=4$, $PC=5$. Now, we reflect $P$ across $AB$, $BC$, and $CA$, call these points $D$, $E$, and $F$, respectively. It is obvious that hexagon $[ADBECF] = 2[ABC]$, so we try to find the area of this hexagon. Note that angle $DAF$ is equal to the sum of angles $DAP$ and $PAF$, which are equal to $2\\angle{BAP}$ and $2\\angle{PAC}$, respectively. Because $\\angle{BAP} + \\angle{PAC} = 60$, we have that $\\angle{DAF} = 120$. By similar reasoning, we know $\\angle{DAF} = \\angle{EBD} = \\angle{FCE} = 120$. Now, split this hexagon into 4 triangles, $DAF$, $EBD$, $FCE$, and $DEF$. It is easy to see that $DAF$, $EBD$, $FCE$ are 30-30-120 triangles with areas $\\dfrac{9\\sqrt{3}}{4}$, $4\\sqrt{3}$, and $\\dfrac{25\\sqrt{3}}{4}$, respectively. We note that $DEF$ is a 3-4-5 right triangle, with side lengths $3\\sqrt{3}$, $4\\sqrt{3}$, and $5\\sqrt{3}$, meaning that $DEF$ has area $18$. Summing, we have $[ADBECF] = 18 + \\dfrac{25\\sqrt{3}}{2}$ Thus, $[ABC] = 9+\\dfrac{25\\sqrt{3}}{4}$, $9+25+3+4 = 41 ~blazing5028"
] |
2012-I-14
| 2,012
| 14
|
Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$
| 375
|
I
|
[
"By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$. Therefore, $\\frac{(a+b+c)}{3}=0$. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$. Without the loss of generality, let $\\overline{ac}$ be the hypotenuse. The magnitudes of $a$, $b$, and $c$ are just $\\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$. So, $|a|^2=\\frac{4}{9}\\cdot((\\frac{x}{2})^2+y^2)=\\frac{x^2}{9}+\\frac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$. Similarly, $|c|^2=\\frac{4}{9}\\cdot(x^2+(\\frac{y}{2})^2)=\\frac{4x^2}{9}+\\frac{y^2}{9}$. The median from $b$ is just half the hypotenuse because the median of any right triangle is just half the hypotenuse. So, $|b|^2=\\frac{4}{9}\\cdot\\frac{x^2+y^2}{4}=\\frac{x^2}{9}+\\frac{y^2}{9}$. Hence, $|a|^2+|b|^2+|c|^2=\\frac{6x^2+6y^2}{9}=\\frac{2x^2+2y^2}{3}=250$. Therefore, $h^2=x^2+y^2=\\frac{3}{2}\\cdot250=375.",
"Note that by vieta's, the sum of the roots is $0,$ so the centroid (at (a+b+c)/3) of the right triangle is at the origin. This is all that is required by the structure of the polynomial. Let $A,B,C$ be the vertices of the triangle corresponding to roots $a,b,c$ with (WLOG) $\\angle ACB=0.$ Let $O$ be the centroid of the triangle, and let $Z$ be the midpoint of $AB.$ By the property of the centroid as the intersection of the medians, $C,O,$ and $Z$ are collinear in that order. Also, by properties of medians and right triangles $AZ=BZ=CZ$ and $OC=2ZO.$ Then, set $AZ=BZ=CZ=3k$ and $OZ=k$ and $CO=2k.$ Stewart's theorem on ACZ and BCZ and rearranging gets \\[3AO^2=12k^2+AC^2,\\] \\[3BO^2=12k^2+BC^2.\\] Summing these up gives us \\[3(AO^2+BO^2)=24k^2+(AC^2+BC^2)\\] Because $AC^2+BC^2=AB^2=36k^2,$ we have \\[AO^2+BO^2=20k^2.\\] Because $CO=2k,$ \\[AO^2+BO^2+CO^2=24k^2.\\] By the second condition, this must be equal to $250.$ Then, $k^2=\\frac{125}{12},$ and the hypotenuse squared is \\[(6k^2)^2=36k^2=36\\left(\\dfrac{125}{12}\\right)=375.\\] ~BS2012",
"Assume $q$ and $r$ are real, so at least one of $a,$ $b,$ and $c$ must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively. By symmetry, the triangle described by $a,$ $b,$ and $c$ must be isosceles and is thus an isosceles right triangle with hypotenuse $\\overline{bc}.$ Now since $P(z)$ has no $z^2$ term, we must have $a+b+c = a + (x + yi) + (x - yi) = 0$ and thus $a = -2x.$ Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, $a-x=y$ and thus $y=-3x.$ We can then solve for $x$: \\begin{align*} |a|^2 + |b|^2 + |c|^2 &= 250\\\\ |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\\\ 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\\\ x^2 &= \\frac{250}{24} \\end{align*} Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and thus $h^2 = 36x^2 = 36 \\cdot \\frac{250}{24} = 375.",
"Let the roots $a$, $b$, and $c$ each be represented by complex numbers $m + ni$, $p + qi$, and $r + ti$. By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get: $m + p + r = 0$ $n + q + t = 0$ And, we know that the sum of the squares of the magnitudes of each is 250, so $m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$ Given the complex plane, we set each of these complex numbers to points: $(m, n)$, $(p, q)$, $(r, t)$. WLOG let $(r, t)$ be the vertex opposite the hypotenuse. If the three points form a right triangle, the vectors from $(r, t)$ to $(m, m)$ and $(p, q)$'s dot product is 0. $mp + r^2 - r(m + p) + nq + t^2 - t(n + q) = 0$ Substituting $m + p + r = 0$ and likewise, simplifying: $mp + 2r^2 + nq + 2t^2 = 0$ Rearranging we get: $r^2 + t^2 = -\\frac{mp + nq}{2}$ The answer is the distance from $(m, n)$ to $(p, q)$ = $m^2 + n^2 + p^2 + q^2 - 2(mp + nq)$. Substituting the equation equal to 250, $= 250 - r^2 - t^2 - 2(mp + nq)$ $= 250 + \\frac{mp + nq}{2} - 2(mp + nq)$ $= 250 - \\frac{3}{2} \\cdot (mp + nq)$ Taking our original equations summing to 0, and squaring each we get: $n + q = -t$; $m + p = -r$ $n^2 + 2nq + q^2 = t^2$; $m^2 + 2mp + p^2 = r^2$ Adding, we get: $m^2 + n^2 + p^2 + q^2 + 2(mp + nq) = r^2 + t^2$ Substituting again we obtain: $250 - r^2 - t^2 + 2(mp + nq) = r^2 + t^2$ $2(r^2 + t^2) = 250 + 2(mp + nq)$ $r^2 + t^2 = 125 + (mp + nq)$ Substituting the equivalence of $r^2 + t^2$: $-\\frac{mp + nq}{2} = 125 + (mp + nq)$ Solving for $mp + nq$, we find it equal to $-\\frac{250}{3}$. Substituting this value into our answer expression, we get: $250 - \\frac{3}{2} \\cdot (-\\frac{250}{3})$, Answer = $375.",
"Let the roots $a$, $b$, and $c$ each be represented by complex numbers $m + ni$, $p + qi$, and $r + ti$. By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get: $m + p + r = 0$ $n + q + t = 0$ And, we know that the sum of the squares of the magnitudes of each is 250, so $m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$ Given the complex plane, we set each of these complex numbers to points: $(m, n)$, $(p, q)$, $(r, t)$. WLOG let $(r, t)$ be the vertex opposite the hypotenuse. If the three points form a right triangle, the vectors from $(r, t)$ to $(m, m)$ and $(p, q)$'s dot product is 0. $mp + r^2 - r(m + p) + nq + t^2 - t(n + q) = 0$ Substituting $m + p + r = 0$ and likewise, simplifying: $mp + 2r^2 + nq + 2t^2 = 0$ Rearranging we get: $r^2 + t^2 = -\\frac{mp + nq}{2}$ The answer is the distance from $(m, n)$ to $(p, q)$ = $m^2 + n^2 + p^2 + q^2 - 2(mp + nq)$. Substituting the equation equal to 250, $= 250 - r^2 - t^2 - 2(mp + nq)$ $= 250 + \\frac{mp + nq}{2} - 2(mp + nq)$ $= 250 - \\frac{3}{2} \\cdot (mp + nq)$ Taking our original equations summing to 0, and squaring each we get: $n + q = -t$; $m + p = -r$ $n^2 + 2nq + q^2 = t^2$; $m^2 + 2mp + p^2 = r^2$ Adding, we get: $m^2 + n^2 + p^2 + q^2 + 2(mp + nq) = r^2 + t^2$ Substituting again we obtain: $250 - r^2 - t^2 + 2(mp + nq) = r^2 + t^2$ $2(r^2 + t^2) = 250 + 2(mp + nq)$ $r^2 + t^2 = 125 + (mp + nq)$ Substituting the equivalence of $r^2 + t^2$: $-\\frac{mp + nq}{2} = 125 + (mp + nq)$ Solving for $mp + nq$, we find it equal to $-\\frac{250}{3}$. Substituting this value into our answer expression, we get: $250 - \\frac{3}{2} \\cdot (-\\frac{250}{3})$, Answer = $375.",
"As noted in the previous solutions, $a+b+c = 0$. Let $a = a_1+a_2 i$, $b = b_1+b_2 i$, $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$. Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem. Thus, we have \\[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\\] \\[= 2 \\left( \\sum_{} ( {a_1}^2 + {a_2}^2 ) \\right) - 2 \\left( \\sum_{cyc} a_1 b_1 + \\sum_{cyc} a_2 b_2 \\right)\\] \\[= 500 - \\left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \\sum_{cyc} ( {a_1}^2 + {a_2}^2 ) \\right)\\] \\[= 500 - (0^2 + 0^2 - 250)\\] so $h^2 = 375 and we may conclude. ~ rzlng",
"As noted in the previous solutions, $a+b+c = 0$. Let $a = a_1+a_2 i$, $b = b_1+b_2 i$, $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$. Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem. Thus, we have \\[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\\] \\[= 2 \\left( \\sum_{} ( {a_1}^2 + {a_2}^2 ) \\right) - 2 \\left( \\sum_{cyc} a_1 b_1 + \\sum_{cyc} a_2 b_2 \\right)\\] \\[= 500 - \\left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \\sum_{cyc} ( {a_1}^2 + {a_2}^2 ) \\right)\\] \\[= 500 - (0^2 + 0^2 - 250)\\] so $h^2 = 375 and we may conclude. ~ rzlng",
"First, note that the roots of this cubic will be $a, b$ and $-(a+b)$ due to Vieta's, which means that the sum of the roots are 0. Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works Set one of the roots as $r$, where $r$ is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as $\\frac{r}{2}+ni$ and $\\frac{r}{2}-ni$, making the roots add up to 0. Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find $n$ out. We use the fact that it is an isosceles right triangle to find that $\\frac{3r}{2}=n$, which means that the 2 other roots are now $\\frac{r}{2}+\\frac{3r}{2}i$ and $\\frac{r}{2}-\\frac{3r}{2}i$ Now we use the fact that $|a|^2+|b|^2+|c|^2=250$. Clearly one of these is $r$ away from the origin, so that gets $r^2$, and then we get $2*\\frac{r}{2}^2+\\frac{3r}{2}^2$ which gets us $5r^2$, getting $r^2+5r^2=250$, so $r=\\sqrt{\\frac{250}{6}}$. So the final answer comes out to \\[(\\frac{250}{6}*\\frac{9}{4}*2)^2=375\\] -dragoon P.S.: The main propose, saying that one root is on the real axis and the right triangle is a right isosceles triangle is not actually only a wishful thinking that came out by luck, but actually is something that must be true due to the complex conjugate root theorem. -SuperDolphin - in this specific problem this logic is incorrect, as the coefficients of the polynomial can be complex. If a question specified real roots, then this would be appropriate. In this case, it is just wishful thinking as stated above",
"First, note that the roots of this cubic will be $a, b$ and $-(a+b)$ due to Vieta's, which means that the sum of the roots are 0. Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works Set one of the roots as $r$, where $r$ is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as $\\frac{r}{2}+ni$ and $\\frac{r}{2}-ni$, making the roots add up to 0. Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find $n$ out. We use the fact that it is an isosceles right triangle to find that $\\frac{3r}{2}=n$, which means that the 2 other roots are now $\\frac{r}{2}+\\frac{3r}{2}i$ and $\\frac{r}{2}-\\frac{3r}{2}i$ Now we use the fact that $|a|^2+|b|^2+|c|^2=250$. Clearly one of these is $r$ away from the origin, so that gets $r^2$, and then we get $2*\\frac{r}{2}^2+\\frac{3r}{2}^2$ which gets us $5r^2$, getting $r^2+5r^2=250$, so $r=\\sqrt{\\frac{250}{6}}$. So the final answer comes out to \\[(\\frac{250}{6}*\\frac{9}{4}*2)^2=375\\] -dragoon P.S.: The main propose, saying that one root is on the real axis and the right triangle is a right isosceles triangle is not actually only a wishful thinking that came out by luck, but actually is something that must be true due to the complex conjugate root theorem. -SuperDolphin - in this specific problem this logic is incorrect, as the coefficients of the polynomial can be complex. If a question specified real roots, then this would be appropriate. In this case, it is just wishful thinking as stated above",
"As shown in the other solutions, $a+b+c = 0$. Without loss of generality, let $b$ be the complex number opposite the hypotenuse. Note that there is an isomorphism between $\\mathbb{C}$ under $+$ and $\\mathbb{R}^2$ under $+$. Let $\\Vec{a}$, $\\Vec{b}$, and $\\Vec{c}$ be the corresponding vectors to $a$, $b$, and $c$. Thus $\\Vec{a} + \\Vec{b} + \\Vec{c} = \\Vec{0}$ $\\Rightarrow 0 = \\Vec{0}\\cdot \\Vec{0} = (\\Vec{a} + \\Vec{b} + \\Vec{c})\\cdot (\\Vec{a} + \\Vec{b} + \\Vec{c}) = \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} + 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c})$ Now $|a|^2 + |b|^2 + |c|^2 = 250$ implies that $\\lVert \\Vec{a}\\rVert^2 + \\lVert \\Vec{b}\\rVert^2 + \\lVert \\Vec{c}\\rVert^2 = 250$ $\\Rightarrow \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} = \\lVert \\Vec{a}\\rVert^2 + \\lVert \\Vec{b}\\rVert^2 + \\lVert \\Vec{c}\\rVert^2 = 250$ Also note that because there is a right angle at $b$, $\\Vec{a} - \\Vec{b}$ and $\\Vec{c} - \\Vec{b}$ are perpendicular. $\\Rightarrow (\\Vec{a} - \\Vec{b})\\cdot (\\Vec{c} - \\Vec{b}) = 0$ $\\Rightarrow 0 = (\\Vec{a} - \\Vec{b})\\cdot (\\Vec{c} - \\Vec{b}) = \\Vec{a}\\cdot \\Vec{c} + \\Vec{b} \\cdot \\Vec{b} - \\Vec{a} \\cdot \\Vec{b} - \\Vec{b} \\cdot \\Vec{c}$ Note that $h^2 = |a-c|^2$ $\\Rightarrow h^2 = \\lVert \\Vec{a} - \\Vec{c} \\rVert^2 = (\\Vec{a} - \\Vec{c})\\cdot (\\Vec{a} - \\Vec{c}) = \\Vec{a} \\cdot \\Vec{a} + \\Vec{c}\\cdot \\Vec{c} - \\Vec{a}\\cdot \\Vec{c} - \\Vec{a}\\cdot \\Vec{c} = \\Vec{a} \\cdot \\Vec{a} + \\Vec{c}\\cdot \\Vec{c} - 2\\Vec{a}\\cdot \\Vec{c}$. $\\Rightarrow 0 = \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} + 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c}) = 250 + 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c})$ $\\Rightarrow -250 = 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c})$ $\\Rightarrow -125 = \\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c}$ $\\Rightarrow -125 = -125 + 0 = (\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c}) + (\\Vec{a}\\cdot \\Vec{c} + \\Vec{b} \\cdot \\Vec{b} - \\Vec{a} \\cdot \\Vec{b} - \\Vec{b} \\cdot \\Vec{c}) = 2\\Vec{a}\\cdot \\Vec{c} + \\Vec{b} \\cdot \\Vec{b}$ $\\Rightarrow 125 = - \\Vec{b} \\cdot \\Vec{b} - 2\\Vec{a}\\cdot \\Vec{c}$ $\\Rightarrow 375 = 250 + 125 = \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} - \\Vec{b} \\cdot \\Vec{b} - 2\\Vec{a}\\cdot \\Vec{c} = \\Vec{a}\\cdot \\Vec{a} + \\Vec{c}\\cdot \\Vec{c} - 2\\Vec{a}\\cdot \\Vec{c} = h^2$ $\\Rightarrow h^2 = 375",
"As shown in the other solutions, $a+b+c = 0$. Without loss of generality, let $b$ be the complex number opposite the hypotenuse. Note that there is an isomorphism between $\\mathbb{C}$ under $+$ and $\\mathbb{R}^2$ under $+$. Let $\\Vec{a}$, $\\Vec{b}$, and $\\Vec{c}$ be the corresponding vectors to $a$, $b$, and $c$. Thus $\\Vec{a} + \\Vec{b} + \\Vec{c} = \\Vec{0}$ $\\Rightarrow 0 = \\Vec{0}\\cdot \\Vec{0} = (\\Vec{a} + \\Vec{b} + \\Vec{c})\\cdot (\\Vec{a} + \\Vec{b} + \\Vec{c}) = \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} + 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c})$ Now $|a|^2 + |b|^2 + |c|^2 = 250$ implies that $\\lVert \\Vec{a}\\rVert^2 + \\lVert \\Vec{b}\\rVert^2 + \\lVert \\Vec{c}\\rVert^2 = 250$ $\\Rightarrow \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} = \\lVert \\Vec{a}\\rVert^2 + \\lVert \\Vec{b}\\rVert^2 + \\lVert \\Vec{c}\\rVert^2 = 250$ Also note that because there is a right angle at $b$, $\\Vec{a} - \\Vec{b}$ and $\\Vec{c} - \\Vec{b}$ are perpendicular. $\\Rightarrow (\\Vec{a} - \\Vec{b})\\cdot (\\Vec{c} - \\Vec{b}) = 0$ $\\Rightarrow 0 = (\\Vec{a} - \\Vec{b})\\cdot (\\Vec{c} - \\Vec{b}) = \\Vec{a}\\cdot \\Vec{c} + \\Vec{b} \\cdot \\Vec{b} - \\Vec{a} \\cdot \\Vec{b} - \\Vec{b} \\cdot \\Vec{c}$ Note that $h^2 = |a-c|^2$ $\\Rightarrow h^2 = \\lVert \\Vec{a} - \\Vec{c} \\rVert^2 = (\\Vec{a} - \\Vec{c})\\cdot (\\Vec{a} - \\Vec{c}) = \\Vec{a} \\cdot \\Vec{a} + \\Vec{c}\\cdot \\Vec{c} - \\Vec{a}\\cdot \\Vec{c} - \\Vec{a}\\cdot \\Vec{c} = \\Vec{a} \\cdot \\Vec{a} + \\Vec{c}\\cdot \\Vec{c} - 2\\Vec{a}\\cdot \\Vec{c}$. $\\Rightarrow 0 = \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} + 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c}) = 250 + 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c})$ $\\Rightarrow -250 = 2(\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c})$ $\\Rightarrow -125 = \\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c}$ $\\Rightarrow -125 = -125 + 0 = (\\Vec{a}\\cdot \\Vec{b} + \\Vec{a}\\cdot \\Vec{c} + \\Vec{b}\\cdot \\Vec{c}) + (\\Vec{a}\\cdot \\Vec{c} + \\Vec{b} \\cdot \\Vec{b} - \\Vec{a} \\cdot \\Vec{b} - \\Vec{b} \\cdot \\Vec{c}) = 2\\Vec{a}\\cdot \\Vec{c} + \\Vec{b} \\cdot \\Vec{b}$ $\\Rightarrow 125 = - \\Vec{b} \\cdot \\Vec{b} - 2\\Vec{a}\\cdot \\Vec{c}$ $\\Rightarrow 375 = 250 + 125 = \\Vec{a}\\cdot \\Vec{a} + \\Vec{b}\\cdot \\Vec{b} + \\Vec{c}\\cdot \\Vec{c} - \\Vec{b} \\cdot \\Vec{b} - 2\\Vec{a}\\cdot \\Vec{c} = \\Vec{a}\\cdot \\Vec{a} + \\Vec{c}\\cdot \\Vec{c} - 2\\Vec{a}\\cdot \\Vec{c} = h^2$ $\\Rightarrow h^2 = 375",
"Note that the roots of the polynomial $(a,b,c)$ must sum to $0$ due to the $z^2$ coefficient equaling $0$ because of Vieta's Formulas. This tells us that $a+b+c = 0$ and $\\overline{a} + \\overline{b} + \\overline{c} = 0$ so $(a+b+c)(\\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\\overline{b} + \\overline{a}b + b\\overline{c} + \\overline{b}c + c\\overline{a} + \\overline{c}a = 250 + a\\overline{b} + \\overline{a}b + b\\overline{c} + \\overline{b}c + c\\overline{a} + \\overline{c}a = 0$ so $a\\overline{b} + \\overline{a}b + b\\overline{c} + \\overline{b}c + c\\overline{a} + \\overline{c}a = -250.$ Note that terms similar to $\\overline{a}b + \\overline{b}a$ appear in $|a-b|^2$ so we decide to sum $|a-b|^2 + |b-c|^2 + |c-a|^2$ out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the $|c-a|^2$ side is the hypotenuse then our sum is equal to $2|c-a|^2 = 2h^2.$ \\[2|a|^2 + 2|b|^2 + 2|c|^2 -a\\overline{b} - \\overline{a}b - b\\overline{c} - \\overline{b}c - c\\overline{a} - \\overline{c}a = 500 + 250 = 2h^2\\] As a result we know that $h^2 = 375 ~SailS",
"Note that the roots of the polynomial $(a,b,c)$ must sum to $0$ due to the $z^2$ coefficient equaling $0$ because of Vieta's Formulas. This tells us that $a+b+c = 0$ and $\\overline{a} + \\overline{b} + \\overline{c} = 0$ so $(a+b+c)(\\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\\overline{b} + \\overline{a}b + b\\overline{c} + \\overline{b}c + c\\overline{a} + \\overline{c}a = 250 + a\\overline{b} + \\overline{a}b + b\\overline{c} + \\overline{b}c + c\\overline{a} + \\overline{c}a = 0$ so $a\\overline{b} + \\overline{a}b + b\\overline{c} + \\overline{b}c + c\\overline{a} + \\overline{c}a = -250.$ Note that terms similar to $\\overline{a}b + \\overline{b}a$ appear in $|a-b|^2$ so we decide to sum $|a-b|^2 + |b-c|^2 + |c-a|^2$ out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the $|c-a|^2$ side is the hypotenuse then our sum is equal to $2|c-a|^2 = 2h^2.$ \\[2|a|^2 + 2|b|^2 + 2|c|^2 -a\\overline{b} - \\overline{a}b - b\\overline{c} - \\overline{b}c - c\\overline{a} - \\overline{c}a = 500 + 250 = 2h^2\\] As a result we know that $h^2 = 375 ~SailS"
] |
2012-II-1
| 2,012
| 1
|
Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$ .
| 34
|
II
|
[
"",
"Solving for $m$ gives us $m = \\frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \\equiv 503 \\mod 5 \\longrightarrow n \\equiv 1 \\mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\\frac{166-1}{5}+1 = 34",
"Dividing by $4$ gives us $5m + 3n = 503$. Thus, we have $503-5m\\equiv 0 \\pmod 3$ since $n$ is an integer. Rearranging then gives $m\\equiv 1\\pmod 3.$ Since $503-5m>0,$ we know that $m<503/5.$ Because $m$ is an integer, we can rewrite this as $m\\le 100.$ Therefore, $m$ ranges from \\[0\\cdot 3+1\\quad \\text{to} \\quad 33\\cdot 3+1,\\] giving $034 values. ~vaporwave",
"Because the x-intercept of the equation is $\\frac{2012}{20}$, and the y-intercept is $\\frac{2012}{12}$, the slope is $\\frac{\\frac{-2012}{12}}{\\frac{2012}{20}} = \\frac{-5}{3}$. Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=34 solutions.",
"Note that a positive integer is divisible by 20 if it ends in 0. Notice that if a multiple of 12 end in 2, 12 must be multiplied by an integer that ends in 1 or 6. So let's start checking because 2012 ends in 2, same as 12. When $n=1$, $m=100$. When $n=6$, $m=97$. What is happening? Why doesn't, say, as values of $n$, do $1, 11, ...$ work, while $6, 16, 26$ work for $m$ to be an integer (as seen in 2022 CEMC Cayley #21 (https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleyContest.pdf, https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleySolution.pdf)?) Notice that we can break $1, 6, ...$ down into $0, 5, 10, ...$. So $20m=2000-12k$, where $k$ is a member of the set ${0, 5, 10, 15...}$. For a number to be divisible by 20, it must be divisible 10, and the quotient (that number divided by 10) must be divisible by 2. This means that the number must end in 0, and its tens digit must be even. So notice that the tens digit cycle in: $0, 4, 8, 2, 6$. Each of this is even (notice that when $k=25$, $2000-12k=1700$, it cycles again). So we know that all values of $n$, which end in 1 or 6, that make $2012-12n\\geq20$ works. So values of $n$ that belong to ${1, 6, ..., 166}$ work. Clearly, that is 34 integer values of $n$. Therefore, the answer is $034. ~hastapasta",
"Note that a positive integer is divisible by 20 if it ends in 0. Notice that if a multiple of 12 end in 2, 12 must be multiplied by an integer that ends in 1 or 6. So let's start checking because 2012 ends in 2, same as 12. When $n=1$, $m=100$. When $n=6$, $m=97$. What is happening? Why doesn't, say, as values of $n$, do $1, 11, ...$ work, while $6, 16, 26$ work for $m$ to be an integer (as seen in 2022 CEMC Cayley #21 (https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleyContest.pdf, https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleySolution.pdf)?) Notice that we can break $1, 6, ...$ down into $0, 5, 10, ...$. So $20m=2000-12k$, where $k$ is a member of the set ${0, 5, 10, 15...}$. For a number to be divisible by 20, it must be divisible 10, and the quotient (that number divided by 10) must be divisible by 2. This means that the number must end in 0, and its tens digit must be even. So notice that the tens digit cycle in: $0, 4, 8, 2, 6$. Each of this is even (notice that when $k=25$, $2000-12k=1700$, it cycles again). So we know that all values of $n$, which end in 1 or 6, that make $2012-12n\\geq20$ works. So values of $n$ that belong to ${1, 6, ..., 166}$ work. Clearly, that is 34 integer values of $n$. Therefore, the answer is $034. ~hastapasta"
] |
2012-II-2
| 2,012
| 2
|
Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$ , $b_1=99$ , and $a_{15}=b_{11}$ . Find $a_9$ .
| 363
|
II
|
[
"Call the common ratio $r.$ Now since the $n$th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \\cdot r^{14} = b_1 \\cdot r^{10} \\implies r^4 = \\frac{99}{27} = \\frac{11}{3}.$ But $a_9$ equals $a_1 \\cdot r^8 = a_1 \\cdot (r^4)^2=27\\cdot {\\left(\\frac{11}{3}\\right)}^2=27\\cdot \\frac{121} 9=363."
] |
2012-II-3
| 2,012
| 3
|
At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements.
| 88
|
II
|
[
"There are two cases: Case 1: One man and one woman is chosen from each department. Case 2: Two men are chosen from one department, two women are chosen from another department, and one man and one woman are chosen from the third department. For the first case, in each department there are ${{2}\\choose{1}} \\times {{2}\\choose{1}} = 4$ ways to choose one man and one woman. Thus there are $4^3 = 64$ total possibilities conforming to case 1. For the second case, there is only ${{2}\\choose{2}} = 1$ way to choose two professors of the same gender from a department, and again there are $4$ ways to choose one man and one woman. Thus there are $1 \\cdot 1 \\cdot 4 = 4$ ways to choose two men from one department, two women from another department, and one man and one woman from the third department. However, there are $3! = 6$ different department orders, so the total number of possibilities conforming to case 2 is $4 \\cdot 6 = 24$. Summing these two values yields the final answer: $64 + 24 = 088.",
"Use generating functions. For each department, there is 1 way to pick 2 males, 4 ways to pick one of each, and 1 way to pick 2 females. Since there are three departments in total, and we wish for three males and three females, the answer will be equal to the coefficient of $x^3y^3$ in the expansion of $(x^2+4xy+y^2)^3$. The requested coefficient is $088 ~sigma"
] |
2012-II-4
| 2,012
| 4
|
Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$ , where $p$ , $q$ , and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$ .
| 61
|
II
|
[
"[asy]draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label(\"$D$\", (1.2,0),dir(-90)); dot((6,1.4)); dot((1.2,0)); label(\"$a$\", (0.6,0),dir(-90)); label(\"$b$\", (3.6,0),dir(-90)); label(\"$c$\", (6,0.7),dir(0));[/asy] Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second. Observe that $\\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$, and $\\dfrac{b+c}{6.2}=1$ or $b+c=6.2$. Subtracting the second equation from the first gives $2a=2.4$, or $a=1.2$. Now, let us solve $b$ and $c$. Note that $\\dfrac{\\sqrt{b^2+c^2}}{5}=1$, or $b^2+c^2=25$. We also have $b+c=6.2$. We have a system of equations: \\[\\left\\{\\begin{array}{l}b+c=6.2\\\\ b^2+c^2=25\\end{array}\\right.\\] Squaring the first equation gives $b^2+2bc+c^2=38.44$, and subtracting the second from this gives $2bc=13.44$. Now subtracting this from $b^2+c^2=25$ gives $b^2-2bc+c^2=(b-c)^2=11.56$, or $b-c=3.4$. Now we have the following two equations: \\[\\left\\{\\begin{array}{l}b+c=6.2\\\\ b-c=3.4\\end{array}\\right.\\] Adding the equations and dividing by two gives $b=4.8$, and it follows that $c=1.4$. The ratios we desire are therefore $1.4:6:4.8=7:30:24$, and our answer is $7+30+24=061 and therefore violating our restriction.",
"[asy]draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label(\"$D$\", (1.2,0),dir(-90)); dot((6,1.4)); dot((1.2,0)); label(\"$a$\", (0.6,0),dir(-90)); label(\"$b$\", (3.6,0),dir(-90)); label(\"$c$\", (6,0.7),dir(0));[/asy] Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG the time taken is $1$ second. Observe that $\\dfrac{2a+b+c}{8.6}=1$ or $2a+b+c=8.6$, and $\\dfrac{b+c}{6.2}=1$ or $b+c=6.2$. Subtracting the second equation from the first gives $2a=2.4$, or $a=1.2$. Now, let us solve $b$ and $c$. Note that $\\dfrac{\\sqrt{b^2+c^2}}{5}=1$, or $b^2+c^2=25$. We also have $b+c=6.2$. We have a system of equations: \\[\\left\\{\\begin{array}{l}b+c=6.2\\\\ b^2+c^2=25\\end{array}\\right.\\] Squaring the first equation gives $b^2+2bc+c^2=38.44$, and subtracting the second from this gives $2bc=13.44$. Now subtracting this from $b^2+c^2=25$ gives $b^2-2bc+c^2=(b-c)^2=11.56$, or $b-c=3.4$. Now we have the following two equations: \\[\\left\\{\\begin{array}{l}b+c=6.2\\\\ b-c=3.4\\end{array}\\right.\\] Adding the equations and dividing by two gives $b=4.8$, and it follows that $c=1.4$. The ratios we desire are therefore $1.4:6:4.8=7:30:24$, and our answer is $7+30+24=061 and therefore violating our restriction.",
"Let P, Q, and R be the east-west distance of the field, the north-south distance, and the distance from the southeast corner to point D, respectively. Ana's distance to point D = $P + Q + (P - R) = 2P + Q - R$ Bob's distance to point D = $Q + R$ Cao's distance to point D = $\\sqrt{Q^2 + R^2}$ Since they arrive at the same time, their distance/speed ratios are equal, so: \\[\\frac{2P + Q - R}{8.6} = \\frac{Q + R}{6.2} = \\frac{\\sqrt{Q^2 + R^2}}{5}\\] \\[\\frac{2P + Q - R}{43} = \\frac{Q + R}{31} = \\frac{\\sqrt{Q^2 + R^2}}{25}\\] Looking at the last two parts of the equation: \\[\\frac{Q + R}{31} = \\frac{\\sqrt{Q^2 + R^2}}{25}\\] \\[25 (Q + R) = 31 \\sqrt{Q^2 + R^2}\\] \\[625 Q^2 + 1250 QR + 625 R^2 = 961 Q^2 + 961 R^2\\] \\[336 Q^2 - 1250 QR + 336 R^2 = 0\\] \\[168 (\\frac{Q}{R})^2 - 625 \\frac{Q}{R}+ 168 = 0\\] \\[\\frac{Q}{R} = \\frac{625 \\pm \\sqrt{625^2 - 4 \\cdot 168^2}}{336}\\] \\[\\frac{Q}{R} = 24/7\\;or\\;7/24\\] Looking at the first two parts of the equation above: \\[\\frac{2 P + Q - R}{43} = \\frac{Q + R}{31}\\] \\[62 P + 31 Q - 31 R = 43 Q + 43 R\\] \\[P = \\frac{6}{31}Q + \\frac{37}{31} R\\] If $\\frac{Q}{R} = \\frac{24}{7}$: \\[R = \\frac{7}{24} Q\\] \\[P = \\frac{6}{31} Q + \\frac{37}{31} \\cdot \\frac{7}{24} Q = \\frac{13}{24} Q\\] However, this makes P < Q, but we are given that P > Q. Therefore, $\\frac{Q}{R} = \\frac{7}{24}$, and: \\[R = \\frac{24}{7} Q\\] \\[P = \\frac{6}{31} Q + \\frac{37}{31} \\cdot \\frac{24}{7} Q = \\frac{30}{7} Q\\] \\[P : Q : R = 30 : 7 : 24\\] The solution is $P + Q + R = 061.",
"Let P, Q, and R be the east-west distance of the field, the north-south distance, and the distance from the southeast corner to point D, respectively. Ana's distance to point D = $P + Q + (P - R) = 2P + Q - R$ Bob's distance to point D = $Q + R$ Cao's distance to point D = $\\sqrt{Q^2 + R^2}$ Since they arrive at the same time, their distance/speed ratios are equal, so: \\[\\frac{2P + Q - R}{8.6} = \\frac{Q + R}{6.2} = \\frac{\\sqrt{Q^2 + R^2}}{5}\\] \\[\\frac{2P + Q - R}{43} = \\frac{Q + R}{31} = \\frac{\\sqrt{Q^2 + R^2}}{25}\\] Looking at the last two parts of the equation: \\[\\frac{Q + R}{31} = \\frac{\\sqrt{Q^2 + R^2}}{25}\\] \\[25 (Q + R) = 31 \\sqrt{Q^2 + R^2}\\] \\[625 Q^2 + 1250 QR + 625 R^2 = 961 Q^2 + 961 R^2\\] \\[336 Q^2 - 1250 QR + 336 R^2 = 0\\] \\[168 (\\frac{Q}{R})^2 - 625 \\frac{Q}{R}+ 168 = 0\\] \\[\\frac{Q}{R} = \\frac{625 \\pm \\sqrt{625^2 - 4 \\cdot 168^2}}{336}\\] \\[\\frac{Q}{R} = 24/7\\;or\\;7/24\\] Looking at the first two parts of the equation above: \\[\\frac{2 P + Q - R}{43} = \\frac{Q + R}{31}\\] \\[62 P + 31 Q - 31 R = 43 Q + 43 R\\] \\[P = \\frac{6}{31}Q + \\frac{37}{31} R\\] If $\\frac{Q}{R} = \\frac{24}{7}$: \\[R = \\frac{7}{24} Q\\] \\[P = \\frac{6}{31} Q + \\frac{37}{31} \\cdot \\frac{7}{24} Q = \\frac{13}{24} Q\\] However, this makes P < Q, but we are given that P > Q. Therefore, $\\frac{Q}{R} = \\frac{7}{24}$, and: \\[R = \\frac{24}{7} Q\\] \\[P = \\frac{6}{31} Q + \\frac{37}{31} \\cdot \\frac{24}{7} Q = \\frac{30}{7} Q\\] \\[P : Q : R = 30 : 7 : 24\\] The solution is $P + Q + R = 061."
] |
2012-II-6
| 2,012
| 6
|
Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$ . Find $c+d$ .
| 125
|
II
|
[
"Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have $|z^3| * |z^2 - (1+2i)|$ $=|z|^3 * |z^2 - (1+2i)|$ $=125|z^2 - (1+2i)|$ Thus we only need to maximize the value of $|z^2 - (1+2i)|$. To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\\frac{-1}{\\sqrt{5}} + \\frac{-2}{\\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \\sqrt{5} (-5 - 10i)$ Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = 125",
"WLOG, let $z_{1}=5(\\cos{\\theta_{1}}+i\\sin{\\theta_{1}})$ and $z_{2}=1+2i=\\sqrt{5}(\\cos{\\theta_{2}+i\\sin{\\theta_{2}}})$ This means that $z_{1}^3=125(\\cos{3\\theta_{1}}+i\\sin{3\\theta_{1}})$ $z_{1}^4=625(\\cos{4\\theta_{1}}+i\\sin{4\\theta_{1}})$ Hence, this means that $z_{2}z_{1}^3=125\\sqrt{5}(\\cos({\\theta_{2}+3\\theta_{1}})+i\\sin({\\theta_{2}+3\\theta_{1}}))$ And $z_{1}^5=3125(\\cos{5\\theta_{1}}+i\\sin{5\\theta_{1}})$ Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line $yi=mx$, or when they are each a $180^{\\circ}$ rotation away from each other. Hence, we must have that $5\\theta_{1}=3\\theta_{1}+\\theta_{2}+180^{\\circ}\\implies\\theta_{1}=\\frac{\\theta_{2}+180^{\\circ}}{2}$ Now, plug this back into $z_{1}^4$(if you want to know why, reread what we want in the problem!) So now, we have that $z_{1}^4=625(\\cos{2\\theta_{2}}+i\\sin{2\\theta_{2}})$ Notice that $\\cos\\theta_{2}=\\frac{1}{\\sqrt{5}}$ and $\\sin\\theta_{2}=\\frac{2}{\\sqrt{5}}$ Then, we have that $\\cos{2\\theta_{2}}=\\cos^2{\\theta_{2}}-\\sin^2{\\theta_{2}}=-\\frac{3}{5}$ and $\\sin{2\\theta_{2}}=2\\sin{\\theta_{2}}\\cos{\\theta_{2}}=\\frac{4}{5}$ Finally, plugging back in, we find that $z_{1}^4=625(-\\frac{3}{5}+\\frac{4i}{5})=-375+500i$ $-375+500=125"
] |
2012-II-7
| 2,012
| 7
|
Let $S$ be the increasing sequence of positive integers whose binary representation has exactly $8$ ones. Let $N$ be the 1000th number in $S$ . Find the remainder when $N$ is divided by $1000$ .
| 32
|
II
|
[
"Okay, an exercise in counting (lots of binomials to calculate!). In base 2, the first number is $11111111$, which is the only way to choose 8 1's out of 8 spaces, or $\\binom{8}{8}$. What about 9 spaces? Well, all told, there are $\\binom{9}{8}=9$, which includes the first 1. Similarly, for 10 spaces, there are $\\binom{10}{8}=45,$ which includes the first 9. For 11 spaces, there are $\\binom{11}{8}=165$, which includes the first 45. You're getting the handle. For 12 spaces, there are $\\binom{12}{8}=495$, which includes the first 165; for 13 spaces, there are $\\binom{13}{8}=13 \\cdot 99 > 1000$, so we now know that $N$ has exactly 13 spaces, so the $2^{12}$ digit is 1. Now we just proceed with the other 12 spaces with 7 1's, and we're looking for the $1000-495=505th$ number. Well, $\\binom{11}{7}=330$, so we know that the $2^{11}$ digit also is 1, and we're left with finding the $505-330=175th$ number with 11 spaces and 6 1's. Now $\\binom{10}{6}=210,$ which is too big, but $\\binom{9}{6}=84.$ Thus, the $2^9$ digit is 1, and we're now looking for the $175-84=91st$ number with 9 spaces and 5 1's. Continuing the same process, $\\binom{8}{5}=56$, so the $2^8$ digit is 1, and we're left to look for the $91-56=35th$ number with 8 spaces and 4 1's. But here $\\binom{7}{4}=35$, so N must be the last or largest 7-digit number with 4 1's. Thus the last 8 digits of $N$ must be $01111000$, and to summarize, $N=1101101111000$ in base $2$. Therefore, $N = 8+16+32+64+256+512+2048+4096 \\equiv 32 \\pmod{1000}$, and the answer is $032."
] |
2012-II-8
| 2,012
| 8
|
The complex numbers $z$ and $w$ satisfy the system \[z + \frac{20i}w = 5+i\] \[w+\frac{12i}z = -4+10i\] Find the smallest possible value of $\vert zw\vert^2$ .
| 40
|
II
|
[
"Multiplying the two equations together gives us \\[zw + 32i - \\frac{240}{zw} = -30 + 46i\\] and multiplying by $zw$ then gives us a quadratic in $zw$: \\[(zw)^2 + (30-14i)zw - 240 =0.\\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \\pm \\sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\\vert zw\\vert^2$ is then obviously $6^2 + 2^2 = 040 and continue from there. mathboy282"
] |
2012-II-9
| 2,012
| 9
|
Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 107
|
II
|
[
"Examine the first term in the expression we want to evaluate, $\\frac{\\sin 2x}{\\sin 2y}$, separately from the second term, $\\frac{\\cos 2x}{\\cos 2y}$. The First Term Using the identity $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, we have: $\\frac{2\\sin x \\cos x}{2\\sin y \\cos y} = \\frac{\\sin x \\cos x}{\\sin y \\cos y} = \\frac{\\sin x}{\\sin y}\\cdot\\frac{\\cos x}{\\cos y}=3\\cdot\\frac{1}{2} = \\frac{3}{2}$ The Second Term Let the equation $\\frac{\\sin x}{\\sin y} = 3$ be equation 1, and let the equation $\\frac{\\cos x}{\\cos y} = \\frac12$ be equation 2. Hungry for the widely-used identity $\\sin^2\\theta + \\cos^2\\theta = 1$, we cross multiply equation 1 by $\\sin y$ and multiply equation 2 by $\\cos y$. Equation 1 then becomes: $\\sin x = 3\\sin y$. Equation 2 then becomes: $\\cos x = \\frac{1}{2} \\cos y$ Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS: $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ Applying the identity $\\cos^2 y = 1 - \\sin^2 y$ (which is similar to $\\sin^2\\theta + \\cos^2\\theta = 1$ but a bit different), we can change $1 = 9\\sin^2 y + \\frac{1}{4} \\cos^2 y$ into: $1 = 9\\sin^2 y + \\frac{1}{4} - \\frac{1}{4} \\sin^2 y$ Rearranging, we get $\\frac{3}{4} = \\frac{35}{4} \\sin^2 y$. So, $\\sin^2 y = \\frac{3}{35}$. Squaring Equation 1 (leading to $\\sin^2 x = 9\\sin^2 y$), we can solve for $\\sin^2 x$: $\\sin^2 x = 9\\left(\\frac{3}{35}\\right) = \\frac{27}{35}$ Using the identity $\\cos 2\\theta = 1 - 2\\sin^2\\theta$, we can solve for $\\frac{\\cos 2x}{\\cos 2y}$. $\\cos 2x = 1 - 2\\sin^2 x = 1 - 2\\cdot\\frac{27}{35} = 1 - \\frac{54}{35} = -\\frac{19}{35}$ $\\cos 2y = 1 - 2\\sin^2 y = 1 - 2\\cdot\\frac{3}{35} = 1 - \\frac{6}{35} = \\frac{29}{35}$ Thus, $\\frac{\\cos 2x}{\\cos 2y} = \\frac{-19/35}{29/35} = -\\frac{19}{29}$. Plugging in the numbers we got back into the original equation : We get $\\frac{\\sin 2x}{\\sin 2y} + \\frac{\\cos 2x}{\\cos 2y} = \\frac32 + \\left(-\\frac{19}{29} \\right) = \\frac{49}{58}$. So, the answer is $49+58=107.",
"As mentioned above, the first term is clearly $\\frac{3}{2}.$ For the second term, we first wish to find $\\frac{\\cos 2x}{\\cos 2y} =\\frac{2\\cos^2 x - 1}{2 \\cos^2y -1}.$ Now we first square the first equation getting $\\frac{\\sin^2x}{\\sin^2y} =\\frac{1-\\cos^2x}{1 - \\cos^2y} =9.$ Squaring the second equation yields $\\frac{\\cos^2x}{\\cos^2y} =\\frac{1}{4}.$ Let $\\cos^2x = a$ and $\\cos^2y = b.$ We have the system of equations \\begin{align*} 1-a &= 9-9b \\\\ 4a &= b \\\\ \\end{align*} Multiplying the first equation by $4$ yields $4-4a = 36 - 36b$ and so $4-b =36 - 36b \\implies b =\\frac{32}{35}.$ We then find $a =\\frac{8}{35}.$ Therefore the second fraction ends up being $\\frac{\\frac{64}{35}-1}{\\frac{16}{35}-1} = -\\frac{19}{29}$ so that means our desired sum is $\\frac{49}{58}$ so the desired sum is $107",
"We draw 2 right triangles with angles x and y that have the same hypotenuse. We get $b^2 + 9a^2 = 4b^2 + a^2$. Then, we find $8a^2 = 3b^2$. Now, we can scale the triangle such that $a = \\sqrt{3}$, $b = \\sqrt{8}$. We find all the side lengths, and we find the hypotenuse of both these triangles to equal $\\sqrt{35}$ This allows us to find sin and cos easily. The first term is $\\frac{3}{2}$, refer to solution 1 for how to find it. The second term is $\\frac{\\cos^2(x) - \\sin^2(x)}{\\cos^2(y) - \\sin^2(y)}$. Using the diagram, we can easily compute this as $\\frac{\\frac{8}{35} - \\frac{27}{35}}{\\frac{32}{35} - \\frac{3}{35}} = \\frac{-19}{29}$ Summing these you get $\\frac{3}{2} + \\frac{-19}{29} = \\frac{49}{58} \\implies 107 -Alexlikemath",
"Let $a = \\sin(x), b = \\sin(y)$ The first equation yields $\\frac{a}{b} = 3.$ Using $sin^2(x) + cos^2(x) = 1$ the second equation yields \\[\\frac{\\sqrt{1-a^2}}{\\sqrt{1-b^2}} = \\frac{1}{2} \\rightarrow \\frac{1-a^2}{1-b^2} = \\frac{1}{4}\\] Solving this yields $\\left(a, b\\right) = \\left(3\\sqrt{\\frac{3}{35}},\\sqrt{\\frac{3}{35}}\\right).$ Finding the first via double angle for sin yields \\[\\frac{\\sin(2x)}{\\sin(2y)} = \\frac{2\\sin{x}\\cos{x}}{2\\sin{y}\\cos{y}} = 3 \\cdot \\frac{1}{2} = \\frac{3}{2}\\] Double angle for cosine is \\[\\cos(2x) = 1-2\\sin^2{x}\\] so $\\frac{\\cos(2x)}{\\sin(2x)} = \\frac{1-2a^2}{1-2b^2} = -\\frac{19}{29}.$ Adding yields $\\frac{49}{58} \\rightarrow 49 + 58 = 107"
] |
2012-II-10
| 2,012
| 10
|
Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$ . Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .
| 496
|
II
|
[
"Solution 1 We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational. Let $x = a + \\frac{b}{c}$ where $a,b,c$ are nonnegative integers and $0 \\le b < c$ (essentially, $x$ is a mixed number). Then, \\[n = \\left(a + \\frac{b}{c}\\right) \\left\\lfloor a +\\frac{b}{c} \\right\\rfloor \\Rightarrow n = \\left(a + \\frac{b}{c}\\right)a = a^2 + \\frac{ab}{c}\\] Here it is sufficient for $\\frac{ab}{c}$ to be an integer. We can use casework to find values of $n$ based on the value of $a$: $a = 0 \\implies$ nothing because n is positive $a = 1 \\implies \\frac{b}{c} = \\frac{0}{1}$ $a = 2 \\implies \\frac{b}{c} = \\frac{0}{2},\\frac{1}{2}$ $a = 3 \\implies\\frac{b}{c} =\\frac{0}{3},\\frac{1}{3},\\frac{2}{3}$ The pattern continues up to $a = 31$. Note that if $a = 32$, then $n > 1000$. However if $a = 31$, the largest possible $x$ is $31 + \\frac{30}{31}$, in which $n$ is still less than $1000$. Therefore the number of positive integers for $n$ is equal to $1+2+3+...+31 = \\frac{31 \\cdot 32}{2} = 496 Video Solution 2012 AIME II #10 ~MathProblemSolvingSkills.com",
"We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational. Let $x = a + \\frac{b}{c}$ where $a,b,c$ are nonnegative integers and $0 \\le b < c$ (essentially, $x$ is a mixed number). Then, \\[n = \\left(a + \\frac{b}{c}\\right) \\left\\lfloor a +\\frac{b}{c} \\right\\rfloor \\Rightarrow n = \\left(a + \\frac{b}{c}\\right)a = a^2 + \\frac{ab}{c}\\] Here it is sufficient for $\\frac{ab}{c}$ to be an integer. We can use casework to find values of $n$ based on the value of $a$: $a = 0 \\implies$ nothing because n is positive $a = 1 \\implies \\frac{b}{c} = \\frac{0}{1}$ $a = 2 \\implies \\frac{b}{c} = \\frac{0}{2},\\frac{1}{2}$ $a = 3 \\implies\\frac{b}{c} =\\frac{0}{3},\\frac{1}{3},\\frac{2}{3}$ The pattern continues up to $a = 31$. Note that if $a = 32$, then $n > 1000$. However if $a = 31$, the largest possible $x$ is $31 + \\frac{30}{31}$, in which $n$ is still less than $1000$. Therefore the number of positive integers for $n$ is equal to $1+2+3+...+31 = \\frac{31 \\cdot 32}{2} = 496 Video Solution 2012 AIME II #10 ~MathProblemSolvingSkills.com",
"Notice that $x\\lfloor x\\rfloor$ is continuous over the region $x \\in [k, k+1)$ for any integer $k$. Therefore, it takes all values in the range $[k\\lfloor k\\rfloor, (k+1)\\lfloor k+1\\rfloor) = [k^2, (k+1)k)$ over that interval. Note that if $k>32$ then $k^2 > 1000$ and if $k=31$, the maximum value attained is $31*32 < 1000$. It follows that the answer is $\\sum_{k=1}^{31} (k+1)k-k^2 = \\sum_{k=1}^{31} k = \\frac{31\\cdot 32}{2} = 496 Video Solution 2012 AIME II #10 ~MathProblemSolvingSkills.com",
"Bounding gives $x^2\\le n<x^2+x$. Thus there are a total of $x$ possible values for $n$, for each value of $x^2$. Checking, we see $31^2+31=992<1000$, so there are \\[1+2+3+...+31= 496 Video Solution 2012 AIME II #10 ~MathProblemSolvingSkills.com",
"After a bit of experimenting, we let $n=l^2+s, s < 2n+1$. We claim that I (the integer part of $x$) = $l$ . (Prove it yourself using contradiction !) so now we get that $x=l+\\frac{s}{l}$. This implies that solutions exist iff $s<l$, or for all natural numbers of the form $l^2+s$ where $s<l$. Hence, 1 solution exists for $l=1$! 2 for $l=2$ and so on. Therefore our final answer is $31+30+\\dots+1= 496 Video Solution 2012 AIME II #10 ~MathProblemSolvingSkills.com"
] |
2012-II-11
| 2,012
| 11
|
Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 8
|
II
|
[
"After evaluating the first few values of $f_k (x)$, we obtain $f_4(x) = f_1(x) = \\frac{2}{3} - \\frac{3}{3x+1} = \\frac{6x-7}{9x+3}$. Since $1001 \\equiv 2 \\mod 3$, $f_{1001}(x) = f_2(x) = \\frac{3x+7}{6-9x}$. We set this equal to $x-3$, i.e. $\\frac{3x+7}{6-9x} = x-3 \\Rightarrow x = \\frac{5}{3}$. The answer is thus $5+3 = 008."
] |
2012-II-12
| 2,012
| 12
|
For a positive integer $p$ , define the positive integer $n$ to be $p$ -safe if $n$ differs in absolute value by more than $2$ from all multiples of $p$ . For example, the set of $10$ -safe numbers is $\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}$ . Find the number of positive integers less than or equal to $10,000$ which are simultaneously $7$ -safe, $11$ -safe, and $13$ -safe.
| 958
|
II
|
[
"We see that a number $n$ is $p$-safe if and only if the residue of $n \\mod p$ is greater than $2$ and less than $p-2$; thus, there are $p-5$ residues $\\mod p$ that a $p$-safe number can have. Therefore, a number $n$ satisfying the conditions of the problem can have $2$ different residues $\\mod 7$, $6$ different residues $\\mod 11$, and $8$ different residues $\\mod 13$. The Chinese Remainder Theorem states that for a number $x$ that is $a$ (mod b) $c$ (mod d) $e$ (mod f) has one solution if $\\gcd(b,d,f)=1$. For example, in our case, the number $n$ can be: 3 (mod 7) 3 (mod 11) 7 (mod 13) so since $\\gcd(7,11,13)$=1, there is 1 solution for n for this case of residues of $n$. This means that by the Chinese Remainder Theorem, $n$ can have $2\\cdot 6 \\cdot 8 = 96$ different residues mod $7 \\cdot 11 \\cdot 13 = 1001$. Thus, there are $960$ values of $n$ satisfying the conditions in the range $0 < n \\le 10010$. However, we must now remove any values greater than $10000$ that satisfy the conditions. By checking residues, we easily see that the only such values are $10006$ and $10007$, so there remain $958 of all numbers are simultaneously 7-safe, 11-safe, and 13-safe."
] |
2012-II-13
| 2,012
| 13
|
Equilateral $\triangle ABC$ has side length $\sqrt{111}$ . There are four distinct triangles $AD_1E_1$ , $AD_1E_2$ , $AD_2E_3$ , and $AD_2E_4$ , each congruent to $\triangle ABC$ ,
with $BD_1 = BD_2 = \sqrt{11}$ . Find $\sum_{k=1}^4(CE_k)^2$ .
| 677
|
II
|
[
"Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\\sqrt{111}$ from point $A$ and $\\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\\sqrt{111}$ and a circle centered at $B$ with radius $\\sqrt{11}$ intersect. Let $D_1$ be the point on the opposite side of $\\overline{AB}$ from $C$, and $D_2$ the point on the same side of $\\overline{AB}$ as $C$. Let $\\theta$ be the measure of angle $BAD_1$ (also the measure of angle $BAD_2$); by the Law of Cosines, \\begin{align*}\\sqrt{11}^2 &= \\sqrt{111}^2 + \\sqrt{111}^2 - 2 \\cdot \\sqrt{111} \\cdot \\sqrt{111} \\cdot \\cos\\theta\\\\ 11 &= 222(1 - \\cos\\theta)\\end{align*} There are two equilateral triangles with $\\overline{AD_1}$ as a side; let $E_1$ be the third vertex that is farthest from $C$, and $E_2$ be the third vertex that is nearest to $C$. Angle $E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \\theta + 60 = 120 + \\theta$; by the Law of Cosines, \\begin{align*}(E_1C)^2 &= (E_1A)^2 + (AC)^2 - 2 (E_1A) (AC)\\cos(120 + \\theta)\\\\ &= 111 + 111 - 222\\cos(120 + \\theta)\\end{align*} Angle $E_2AC = \\theta$; by the Law of Cosines, \\begin{align*}(E_2C)^2 &= (E_2A)^2 + (AC)^2 - 2 (E_2A) (AC)\\cos\\theta\\\\ &= 111 + 111 - 222\\,\\cos\\theta\\end{align*} There are two equilateral triangles with $\\overline{AD_2}$ as a side; let $E_3$ be the third vertex that is farthest from $C$, and $E_4$ be the third vertex that is nearest to $C$. Angle $E_3AC = E_3AB + BAC = (60 - \\theta) + 60 = 120 - \\theta$; by the Law of Cosines, \\begin{align*}(E_3C)^2 &= (E_3A)^2 + (AC)^2 - 2 (E_3A) (AC)\\cos(120 - \\theta)\\\\ &= 111 + 111 - 222\\cos(120 - \\theta)\\end{align*} Angle $E_4AC = \\theta$; by the Law of Cosines, \\begin{align*}(E_4C)^2 &= (E_4A)^2 + (AC)^2 - 2 (E_4A) (AC)\\cos\\theta \\\\ &= 111 + 111 - 222\\cos\\theta\\end{align*} The solution is: \\begin{align*} \\sum_{k=1}^4(CE_k)^2 &= (E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2\\\\ &= 222(1 - \\cos(120 + \\theta)) + 222(1 - \\cos(120 - \\theta)) + 222(1 - \\cos\\theta) + 222(1 - \\cos\\theta)\\\\ &= 222((1 - (\\cos120\\cos\\theta - \\sin120\\sin\\theta)) + (1 - (\\cos120\\cos\\theta + \\sin120\\sin\\theta)) + 2(1 -\\cos\\theta))\\\\ &= 222(1 - \\cos120\\cos\\theta + \\sin120\\sin\\theta + 1 - \\cos120\\cos\\theta - \\sin120\\sin\\theta + 2 - 2\\cos\\theta)\\\\ &= 222(1 + \\frac{1}{2}\\cos\\theta + 1 + \\frac{1}{2}\\cos\\theta + 2 - 2\\cos\\theta)\\\\ &= 222(4 - \\cos\\theta)\\\\ &= 666 + 222(1 - \\cos\\theta) \\end{align*} Substituting $11$ for $222(1 - \\cos\\theta)$ gives the solution $666 + 11 = 677",
"This problem is pretty much destroyed by complex plane geometry, which is similar to vector geometry only with the power of easy rotation. Place the triangle in the complex plane by letting $C$ be the origin, placing $B$ along the x-axis, and $A$ in the first quadrant. Let $r=\\sqrt{111}$. If $\\omega$ denotes the primative sixth root of unity, $e^{i\\pi/3}$, then we have $C=0$, $B=r$, and $A=r\\omega.$ Recall that counter-clockwise rotation in the complex plane by an angle $\\theta$ is accomplished by multiplication by $e^{i\\theta}$ (and clockwise rotation is multiplication by its conjugate). So, we can find $D_1$ and $D_2$ by rotating $B$ around $A$ by angles of $\\theta$ and $-\\theta$, where $\\theta$ is the apex angle in the isoceles triangle with sides $\\sqrt{111}$, $\\sqrt{111}$, and $\\sqrt{11}$. That is, let $z=e^{i\\theta}$, and then: $D_1=A+z(B-A)$, and $D_2=A+\\overline{z}(B-A)$. Now notice that $B-A=\\overline{A}$, so this simplifies further to: $D_1=A+z\\overline{A}$, and $D_2=A+\\overline{z}\\overline{A}$. Similarly, we can write $E_1$, $E_2$, $E_3$, and $E_4$ by rotating $D_1$ and $D_2$ around $A$ by $\\pm\\pi/3$: $E_1=A+\\omega(D_1-A)$, $E_2=A+\\overline{\\omega}(D_1-A)$, $E_3=A+\\omega(D_2-A)$, $E_4=A+\\overline{\\omega}(D_2-A)$. Thus: $E_1=A+\\omega z \\overline{A}$, $E_2=A+\\overline{\\omega} z \\overline{A}$, $E_3=A+\\omega\\overline{z}\\overline{A}$, $E_4=A+\\overline{\\omega}\\overline{z}\\overline{A}$. Now to find some magnitudes, which is easy since we chose $C$ as the origin: $\\|E_1\\|^2=(A+\\omega z \\overline{A})(\\overline{A}+\\overline{\\omega z}A)=2\\|A\\|^2+\\omega z \\overline{A}^2 + \\overline{\\omega}\\overline{z}A^2$, $\\|E_2\\|^2=(A+\\overline{\\omega} z \\overline{A})(\\overline{A}+\\omega \\overline{z}A)=2\\|A\\|^2+\\overline{\\omega} z \\overline{A}^2 + \\omega\\overline{z}A^2$, $\\|E_3\\|^2=(A+\\omega \\overline{z} \\overline{A})(\\overline{A}+\\overline{\\omega} z A)=2\\|A\\|^2+\\omega \\overline{z} \\overline{A}^2 + \\overline{\\omega}zA^2$, $\\|E_4\\|^2=(A+\\overline{\\omega z} \\overline{A})(\\overline{A}+\\omega zA)=2\\|A\\|^2+\\overline{\\omega z} \\overline{A}^2 + \\omega zA^2$. Adding these up, the sum equals $8\\|A\\|^2+(\\overline{A}^2+A^2)(\\omega z + \\omega \\overline{z} + \\overline{\\omega}z + \\overline{\\omega}\\overline{z}) = 8\\|A\\|^2+(\\overline{A}^2+A^2)(z + \\overline{z})( \\omega+ \\overline{\\omega})$. (Isn't that nice?) Notice that $\\overline{A}^2+A^2 = r^2(\\overline{\\omega}^2+\\omega^2) = -r^2$, and $\\omega+ \\overline{\\omega}=1$, so that this sum simplifies further to $888-111(z + \\overline{z})$. Finally, $z + \\overline{z} = 2\\cos{\\theta}$, which is found using the law of cosines on that isoceles triangle: $11=111+111-222\\cos{\\theta}$, so $2\\cos{\\theta}=211/111$. Thus, the sum equals $888-211=677.",
"This method uses complex numbers with $A$ as the origin. Let $A=0$, $B=\\sqrt{111}$, $C = \\sqrt{111}\\theta$, where $\\theta = e^{i \\pi/3} = \\frac{1}{2} + \\frac{\\sqrt{3}}{2}i$. Also, let $x$ be $D_1$ or $D_2$. Then $|x|=\\sqrt{111}, |x-\\sqrt{111}|=\\sqrt{11}$ Therefore, $11 = |x-\\sqrt{111}|^2 = |x|^2 + 111 -2\\sqrt{111}Re(x) = 222 - 2\\sqrt{111}Re(x)$, so \\[2\\sqrt{111}Re(x) = 211.\\] Since $E_1$, $E_2$ are one of $D_1\\theta$ or $D_1\\theta^{-1}$, without loss of generality, let $E_1=D_1\\theta$ and $E_2 = D_1\\theta^{-1}$. Then \\[|CE_1|^2 = |\\sqrt{111}-D_1|^2 = |\\sqrt{111}-x|^2=11\\] \\[|CE_2|^2 = |\\theta^2 \\sqrt{111} - D_1|^2 = 222-2\\sqrt{111} Re(D_1 \\theta^2)\\] One can similarly get $|CE_3|=11$ and $|CE_4|=222-2\\sqrt{111} Re(D_2 \\theta^2)$, so the desired sum is equal to \\[22+444-2\\sqrt{111}(Re(D_1 \\theta^2)+Re(D_1 \\theta^2))\\] Note that $Re(D_{1,2} \\theta^2) = Re((Re(x) \\pm Im(x)i)(-1/2 + \\sqrt{3}/2i)) = - (Re(x)/2 \\pm Im(x)\\sqrt{3}/2)$, so the sum of these two is just $-Re(x)$. Therefore the desired sum is equal to \\[22+444+2\\sqrt{111}Re(x) = 22+444+211 = 677.\\]",
"This method uses the observation that every point is equidistant from $A$. Without loss of generality, we can assume $C$ is on the same side of $AB$ as $D_1$. We can start off by angle chasing the angles around $A$. We let $\\angle BAD_1 = \\alpha$. Then, we note that $BD_1 = BD_2$ and $AD_1 = AB = AD_2$. Thus, $\\bigtriangleup ABD_1 \\cong \\bigtriangleup ABD_2$. Thus, $\\angle BAD_2 = \\alpha$ also. We can now angle chase the angles about $A$. Because $\\angle D_2AE_3 = 60$, $\\angle D_1AE_3 = 60- 2 \\alpha$. We can use all the congruent equilateral triangles in a similar manner obtaining: \\[\\angle D_1AE_3 = 60- 2 \\alpha\\] \\[\\angle E_3AC = \\alpha\\] \\[\\angle CAE_1 = \\alpha\\] \\[\\angle D_2AE_2 = 60- 2 \\alpha\\] \\[\\angle E_2AE_4 = 2 \\alpha\\] Now, $AE_3 = AC = \\sqrt{111}$ and $\\angle E_3AC = \\alpha$. Thus, $\\bigtriangleup E_3AC \\cong \\bigtriangleup BAD_1$. Thus, $CE_3^2 = BD_1^2 = 11$. Similarly, $AC = AE_1 = \\sqrt{111}$ and $\\angle CAE_1 = \\alpha$. Thus, $\\bigtriangleup CAE_1 \\cong \\bigtriangleup BAD_1$. Thus, $CE_1^2 = BD_1^2 = 11$. We can use $\\bigtriangleup CAE_2$ to find $CE_2^2$. Law of Cosines yields \\[CE_2^2 = AE_2^2 + AC^2 - 2 \\cdot AE_2 \\cdot AC \\cdot \\cos(\\angle E_2AC).\\] Substituting the known lengths and angles gives \\[CE_2^2 = 222 - 222 \\cdot \\cos(120- \\alpha).\\] Expanding this with the Cosine Subtraction Identity we get \\[CE_2^2 = 222 - 222(\\cos120 \\cos \\alpha + \\sin120 \\sin \\alpha).\\] We could attempt to calculate this but we can clear it up by simultaneously finding $CE_4^2$ too. We use Law of Cosines on $\\bigtriangleup CAE_4$ to get \\[CE_4^2 = AE_4^2 + AC^2 - 2 \\cdot AE_4 \\cdot AC \\cdot \\cos(\\angle E_4AC).\\] Substituting the known lengths and angles gives \\[CE_4^2 = 222 - 222 \\cdot \\cos(120 + \\alpha).\\] Expanding this with the Cosine Addition Identity we get \\[CE_4^2 = 222 - 222(\\cos120 \\cos \\alpha - \\sin120 \\sin \\alpha).\\] Adding this to our equation for $CE_2^2$, we get \\[CE_2^2 + CE_4^2 = 444 - 222(\\cos120 \\cos \\alpha - \\sin120 \\sin \\alpha) - 222(\\cos120 \\cos \\alpha + \\sin120 \\sin \\alpha).\\] Simplifying we get \\[CE_2^2 + CE_4^2 = 444 - 444(\\cos120 \\cos \\alpha)\\] We can find $\\cos \\alpha$ by using Law of Cosines on $\\bigtriangleup BAD_1$. This gives \\[11 = 222 - 222 \\cos \\alpha.\\] Thus $\\cos \\alpha = \\frac{211}{222}$. Substituting it in gives \\[CE_2^2 + CE_4^2 = 444 - 444(\\cos120 \\cdot \\frac{211}{222}).\\] Thus \\[CE_2^2 + CE_4^2 = 444 + 211 = 655.\\] Therefore the desired sum is equal to \\[11+11+655 = 677.\\]",
"We create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that $\\overline{C E_1} = \\sqrt{11}$, and $\\overline{C E_3} = \\sqrt{11}$. If we set $\\angle B A D_2 = \\theta$, we can start angle chasing. In particular, we will like to find $\\angle D E_4 C$, and $\\angle D E_2 C$, since then we will be able to set up some Law Of Cosines. $\\angle D E_4 C = \\angle D E_4 A + \\angle A E_4 C = 90 - \\frac{\\theta}{2} + 30 + \\frac{\\theta}{2} = 120^{\\circ}$ That was convenient! We can do it with the other angle as well. $\\angle D E_2 C = \\angle D E_2 A - \\angle C E_2 A = 90 - \\frac{\\theta}{2} - (30 - \\frac{\\theta}{2}) = 60^{\\circ}$. That means we are able to set up Law of Cosines, on triangles $\\triangle D E_4 C$ and $\\triangle D E_2 C$, with some really convenient angles. Let $CE_2 = x$, and $CE_4 = y$. \\[333 = 11 + x^2 - \\sqrt{11} x\\] \\[333 = 11 + y^2 + \\sqrt{11} y\\] We subtract and get: \\[0 = (x+y)(x-y-\\sqrt{11})\\] $x+y$ obviously can't be 0, so $x-y = \\sqrt{11}$ We add and get: \\[666 = 22 + x^2 + y^2 + \\sqrt{11} (y-x)\\]. $y-x = -\\sqrt{11}$. Thus, we can fill in and solve. \\[666 = 22 + x^2 + y^2 - 11\\] \\[655 = x^2 + y^2\\] Thus our answer is $C E_1^2 + C E_2^2 + C E_2^2 + C E_4^2 = 11 + 11 + C E_2^2 + C E_4^2 = 11 + 11 + x^2 + y^2 = 11 + 11 + 655 = 677. -Alexlikemath"
] |
2012-II-14
| 2,012
| 14
|
In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$ .
| 16
|
II
|
[
"Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.) Case 1: To create our groups of three, there are $\\dfrac{\\dbinom{9}{3}\\dbinom{6}{3}\\dbinom{3}{3}}{3!}$. In general, the number of ways we can arrange people within the rings to count properly is $\\dfrac{(n-1)!}{2}$, since there are $(n-1)!$ ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has $\\dfrac{(3-1)!}{2}=1$ arrangements. Therefore, for this case, there are $\\left(\\dfrac{\\dbinom{9}{3}\\dbinom{6}{3}\\dbinom{3}{3}}{3!}\\right)(1)^3=280$ Case 2: For three and six, there are $\\dbinom{9}{6}=84$ sets for the rings. For organization within the ring, as before, there is only one way (i.e. $(3-1)!/2$)to arrange the ring of three. For six, there is $\\dfrac{(6-1)!}{2}=60$. This means there are $(84)(1)(60)=5040$ arrangements. Case 3: For four and five, there are $\\dbinom{9}{5}=126$ sets for the rings. Within the five, there are $\\dfrac{4!}{2}=12$, and within the group of four there are $\\dfrac{3!}{2}=3$ arrangements. This means the total is $(126)(12)(3)=4536$. Case 4: For the nine case, there is $\\dbinom{9}{9}=1$ arrangement for the ring. Within it, there are $\\dfrac{8!}{2}=20160$ arrangements. Summing the cases, we have $280+5040+4536+20160=30016 \\to 016 or like would be incorrect. It is not possible to configure A, B, C such that they can be both a group of 3 and a group of 6, since {A, B, C} only has 3 elements. ~mathboy282",
"Let $f_N$ be the number of ways a group of $N$ people can shake hands with exactly two of the other people from the group, where $N \\ge 3$. We can easily find that $f_3=1$. Continuing on, we will label the people as $A$,$B$,$C$,... corresponding with $N$. $f_4$: There are $\\dbinom{3}{2}=3$ possible ways for person $A$ to shake hands with two others (WLOG assume $A$ shakes hands with $B$ and $C$), and there is only one possible outcome thereon ($B$ and $C$ both shakes hands with $D$). Therefore, we can conclude that $f_4=3$ $f_5$: There are $\\dbinom{4}{2}=6$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$). Then, $B$ and $C$ must also shake hands with the remaining two people ($2$ ways), and those last $2$ people must shake hands with each other ($1$ way). Therefore, $f_5=\\dbinom{4}{2}\\ast(2)\\ast(1)=12$ $f_6$: There are $\\dbinom{5}{2}=10$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$). However, $B$ and $C$ shake hands with each other, then there are $f_3$ ways for the rest of the people to shake hands. If $B$ and $C$ shake hands with two others out of the three remaining people (3$\\ast$2 ways), those $2$ people must shake hands with the last person ($1$ way). Therefore, $f_6=\\dbinom{5}{2}\\ast(f_3)+3\\ast(2)\\ast(1)=70$ Now we have enough information to find $f_9$ $f_9$: There are $\\dbinom{8}{2}=28$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$). Case 1: $B$ and $C$ shake hands with each other There are $f_6$ ways for $D$,$E$,$F$,$G$,$H$, and $I$ to shake hands Case 2: $B$ and $C$ shake hands with one of the others out of the $6$ remaining people ($6$ ways) There are $f_5$ ways for $E$,$F$,$G$,$H$, and $I$ to shake hands Case 3: If $B$ and $C$ shake hands with two different people (there are 6$\\ast$5 ways to choose the people but WLOG assume they are $D$ and $E$). Calculations of Case 3 are in its subcases. Subcase 3.1: $D$ and $E$ shake hands with each other There are $f_4$ ways for $F$,$G$,$H$, and $I$ to shake hands Subcase 3.2: $D$ and $E$ each shake hands with one other person (there are $4$ ways to choose the person, WLOG let that be $F$) There are $f_3$ ways for $G$,$H$, and $I$ to shake hands Subcase 3.3: $D$ and $E$ each shake hands with two different people (there are 4$\\ast$3 ways, WLOG let that be $F$ and $G$) There are $2\\ast1$ ways for $F$ and $G$ to then shake hands with $H$ and $I$, and $H$ and $I$ will shake hands with each other. We have \\begin{align*} 28(f_6+6\\ast(f_5)+6\\ast(5)\\ast(f_4)+4\\ast(f_3)+4\\ast(3)\\ast(2)\\ast(1) &= 28(70+6\\ast(12)+6\\ast(5)\\ast(3+4\\ast(1)+4\\ast(3)\\ast(2)\\ast(1) \\\\ &= 30016 \\end{align*} which gives us the answer of $016 ~Danielzh"
] |
2012-II-15
| 2,012
| 15
|
Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ , $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 919
|
II
|
[
"Use the angle bisector theorem to find $CD=\\tfrac{21}{8}$, $BD=\\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\\angle CAD = \\tfrac{\\pi} {3}$, hence $\\angle BAD = \\tfrac{\\pi}{3}$ as well, and $\\triangle BCE$ is equilateral, so $BC=CE=BE=7$. [asy] size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot(\"$W$\",circumcenter(A,B,C),dir(180)); label(\"$\\gamma$\",gamma,dir(180)); [/asy] In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \\cdot \\cos \\angle AEF$ and $XF=AF \\cdot \\cos \\angle AFE$, we get \\begin{align}\\tag{1} EF = AE \\cdot \\cos \\angle AEF + AF \\cdot \\cos \\angle AFE. \\end{align} Note $\\angle AFE = \\angle ACE$, and the Law of Cosines in $\\triangle ACE$ gives $\\cos \\angle ACE = -\\tfrac 17$. Also, $\\angle AEF = \\angle DEF$, and $\\angle DFE = \\tfrac{\\pi}{2}$ ($DE$ is a diameter), so $\\cos \\angle AEF = \\tfrac{EF}{DE} = \\tfrac{8}{49}\\cdot EF$. Plugging in all our values into equation $(1)$, we get: \\[EF = \\tfrac{64}{49} EF -\\tfrac{1}{7} AF \\quad \\Longrightarrow \\quad EF = \\tfrac{7}{15} AF.\\] The Law of Cosines in $\\triangle AEF$, with $EF=\\tfrac 7{15}AF$ and $\\cos\\angle AFE = -\\tfrac 17$ gives \\[8^2 = AF^2 + \\tfrac{49}{225} AF^2 + \\tfrac 2{15} AF^2 = \\tfrac{225+49+30}{225}\\cdot AF^2\\] Thus $AF^2 = \\frac{900}{19}$. The answer is $919. ~Shen Kislay Kai",
"Let $a = BC$, $b = CA$, $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$. We claim that $\\angle MAD=\\angle DAF$. $\\textit{Proof}$. Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\\perp BC$. Therefore, $M\\in \\gamma$. Now let $X = FD\\cap \\omega$. Since $\\angle EFX=90^\\circ$, $EX$ is a diameter, so $X$ lies on the perpendicular bisector of $BC$; hence $E$, $M$, $X$ are collinear. From $\\angle DAG = \\angle DMX = 90^\\circ$, quadrilateral $ADMX$ is cyclic. Therefore, $\\angle MAD = \\angle MXD$. But $\\angle MXD$ and $\\angle EAF$ are both subtended by arc $EF$ in $\\omega$, so they are equal. Thus $\\angle MAD=\\angle DAF$, as claimed. [asy] size(175); defaultpen(fontsize(10pt)); picture pic; pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); pair M=MP(\"M\",midpoint(B--C),dir(220)); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); dot(\"$W$\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"$\\gamma$\",gamma,dir(180)); draw(A--B--F--cycle, black+1); [/asy] As a result, $\\angle CAM = \\angle FAB$. Combined with $\\angle BFA=\\angle MCA$, we get $\\triangle ABF\\sim\\triangle AMC$ and therefore\\[\\frac c{AM}=\\frac {AF}b\\qquad \\Longrightarrow \\qquad AF^2=\\frac{b^2c^2}{AM^2} = \\frac{15^2}{AM^2}\\] By Stewart's Theorem on $\\triangle ABC$ (with cevian $AM$), we get \\[AM^2 = \\tfrac 12 (b^2+c^2)-\\tfrac 14 a^2 = \\tfrac{19}{4},\\] so $AF^2 = \\tfrac{900}{19}$, so the answer is $900+19=919. -Solution by thecmd999",
"Use the angle bisector theorem to find $CD=\\tfrac{21}{8}$, $BD=\\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\\tfrac{49}{8}$, and so $AE=8$. Then use the Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. [asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"$W$\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"$\\gamma$\",gamma,dir(180)); label(\"$u$\",X--D,dir(60)); label(\"$v$\",D--F,dir(70)); [/asy] Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get \\[(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}\\] which simplifies to \\[u^2+2uv = \\frac{5341}{192}.\\] By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get \\[XD^2=u^2=\\frac{931}{192}.\\] Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and \\[AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.\\] The answer is $900+19=919. -Solution by TheBoomBox77",
"Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, \\[\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.\\] Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get\\[25k^2+9k^2-15k^2 = 49 \\qquad \\Longrightarrow\\qquad k=\\frac 7{\\sqrt{19}}\\] Use Ptolemy's Theorem on $BFCA$, to get \\[15k+15k=7\\cdot AF, \\qquad \\Longrightarrow\\qquad AF= \\frac{30}{\\sqrt{19}},\\] so $AF^2=\\frac{900}{19}$ and the answer is $900+19=919$ ~Shen Kislay Kai",
"Denote $AB = c, BC = a, AC = b, \\angle A = 2 \\alpha.$ Let M be midpoint BC. Let $\\theta$ be the circle centered at $A$ with radius $\\sqrt{AB \\cdot AC} =\\sqrt{bc}.$ We calculate the length of some segments. The median $AM = \\sqrt{\\frac {b^2}{2} + \\frac {c^2}{2} - \\frac {a^2}{4}}.$ The bisector $AD = \\frac {2 b c \\cos \\alpha}{b+c}.$ One can use Stewart's Theorem in both cases. $AD$ is bisector of $\\angle A \\implies BD = \\frac {a c}{b + c}, CD = \\frac {a b}{b + c} \\implies$ \\[BD \\cdot CD = \\frac {a^2 bc }{(b+c)^2}.\\] We use Power of Point $D$ and get $AD \\cdot DE = BD \\cdot CD.$ \\[AE = AD + DE = AD + \\frac {BD \\cdot CD}{AD},\\] \\[AE =\\frac {2 b c \\cos \\alpha}{b+c} + \\frac {a^2 bc \\cdot (b+c) }{(b+c)^2 \\cdot 2 b c \\cos \\alpha} =\\] \\[= \\frac {b c \\cos^2 \\alpha + a^2}{2(b+c)\\cos \\alpha} =\\frac {4bc \\cos^2 \\alpha + b^2 +c^2 -2 b c \\cos 2\\alpha}{2(b+c) \\cos \\alpha} = \\frac {b+c}{2} \\implies AD \\cdot AE = 2 bc \\cos \\alpha.\\] We consider the inversion with respect $\\theta.$ $B$ swap $B' \\implies AB' = AC, B' \\in AB \\implies B'$ is symmetric to $C$ with respect to $AE.$ $C$ swap $C' \\implies AC' = AB, C'$ lies on line $AC \\implies C'$ is symmetric to $B$ with respect to $AE.$ $BC^2 = AB^2 + AC^2 + AB \\cdot BC \\implies \\alpha = 60^\\circ \\implies AD \\cdot AE = bc \\implies D$ swap $E.$ Points $D$ and $E$ lies on $\\Gamma \\implies \\Gamma$ swap $\\Gamma.$ $DE$ is diameter $\\Gamma, \\angle DME = 90^\\circ \\implies M \\in \\Gamma.$ Therefore $M$ is crosspoint of $BC$ and $\\Gamma.$ Let $\\Omega$ be circumcircle $AB'C'. \\Omega$ is image of line $BC.$ Point $M$ maps into $M' \\implies M' = \\Gamma \\cap \\Omega.$ Points $A, B',$ and $C'$ are symmetric to $A, C,$ and $B,$ respectively. Point $M'$ lies on $\\Gamma$ which is symmetric with respect to $AE$ and on $\\Omega$ which is symmetric to $\\omega$ with respect to $AE \\implies$ $M'$ is symmetric $F$ with respect to $AE \\implies AM' = AF.$ We use Power of Point $A$ and get \\[AF = AM' = \\frac {AD \\cdot AE}{AM} = \\frac {4b c}{\\sqrt{2 b^2 + 2 c^2 – a^2}} = \\frac {4 \\cdot 3 \\cdot 5}{\\sqrt{ 50 + 18 – 49}} = \\frac {30}{\\sqrt{19}} \\implies \\textbf{919}.\\] [email protected], vvsss",
"To do this, we first define the intersection of $EF$ and $BC$ to be $K$. Lemma 1: $(K, C, D, B)$ are harmonic. First of all, define the midpoint of $BC$ to be $M$. Then, we have that angle FMD is $90$ degrees, and as a result, $M$ lies on this circle. By Power of a Point, $(KD)(KM) = (KE)(KF) = (KB)(KC)$. As a result, $(K, C, D, B)$ are harmonic from another famous harmonic lemma. As a result, since $\\angle EFD = 90^{\\circ}$, by another Harmonic Lemma, $FD$ is the angle bisector of $BFC$. Since $\\frac{BD}{CD} = \\frac{5}{3}$ by angle bisector theorem, $\\frac{BF}{CF} = \\frac{5}{3}$. Since $\\angle BAC$ is $120^{\\circ}$ by Law of Cosines (LOC), we can use LOC to finish off. Call $BF = 5a$, and $CF = 3a$, $(5a)^2+(3a)^2-15a^2 = 49$, so $a = \\frac{7}{\\sqrt{19}}$. We do Ptolemy's Theorem on $ABFC$. Our answer is: \\[\\frac{(3)(35)+(5)(21)}{7\\sqrt{19}} = \\frac{30}{\\sqrt{19}}.\\] As a result, the final answer is $919. -sepehr2010 Minor edits ~Zhenghua"
] |
2013-I-1
| 2,013
| 1
|
The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling?
| 150
|
I
|
[
"Let $r$ represent the rate Tom swims in miles per minute. Then we have $\\frac{1/2}{r} + \\frac{8}{5r} + \\frac{30}{10r} = 255$ Solving for $r$, we find $r = 1/50$, so the time Tom spends biking is $\\frac{30}{(10)(1/50)} = 150 minutes."
] |
2013-I-3
| 2,013
| 3
|
Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$
| 18
|
I
|
[
"It's important to note that $\\dfrac{AE}{EB} + \\dfrac{EB}{AE}$ is equivalent to $\\dfrac{AE^2 + EB^2}{(AE)(EB)}$ We define $a$ as the length of the side of larger inner square, which is also $EB$, $b$ as the length of the side of the smaller inner square which is also $AE$, and $s$ as the side length of $ABCD$. Since we are given that the sum of the areas of the two squares is$\\frac{9}{10}$ of the the area of ABCD, we can represent that as $a^2 + b^2 = \\frac{9s^2}{10}$. The sum of the two nonsquare rectangles can then be represented as $2ab = \\frac{s^2}{10}$. Looking back at what we need to find, we can represent $\\dfrac{AE^2 + EB^2}{(AE)(EB)}$ as $\\dfrac{a^2 + b^2}{ab}$. We have the numerator, and dividing$\\frac{s^2}{10}$ by two gives us the denominator $\\frac{s^2}{20}$. Dividing $\\dfrac{\\frac{9s^2}{10}}{\\frac{s^2}{20}}$ gives us an answer of $018.",
"Let the side of the square be $1$. Therefore the area of the square is also $1$. We label $AE$ as $a$ and $EB$ as $b$. Notice that what we need to find is equivalent to: $\\frac{a^2+b^2}{ab}$. Since the sum of the two squares ($a^2+b^2$) is $\\frac{9}{10}$ (as stated in the problem) the area of the whole square, it is clear that the sum of the two rectangles is $1-\\frac{9}{10} \\implies \\frac{1}{10}$. Since these two rectangles are congruent, they each have area: $\\frac{1}{20}$. Also note that the area of this is $ab$. Plugging this into our equation we get: $\\frac{\\frac{9}{10}}{\\frac{1}{20}} \\implies 018",
"Let $AE$ be $x$, and $EB$ be $1$. Then we are looking for the value $x+\\frac{1}{x}$. The areas of the smaller squares add up to $9/10$ of the area of the large square, $(x+1)^2$. Cross multiplying and simplifying we get $x^2-18x+1=0$. Rearranging, we get $x+\\frac{1}{x}=018",
"As before, $\\dfrac{AE}{EB} + \\dfrac{EB}{AE}$ is equivalent to $\\dfrac{AE^2 + EB^2}{(AE)(EB)}$. Let $x$ represent the value of $AE=CF$. Since $EB=FB=1-x,$ the area of the two rectangles is $2x(1-x)=-2x^2+2x=\\frac1{10}$. Adding $2x^2-2x$ to both sides and dividing by $2$ gives $x^2-x+\\frac1{20}=0.$ Note that the two possible values of $x$ in the quadratic both sum to $1,$ like how $AE$ and $EB$ does. Therefore, $EB$ must be the other root of the quadratic that $AE$ isn't. Applying Vietas and manipulating the numerator, we get $\\frac{x_1^2+x_2^2}{x_1x_2}=\\frac{(x_1+x_2)^2-2x_1x_2}{\\frac{1}{20}}=\\frac{1^2-\\frac1{10}}{\\frac1{20}}=\\frac{\\frac9{10}}{\\frac{1}{20}}=018.",
"As before, $\\dfrac{AE}{EB} + \\dfrac{EB}{AE}$ is equivalent to $\\dfrac{AE^2 + EB^2}{(AE)(EB)}$. Let $x$ represent the value of $AE=CF$. Since $EB=FB=1-x,$ the area of the two rectangles is $2x(1-x)=-2x^2+2x=\\frac1{10}$. Adding $2x^2-2x$ to both sides and dividing by $2$ gives $x^2-x+\\frac1{20}=0.$ Note that the two possible values of $x$ in the quadratic both sum to $1,$ like how $AE$ and $EB$ does. Therefore, $EB$ must be the other root of the quadratic that $AE$ isn't. Applying Vietas and manipulating the numerator, we get $\\frac{x_1^2+x_2^2}{x_1x_2}=\\frac{(x_1+x_2)^2-2x_1x_2}{\\frac{1}{20}}=\\frac{1^2-\\frac1{10}}{\\frac1{20}}=\\frac{\\frac9{10}}{\\frac{1}{20}}=018.",
"Let $AE = x$ and $BE = y$. From this, we get $AB = x + y$. The problem is asking for $\\frac{x}{y} + \\frac{y}{x}$, which can be rearranged to give $\\frac{x^2 + y^2}{xy}$. The problem tells us that $x^2 + y^2 = \\frac{9(x+y)^2}{10}$. We simplify to get $x^2 + y^2 = 18xy$. Finally, we divide both sides by $xy$ to get $\\frac{x^2 + y^2}{xy} = 018. - Spacesam",
"Let $AE = x$ and $BE = y$. From this, we get $AB = x + y$. The problem is asking for $\\frac{x}{y} + \\frac{y}{x}$, which can be rearranged to give $\\frac{x^2 + y^2}{xy}$. The problem tells us that $x^2 + y^2 = \\frac{9(x+y)^2}{10}$. We simplify to get $x^2 + y^2 = 18xy$. Finally, we divide both sides by $xy$ to get $\\frac{x^2 + y^2}{xy} = 018. - Spacesam",
"After we get the polynomial $x^2 - 18x + 1,$ we want to find $x + \\frac 1 {x}.$ Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \\frac 1 {x}.$ Hence $x + \\frac 1 {x}$ is just $18$ by Vieta's formula, or $018",
"After we get the polynomial $x^2 - 18x + 1,$ we want to find $x + \\frac 1 {x}.$ Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \\frac 1 {x}.$ Hence $x + \\frac 1 {x}$ is just $18$ by Vieta's formula, or $018",
"We have the equation $x^2 + y^2$ = $\\frac {9}{10} \\cdot (x+y)^2$. We get $x^2 + y^2 = 18xy$. We rearrange to get $x^2 + y^2 - 18xy = 0$. Since the problem only asks us for a ratio, we assume $x$ = $1$. We have $y^2 - 18y + 1$ = $0$. Solving the quadratic yields $9 + 4 \\sqrt 5$ and $9 - 4 \\sqrt 5$. It doesn't really matter which one it is, since both of them are positive. We will use $9 + 4 \\sqrt 5$. We have $9 + 4 \\sqrt 5 + \\frac {1}{9+4 \\sqrt 5}$. Rationalizing the denominator gives us $9 + 4 \\sqrt 5 + \\frac {9 - 4 \\sqrt 5}{81-80} = (9 + 4 \\sqrt 5) + (9 - 4 \\sqrt 5) = 18$. Our answer is $018 ~Arcticturn",
"Set side length of square to be $10$, $AE = x$ and $EB = y$. From this, we get $y+x=10$, and since the area of the square will be 100, the area of the two rectangles will be $2xy = 10$. We can substitute and say that $2xy = x+y$, and subtract $y$ from both sides, and then divide by $y$, getting the equation $\\frac {x}{y} = 2x-1$, and doing the same thing with $x$ to get $\\frac {y}{x} = 2y-1$. Adding these equations, we get the desired sum to be $2(x+y) - 2$, or $20-2$ which is equal to $018. ~ E___",
"You can call the side length of the square 1. Call length AE x and solve the quadratic equation x^2+1-2x+x^2=9/10 with the quadratic formula.",
"You can call the side length of the square 1. Call length AE x and solve the quadratic equation x^2+1-2x+x^2=9/10 with the quadratic formula."
] |
2013-I-4
| 2,013
| 4
|
In the array of $13$ squares shown below, $8$ squares are colored red, and the remaining $5$ squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated $90^{\circ}$ around the central square is $\frac{1}{n}$ , where $n$ is a positive integer. Find $n$ . [asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); draw((2,0)--(2,2)--(3,2)--(3,0)--(3,1)--(2,1)--(4,1)--(4,0)--(2,0)); draw((1,2)--(1,4)--(0,4)--(0,2)--(0,3)--(1,3)--(-1,3)--(-1,2)--(1,2)); draw((-1,1)--(-3,1)--(-3,0)--(-1,0)--(-2,0)--(-2,1)--(-2,-1)--(-1,-1)--(-1,1)); draw((0,-1)--(0,-3)--(1,-3)--(1,-1)--(1,-2)--(0,-2)--(2,-2)--(2,-1)--(0,-1)); size(100);[/asy]
| 429
|
I
|
[
"When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there enough blue squares for there to be more than one blue square in each three-square formation. Thus there are 2 reds and 1 blue in each, and a blue in the center. There are 3 ways to choose which of the squares in the formation will be blue, leaving the other two red. There are $\\binom{13}{5}$ ways to have 5 blue squares in an array of 13. $\\frac{3}{\\binom{13}{5}}$ = $\\frac{1}{429}$ , so $n$ = $429",
"By the Pigeonhole Principle, if the middle square isn't blue, than any 90 degree rotation won't map onto itself because one extra blue square will be mapped onto a red square. Therefore, we have to have 1 blue square in the middle and 1 blue square in each of the 4 seperate containers(think if there is two in each a 90 degree rotation wouldn't move it to the other side). There are 3 places in each of the seperate tilings for it and everything else will be red. Therefore, there are only 3 good combinations. Now the total amount is 13choose5, so we lead to the answer: $\\frac{3}{\\binom{13}{5}}$ = $\\frac{1}{429}$ , so $n$ = $429 ~MathCosine",
"We use the same idea as the other solutions, knowing the middle square must be blue, which has probability $\\frac{5}{13}$, the next blue square can go anywhere; then the last three must go in the three spots they rotate in, giving us $\\frac{3!}{11*10*9}$. Multiplying yields $429. ~boppitybobii"
] |
2013-I-5
| 2,013
| 5
|
The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$ , where $a$ , $b$ , and $c$ are positive integers. Find $a+b+c$ .
| 98
|
I
|
[
"We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$. Therefore, we have that $9x^3 = (x+1)^3$, so it follows that $x\\sqrt[3]{9} = x+1$. Solving for $x$ yields $\\frac{1}{\\sqrt[3]{9}-1} = \\frac{\\sqrt[3]{81}+\\sqrt[3]{9}+1}{8}$, so the answer is $98.",
"Let $r$ be the real root of the given polynomial. Now define the cubic polynomial $Q(x)=-x^3-3x^2-3x+8$. Note that $1/r$ must be a root of $Q$. However we can simplify $Q$ as $Q(x)=9-(x+1)^3$, so we must have that $(\\frac{1}{r}+1)^3=9$. Thus $\\frac{1}{r}=\\sqrt[3]{9}-1$, and $r=\\frac{1}{\\sqrt[3]{9}-1}$. We can then multiply the numerator and denominator of $r$ by $\\sqrt[3]{81}+\\sqrt[3]{9}+1$ to rationalize the denominator, and we therefore have $r=\\frac{\\sqrt[3]{81}+\\sqrt[3]{9}+1}{8}$, and the answer is $98.",
"It is clear that for the algebraic degree of $x$ to be $3$ that there exists some cubefree integer $p$ and positive integers $m,n$ such that $a = m^3p$ and $b = n^3p^2$ (it is possible that $b = n^3p$, but then the problem wouldn't ask for both an $a$ and $b$). Let $f_1$ be the automorphism over $\\mathbb{Q}[\\sqrt[3]{a}][\\omega]$ which sends $\\sqrt[3]{a} \\to \\omega \\sqrt[3]{a}$ and $f_2$ which sends $\\sqrt[3]{a} \\to \\omega^2 \\sqrt[3]{a}$ (note : $\\omega$ is a cubic root of unity). Letting $r$ be the root, we clearly we have $r + f_1(r) + f_2(r) = \\frac{3}{8}$ by Vieta's formulas. Thus it follows $c=8$. Now, note that $\\sqrt[3]{a} + \\sqrt[3]{b} + 1$ is a root of $x^3 - 3x^2 - 24x - 64 = 0$. Thus $(x-1)^3 = 27x + 63$ so $(\\sqrt[3]{a} + \\sqrt[3]{b})^3 = 27(\\sqrt[3]{a} + \\sqrt[3]{b}) + 90$. Checking the non-cubicroot dimension part, we get $a + b = 90$ so it follows that $a + b + c = 98.",
"We have $cx-1=\\sqrt[3]{a}+\\sqrt[3]{b}.$ Therefore $(cx-1)^3=(\\sqrt[3]{a}+\\sqrt[3]{b})^3=a+b+3\\sqrt[3]{ab}(\\sqrt[3]{a}+\\sqrt[3]{b})=a+b+3\\sqrt[3]{ab}(cx-1).$ We have \\[c^3x^3-3c^2x^2-(3c\\sqrt[3]{ab}-3c)x-(a+b+1-3\\sqrt[3]{ab})=0.\\] We will find $a,b,c$ so that the equation is equivalent to the original one. Let $\\dfrac{3c^2}{c^3}=\\dfrac{3}{8}, \\dfrac{3c\\sqrt[3]{ab}-3c}{c^3}=\\dfrac{3}{8}, \\dfrac{a+b+1-3\\sqrt[3]{ab}}{c^3}=\\dfrac{1}{8}.$ Easily, $c=8, \\sqrt[3]{ab}=9,$ and $a+b=90.$ So $a + b + c = 90+8=98."
] |
2013-I-6
| 2,013
| 6
|
Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 47
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I
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[
"The total ways the textbooks can be arranged in the 3 boxes is $12\\textbf{C}3\\cdot 9\\textbf{C}4$, which is equivalent to $\\frac{12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7\\cdot 6}{144}=12\\cdot11\\cdot10\\cdot7\\cdot3$. If all of the math textbooks are put into the box that can hold $3$ textbooks, there are $9!/(4!\\cdot 5!)=9\\textbf{C}4$ ways for the other textbooks to be arranged. If all of the math textbooks are put into the box that can hold $4$ textbooks, there are $9$ ways to choose the other book in that box, times $8\\textbf{C}3$ ways for the other books to be arranged. If all of the math textbooks are put into the box with the capability of holding $5$ textbooks, there are $9\\textbf{C}2$ ways to choose the other 2 textbooks in that box, times $7\\textbf{C}3$ ways to arrange the other 7 textbooks. $9\\textbf{C}4=9\\cdot7\\cdot2=126$, $9\\cdot 8\\textbf{C}3=9\\cdot8\\cdot7=504$, and $9\\textbf{C}2\\cdot 7\\textbf{C}3=9\\cdot7\\cdot5\\cdot4=1260$, so the total number of ways the math textbooks can all be placed into the same box is $126+504+1260=1890$. So, the probability of this occurring is $\\frac{(9\\cdot7)(2+8+(4\\cdot5))}{12\\cdot11\\cdot10\\cdot7\\cdot3}=\\frac{1890}{27720}$. If the numerator and denominator are both divided by $9\\cdot7$, we have $\\frac{(2+8+(4\\cdot5))}{4\\cdot11\\cdot10}=\\frac{30}{440}$. Simplifying the numerator yields $\\frac{30}{10\\cdot4\\cdot11}$, and dividing both numerator and denominator by $10$ results in $\\frac{3}{44}$. This fraction cannot be simplified any further, so $m=3$ and $n=44$. Therefore, $m+n=3+44=047.",
"Consider the books as either math or not-math where books in each category are indistiguishable from one another. Then, there are $\\,_{12}C_{3}$ total distinguishable ways to pack the books. Now, in order to determine the desired propability, we must find the total number of ways the condition that all math books are in the same box can be satisfied. We proceed with casework for each box: Case 1: The math books are placed into the smallest box. This can be done in $\\binom{3}{3}$ ways. Case 2: The math books are placed into the middle box. This can be done in $\\binom{4}{3}$ ways. Case 3: The math books are placed into the largest box. This can be done in $\\binom{5}{3}$ ways. So, the total ways the condition can be satisfied is $\\binom{3}{3} + \\binom{4}{3} + \\binom{5}{3}$. This can be simplified to $\\binom{6}{4} = \\binom{6}{2}$ by the Hockey Stick Identity. Therefore, the desired probability is $\\dfrac{\\dbinom{6}{2} }{\\dbinom{12}{3}}$ = $\\dfrac{3}{44}$, and $m+n=3+44=047.",
"There are three cases as follows. Note these are PERMUTATIONS, as the books are distinct! 1. Math books in the 3-size box. Probability is $\\frac{3\\cdot2\\cdot1}{12\\cdot11\\cdot10}$, because we choose one of the $3$ places for math book 1, then one of the $2$ for math book 2, and the last one. Total number of orders: $12\\cdot11\\cdot10=1320$. 2. In the 4-size: same logic gets you $\\frac{1}{55}$, since we have $4$ places for math book 1, and so on. 3. In the 5-size: you get $\\frac{1}{22}$, for a sum of $\\frac{3}{44}$ so your answer is $047.",
"There are three cases as follows. Note these are PERMUTATIONS, as the books are distinct! 1. Math books in the 3-size box. Probability is $\\frac{3\\cdot2\\cdot1}{12\\cdot11\\cdot10}$, because we choose one of the $3$ places for math book 1, then one of the $2$ for math book 2, and the last one. Total number of orders: $12\\cdot11\\cdot10=1320$. 2. In the 4-size: same logic gets you $\\frac{1}{55}$, since we have $4$ places for math book 1, and so on. 3. In the 5-size: you get $\\frac{1}{22}$, for a sum of $\\frac{3}{44}$ so your answer is $047.",
"Assume that the $9$ other books are all distinct. The number of ways to place the other $9$ books and the $3$ is $\\frac {12!}{3!}$. The number of ways to put all the $3$ math books into the box that holds $3$ books is $9!$. The number of ways to put all the $3$ math books into the box that holds $4$ books is $\\binom {4}{3} \\cdot 9!$, and the number of ways to put all the $3$ math books into the box that holds $5$ books is $\\binom {5}{3} \\cdot 9!$. The number of desired outcomes is $15 \\cdot 9!$, and the total number of outcomes is $\\frac {12!}{6}$. Simplifying, we get the answer is $\\frac {3}{44}$, so our answer is $047. ~Arcticturn"
] |
2013-I-7
| 2,013
| 7
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A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ .
| 41
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I
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[
"Let the height of the box be $x$. After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{\\left(\\frac{x}{2}\\right)^2 + 64}$, and $\\sqrt{\\left(\\frac{x}{2}\\right)^2 + 36}$. Since the area of the triangle is $30$, the altitude of the triangle from the base with length $10$ is $6$. Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of $10$. We find: \\[10 = \\sqrt{\\left(28+x^2/4\\right)}+x/2\\] Solving for $x$ gives us $x=\\frac{36}{5}$. Since this fraction is simplified: \\[m+n=041\\] Small Note The 3D version of the Pythagorean theorem can also be applied here. ~MC",
"We may use vectors. Let the height of the box be $2h$. Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$. Let the center point of the bottom face be $P_1$, the center of the left face be $P_2$ and the center of the front face be $P_3$. We are given that the area of the triangle $\\triangle P_1 P_2 P_3$ is $30$. Thus, by a well known formula, we note that $\\frac{1}{2}|\\vec{P_1P_2} \\text{x} \\vec{P_1P_3}|=30$ We quickly attain that $\\vec{P_1P_2}=<-6,0,h>$ and $\\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry) Computing the cross product, we find: \\[\\vec{P_1P_2} x \\vec{P_1P_3}=-<6h,8h,48>\\] Thus: \\[\\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60\\] \\[h=3.6\\] \\[2h=7.2\\] \\[2h=36/5\\] \\[m+n=041\\]",
"We may use vectors. Let the height of the box be $2h$. Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$. Let the center point of the bottom face be $P_1$, the center of the left face be $P_2$ and the center of the front face be $P_3$. We are given that the area of the triangle $\\triangle P_1 P_2 P_3$ is $30$. Thus, by a well known formula, we note that $\\frac{1}{2}|\\vec{P_1P_2} \\text{x} \\vec{P_1P_3}|=30$ We quickly attain that $\\vec{P_1P_2}=<-6,0,h>$ and $\\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry) Computing the cross product, we find: \\[\\vec{P_1P_2} x \\vec{P_1P_3}=-<6h,8h,48>\\] Thus: \\[\\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60\\] \\[h=3.6\\] \\[2h=7.2\\] \\[2h=36/5\\] \\[m+n=041\\]",
"Let the height of the box be $x$. After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{(x/2)^2 + 64}$, and $\\sqrt{(x/2)^2 + 36}$. Therefore, we can use Heron's formula to set up an equation for the area of the triangle. The semiperimeter is $\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2$. Therefore, when we square Heron's formula, we find \\begin{align*}900 &= \\frac{1}{2}\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2\\right)\\times\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2 - 10\\right)\\\\&\\qquad\\times\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2 - \\sqrt{(x/2)^2 + 64}\\right)\\\\&\\qquad\\times\\left(\\left(10 + \\sqrt{(x/2)^2 + 64} + \\sqrt{(x/2)^2 + 36}\\right)/2 - \\sqrt{(x/2)^2 + 36}\\right).\\end{align*} Solving, we get $041.",
"It isn't hard to see that the triangle connecting the centers of the faces of the rectangular prism is congruent to the triangle connecting the midpoints of three edges that concur. So we can now apply de Gua's theorem (https://en.wikipedia.org/wiki/De_Gua%27s_theorem) to see that: $30^2=24^2+(3x)^2+(4x)^2$ Where $x$ is half the desired length of the height. Solving yields $2x=\\frac{36}{5}$ And thus $36+5=041 ---Solution 4 contributed by Siddharth Namachivayam",
"Let half the height be $a$ (we want to find $2a$), then we see that the three sides of the triangle are (by Pyth Theorem) $10, \\sqrt{a^2+36}, \\sqrt{a^2+64}$. Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is $\\frac{a^2}{\\sqrt{(a^2+36)(a^2+64)}}$ by the Law of Cosines, meaning that its sine is $\\frac{\\sqrt{100a^2+2304}}{\\sqrt{(a^2+36)(a^2+64)}}$. Finally, multiply the two square-rooted sides by this sine and one-half, and equate to 30. You get $\\sqrt{25a^2+576} = 30$, giving $a=\\frac{18}{5}$, so our answer is $041",
"Let half the height be $a$ (we want to find $2a$), then we see that the three sides of the triangle are (by Pyth Theorem) $10, \\sqrt{a^2+36}, \\sqrt{a^2+64}$. Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is $\\frac{a^2}{\\sqrt{(a^2+36)(a^2+64)}}$ by the Law of Cosines, meaning that its sine is $\\frac{\\sqrt{100a^2+2304}}{\\sqrt{(a^2+36)(a^2+64)}}$. Finally, multiply the two square-rooted sides by this sine and one-half, and equate to 30. You get $\\sqrt{25a^2+576} = 30$, giving $a=\\frac{18}{5}$, so our answer is $041",
"Let $x$ be $\\frac12$ the height of the box. We will solve for $x$ and then multiply by $2$ at the end. The three side lengths of the triangle, by the Pythagorean Theorem, are $\\sqrt{x^2+6^2},\\sqrt{x^2+8^2},$ and $10$. Heron's formula states that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter. Applying difference of squares to make the formula less computational, we get $\\frac{\\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}=\\frac{\\sqrt{(a^2+b^2+2ab-c^2)(c^2-b^2-a^2+2ab)}}4=\\frac{\\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}4$. Now, we plug in $\\sqrt{x^2+6^2}$ for $a$, $\\sqrt{x^2+8^2}$ for $b$, and $10$ for $c$. This gives us \\[\\frac{\\sqrt{4(x^2+6^2)(x^2+8^2)-(x^2+6^2+x^2+8^2-10^2)^2}}4=30\\] \\[\\sqrt{4x^4+400x^2+4\\cdot48^2-4x^4}=120\\] \\[400x^2+4\\cdot48^2=120^2\\] \\[25x^2+24^2=30^2\\] \\[25x^2=6^2(5^2-4^2)\\] \\[25x^2=18^2\\] \\[5x=18\\] \\[x=\\frac{18}5\\] Now multiplying by $2$, we get that the height is $\\frac{36}5$, and $m+n=36+5=041",
"Let $x$ be $\\frac12$ the height of the box. We will solve for $x$ and then multiply by $2$ at the end. The three side lengths of the triangle, by the Pythagorean Theorem, are $\\sqrt{x^2+6^2},\\sqrt{x^2+8^2},$ and $10$. Heron's formula states that the area of the triangle is $\\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter. Applying difference of squares to make the formula less computational, we get $\\frac{\\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}=\\frac{\\sqrt{(a^2+b^2+2ab-c^2)(c^2-b^2-a^2+2ab)}}4=\\frac{\\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}4$. Now, we plug in $\\sqrt{x^2+6^2}$ for $a$, $\\sqrt{x^2+8^2}$ for $b$, and $10$ for $c$. This gives us \\[\\frac{\\sqrt{4(x^2+6^2)(x^2+8^2)-(x^2+6^2+x^2+8^2-10^2)^2}}4=30\\] \\[\\sqrt{4x^4+400x^2+4\\cdot48^2-4x^4}=120\\] \\[400x^2+4\\cdot48^2=120^2\\] \\[25x^2+24^2=30^2\\] \\[25x^2=6^2(5^2-4^2)\\] \\[25x^2=18^2\\] \\[5x=18\\] \\[x=\\frac{18}5\\] Now multiplying by $2$, we get that the height is $\\frac{36}5$, and $m+n=36+5=041",
"Let the rectangular prism be $ABCDEFGH$ with $AB=12$, $AC=16$, and $AD=\\alpha$ edges of the cube. It is easy to notice that the triangle given by the problem is similar to the triangle $BCD$ by a ratio of $1:2$. Thus $[BCD]=120$. We know that the volume of the triangular pyramid defined by $A,B,C,D$ is equal to $\\frac{1}{3}\\cdot12\\cdot16\\cdot\\alpha\\cdot\\frac{1}{2}=32\\alpha$, and we can find the volume by using the triangle $BCD$ as the base. Letting $h$ be the distance from $A$ to $\\triangle BCD$, we find that the volume is $\\frac{1}{3}\\cdot h\\cdot120=40h$. Thus $h=\\frac{4}{5}\\alpha$, and all that remains is to find an equation for $h$. Let $A$ be the origin of a three-dimensional plane, and let the line segments $AB,AC,AD$ lie on the $x,y,z$ axes respectively. We know that the equation of the plane in which $\\triangle BCD$ lies can be represented by the equation $ax+by+cz=d$, where $a,b,c,d$ are real values, and we know that $B(12,0,0),C(0,16,0),D(0,0,\\alpha)$ lie on $\\triangle BCD$ which in turn lies on this plane. Thus the three points are solutions to the equation. Plugging in the values yields $d=12a=16b=\\alpha c$, and inserting these values for $a,b,c$ in terms of $d$ and simplifying yields $\\frac{1}{12}x+\\frac{1}{16}y+\\frac{1}{\\alpha}z=1$. This is the equation of the plane. By the formula, the value of $h$ (or the distance between $A$ and $\\triangle BCD$) is equivalent to $\\frac{1}{\\sqrt{\\frac{1}{12^2}+\\frac{1}{16^2}+\\frac{1}{\\alpha^2}}}$. Plugging this value in, we find that: \\[\\frac{4}{5}\\alpha=\\frac{1}{\\sqrt{\\frac{1}{12^2}+\\frac{1}{16^2}+\\frac{1}{\\alpha^2}}}\\] \\[\\sqrt{\\frac{1}{12^2}+\\frac{1}{16^2}+\\frac{1}{\\alpha^2}}=\\frac{5}{4\\alpha}\\] \\[\\frac{12^2+16^2}{12^2\\cdot16^2}+\\frac{1}{\\alpha^2}=\\frac{25}{16\\alpha^2}\\] \\[\\frac{20^2}{192^2}=\\frac{9}{16\\alpha^2}\\] \\[\\frac{20}{192}=\\frac{3}{4\\alpha}\\] \\[\\alpha=\\frac{3}{4}\\cdot\\frac{192}{20}=\\frac{36}{5}\\] Thus the answer is $36+5=041. ~eevee9406",
"Let the rectangular prism be $ABCDEFGH$ with $AB=12$, $AC=16$, and $AD=\\alpha$ edges of the cube. It is easy to notice that the triangle given by the problem is similar to the triangle $BCD$ by a ratio of $1:2$. Thus $[BCD]=120$. We know that the volume of the triangular pyramid defined by $A,B,C,D$ is equal to $\\frac{1}{3}\\cdot12\\cdot16\\cdot\\alpha\\cdot\\frac{1}{2}=32\\alpha$, and we can find the volume by using the triangle $BCD$ as the base. Letting $h$ be the distance from $A$ to $\\triangle BCD$, we find that the volume is $\\frac{1}{3}\\cdot h\\cdot120=40h$. Thus $h=\\frac{4}{5}\\alpha$, and all that remains is to find an equation for $h$. Let $A$ be the origin of a three-dimensional plane, and let the line segments $AB,AC,AD$ lie on the $x,y,z$ axes respectively. We know that the equation of the plane in which $\\triangle BCD$ lies can be represented by the equation $ax+by+cz=d$, where $a,b,c,d$ are real values, and we know that $B(12,0,0),C(0,16,0),D(0,0,\\alpha)$ lie on $\\triangle BCD$ which in turn lies on this plane. Thus the three points are solutions to the equation. Plugging in the values yields $d=12a=16b=\\alpha c$, and inserting these values for $a,b,c$ in terms of $d$ and simplifying yields $\\frac{1}{12}x+\\frac{1}{16}y+\\frac{1}{\\alpha}z=1$. This is the equation of the plane. By the formula, the value of $h$ (or the distance between $A$ and $\\triangle BCD$) is equivalent to $\\frac{1}{\\sqrt{\\frac{1}{12^2}+\\frac{1}{16^2}+\\frac{1}{\\alpha^2}}}$. Plugging this value in, we find that: \\[\\frac{4}{5}\\alpha=\\frac{1}{\\sqrt{\\frac{1}{12^2}+\\frac{1}{16^2}+\\frac{1}{\\alpha^2}}}\\] \\[\\sqrt{\\frac{1}{12^2}+\\frac{1}{16^2}+\\frac{1}{\\alpha^2}}=\\frac{5}{4\\alpha}\\] \\[\\frac{12^2+16^2}{12^2\\cdot16^2}+\\frac{1}{\\alpha^2}=\\frac{25}{16\\alpha^2}\\] \\[\\frac{20^2}{192^2}=\\frac{9}{16\\alpha^2}\\] \\[\\frac{20}{192}=\\frac{3}{4\\alpha}\\] \\[\\alpha=\\frac{3}{4}\\cdot\\frac{192}{20}=\\frac{36}{5}\\] Thus the answer is $36+5=041. ~eevee9406"
] |
2013-I-8
| 2,013
| 8
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The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by $1000$ .
| 371
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I
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[
"We know that the domain of $\\text{arcsin}$ is $[-1, 1]$, so $-1 \\le \\log_m nx \\le 1$. Now we can apply the definition of logarithms: \\[m^{-1} = \\frac1m \\le nx \\le m\\] \\[\\implies \\frac{1}{mn} \\le x \\le \\frac{m}{n}\\] Since the domain of $f(x)$ has length $\\frac{1}{2013}$, we have that \\[\\frac{m}{n} - \\frac{1}{mn} = \\frac{1}{2013}\\] \\[\\implies \\frac{m^2 - 1}{mn} = \\frac{1}{2013}\\] A larger value of $m$ will also result in a larger value of $n$ since $\\frac{m^2 - 1}{mn} \\approx \\frac{m^2}{mn}=\\frac{m}{n}$ meaning $m$ and $n$ increase about linearly for large $m$ and $n$. So we want to find the smallest value of $m$ that also results in an integer value of $n$. The problem states that $m > 1$. Thus, first we try $m = 2$: \\[\\frac{3}{2n} = \\frac{1}{2013} \\implies 2n = 3 \\cdot 2013 \\implies n \\notin \\mathbb{Z}\\] Now, we try $m=3$: \\[\\frac{8}{3n} = \\frac{1}{2013} \\implies 3n = 8 \\cdot 2013 \\implies n = 8 \\cdot 671 = 5368\\] Since $m=3$ is the smallest value of $m$ that results in an integral $n$ value, we have minimized $m+n$, which is $5368 + 3 = 5371 \\equiv 371.",
"We start with the same method as above. The domain of the arcsin function is $[-1, 1]$, so $-1 \\le \\log_{m}(nx) \\le 1$. \\[\\frac{1}{m} \\le nx \\le m\\] \\[\\frac{1}{mn} \\le x \\le \\frac{m}{n}\\] \\[\\frac{m}{n} - \\frac{1}{mn} = \\frac{1}{2013}\\] \\[n = 2013m - \\frac{2013}{m}\\] For $n$ to be an integer, $m$ must divide $2013$, and $m > 1$. To minimize $n$, $m$ should be as small as possible because increasing $m$ will decrease $\\frac{2013}{m}$, the amount you are subtracting, and increase $2013m$, the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$. We let $m$ equal $3$, the smallest factor of $2013$ that isn't $1$. Then we have $n = 2013*3 - \\frac{2013}{3} = 6039 - 671 = 5368$ $m + n = 5371$, so the answer is $371.",
"Note that we need $-1\\le f(x)\\le 1$, and this eventually gets to $\\frac{m^2-1}{mn}=\\frac{1}{2013}$. From there, break out the quadratic formula and note that \\[m= \\frac{n+\\sqrt{n^2+4026^2}}{2013\\times 2}.\\] Then we realize that the square root, call it $a$, must be an integer. Then $(a-n)(a+n)=4026^2.$ Observe carefully that $4026^2 = 2\\times 2\\times 3\\times 3\\times 11\\times 11\\times 61\\times 61$! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$, in which $a=6710$ and $n=5368$. That is our best solution, upon which we see that $m=3$, thus $371.",
"Note that we need $-1\\le f(x)\\le 1$, and this eventually gets to $\\frac{m^2-1}{mn}=\\frac{1}{2013}$. From there, break out the quadratic formula and note that \\[m= \\frac{n+\\sqrt{n^2+4026^2}}{2013\\times 2}.\\] Then we realize that the square root, call it $a$, must be an integer. Then $(a-n)(a+n)=4026^2.$ Observe carefully that $4026^2 = 2\\times 2\\times 3\\times 3\\times 11\\times 11\\times 61\\times 61$! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$, in which $a=6710$ and $n=5368$. That is our best solution, upon which we see that $m=3$, thus $371."
] |
2013-I-9
| 2,013
| 9
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A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ , $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$ . [asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]
| 113
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[
"Let $M$ and $N$ be the points on $\\overline{AB}$ and $\\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\\overline{BC}$ where the folded $A$ touches it. [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP(\"B\", (0,0), dir(200)); pair A = MP(\"A\", 12*dir(60), dir(90)); pair C = MP(\"C\", (12,0), dir(-20)); pair D = MP(\"D\", (9,0), dir(-80)); pair Y = MP(\"Y\", midpoint(A--D), dir(-50)); pair M = MP(\"M\", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP(\"N\", extension(A,C,M,Y), dir(20)); pair F = MP(\"F\", foot(A,B,C), dir(-90)); pair X = MP(\"X\", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA(\"\\theta\",F,A,D,1.8); [/asy] We have $AF=6\\sqrt{3}$ and $FD=3$, so $AD=3\\sqrt{13}$. Denote $\\angle DAF = \\theta$; we get $\\cos\\theta = 2\\sqrt{3}/\\sqrt{13}$. In triangle $AXY$, $AY=\\tfrac 12 AD = \\tfrac 32 \\sqrt{13}$, and $AX=AY\\sec\\theta =\\tfrac{13}{4}\\sqrt{3}$. In triangle $AMX$, we get $\\angle AMX=60^\\circ-\\theta$ and then use sine-law to get $MX=\\tfrac 12 AX\\csc(60^\\circ-\\theta)$; similarly, from triangle $ANX$ we get $NX=\\tfrac 12 AX\\csc(60^\\circ+\\theta)$. Thus \\[MN=\\tfrac 12 AX(\\csc(60^\\circ-\\theta) +\\csc(60^\\circ+\\theta)).\\] Since $\\sin(60^\\circ\\pm \\theta) = \\tfrac 12 (\\sqrt{3}\\cos\\theta \\pm \\sin\\theta)$, we get \\begin{align*} \\csc(60^\\circ-\\theta) +\\csc(60^\\circ+\\theta) &= \\frac{\\sqrt{3}\\cos\\theta}{\\cos^2\\theta - \\tfrac 14} = \\frac{24 \\cdot \\sqrt{13}}{35} \\end{align*} Then \\[MN = \\frac 12 AX \\cdot \\frac{24 \\cdot \\sqrt{13}}{35} = \\frac{39\\sqrt{39}}{35}\\] The answer is $39 + 39 + 35 = 113.",
"Let $P$ and $Q$ be the points on $\\overline{AB}$ and $\\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\\overline{BC}$ where the folded $A$ touches it. Let $a$, $b$, and $x$ be the lengths $AP$, $AQ$, and $PQ$, respectively. We have $PD = a$, $QD = b$, $BP = 12 - a$, $CQ = 12 - b$, $BD = 9$, and $CD = 3$. Using the Law of Cosines on $BPD$: $a^{2} = (12 - a)^{2} + 9^{2} - 2 \\times (12 - a) \\times 9 \\times \\cos{60}$ $a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a$ $a = \\frac{39}{5}$ Using the Law of Cosines on $CQD$: $b^{2} = (12 - b)^{2} +3^{2} - 2 \\times (12 - b) \\times 3 \\times \\cos{60}$ $b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b$ $b = \\frac{39}{7}$ Using the Law of Cosines on $DPQ$: $x^{2} = a^{2} + b^{2} - 2ab \\cos{60}$ $x^{2} = (\\frac{39}{5})^2 + (\\frac{39}{7})^2 - (\\frac{39}{5} \\times \\frac{39}{7})$ $x = \\frac{39 \\sqrt{39}}{35}$ The solution is $39 + 39 + 35 = 113.",
"Proceed with the same labeling as in Solution 1. $\\angle B = \\angle C = \\angle A = \\angle PDQ = 60^\\circ$ $\\angle PDB + \\angle PDQ + \\angle QDC = \\angle QDC + \\angle CQD + \\angle C = 180^\\circ$ Therefore, $\\angle PDB = \\angle CQD$. Similarly, $\\angle BPD = \\angle QDC$. Now, $\\bigtriangleup BPD$ and $\\bigtriangleup CDQ$ are similar triangles, so $\\frac{3}{12-a} = \\frac{12-b}{9} = \\frac{b}{a}$. Solving this system of equations yields $a = \\frac{39}{5}$ and $b = \\frac{39}{7}$. Using the Law of Cosines on $APQ$: $x^{2} = a^{2} + b^{2} - 2ab \\cos{60}$ $x^{2} = (\\frac{39}{5})^2 + (\\frac{39}{7})^2 - (\\frac{39}{5} \\times \\frac{39}{7})$ $x = \\frac{39 \\sqrt{39}}{35}$ The solution is $39 + 39 + 35 = 113 so that all sides are integers. Applying Law of cosines becomes easier, you just need to remember to scale back up.",
"We let the original position of $A$ be $A$, and the position of $A$ after folding be $D$. Also, we put the triangle on the coordinate plane such that $A=(0,0)$, $B=(-6,-6\\sqrt3)$, $C=(6,-6\\sqrt3)$, and $D=(3,-6\\sqrt3)$. [asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP(\"B\", (0,0), dir(200)); pair A = (9,0); pair C = MP(\"C\", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label(\"$D$\", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label(\"$A$\",X,dir(90)); draw(A--X); [/asy] Note that since $A$ is reflected over the fold line to $D$, the fold line is the perpendicular bisector of $AD$. We know $A=(0,0)$ and $D=(3,-6\\sqrt3)$. The midpoint of $AD$ (which is a point on the fold line) is $(\\tfrac32, -3\\sqrt3)$. Also, the slope of $AD$ is $\\frac{-6\\sqrt3}{3}=-2\\sqrt3$, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$, or $\\frac{1}{2\\sqrt3}=\\frac{\\sqrt3}{6}$. Then, using point slope form, the equation of the fold line is \\[y+3\\sqrt3=\\frac{\\sqrt3}{6}\\left(x-\\frac32\\right)\\]\\[y=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\] Note that the equations of lines $AB$ and $AC$ are $y=\\sqrt3x$ and $y=-\\sqrt3x$, respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$: \\[\\sqrt3 x=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\]\\[\\frac{5\\sqrt3}{6}x=-\\frac{13\\sqrt3}{4} \\implies x=-\\frac{39}{10}\\] Therefore, the point of intersection is $\\left(-\\tfrac{39}{10},-\\tfrac{39\\sqrt3}{10}\\right)$. Now, lets find the intersection with $AC$. Substituting for $y$ yields \\[-\\sqrt3 x=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\]\\[\\frac{-7\\sqrt3}{6}x=-\\frac{13\\sqrt3}{4} \\implies x=\\frac{39}{14}\\] Therefore, the point of intersection is $\\left(\\tfrac{39}{14},-\\tfrac{39\\sqrt3}{14}\\right)$. Now, we just need to use the distance formula to find the distance between $\\left(-\\tfrac{39}{10},-\\tfrac{39\\sqrt3}{10}\\right)$ and $\\left(\\tfrac{39}{14},-\\tfrac{39\\sqrt3}{14}\\right)$. \\[\\sqrt{\\left(\\frac{39}{14}+\\frac{39}{10}\\right)^2+\\left(-\\frac{39\\sqrt3}{14}+\\frac{39\\sqrt3}{10}\\right)^2}\\] The number 39 is in all of the terms, so let's factor it out: \\[39\\sqrt{\\left(\\frac{1}{14}+\\frac{1}{10}\\right)^2+\\left(-\\frac{\\sqrt3}{14}+\\frac{\\sqrt3}{10}\\right)^2}=39\\sqrt{\\left(\\frac{6}{35}\\right)^2+\\left(\\frac{\\sqrt3}{35}\\right)^2}\\]\\[\\frac{39}{35}\\sqrt{6^2+\\sqrt3^2}=\\frac{39\\sqrt{39}}{35}\\] Therefore, our answer is $39+39+35=113, and we are done. Solution by nosaj.",
"We let the original position of $A$ be $A$, and the position of $A$ after folding be $D$. Also, we put the triangle on the coordinate plane such that $A=(0,0)$, $B=(-6,-6\\sqrt3)$, $C=(6,-6\\sqrt3)$, and $D=(3,-6\\sqrt3)$. [asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP(\"B\", (0,0), dir(200)); pair A = (9,0); pair C = MP(\"C\", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label(\"$D$\", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label(\"$A$\",X,dir(90)); draw(A--X); [/asy] Note that since $A$ is reflected over the fold line to $D$, the fold line is the perpendicular bisector of $AD$. We know $A=(0,0)$ and $D=(3,-6\\sqrt3)$. The midpoint of $AD$ (which is a point on the fold line) is $(\\tfrac32, -3\\sqrt3)$. Also, the slope of $AD$ is $\\frac{-6\\sqrt3}{3}=-2\\sqrt3$, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$, or $\\frac{1}{2\\sqrt3}=\\frac{\\sqrt3}{6}$. Then, using point slope form, the equation of the fold line is \\[y+3\\sqrt3=\\frac{\\sqrt3}{6}\\left(x-\\frac32\\right)\\]\\[y=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\] Note that the equations of lines $AB$ and $AC$ are $y=\\sqrt3x$ and $y=-\\sqrt3x$, respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$: \\[\\sqrt3 x=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\]\\[\\frac{5\\sqrt3}{6}x=-\\frac{13\\sqrt3}{4} \\implies x=-\\frac{39}{10}\\] Therefore, the point of intersection is $\\left(-\\tfrac{39}{10},-\\tfrac{39\\sqrt3}{10}\\right)$. Now, lets find the intersection with $AC$. Substituting for $y$ yields \\[-\\sqrt3 x=\\frac{\\sqrt3}{6}x-\\frac{13\\sqrt3}{4}\\]\\[\\frac{-7\\sqrt3}{6}x=-\\frac{13\\sqrt3}{4} \\implies x=\\frac{39}{14}\\] Therefore, the point of intersection is $\\left(\\tfrac{39}{14},-\\tfrac{39\\sqrt3}{14}\\right)$. Now, we just need to use the distance formula to find the distance between $\\left(-\\tfrac{39}{10},-\\tfrac{39\\sqrt3}{10}\\right)$ and $\\left(\\tfrac{39}{14},-\\tfrac{39\\sqrt3}{14}\\right)$. \\[\\sqrt{\\left(\\frac{39}{14}+\\frac{39}{10}\\right)^2+\\left(-\\frac{39\\sqrt3}{14}+\\frac{39\\sqrt3}{10}\\right)^2}\\] The number 39 is in all of the terms, so let's factor it out: \\[39\\sqrt{\\left(\\frac{1}{14}+\\frac{1}{10}\\right)^2+\\left(-\\frac{\\sqrt3}{14}+\\frac{\\sqrt3}{10}\\right)^2}=39\\sqrt{\\left(\\frac{6}{35}\\right)^2+\\left(\\frac{\\sqrt3}{35}\\right)^2}\\]\\[\\frac{39}{35}\\sqrt{6^2+\\sqrt3^2}=\\frac{39\\sqrt{39}}{35}\\] Therefore, our answer is $39+39+35=113, and we are done. Solution by nosaj.",
"Note: this requires lots of calculations that increase your chance of errors, but it only requires simple understanding of areas, similar triangles, and Heron's formula. I'll just put the strategy here because I am too lazy to calculate it myself right now. 1. notice that the two triangles on the sides of the folded corner are similar. using this, we can find that the side lengths of them are 9,7.8,4.2 and 3, 45/7, 39/7 2. use heron's formula to find the areas of those two triangles. remember that it is sqrt[s(s-a)(s-b)(s-c)] 3. using the area of these triangles, we can find the area of the triangle with the length we need. 4. use heron's formula again, with the unknown length as x, and since we know the area and the other two side lengths, we can just solve for x with this equation. -EmilyQ",
"Thanks to Solution 1 for the diagram below: [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP(\"B\", (0,0), dir(200)); pair A = MP(\"A\", 12*dir(60), dir(90)); pair C = MP(\"C\", (12,0), dir(-20)); pair D = MP(\"D\", (9,0), dir(-80)); pair Y = MP(\"Y\", midpoint(A--D), dir(-50)); pair M = MP(\"M\", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP(\"N\", extension(A,C,M,Y), dir(20)); pair F = MP(\"F\", foot(A,B,C), dir(-90)); pair X = MP(\"X\", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA(\"\\theta\",F,A,D,1.8); [/asy] We will use the notation already on the diagram, but our solution is slightly different. We will only need M and N. Let NC be length a, which implies NA be 12-a. Also, AC = 3 because AB = 9 By the Law of Cosines on NCA, $(12-a)^2=a^2+3^2-2(a)(3)(cos60)$ which simplifies to: $a=\\frac{45}{7}$ Which means that NC = 45/7 and NA = 39/7. We can do the same thing for MBA. This time, MB = b. $(12-b)^2=b^2+81-2(9)(b)(cos60)$ Which gives: $b=\\frac{21}{5}$ Which implies that MA = $\\frac{39}{5}$ Now, since MAN is 60 degrees, we can apply the Law of Cosines again(I know, I don't like bashy things too) to get: $c^2=(\\frac{39}{5})^2+39/7^2-2(39/7)(39/5)(cos60)$ Which leads us to our answer => 113 ~MC",
"Thanks to Solution 1 for the diagram below: [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP(\"B\", (0,0), dir(200)); pair A = MP(\"A\", 12*dir(60), dir(90)); pair C = MP(\"C\", (12,0), dir(-20)); pair D = MP(\"D\", (9,0), dir(-80)); pair Y = MP(\"Y\", midpoint(A--D), dir(-50)); pair M = MP(\"M\", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP(\"N\", extension(A,C,M,Y), dir(20)); pair F = MP(\"F\", foot(A,B,C), dir(-90)); pair X = MP(\"X\", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA(\"\\theta\",F,A,D,1.8); [/asy] We will use the notation already on the diagram, but our solution is slightly different. We will only need M and N. Let NC be length a, which implies NA be 12-a. Also, AC = 3 because AB = 9 By the Law of Cosines on NCA, $(12-a)^2=a^2+3^2-2(a)(3)(cos60)$ which simplifies to: $a=\\frac{45}{7}$ Which means that NC = 45/7 and NA = 39/7. We can do the same thing for MBA. This time, MB = b. $(12-b)^2=b^2+81-2(9)(b)(cos60)$ Which gives: $b=\\frac{21}{5}$ Which implies that MA = $\\frac{39}{5}$ Now, since MAN is 60 degrees, we can apply the Law of Cosines again(I know, I don't like bashy things too) to get: $c^2=(\\frac{39}{5})^2+39/7^2-2(39/7)(39/5)(cos60)$ Which leads us to our answer => 113 ~MC"
] |
2013-I-10
| 2,013
| 10
|
There are nonzero integers $a$ , $b$ , $r$ , and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$ . For each possible combination of $a$ and $b$ , let ${p}_{a,b}$ be the sum of the zeros of $P(x)$ . Find the sum of the ${p}_{a,b}$ 's for all possible combinations of $a$ and $b$ .
| 80
|
I
|
[
"Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have $(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$ Applying difference of squares, and regrouping, we have $(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$ So matching coefficients, we obtain $q(r^2 + s^2) = 65$ $b = r^2 + s^2 + 2rq$ $a = q + 2r$ By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair. We proceed by determining possible values for $q$, $r$, and $s$ and using these to determine $a$ and $b$. If $q = 1$, $r^2 + s^2 = 65$ so (r, s) = $(\\pm1, \\pm 8), (\\pm8, \\pm 1), (\\pm4, \\pm 7), (\\pm7, \\pm 4)$ Similarly, for $q = 5$, $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\\pm2, \\pm 3), (\\pm3, \\pm 2)$ For $q = 13$, $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\\pm2, \\pm 1), (\\pm1, \\pm 2)$ Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$). Our answer is then $(1)(8) + (5)(4) + (13)(4) = 080."
] |
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