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1998-I-7
| 1,998 | 7 |
Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$
| 196 | null |
[
"We want $x_1 +x_2+x_3+x_4 =98$. This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$, as for all integers $y_i$, $2y_i + 1$ will be odd. Substituting we get \\[2y_1+2y_2+2y_3+2y_4 +4 = 98 \\implies y_1+y_2+y_3+y_4 =47\\] Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\\choose3}$. Computing this and dividing by 100 gives us an answer of $196. ~smartguy888",
"We want $x_1 +x_2+x_3+x_4 =98$. This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$, as for all integers $y_i$, $2y_i + 1$ will be odd. Substituting we get \\[2y_1+2y_2+2y_3+2y_4 +4 = 98 \\implies y_1+y_2+y_3+y_4 =47\\] Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\\choose3}$. Computing this and dividing by 100 gives us an answer of $196. ~smartguy888",
"Define $x_i = 2y_i - 1$. Then $2\\left(\\sum_{i = 1}^4 y_i\\right) - 4 = 98$, so $\\sum_{i = 1}^4 y_i = 51$. So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50\\choose3} = \\frac{50 * 49 * 48}{3 * 2} = 19600$, and $\\frac n{100} = 196.",
"Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into $4$ boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have $94$ stones left. Because we want an odd number in each box, we pair the stones, creating $47$ sets of $2$. Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #). Our problem can now be restated: how many ways are there to partition a line of $47$ stones? We can easily solve this by using $3$ sticks to separate the stones into $4$ groups, and this is the same as arranging a line of $3$ sticks and $47$ stones. \\[\\frac{50!}{47! \\cdot 3!} = 19600\\] \\[\\frac{50 * 49 * 48}{3 * 2} = 19600\\] Our answer is therefore $\\frac{19600}{100} = 196",
"Let $x = a + b$ and $y = c + d$. Then $x + y = 98$, where $x, y$ are positive even integers ranging from $2-98$. -When $(x, y) = (2, 96)$, $(a, b) = (1, 1)$ and $(c, d) = (1, 95), (3, 93),...,(95, 1)$. This accounts for $48$ solutions. -When $(x, y) = (4, 94)$, $(a, b) = (1, 3), (3, 1)$ and $(c, d) = (1, 93), (3, 91),...,(93, 1)$. This accounts for $94$ solutions. We quickly see that the total number of acceptable ordered pairs $(a, b, c, d)$ is: \\begin{align*} &\\mathrel{\\phantom{=}} 1 \\cdot 48 + 2 \\cdot 47 + 3 \\cdot 46 + ... + 48 \\cdot 1\\\\ &= (24.5 - 23.5)(24.5 + 23.5) + (24.5 - 22.5)(24.5 + 22.5) + ... + (24.5 + 23.5)(24.5 - 23.5)\\\\ &= 48(24.5)^2 - 2(0.5^2 + 1.5^2 + ... + 23.5^2)\\\\ &= 28812 - \\frac{1^2 + 3^2 + ... + 47^2}{2}\\\\ &= 28812 - \\frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\\\ &= 28812 - \\frac{\\frac{47(47 + 1)(2(47) + 1)}{6} - \\frac{4(23)(23 + 1)(2(23) + 1)}{6}}{2}\\\\ &= 19600 \\end{align*} Therefore, $\\frac{n}{100} = \\frac{19600}{100} = 196.) -baker77",
"We write the generating functions for each of the terms, and obtain $(x+x^3+x^5\\cdots)^4$ as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\\left(\\frac{1}{1-x^2}\\right)^4=\\binom{3}{3} +\\binom{4}{3}x^2+\\binom{5}{3}x^4 \\cdots$ and in general we want that the coefficient of $x^{2k}$ is $\\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\\binom{50}{3}=19600$ in which $\\frac{n}{100}=196"
] |
1998-I-8
| 1,998 | 8 |
Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encountered. What positive integer $x$ produces a sequence of maximum length?
| 618 | null |
[
"The best way to start is to just write out some terms. 0 1 2 3 4 5 6 $\\quad 1000 \\quad$aa $\\quad x \\quad$aaa $1000 - x$ $2x - 1000$a $2000 - 3x$ $5x - 3000$ $5000 - 8x$ It is now apparent that each term can be written as $n \\equiv 0 \\pmod{2}\\quad\\quad F_{n-1}\\cdot 1000-F_n\\cdot x$ $n \\equiv 1 \\pmod{2}\\quad\\quad F_{n}\\cdot x-F_{n-1}\\cdot 1000$ where the $F_{n}$ are Fibonacci numbers. This can be proven through induction. Solution 1 We can start to write out some of the inequalities now: $x > 0$ $1000 - x > 0 \\Longrightarrow x < 1000$ $2x - 1000 > 0 \\Longrightarrow x > 500$ $2000 - 3x > 0 \\Longrightarrow x < 666.\\overline{6}$ $5x - 3000 > 0 \\Longrightarrow x > 600$ And in general, $n \\equiv 0 \\pmod{2}\\quad\\quad x < \\frac{F_{n-1}}{F_n} \\cdot 1000$ $n \\equiv 1 \\pmod{2}\\quad\\quad x > \\frac{F_{n-1}}{F_{n}} \\cdot 1000$ It is apparent that the bounds are slowly closing in on $x$, so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11): $x < \\frac{F_{9}}{F_{10}} \\cdot 1000 = \\frac{34}{55} \\cdot 1000 = 618.\\overline{18}$ $x > \\frac{F_{10}}{F_{11}} \\cdot 1000 = \\frac{55}{89} \\cdot 1000 \\approx 617.977$ Thus the sequence is maximized when $x = 618",
"We can start to write out some of the inequalities now: $x > 0$ $1000 - x > 0 \\Longrightarrow x < 1000$ $2x - 1000 > 0 \\Longrightarrow x > 500$ $2000 - 3x > 0 \\Longrightarrow x < 666.\\overline{6}$ $5x - 3000 > 0 \\Longrightarrow x > 600$ And in general, $n \\equiv 0 \\pmod{2}\\quad\\quad x < \\frac{F_{n-1}}{F_n} \\cdot 1000$ $n \\equiv 1 \\pmod{2}\\quad\\quad x > \\frac{F_{n-1}}{F_{n}} \\cdot 1000$ It is apparent that the bounds are slowly closing in on $x$, so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11): $x < \\frac{F_{9}}{F_{10}} \\cdot 1000 = \\frac{34}{55} \\cdot 1000 = 618.\\overline{18}$ $x > \\frac{F_{10}}{F_{11}} \\cdot 1000 = \\frac{55}{89} \\cdot 1000 \\approx 617.977$ Thus the sequence is maximized when $x = 618",
"It is well known that $\\lim_{n\\rightarrow\\infty} \\frac{F_{n-1}}{F_n} = \\phi - 1 =\\frac{1 + \\sqrt{5}}{2} - 1 \\approx .61803$, so $1000 \\cdot \\frac{F_{n-1}}{F_n}$ approaches $x = 618"
] |
1998-I-9
| 1,998 | 9 |
Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly $m$ minutes. The probability that either one arrives while the other is in the cafeteria is $40 \%,$ and $m = a - b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime . Find $a + b + c.$
| 87 | null |
[
"Solution 1 Let the two mathematicians be $M_1$ and $M_2$. Consider plotting the times that they are on break on a coordinate plane with one axis being the time $M_1$ arrives and the second axis being the time $M_2$ arrives (in minutes past 9 a.m.). The two mathematicians meet each other when $|M_1-M_2| \\leq m$. Also because the mathematicians arrive between 9 and 10, $0 \\leq M_1,M_2 \\leq 60$. Therefore, $60\\times 60$ square represents the possible arrival times of the mathematicians, while the shaded region represents the arrival times where they meet. [asy] import graph; size(180); real m=60-12*sqrt(15); draw((0,0)--(60,0)--(60,60)--(0,60)--cycle); fill((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle,lightgray); draw((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle); xaxis(\"$M_1$\",-10,80); yaxis(\"$M_2$\",-10,80); label(rotate(45)*\"$M_1-M_2\\le m$\",((m+60)/2,(60-m)/2),NW,fontsize(9)); label(rotate(45)*\"$M_1-M_2\\ge -m$\",((60-m)/2,(m+60)/2),SE,fontsize(9)); label(\"$m$\",(m,0),S); label(\"$m$\",(0,m),W); label(\"$60$\",(60,0),S); label(\"$60$\",(0,60),W); [/asy] It's easier to compute the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet: $\\frac{(60-m)^2}{60^2} = 0.6$ $(60-m)^2 = 36\\cdot 60$ $60 - m = 12\\sqrt{15}$ $\\Rightarrow m = 60-12\\sqrt{15}$ So the answer is $60 + 12 + 15 = 087$. Solution 2 Case 1: Case 2: We draw a number line representing the time interval. If mathematician $M_1$ comes in at the center of the time period, then the two mathematicions will meet if $M_2$ comes in somewhere between $m$ minutes before and after $M_1$ comes (a total range of $2m$ minutes). However, if $M_1$ comes into the cafeteria in the first or last $m$ minutes, then the range in which $M_2$ is reduced to somewhere in between $m$ and $2m$. We now try to find the weighted average of the chance that the two meet. In the central $60-2m$ minutes, $M_1$ and $M_2$ have to enter the cafeteria within $m$ minutes of each other; so if we fix point $M_1$ then $M_2$ has a $\\frac{2m}{60} = \\frac{m}{30}$ probability of meeting. In the first and last $2m$ minutes, the probability that the two meet ranges from $\\frac{m}{60}$ to $\\frac{2m}{60}$, depending upon the location of $M_1$ with respect to the endpoints. Intuitively, the average probability will occur at $\\frac{\\frac{3}{2}m}{60} = \\frac{m}{40}$. So the weighted average is: $\\frac{\\frac{m}{30}(60-2m) + \\frac{m}{40}(2m)}{60} = \\frac{40}{100}$ $0 = \\frac{m^2}{60} - 2m + 24$ $0 = m^2 - 120m + 1440$ Solving this quadratic, we get two roots, $60 \\pm 12\\sqrt{15}$. However, $m < 60$, so we discard the greater root; and thus our answer $60 + 12 + 15 = 087$.",
"Let the two mathematicians be $M_1$ and $M_2$. Consider plotting the times that they are on break on a coordinate plane with one axis being the time $M_1$ arrives and the second axis being the time $M_2$ arrives (in minutes past 9 a.m.). The two mathematicians meet each other when $|M_1-M_2| \\leq m$. Also because the mathematicians arrive between 9 and 10, $0 \\leq M_1,M_2 \\leq 60$. Therefore, $60\\times 60$ square represents the possible arrival times of the mathematicians, while the shaded region represents the arrival times where they meet. [asy] import graph; size(180); real m=60-12*sqrt(15); draw((0,0)--(60,0)--(60,60)--(0,60)--cycle); fill((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle,lightgray); draw((m,0)--(60,60-m)--(60,60)--(60-m,60)--(0,m)--(0,0)--cycle); xaxis(\"$M_1$\",-10,80); yaxis(\"$M_2$\",-10,80); label(rotate(45)*\"$M_1-M_2\\le m$\",((m+60)/2,(60-m)/2),NW,fontsize(9)); label(rotate(45)*\"$M_1-M_2\\ge -m$\",((60-m)/2,(m+60)/2),SE,fontsize(9)); label(\"$m$\",(m,0),S); label(\"$m$\",(0,m),W); label(\"$60$\",(60,0),S); label(\"$60$\",(0,60),W); [/asy] It's easier to compute the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet: $\\frac{(60-m)^2}{60^2} = 0.6$ $(60-m)^2 = 36\\cdot 60$ $60 - m = 12\\sqrt{15}$ $\\Rightarrow m = 60-12\\sqrt{15}$ So the answer is $60 + 12 + 15 = 087$. Solution 2 Case 1: Case 2: We draw a number line representing the time interval. If mathematician $M_1$ comes in at the center of the time period, then the two mathematicions will meet if $M_2$ comes in somewhere between $m$ minutes before and after $M_1$ comes (a total range of $2m$ minutes). However, if $M_1$ comes into the cafeteria in the first or last $m$ minutes, then the range in which $M_2$ is reduced to somewhere in between $m$ and $2m$. We now try to find the weighted average of the chance that the two meet. In the central $60-2m$ minutes, $M_1$ and $M_2$ have to enter the cafeteria within $m$ minutes of each other; so if we fix point $M_1$ then $M_2$ has a $\\frac{2m}{60} = \\frac{m}{30}$ probability of meeting. In the first and last $2m$ minutes, the probability that the two meet ranges from $\\frac{m}{60}$ to $\\frac{2m}{60}$, depending upon the location of $M_1$ with respect to the endpoints. Intuitively, the average probability will occur at $\\frac{\\frac{3}{2}m}{60} = \\frac{m}{40}$. So the weighted average is: $\\frac{\\frac{m}{30}(60-2m) + \\frac{m}{40}(2m)}{60} = \\frac{40}{100}$ $0 = \\frac{m^2}{60} - 2m + 24$ $0 = m^2 - 120m + 1440$ Solving this quadratic, we get two roots, $60 \\pm 12\\sqrt{15}$. However, $m < 60$, so we discard the greater root; and thus our answer $60 + 12 + 15 = 087$.",
"Case 1: Case 2: We draw a number line representing the time interval. If mathematician $M_1$ comes in at the center of the time period, then the two mathematicions will meet if $M_2$ comes in somewhere between $m$ minutes before and after $M_1$ comes (a total range of $2m$ minutes). However, if $M_1$ comes into the cafeteria in the first or last $m$ minutes, then the range in which $M_2$ is reduced to somewhere in between $m$ and $2m$. We now try to find the weighted average of the chance that the two meet. In the central $60-2m$ minutes, $M_1$ and $M_2$ have to enter the cafeteria within $m$ minutes of each other; so if we fix point $M_1$ then $M_2$ has a $\\frac{2m}{60} = \\frac{m}{30}$ probability of meeting. In the first and last $2m$ minutes, the probability that the two meet ranges from $\\frac{m}{60}$ to $\\frac{2m}{60}$, depending upon the location of $M_1$ with respect to the endpoints. Intuitively, the average probability will occur at $\\frac{\\frac{3}{2}m}{60} = \\frac{m}{40}$. So the weighted average is: $\\frac{\\frac{m}{30}(60-2m) + \\frac{m}{40}(2m)}{60} = \\frac{40}{100}$ $0 = \\frac{m^2}{60} - 2m + 24$ $0 = m^2 - 120m + 1440$ Solving this quadratic, we get two roots, $60 \\pm 12\\sqrt{15}$. However, $m < 60$, so we discard the greater root; and thus our answer $60 + 12 + 15 = 087$."
] |
1998-I-10
| 1,998 | 10 |
Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon . A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime . Find $a + b + c$ .
| 152 | null |
[
"The key is to realize the significance that the figures are spheres, not circles. The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out. Let us examine the relation between one of the outside 8 spheres and the center one (with radius $r$): If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$. Then by the Pythagorean Theorem: \\[x^2 + (r-100)^2 = (r+100)^2 \\Longrightarrow x = 20\\sqrt{r}\\] $x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\\sqrt{r}$. We can draw another right triangle as shown above. One leg has a length of $200$. The other can be found by partitioning the leg into three sections and using $45-45-90 \\triangle$s to see that the leg is $100\\sqrt{2} + 200 + 100\\sqrt{2} = 200(\\sqrt{2} + 1)$. Pythagorean Theorem: \\begin{eqnarray*} (40\\sqrt{r})^2 &=& 200^2 + [200(\\sqrt{2}+1)]^2\\\\ 1600r &=& 200^2[(1 + \\sqrt{2})^2 + 1] \\\\ r &=& 100 + 50\\sqrt{2} \\end{eqnarray*} Thus $a + b + c = 100 + 50 + 2 = 152.",
"Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$. Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \\begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*cos(45^\\circ) \\\\ &=& 2x^2 - 2x^2*\\frac{\\sqrt{2}}{2} \\\\ &=& (2-\\sqrt{2})x^2 \\end{eqnarray*} And thus \\[x = \\frac{200}{\\sqrt{2-\\sqrt{2}}}\\] From the above, $x = 20\\sqrt{r}$, so we get \\begin{eqnarray*} r &=& (\\frac{200}{20(\\sqrt{2-\\sqrt{2}})})^2 \\\\ &=& (\\frac{10}{\\sqrt{2-\\sqrt{2}})})^2 \\cdot \\frac{2+\\sqrt{2}}{2+\\sqrt{2}} \\\\ &=& \\frac{200 + 100\\sqrt{2}}{2} \\\\ &=& 100 + 50\\sqrt{2} \\end{eqnarray*} And hence the answer is $100 + 50 + 2 \\Rightarrow 152 ~hashbrown2009"
] |
1998-I-11
| 1,998 | 11 |
Three of the edges of a cube are $\overline{AB}, \overline{BC},$ and $\overline{CD},$ and $\overline{AD}$ is an interior diagonal . Points $P, Q,$ and $R$ are on $\overline{AB}, \overline{BC},$ and $\overline{CD},$ respectively, so that $AP = 5, PB = 15, BQ = 15,$ and $CR = 10.$ What is the area of the polygon that is the intersection of plane $PQR$ and the cube?
| 525 | null |
[
"\"For non-asymptote version of image, see Image:1998_AIME-11.png\" [asy] import three; size(280); defaultpen(linewidth(0.6)+fontsize(9)); currentprojection=perspective(30,-60,40); triple A=(0,0,0),B=(20,0,0),C=(20,0,20),D=(20,20,20); triple P=(5,0,0),Q=(20,0,15),R=(20,10,20),Pa=(15,20,20),Qa=(0,20,5),Ra=(0,10,0); draw(box((0,0,0),(20,20,20))); draw(P--Q--R--Pa--Qa--Ra--cycle,linewidth(0.7)); label(\"\\(A\\,(0,0,0)\\)\",A,SW); label(\"\\(B\\,(20,0,0)\\)\",B,S); label(\"\\(C\\,(20,0,20)\\)\",C,SW); label(\"\\(D\\,(20,20,20)\\)\",D,E); label(\"\\(P\\,(5,0,0)\\)\",P,SW); label(\"\\(Q\\,(20,0,15)\\)\",Q,E); label(\"\\(R\\,(20,10,20)\\)\",R,E); label(\"\\((15,20,20)\\)\",Pa,N); label(\"\\((0,20,5)\\)\",Qa,W); label(\"\\((0,10,0)\\)\",Ra,W); [/asy] This approach uses analytic geometry. Let $A$ be at the origin, $B$ at $(20,0,0)$, $C$ at $(20,0,20)$, and $D$ at $(20,20,20)$. Thus, $P$ is at $(5,0,0)$, $Q$ is at $(20,0,15)$, and $R$ is at $(20,10,20)$. Let the plane $PQR$ have the equation $ax + by + cz = d$. Using point $P$, we get that $5a = d$. Using point $Q$, we get \\[20a + 15c = d \\Longrightarrow 4d + 15c = d \\Longrightarrow d = -5c\\]Using point $R$, we get \\[20a + 10b + 20c = d \\Longrightarrow 4d + 10b - 4d = d \\Longrightarrow d = 10b\\]Thus plane $PQR$’s equation reduces to \\[\\frac{d}{5}x + \\frac{d}{10}y - \\frac{d}{5}z = d \\Longrightarrow 2x + y - 2z = 10\\] We know need to find the intersection of this plane with that of $z = 0$, $z = 20$, $x = 0$, and $y = 20$. After doing a little bit of algebra, the intersections are the lines \\[y = -2x + 10\\] \\[y = -2x + 50\\] \\[y = 2z + 10\\] \\[z = x + 5\\]Thus, there are three more vertices on the polygon, which are at $(0,10,0)(0,20,5)(15,20,20)$. We can find the lengths of the sides of the polygons now. There are 4 right triangles with legs of length 5 and 10, so their hypotenuses are $5\\sqrt{5}$. The other two are of $45-45-90 \\triangle$s with legs of length 15, so their hypotenuses are $15\\sqrt{2}$. So we have a hexagon with sides $15\\sqrt{2},5\\sqrt{5}, 5\\sqrt{5},15\\sqrt{2}, 5\\sqrt{5},5\\sqrt{5}$ By symmetry, we know that opposite angles of the polygon are congruent. We can also calculate the length of the long diagonal by noting that it is of the same length of a face diagonal, making it $20\\sqrt{2}$. [asy] size(190); pointpen=black;pathpen=black; real s=2^.5; pair P=(0,0),Q=(7.5*s,2.5*s),R=Q+(0,15*s),Pa=(0,20*s),Qa=(-Q.x,Q.y),Ra=(-R.x,R.y); D(P--Q--R--Pa--Ra--Qa--cycle);D(R--Ra);D(Q--Qa);D(P--Pa); MP(\"15\\sqrt{2}\",(Q+R)/2,E); MP(\"5\\sqrt{5}\",(P+Q)/2,SE); MP(\"5\\sqrt{5}\",(R+Pa)/2,NE); MP(\"20\\sqrt{2}\",(P+Pa)/2,W); [/asy] The height of the triangles at the top/bottom is $\\frac{20\\sqrt{2} - 15\\sqrt{2}}{2} = \\frac{5}{2}\\sqrt{2}$. The Pythagorean Theorem gives that half of the base of the triangles is $\\frac{15}{\\sqrt{2}}$. We find that the middle rectangle is actually a square, so the total area is $(15\\sqrt{2})^2 + 4\\left(\\frac 12\\right)\\left(\\frac 52\\sqrt{2}\\right)\\left(\\frac{15}{\\sqrt{2}}\\right) = 525$.",
"First, note that whenever the plane intersects two opposite faces of the cube, the resulting line segments must be parallel. Because they are part of parallel planes (the faces), they must be either parallel or skew; they are both part of plane $PQR$, so they cannot be skew. Therefore, they are parallel. Let the cube's vertices be $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$, with $A$, $B$, and $C$ on the bottom face as before, $H$ being the other bottom vertex, $D$ directly above $C$, $E$ above $B$, $F$ above $A$, and $G$ above $H$. Clearly, the next vertex of the intersection (starting with $P$, $Q$, $R$) will be somewhere on $DG$. Let it be $X$, and have a distance of $x$ from D, and a distance of $20 - x$ from $G$. Then, the next vertex will be somewhere on $FG$. It must be parallel to $PQ$, so this implies that it has a distance of $20 - x$ from $G$, and thus a distance of $x$ from $F$. Now, the next vertex (call it $Y$) will be somewhere on $AF$. The segment must be parallel to $QR$, so $FY$ must have length $2x$, and $AY$ must be $20 - 2x$. Since $DX \\parallel AP$, $DR \\parallel AY$, and $RX \\parallel PY$, we must have $\\triangle{APY} \\sim \\triangle{DXR}$; therefore, \\[\\frac{AP}{DX}=\\frac{AY}{DR}\\] \\[\\frac{5}{x}=\\frac{20-2x}{10}\\] \\[x^{2}-10x+25=0\\] \\[x=5\\] We can now find that the hexagon has side lengths $15\\sqrt {2}$, $5\\sqrt {5}$, $5\\sqrt {5}$, $15\\sqrt {2}$, $5\\sqrt {5}$, and $5\\sqrt {5}$. Moreover, opposite angles of this must be equal (by symmetry), so segment $RY$ divides the hexagon into two isosceles trapezoids. It is easy to find the length of $RY$ (they're midpoints of opposite edges, so the distance between the two points is equal to a face diagonal of the cube, or $20\\sqrt {2}$), so it is now easy to finish the problem. From here, we can continue as in the first solution."
] |
1998-I-12
| 1,998 | 12 |
Let $ABC$ be equilateral , and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime . What is $a^{2} + b^{2} + c^{2}$ ?
| 83 | null |
[
"WLOG, assume that $AB = BC = AC = 2$. We let $x = EP = FQ$, $y = EQ$, $k = PQ$. Since $AE = \\frac {1}{2}AB$ and $AD = \\frac {1}{2}AC$, $\\triangle AED \\sim \\triangle ABC$ and $ED \\parallel BC$. By alternate interior angles, we have $\\angle PEQ = \\angle BFQ$ and $\\angle EPQ = \\angle FBQ$. By vertical angles, $\\angle EQP = \\angle FQB$. Thus $\\triangle EQP \\sim \\triangle FQB$, so $\\frac {EP}{EQ} = \\frac {FB}{FQ}\\Longrightarrow\\frac {x}{y} = \\frac {1}{x}\\Longrightarrow x^{2} = y$. Since $\\triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1\\Longrightarrow x + y = 1$. Solving for $x$ and $y$ using $x^{2} = y$ and $x + y = 1$ gives $x = \\frac {\\sqrt {5} - 1}{2}$ and $y = \\frac {3 - \\sqrt {5}}{2}$. Using the Law of Cosines, we get $k^{2} = x^{2} + y^{2} - 2xy\\cos{\\frac {\\pi}{3}}$ $= \\left(\\frac {\\sqrt {5} - 1}{2}\\right)^{2} + \\left(\\frac {3 - \\sqrt {5}}{2}\\right)^{2} - 2\\left(\\frac {\\sqrt {5} - 1}{2}\\right)\\left(\\frac {3 - \\sqrt {5}}{2}\\right)\\cos{\\frac {\\pi}{3}}$ $= 7 - 3\\sqrt {5}$ We want the ratio of the squares of the sides, so $\\frac {(2)^{2}}{k^{2}} = \\frac {4}{7 - 3\\sqrt {5}} = 7 + 3\\sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = 083.",
"WLOG, let $\\Delta ABC$ have side length $2.$ Then, $DE = EF = FD = 1.$ We also notice that $\\angle CEP = \\angle DEF = 60^{\\circ},$ meaning $\\angle CEF = \\angle CEP + \\angle DEF = 120^{\\circ}.$ Let $EP = x.$ Since $FQ = x$ by congruent triangles $\\Delta EPC$ and $\\Delta FQA,$ $EQ = EF - FQ = 1-x.$ We can now apply Law of Cosines to $\\Delta CEP, \\Delta PEQ,$ and $\\Delta CEQ.$ By LoC on $\\Delta CEP,$ we get \\[CP^2 = 1^2 + x^2 - 2\\cdot 1\\cdot x\\cdot \\left(\\frac{1}{2}\\right) = x^2 - x + 1.\\] In a similar vein, using LoC on $\\Delta PEQ$ and $\\Delta CEQ,$ respectively, earns \\[PQ^2 = x^2 + (1-x)^2 - 2\\cdot x\\cdot (1-x)\\cdot \\left(\\frac{1}{2}\\right) = 3x^2 - 3x + 1\\] \\[CQ^2 = 1^2 + (1-x)^2 - 2\\cdot 1\\cdot (1-x)\\cdot \\left(-\\frac{1}{2}\\right) = x^2 - 3x + 3\\] We have $CP^2, PQ^2,$ and $CQ^2.$ Additionally, by segment addition, $CP + PQ = CQ.$ Solving for $CP, PQ,$ and $CQ$ from the Law of Cosines expressions and plugging them into the segment addition gets the (admittedly-ugly) equation \\[\\sqrt{x^2-3x+3} = \\sqrt{x^2-x+1} + \\sqrt{3x^2-3x+1}.\\] Since the equation is ugly, we look at what the problem is asking for us to solve. We want $\\frac{[ABC]}{[PQR]}.$ We see that $[ABC] = \\sqrt{3}$ and $[PQR] = [DEF] - [PDR] - [RFQ] - [QEP] = \\frac{\\sqrt{3}}{4} - \\frac{3}{2}\\left(\\frac{\\sqrt{3}}{2}x(1-x)\\right),$ since $[PDR] = [RFQ] = [QEP] = \\frac{1}{2}x(1-x)\\frac{\\sqrt{3}}{2}$ from the sine area formula. Simplifying $\\frac{[ABC]}{[PQR]}$ gets us wanting to find $\\frac{4}{3x^2-3x+1}.$ We see $3x^2-3x+1$ in both the denominator of what we want and under a radicand in our algebraic expression, which leads us to think the calculations may not be that bad. Isolate $\\sqrt{3x^2-3x+1}$ and square to get \\[3x^2-3x+1 = 2x^2-4x+4-2\\sqrt{(x^2-3x+3)(x^2-x+1)}\\] Isolate the radicand and square and expand to get $x^4+2x^3-5x^2-6x+9=4x^4-16x^3+28x^2-24x+12,$ and moving terms to one side and dividing by $3,$ we get \\[x^4-6x^3+11x^2-6x+1=0.\\] This can be factored into $(x^2-3x+1)^2 = 0 \\rightarrow x^2-3x+1 = 0 \\rightarrow x = \\frac{3 \\pm \\sqrt{5}}{2}.$ From the equation $x^2-3x+1=0,$ we have $x^2=3x+1,$ so plugging that value into the expression we want to find, we get $\\frac{4}{3(3x+1)-3x+1} = \\frac{4}{6x+2}.$ Substituting $x = \\frac{3-\\sqrt{5}}{2}$ into $\\frac{4}{6x+2}$ gets an expression of $7+3\\sqrt{5},$ so $a^2+b^2+c^2 = 083. -PureSwag"
] |
1998-I-13
| 1,998 | 13 |
If $\{a_1,a_2,a_3,\ldots,a_n\}$ is a set of real numbers , indexed so that $a_1 < a_2 < a_3 < \cdots < a_n,$ its complex power sum is defined to be $a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,$ where $i^2 = - 1.$ Let $S_n$ be the sum of the complex power sums of all nonempty subsets of $\{1,2,\ldots,n\}.$ Given that $S_8 = - 176 - 64i$ and $S_9 = p + qi,$ where $p$ and $q$ are integers, find $|p| + |q|.$
| 368 | null |
[
"We note that the number of subsets (for now, including the empty subset, which we will just define to have a power sum of zero) with $9$ in it is equal to the number of subsets without a $9$. To easily see this, take all possible subsets of $\\{1,2,\\ldots,8\\}$. Since the sets are ordered, a $9$ must go at the end; hence we can just append a $9$ to any of those subsets to get a new one. Now that we have drawn that bijection, we can calculate the complex power sum recursively. Since appending a $9$ to a subset doesn't change anything about that subset's complex power sum besides adding an additional term, we have that $S_9 = 2S_8 + T_9$, where $T_9$ refers to the sum of all of the $9i^x$. If a subset of size 1 has a 9, then its power sum must be $9i$, and there is only $1$ of these such subsets. There are ${8\\choose1}$ with $9\\cdot i^2$, ${8\\choose2}$ with $9\\cdot i^3$, and so forth. So $T_9 =\\sum_{k=0}^{8} 9{8\\choose{k}}i^{k+1}$. This is exactly the binomial expansion of $9i \\cdot (1+i)^8$. We can use De Moivre's Theorem to calculate the power: $(\\sqrt{2})^8\\cos{8\\cdot45} = 16$. Hence $T_9 = 16\\cdot9i = 144i$, and $S_9 = 2S_8 + 144i = 2(-176 -64i) + 144i = -352 + 16i$. Thus, $|p| + |q| = |-352| + |16| = 368."
] |
1998-I-14
| 1,998 | 14 |
An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$ ?
| 130 | null |
[
"\\[2mnp = (m+2)(n+2)(p+2)\\] Let’s solve for $p$: \\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\\] \\[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\\] \\[p = \\frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \\frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\\] Clearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \\Longrightarrow (m-2)(n-2) = 9$. The possible pairs of factors of $9$ are $(1,9)(3,3)$. These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$, while the second pair gives $98$. We now check that $130$ is optimal, setting $a=m-2$, $b=n-2$ in order to simplify calculations. Since \\[0 \\le (a-1)(b-1) \\implies a+b \\le ab+1\\] We have \\[p = \\frac{2(a+4)(b+4)}{ab-8} = \\frac{2ab+8(a+b)+32}{ab-8} \\le \\frac{2ab+8(ab+1)+32}{ab-8} = 10 + \\frac{120}{ab-8} \\le 130\\] Where we see $(m,n)=(3,11)$ gives us our maximum value of $130.",
"Similarly as above, we solve for $p,$ but we express the denominator differently: \\[p=\\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \\implies \\dfrac{1}{p}=\\dfrac{1}{2}-\\dfrac{2(m+n+2)}{(m+2)(n+2)}.\\] Hence, it suffices to maximize $\\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer. Then since $\\dfrac{m+n+2}{(m+2)(n+2)}>\\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ \\[\\implies \\dfrac{1}{p}=\\dfrac{1}{2}-\\dfrac{2(n+5)}{5(n+2)}=\\dfrac{n-10}{10(n+2)},\\] where we simply let $n=11$ to achieve $p=130 ~Generic_Username"
] |
1998-I-15
| 1,998 | 15 |
Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j$ . Let $D_{40}$ be the set of all dominos whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$
| 761 | null |
[
"We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes. You need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment. But you can get up to 38 on each segment because you go in to the point then out on a different segment. Counting going out from the starting and ending at the ending point we have: $\\frac{38\\cdot 38 + 2\\cdot 39}2 = 761 Clarification To clarify the above solution, every time you move to a new vertex, you take one path in and must leave by another path. Therefore every vertex needs to have an even number of segments leaving it (with the exception of the first and last), because the \"in\" and \"out\" segments must make a pair.",
"A proper sequence can be represented by writing the common coordinates of adjacent ordered pairs once. For example, represent (4,7),(7,3),(3,5) as $4,7,3,5 .$ Label the vertices of a regular $n$ -gon $1,2,3, \\ldots, n .$ Each domino is thereby represented by a directed segment from one vertex of the $n$ -gon to another, and a proper sequence is represented as a path that retraces no segment. Each time that such a path reaches a non-terminal vertex, it must leave it. Thus, when $n$ is even, it is not possible for such a path to trace every segment, for an odd number of segments emanate from each vertex. By removing $\\frac{1}{2}(n-2)$ suitable segments, however, it can be arranged that $n-2$ segments will emanate from $n-2$ of the vertices and that an odd number of segments will emanate from exactly two of the vertices. In this situation, a path can be found that traces every remaining segment exactly once, starting at one of the two exceptional vertices and finishing at the other. This path will have length $\\binom{n}{2}-\\frac{1}{2}(n-2),$ which is 761 when $n=40$. ~phoenixfire Note 1 When $n$ is odd, a proper sequence of length $\\binom{n}{2}$ can be found using the dominos of $D_{n}$. In this case, the second coordinate of the final domino equals the first coordinate of the first domino. ~phoenixfire",
"Let $A_{n}=\\{1,2,3, \\ldots, n\\}$ and $D_{n}$ be the set of dominos that can be formed using integers in $A_{n} .$ Each $k$ in $A_{n}$ appears in $2(n-1)$ dominos in $D_{n},$ hence appears at most $n-1$ times in a proper sequence from $D_{n}.$ Except possibly for the integers $i$ and $j$ that begin and end a proper sequence, every integer appears an even number of times in the sequence. Thus, if $n$ is even, each integer different from $i$ and $j$ appears on at most $n-2$ dominos in the sequence, because $n-2$ is even, and $i$ and $j$ themselves appear on at most $n-1$ dominos each. This gives an upper bound of \\[\\frac{1}{2}\\left[(n-2)^{2}+2(n-1)\\right]=\\frac{n^{2}-2 n+2}{2}\\] dominos in the longest proper sequence in $D_{n}.$ This bound is in fact attained for every even $n.$ It is easy to verify this for $n=2$, so assume inductively that a sequence of this length has been found for a particular value of $n$. Without loss of generality, assume $i=1$ and $j=2,$ and let $_{p} X_{p+2}$ denote a four-domino sequence of the form $(p, n+1)(n+1, p+1)(p+1, n+2)(n+2, p+2) .$ By appending \\[{ }_{2} X_{4},{ }_{4} X_{6}, \\ldots,{ }_{n-2} X_{n},(n, n+1)(n+1,1)(1, n+2)(n+2,2)\\] to the given proper sequence, a proper sequence of length \\[\\frac{n^{2}-2 n+2}{2}+4 \\cdot \\frac{n-2}{2}+4=\\frac{n^{2}+2 n+2}{2}=\\frac{(n+2)^{2}-2(n+2)+2}{2}\\] is obtained that starts at $i=1$ and ends at $j=2 .$ This completes the inductive proof. In particular, the longest proper sequence when $n=40$ is 761. ~phoenixfire Note 2 In the language of graph theory, this is an example of an Eulerian circuit. ~phoenixfire",
"Consider the segments joining the vertices of a regular $n$-gon. For odd $n$, we see that the number of segments is quite easily $\\binom{n-1}{2}$. This is because every vertex touches every other vertex the same number of times. ($\\frac{n-1}{2}$ times to be exact). Hence the answer for odd cases is $n\\frac{n-1}{2}=\\binom{n-1}{2}$. (This is because a segment that starts at the first vertex also ends at the first vertex). For even $n$ however, every vertex touches $\\frac{n-2}{2}$ vertices. However, one may be motivated to say that the answer (as in the odd case) is $n\\frac{n-2}{2}$. But this is incorrect, because for the even case, it never ends at the vertex (the first vertex) you started at. So it must end at another vertex. But that vertex has already $\\frac{n-2}{2}$ other segments touching it. So we have that the final answer is $1$ plus $n\\frac{n-2}{2}$. The case for $n=40$, is $761$. ~th1nq3r Note The reason it touches every single other vertex $\\frac{n-1}{2}$ for odd $n$ is because there are a total of $\\binom{n-1}{2}$ segments, and once dividing $\\binom{n-1}{2}$ by $n$, you will then have the number of segments that are connected to each vertex. For the even case every vertex has at least $\\frac{n-2}{2}$ other segments touching it. This is because (you CAN convince yourself through a painful induction/observation argument, but you probably shouldn't) $n$ is even, and if you try to apply the odd case to the even case, (namely that there is $\\frac{n-1}{2}$ segments touching each vertex), there would have to then be $n\\frac{n-1}{2}$ total segments, which never works since it never loops back to the vertex you started on). So at LEAST, there should be $\\frac{n-2}{2}$ segments touching each vertex. However, there is also at most $\\frac{n-2}{2}$ segments touching each vertex. Hence by the (what I call) the \"less-than-or-greater-than\" argument, there must be $\\frac{n-2}{2}$ segments out of each vertex. (Mind the plus one extra vertex since once again, the vertex you started on, it doesn't loop around, so it must end at another vertex). Hence the answer is $n\\frac{n-2}{2}+1$. (I am not very good at explaining things. Sorry if it didn't make sense. Maybe if you find some way to contact me on aops, I could try and help). (For instance, I could see possible confusion at the part where I claim that the minimum lines be $\\frac{n-2}{2}$. You might think, \"Oh why not $\\frac{n-1}{2}-1$, or even $\\frac{n-1}{2}-2$ or even subtract $3$?\" Well if you actually start off with the fact that there should be at MOST $\\frac{n-2}{2}$ segments from each vertex, then it is obvious, since then there must be a minimum of $\\frac{n-2}{2}$ segments from each vertex). ~th1nq3r",
"We can see that $|D_{40}| = 780$, since we are choosing 2 integers $[1,40]$, and order doesn't matter because $(i,j)$ and $(j,i)$ aren't both in the set. Then from doing a smaller example of $D_4$, we can note that non-endpoints must have an even number of pairs in $D$ in order for one domino's end to match another's beginning. Then, in order to maximize the length we want to minimize the number of dominoes we remove to make all pairs be even (except endpoints). Then we can see that for every two pairs with an odd number of pairs, we can connect single ends to form a circular chain (i.e. if we want to ensure that 2,3 have even pairs then we can imagine (3,1)(1,2)). Thus we can divide 40 by 4 to get the number of numbers to remove. Thus we need to remove 10 numbers. But each number is connected to another, so we remove 20 dominoes. Thus there are 780-20 = 760 dominoes not including endpoints. Finally, we can add one more domino at the end since it doesn't need to match up with the first of another to make it 761."
] |
1999-I-1
| 1,999 | 1 |
Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding terms also being prime.
| 29 | null |
[
"Obviously, all of the terms must be odd. The common difference between the terms cannot be $2$ or $4$, since otherwise there would be a number in the sequence that is divisible by $3$. However, if the common difference is $6$, we find that $5,11,17,23$, and $29$ form an arithmetic sequence. Thus, the answer is $029$."
] |
1999-I-2
| 1,999 | 2 |
Consider the parallelogram with vertices $(10,45),$ $(10,114),$ $(28,153),$ and $(28,84).$ A line through the origin cuts this figure into two congruent polygons. The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
| 118 | null |
[
"Solution 1 Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$. Let the second point on the line $x=28$ be $(28, 153-a)$. For two given points, the line will pass the origin if the coordinates are proportional (such that $\\frac{y_1}{x_1} = \\frac{y_2}{x_2}$). Then, we can write that $\\frac{45 + a}{10} = \\frac{153 - a}{28}$. Solving for $a$ yields that $1530 - 10a = 1260 + 28a$, so $a=\\frac{270}{38}=\\frac{135}{19}$. The slope of the line (since it passes through the origin) is $\\frac{45 + \\frac{135}{19}}{10} = \\frac{99}{19}$, and the solution is $m + n = 118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$. Let the second point on the line $x=28$ be $(28, 153-a)$. For two given points, the line will pass the origin if the coordinates are proportional (such that $\\frac{y_1}{x_1} = \\frac{y_2}{x_2}$). Then, we can write that $\\frac{45 + a}{10} = \\frac{153 - a}{28}$. Solving for $a$ yields that $1530 - 10a = 1260 + 28a$, so $a=\\frac{270}{38}=\\frac{135}{19}$. The slope of the line (since it passes through the origin) is $\\frac{45 + \\frac{135}{19}}{10} = \\frac{99}{19}$, and the solution is $m + n = 118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$, and $(28,153)$ gives $(19,99)$, which is the center of the parallelogram. Thus the slope of the line must be $\\frac{99}{19}$, and the solution is $118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$, and $(28,153)$ gives $(19,99)$, which is the center of the parallelogram. Thus the slope of the line must be $\\frac{99}{19}$, and the solution is $118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"Note that the area of the parallelogram is equivalent to $69 \\cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$ The points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \\frac{10m}{n})$ and $(28, \\frac{28m}{n}),$ respectively. We see that $\\frac{10m}{n} - 45 + \\frac{28m}{n} - 84 = 69,$ so $\\frac{38m}{n} = 198 \\implies \\frac{m}{n} = \\frac{99}{19} \\implies 118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"Note that the area of the parallelogram is equivalent to $69 \\cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$ The points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \\frac{10m}{n})$ and $(28, \\frac{28m}{n}),$ respectively. We see that $\\frac{10m}{n} - 45 + \\frac{28m}{n} - 84 = 69,$ so $\\frac{38m}{n} = 198 \\implies \\frac{m}{n} = \\frac{99}{19} \\implies 118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"$(\\Sigma x_i /4, \\Sigma y_i /4)$ is the centroid, which generates $(19,99)$, so the answer is $118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc",
"$(\\Sigma x_i /4, \\Sigma y_i /4)$ is the centroid, which generates $(19,99)$, so the answer is $118. This is the fastest way because you do not need to find the opposite vertices by drawing. Solution by maxamc"
] |
1999-I-3
| 1,999 | 3 |
Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square.
| 38 | null |
[
"If $n^2-19n+99=x^2$ for some positive integer $x$, then rearranging we get $n^2-19n+99-x^2=0$. Now from the quadratic formula, $n=\\frac{19\\pm \\sqrt{4x^2-35}}{2}$ Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$. Rearranging gives $(2x+q)(2x-q)=35$. Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$, giving $x=3$ or $9$. This gives $n=1, 9, 10,$ or $18$, and the sum is $1+9+10+18=38.",
"Suppose there is some $k$ such that $x^2 - 19x + 99 = k^2$. Completing the square, we have that $(x - 19/2)^2 + 99 - (19/2)^2 = k^2$, that is, $(x - 19/2)^2 + 35/4 = k^2$. Multiplying both sides by 4 and rearranging, we see that $(2k)^2 - (2x - 19)^2 = 37$. Thus, $(2k - 2x + 19)(2k + 2x - 19) = 35$. We then proceed as we did in the previous solution.",
"When $n \\geq 12$, we have \\[(n-10)^2 < n^2 -19n + 99 < (n-8)^2.\\] So if $n \\geq 12$ and $n^2 -19n + 99$ is a perfect square, then \\[n^2 -19n + 99 = (n-9)^2\\] or $n = 18$. For $1 \\leq n < 12$, it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$ We conclude that the answer is $1 + 9 + 10 + 18 = 38"
] |
1999-I-4
| 1,999 | 4 |
The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $\frac{43}{99}$ and the area of octagon $ABCDEFGH$ is $\frac{m}{n}$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ AIME 1999 Problem 4.png
| 185 | null |
[
"Triangles $AOB$, $BOC$, $COD$, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$, etc.), and each area is $\\frac{\\frac{43}{99}\\cdot\\frac{1}{2}}{2}$. Since the area of a triangle is $bh/2$, the area of all $8$ of them is $\\frac{86}{99}$ and the answer is $185.",
"Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$. The area of the octagon (by subtraction of areas) is $1 - 4\\left(\\frac{1}{2}xy\\right) = 1 - 2xy$. By the Pythagorean theorem, \\[x^2 + y^2 = \\left(\\frac{43}{99}\\right)^2\\] Also, \\begin{align*}x + y + \\frac{43}{99} &= 1\\\\ x^2 + 2xy + y^2 &= \\left(\\frac{56}{99}\\right)^2\\end{align*} Substituting, \\begin{align*}\\left(\\frac{43}{99}\\right)^2 + 2xy &= \\left(\\frac{56}{99}\\right)^2 \\\\ 2xy = \\frac{(56 + 43)(56 - 43)}{99^2} &= \\frac{13}{99} \\end{align*} Thus, the area of the octagon is $1 - \\frac{13}{99} = \\frac{86}{99}$, so $m + n = 185."
] |
1999-I-5
| 1,999 | 5 |
For any positive integer $x_{}$ , let $S(x)$ be the sum of the digits of $x_{}$ , and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999?
| 223 | null |
[
"For most values of $x$, $T(x)$ will equal $2$. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, \\[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\\] And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2$. From $7$ to $1999$, there are $\\left\\lceil \\frac{1999 - 7}{9}\\right\\rceil = 222$ solutions; including $2$ and there are a total of $223 solutions."
] |
1999-I-6
| 1,999 | 6 |
A transformation of the first quadrant of the coordinate plane maps each point $(x,y)$ to the point $(\sqrt{x},\sqrt{y}).$ The vertices of quadrilateral $ABCD$ are $A=(900,300), B=(1800,600), C=(600,1800),$ and $D=(300,900).$ Let $k_{}$ be the area of the region enclosed by the image of quadrilateral $ABCD.$ Find the greatest integer that does not exceed $k_{}.$
| 314 | null |
[
"\\begin{eqnarray*}A' = & (\\sqrt {900}, \\sqrt {300})\\\\ B' = & (\\sqrt {1800}, \\sqrt {600})\\\\ C' = & (\\sqrt {600}, \\sqrt {1800})\\\\ D' = & (\\sqrt {300}, \\sqrt {900}) \\end{eqnarray*} First we see that lines passing through $AB$ and $CD$ have equations $y = \\frac {1}{3}x$ and $y = 3x$, respectively. Looking at the points above, we see the equations for $A'B'$ and $C'D'$ are $y^2 = \\frac {1}{3}x^2$ and $y^2 = 3x^2$, or, after manipulation $y = \\frac {x}{\\sqrt {3}}$ and $y = \\sqrt {3}x$, respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines. Now take a look at $BC$ and $AD$, which have the equations $y = - x + 2400$ and $y = - x + 1200$. The image equations hence are $x^2 + y^2 = 2400$ and $x^2 + y^2 = 1200$, respectively, which are the equations for circles. To find the area between the circles (actually, parts of the circles), we need to figure out the angle of the arc. This could be done by $\\arctan \\sqrt {3} - \\arctan \\frac {1}{\\sqrt {3}} = 60^\\circ - 30^\\circ = 30^\\circ$. So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = $\\frac {30^\\circ}{360^\\circ}(R^2\\pi - r^2\\pi) = \\frac {1}{12}(2400\\pi - 1200\\pi) = 100\\pi$. Hence the answer is $314."
] |
1999-I-7
| 1,999 | 7 |
There is a set of 1000 switches, each of which has four positions, called $A, B, C$ , and $D$ . When the position of any switch changes, it is only from $A$ to $B$ , from $B$ to $C$ , from $C$ to $D$ , or from $D$ to $A$ . Initially each switch is in position $A$ . The switches are labeled with the 1000 different integers $(2^{x})(3^{y})(5^{z})$ , where $x, y$ , and $z$ take on the values $0, 1, \ldots, 9$ . At step $i$ of a 1000-step process, the $i$ -th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$ -th switch. After step 1000 has been completed, how many switches will be in position $A$ ?
| 650 | null |
[
"For each $i$th switch (designated by $x_{i},y_{i},z_{i}$), it advances itself only one time at the $i$th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$. Let $N = 2^{9}3^{9}5^{9}$ be the max switch label. To find the divisor multiples in the range of $d_{i}$ to $N$, we consider the exponents of the number $\\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}$. In general, the divisor-count of $\\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A: $4n = [(9-x)+1] [(9-y)+1] [(9-z)+1] = (10-x)(10-y)(10-z)$, where $0 \\le x,y,z \\le 9.$ We consider the cases where the 3 factors above do not contribute multiples of 4. Case of no 2's: The switches must be $(\\mathrm{odd})(\\mathrm{odd})(\\mathrm{odd})$. There are $5$ odd integers in $0$ to $9$, so we have $5 \\times 5 \\times 5 = 125$ ways. Case of a single 2: The switches must be one of $(2\\cdot \\mathrm{odd})(\\mathrm{odd})(\\mathrm{odd})$ or $(\\mathrm{odd})(2 \\cdot \\mathrm{odd})(\\mathrm{odd})$ or $(\\mathrm{odd})(\\mathrm{odd})(2 \\cdot \\mathrm{odd})$. Since $0 \\le x,y,z \\le 9,$ the terms $2\\cdot 1, 2 \\cdot 3,$ and $2 \\cdot 5$ are three valid choices for the $(2 \\cdot odd)$ factor above. We have ${3\\choose{1}} \\cdot 3 \\cdot 5^{2}= 225$ ways. The number of switches in position A is $1000-125-225 = 650."
] |
1999-I-8
| 1,999 | 8 |
Let $\mathcal{T}$ be the set of ordered triples $(x,y,z)$ of nonnegative real numbers that lie in the plane $x+y+z=1.$ Let us say that $(x,y,z)$ supports $(a,b,c)$ when exactly two of the following are true: $x\ge a, y\ge b, z\ge c.$ Let $\mathcal{S}$ consist of those triples in $\mathcal{T}$ that support $\left(\frac 12,\frac 13,\frac 16\right).$ The area of $\mathcal{S}$ divided by the area of $\mathcal{T}$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$
| 25 | null |
[
"This problem just requires a good diagram and strong 3D visualization. The region in $(x,y,z)$ where $x \\ge \\frac{1}{2}, y \\ge \\frac{1}{3}$ is that of a little triangle on the bottom of the above diagram, of $y \\ge \\frac{1}{3}, z \\ge \\frac{1}{6}$ is the triangle at the right, and $x \\ge \\frac 12, z \\ge \\frac 16$ the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by $x+y+z=1,\\ x,y,z \\ge 0$. We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third. The side length of the large equilateral triangle is $\\sqrt{2}$, which we can find using 45-45-90 $\\triangle$ with the axes. Using the formula $A = \\frac{s^2\\sqrt{3}}{4}$ for equilateral triangles, the area of the large triangle is $\\frac{(\\sqrt{2})^2\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{2}$. Since the lines of the smaller triangles are parallel to those of the large triangle, by corresponding angles we see that all of the triangles are similar, so they are all equilateral triangles. We can solve for their side lengths easily by subtraction, and we get $\\frac{\\sqrt{2}}{6}, \\frac{\\sqrt{2}}{3}, \\frac{\\sqrt{2}}{2}$. Calculating their areas, we get $\\frac{\\sqrt{3}}{8}, \\frac{\\sqrt{3}}{18}, \\frac{\\sqrt{3}}{72}$. The ratio $\\frac{\\mathcal{S}}{\\mathcal{T}} = \\frac{\\frac{9\\sqrt{3} + 4\\sqrt{3} + \\sqrt{3}}{72}}{\\frac{\\sqrt{3}}{2}} = \\frac{14}{36} = \\frac{7}{18}$, and the answer is $m + n = 025."
] |
1999-I-9
| 1,999 | 9 |
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
| 259 | null |
[
"Suppose we pick an arbitrary point on the complex plane, say $(1,1)$. According to the definition of $f(z)$, \\[f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,\\] this image must be equidistant to $(1,1)$ and $(0,0)$. Thus the image must lie on the line with slope $-1$ and which passes through $\\left(\\frac 12, \\frac12\\right)$, so its graph is $x + y = 1$. Substituting $x = (a-b)$ and $y = (a+b)$, we get $2a = 1 \\Rightarrow a = \\frac 12$. By the Pythagorean Theorem, we have $\\left(\\frac{1}{2}\\right)^2 + b^2 = 8^2 \\Longrightarrow b^2 = \\frac{255}{4}$, and the answer is $259.",
"Plugging in $z=1$ yields $f(1) = a+bi$. This implies that $a+bi$ must fall on the line $Re(z)=a=\\frac{1}{2}$, given the equidistant rule. By $|a+bi|=8$, we get $a^2 + b^2 = 64$, and plugging in $a=\\frac{1}{2}$ yields $b^2=\\frac{255}{4}$. The answer is thus $259.",
"We are given that $(a + bi)z$ is equidistant from the origin and $z.$ This translates to \\begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\\\ |z(a - 1 + bi)| & = & |z(a + bi)| \\\\ |z|\\cdot|(a - 1) + bi| & = & |z|\\cdot|a + bi| \\\\ |(a - 1) + bi| & = & |a + bi| \\\\ (a - 1)^2 + b^2 & = & a^2 + b^2 \\\\ & \\Rightarrow & a = \\frac 12 \\end{eqnarray*} Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ Because $a = \\frac 12,$ thus $b^2 = \\frac {255}4.$ So the answer is $259.",
"Let $P$ and $Q$ be the points in the complex plane represented by $z$ and $(a+bi)z$, respectively. $|a+bi| = 8$ implies $OQ = 8OP$. Also, we are given $OQ = PQ$, so $OPQ$ is isosceles with base $OP$. Notice that the base angle of this isosceles triangle is equal to the argument $\\theta$ of the complex number $a + bi$, because $(a+bi)z$ forms an angle of $\\theta$ with $z$. Drop the altitude/median from $Q$ to base $OP$, and you end up with a right triangle that shows $\\cos \\theta = \\frac{\\frac{1}{2}OP}{8OQ} = \\frac{\\frac{1}{2}|z|}{8|z|} = \\frac{1}{16}$. Since $a$ and $b$ are positive, $z$ lies in the first quadrant and $\\theta < \\pi/2$; hence by right triangle trigonometry $\\sin \\theta = \\frac{\\sqrt{255}}{16}$. Finally, $b = |a+bi|\\sin\\theta = 8\\frac{\\sqrt{255}}{16} = \\frac{\\sqrt{255}}{2}$, and $b^2 = \\frac{255}{4}$, so the answer is $259$.",
"Similarly to in Solution 3, we see that $|(a + bi)z - z| = |(a + bi)z|$. Letting the point $z = c + di$, we have $\\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \\sqrt{(ac-bd)^2+(ad+bc)^2}$. Expanding both sides of this equation (after squaring, of course) and canceling terms, we get $(d^2+c^2)(-2a+1) = 0$. Of course, $(d^2+c^2)$ can't be zero because this property of the function holds for all complex $z$. Therefore, $a = \\frac{1}{2}$ and we proceed as above to get $259. ~ anellipticcurveoverq",
"This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem. Consider any complex number $z=c+di$. Let $z$ denote point $P$ on the complex plane. Then $P=(c,d)$ on the complex plane. The equation for the line $OP$ is $y=\\frac{d}{c}x$. Let the image of point $P$ be $Q$, after the point undergoes the function. Since each image is equidistant from the preimage and the origin, $Q$ must be on the perpendicular bisector of $OP$.Given $z=c+di$, $f(z)=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. Then $Q=(ac-bd,ad+bc)$. The midpoint of $OP$ is $(0.5c, 0.5d)$. Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of $-1$, using the point-slope form, the equation of the perpendicular line to $OP$ is $y-0.5d=-\\frac{c}{d}(x-0.5c)$. Rearranging, we have $y=-\\frac{cx}{d}+\\frac{c^2}{2d}+\\frac{d}{2}$. Since we know that $Q=(ac-bd,ad+bc)$, thus we plug in $Q$ into the line: $ad+bc=-\\frac{ac^2-bcd}{d}+\\frac{c^2}{2d}+\\frac{d}{2}$. Let's start canceling. $2ad^2+2bcd=-2ac^2+2bcd+c^2+d^2$. Subtracting, $c^2+d^2-2ac^2=2ad^2$. Thus $c^2+d^2=2ac^2+2ad^2$. Since this is an identity for any $(c,d)$, thus $2a=1$. $a=\\frac{1}{2}$. Since $|a+bi|=8$, thus $a^2+b^2=64$ (or simply think of $a+bi$ as the point $(a,b)$, and $|a+bi|$ being the distance of $(a,b)$ to the origin). Thus plug in $a=\\frac{1}{2}, b^2=\\frac{255}{4}$. Since $255$ and $4$ are relatively prime, the final result is $255+4=259. ~hastapasta"
] |
1999-I-10
| 1,999 | 10 |
Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
| 489 | null |
[
"First, let us find the number of triangles that can be formed from the 10 points. Since none of the points are collinear, it is possible to pick ${10\\choose3}$ sets of 3 points which form triangles. However, a fourth distinct segment must also be picked. Since the triangle accounts for 3 segments, there are $45 - 3 = 42$ segments remaining. The total number of ways of picking four distinct segments is ${45\\choose4}$. Thus, the requested probability is $\\frac{{10\\choose3} \\cdot 42}{{45\\choose4}} = \\frac{10 \\cdot 9 \\cdot 8 \\cdot 42 \\cdot 4!}{45 \\cdot 44 \\cdot 43 \\cdot 42 \\cdot 3!} = \\frac{16}{473}$. The solution is $m + n = 489$.",
"Note that 4 segments can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the first one, then the one segment that completes the triangle. Note that the fourth segment doesn't matter in this case. Note that there are $(9 - 1) \\times 2 = 16$ segments that share an endpoint with the first segment. The answer is then $4 \\times \\frac{16}{44} \\times \\frac{1}{43} = \\frac{16}{11} \\times \\frac{1}{43} = \\frac{16}{473} \\implies m + n = 489 -whatRthose",
"Instead of working with the four segments, let's focus on their endpoints. When we select these segments, we are working with $4, 5, 6, 7,$ or $8$ endpoints in total. If we have $6, 7,$ or $8$ endpoints, it is easy to see that we cannot form a triangle by drawing four segments between them, because at least one point will be \"left out\". However, if we have $5$ endpoints, we can use three segments form a triangle with three of the points and connect the remaining two points with the last segment. There are ${10\\choose5}$ ways to select these $5$ points from the original $10,$ and ${5\\choose3}$ ways to decide which three points are in the triangle. Finally, if we have $4$ endpoints, we can also form a triangle with three of the points, then use the remaining segment to connect the last point to either of the previous three. We have ${10\\choose4}$ ways to select the $4$ points and ${4\\choose3}$ ways to choose three points for the triangle. Finally, we must connect the last point to one vertex of the triangle; we can do this in $3$ ways. As in Solution 1, there are ${45\\choose4}$ total ways to select four segments. So, our desired probability is \\[\\dfrac{{10\\choose5}{5\\choose3}+{10\\choose4}{4\\choose3}\\cdot 3}{{45\\choose4}}=\\dfrac{16}{473} \\implies m + n = 489.\\] --vaporwave"
] |
1999-I-11
| 1,999 | 11 |
Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$
| 177 | null |
[
"Let $s = \\sum_{k=1}^{35}\\sin 5k = \\sin 5 + \\sin 10 + \\ldots + \\sin 175$. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity $\\sin a \\sin b = \\frac 12(\\cos (a-b) - \\cos (a+b))$, we can rewrite $s$ as \\begin{align*} s \\cdot \\sin 5 = \\sum_{k=1}^{35} \\sin 5k \\sin 5 &= \\sum_{k=1}^{35} \\frac{1}{2}(\\cos (5k - 5)- \\cos (5k + 5))\\\\ &= 0.5(\\cos 0 - \\cos 10 + \\cos 5 - \\cos 15 + \\cos 10 \\ldots + \\cos 165 - \\cos 175+ \\cos 170 - \\cos 180) \\end{align*} This telescopes to \\[s = \\frac{\\cos 0 + \\cos 5 - \\cos 175 - \\cos 180}{2 \\sin 5} = \\frac{1 + \\cos 5}{\\sin 5}.\\] Manipulating this to use the identity $\\tan x = \\frac{1 - \\cos 2x}{\\sin 2x}$, we get \\[s = \\frac{1 - \\cos 175}{\\sin 175} \\Longrightarrow s = \\tan \\frac{175}{2},\\] and our answer is $177.",
"Let $x=e^{\\frac{i\\pi}{36}}$. By Euler's Formula, $\\sin{5k^\\circ}=\\frac{x^k-\\frac{1}{x^{k}}}{2i}$. The sum we want is thus $\\frac{x-\\frac{1}{x}}{2i}+\\frac{x^2-\\frac{1}{x^{2}}}{2i}+\\cdots+\\frac{x^{35}-\\frac{1}{x^{35}}}{2i}$ We factor the $\\frac{1}{2i}$ and split into two geometric series to get $\\frac{1}{2i}\\left(\\frac{-\\frac{1}{x^{35}}(x^{35}-1)}{x-1}+\\frac{x(x^{35}-1)}{x-1}\\right)$ However, we note that $x^{36}=-1$, so $-\\frac{1}{x^{35}}=x$, so our two geometric series are actually the same. We combine the terms and simplify to get $\\frac{1}{i}\\left(\\frac{x^{36}-x}{x-1}\\right)$ Apply Euler's identity and simplify again to get $\\frac{1}{i}\\left(\\frac{-x-1}{x-1}\\right)$ Now, we need to figure out how to express this as the tangent of something. We note that $\\tan(5k^\\circ)=\\frac{\\sin(5k^\\circ)}{\\cos(5k^\\circ)}=\\frac{\\frac{x^k-\\frac{1}{x^k}}{2i}}{\\frac{x^k+\\frac{1}{x^k}}{2}}=\\frac{1}{i}\\frac{x^{2k}-1}{x^{2k}+1}$. So, we set the two equal to each other to solve for $k$. Cross multiplying gets $(-x-1)(x^{2k}+1)=(x-1)(x^{2k}-1)$. Expanding yields $-x^{2k+1}-x-x^{2k}-1=x^{2k+1}-x-x^{2k}+1$. Simplifying yields $x^{2k+1}=-1$. Since $2k+1=36$ is the smallest solution, we have $k=\\frac{35}{2}$, and the argument of tangent is $5k=\\frac{175}{2}$. The requested sum is $175+2=177.",
"This solution is inspired by 1997 AIME Problems/Problem 11. First of all, $\\sum_{k=1}^{35}\\sin 5k=\\sum_{k=0}^{35}\\sin 5k$ because the difference is $\\sin 0 = 0$. Now consider the sum $\\sum_{k=0}^{35}\\cos 5k$. Since $\\cos 5=-\\cos 175$, $\\cos 10=-\\cos 170$ and so on, this simplifies to $\\cos 0=1$. Now, consider the ratio \\[\\dfrac{\\sum\\limits_{n=0}^{35}\\sin 5k}{\\sum\\limits_{n=0}^{35}\\cos 5k.}\\] Using the sum-to-product identities, the numerator can be written as \\[\\sin \\frac{175}{2} (\\cos \\frac{175}{2} + \\cos \\frac {165}{2}+\\cdots + \\cos \\frac{5}{2}.)\\] Similarly (via sum-to-product identities), the denominator is equivalent to \\[\\cos \\frac{175}{2} (\\cos \\frac{175}{2} + \\cos \\frac {165}{2}+\\cdots + \\cos \\frac{5}{2}.)\\] If we substitute these equivalent expressions into the original ratio, many parts cancel. We are left with \\[\\frac{\\sin \\frac{175}{2}}{\\cos \\frac{175}{2}} = \\tan{\\frac{175}{2}}.\\] Aha! Remembering that the original denominator $\\sum_{k=0}^{35}\\cos 5k$ was equal to 1, we realize that the numerator, $\\sum_{k=0}^{35}\\sin 5k=\\tan{\\frac{175}{2}}$. Since $\\sum_{k=0}^{35}\\sin 5k=\\sum_{k=1}^{35}\\sin 5k$, we get $\\tan \\frac{175}{2}$, or $m+n=177 ~ewei12"
] |
1999-I-12
| 1,999 | 12 |
The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is 21. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
| 345 | null |
[
"[asy] pathpen = black + linewidth(0.65); pointpen = black; pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C); path P = incircle(A,B,C); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--cycle);D(P); D(MP(\"P\",IP(A--B,P))); pair Q=IP(C--A,P),R=IP(B--C,P); D(MP(\"R\",R,NE));D(MP(\"Q\",Q,NW)); MP(\"23\",(A+Q)/2,W);MP(\"27\",(B+R)/2,E); [/asy] Solution 1 Let $Q$ be the tangency point on $\\overline{AC}$, and $R$ on $\\overline{BC}$. By the Two Tangent Theorem, $AP = AQ = 23$, $BP = BR = 27$, and $CQ = CR = x$. Using $rs = A$, where $s = \\frac{27 \\cdot 2 + 23 \\cdot 2 + x \\cdot 2}{2} = 50 + x$, we get $(21)(50 + x) = A$. By Heron's formula, $A = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{(50+x)(x)(23)(27)}$. Equating and squaring both sides, \\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\\\ 441(50+x) &=& 621x\\\\ 180x = 441 \\cdot 50 &\\Longrightarrow & x = \\frac{245}{2} \\end{eqnarray*} We want the perimeter, which is $2s = 2\\left(50 + \\frac{245}{2}\\right) = 345",
"Let $Q$ be the tangency point on $\\overline{AC}$, and $R$ on $\\overline{BC}$. By the Two Tangent Theorem, $AP = AQ = 23$, $BP = BR = 27$, and $CQ = CR = x$. Using $rs = A$, where $s = \\frac{27 \\cdot 2 + 23 \\cdot 2 + x \\cdot 2}{2} = 50 + x$, we get $(21)(50 + x) = A$. By Heron's formula, $A = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{(50+x)(x)(23)(27)}$. Equating and squaring both sides, \\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\\\ 441(50+x) &=& 621x\\\\ 180x = 441 \\cdot 50 &\\Longrightarrow & x = \\frac{245}{2} \\end{eqnarray*} We want the perimeter, which is $2s = 2\\left(50 + \\frac{245}{2}\\right) = 345",
"Let the incenter be denoted $I$. It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\\angle ABI = \\angle CBI = \\alpha, \\angle BAI = \\angle CAI = \\beta,$ and $\\angle BCI = \\angle ACI = \\gamma.$ We have that \\begin{eqnarray*} \\tan \\alpha & = & \\frac {21}{27} \\\\ \\tan \\beta & = & \\frac {21}{23} \\\\ \\tan \\gamma & = & \\frac {21}x. \\end{eqnarray*} So naturally we look at $\\tan \\gamma.$ But since $\\gamma = \\frac \\pi2 - (\\beta + \\alpha)$ we have \\begin{eqnarray*} \\tan \\gamma & = & \\tan\\left(\\frac \\pi2 - (\\beta + \\alpha)\\right) \\\\ & = & \\frac 1{\\tan(\\alpha + \\beta)} \\\\ \\Rightarrow \\frac {21}x & = & \\frac {1 - \\frac {21\\cdot 21}{23\\cdot 27}}{\\frac {21}{27} + \\frac {21}{23}} \\end{eqnarray*} Doing the algebra, we get $x = \\frac {245}2.$ The perimeter is therefore $2\\cdot\\frac {245}2 + 2\\cdot 23 + 2\\cdot 27 = 345",
"Let unknown side has length as $x$, Assume three sides of triangles are $a,b,c$, the area of the triangle is $S$. Note that $r=\\frac{2S}{a+b+c}=21,S=1050+21x$ $\\tan\\angle{\\frac{B}{2}}=\\frac{7}{9}, \\tan\\angle{B}=\\frac{63}{16}$. Use trig identity, knowing that $1+\\cot^2\\angle{B}=\\csc^2\\angle{B}$, getting that $\\sin\\angle{B}=\\frac{63}{65}$ Now equation $(x+27)*50*\\frac{63}{65}*\\frac{1}{2}=1050+21x; x=\\frac{245}{2}$, the final answer is $245+100=345$ ~bluesoul"
] |
1999-I-13
| 1,999 | 13 |
Forty teams play a tournament in which every team plays every other( $39$ different opponents) team exactly once. No ties occur, and each team has a $50 \%$ chance of winning any game it plays. The probability that no two teams win the same number of games is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $\log_2 n.$
| 742 | null |
[
"There are ${40 \\choose 2} = 780$ total pairings of teams, and thus $2^{780}$ possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some $k$, with $0 \\leq k \\leq 39$, where $k$ represents the number of games the team won. With this in mind, we see that there are a total of $40!$ outcomes in which no two teams win the same number of games. Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination. The desired probability is thus $\\frac{40!}{2^{780}}$. We wish to simplify this into the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime. The only necessary step is to factor out all the powers of 2 from $40!$; the remaining number is clearly relatively prime to all powers of 2. The number of powers of 2 in $40!$ is $\\left \\lfloor \\frac{40}{2} \\right \\rfloor + \\left \\lfloor \\frac{40}{4} \\right \\rfloor + \\left \\lfloor \\frac{40}{8} \\right \\rfloor + \\left \\lfloor \\frac{40}{16} \\right \\rfloor + \\left \\lfloor \\frac{40}{32} \\right \\rfloor = 20 + 10 + 5 + 2 + 1 = 38.$ $780-38 = 742."
] |
1999-I-14
| 1,999 | 14 |
Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
| 463 | null |
[
"[asy] real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP(\"P\",P,NW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP(\"13\",(A+B)/2,S);MP(\"15\",(A+C)/2,NW);MP(\"14\",(C+B)/2,NE); [/asy] Solution 1 Drop perpendiculars from $P$ to the three sides of $\\triangle ABC$ and let them meet $\\overline{AB}, \\overline{BC},$ and $\\overline{CA}$ at $D, E,$ and $F$ respectively. [asy] import olympiad; real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP(\"P\",P,SSW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP(\"13\",(A+B)/2,S);MP(\"15\",(A+C)/2,NW);MP(\"14\",(C+B)/2,NE); /* constructing D,E,F as foot of perps from P */ pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); D(MP(\"D\",D,NE)--P--MP(\"E\",E,SSW),dashed);D(P--MP(\"F\",F),dashed); D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); [/asy] Let $BE = x, CF = y,$ and $AD = z$. We have that \\begin{align*}DP&=z\\tan\\theta\\\\ EP&=x\\tan\\theta\\\\ FP&=y\\tan\\theta\\end{align*} We can then use the tool of calculating area in two ways \\begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\\\ &=\\frac{1}{2}(13)(z\\tan\\theta)+\\frac{1}{2}(14)(x\\tan\\theta)+\\frac{1}{2}(15)(y\\tan\\theta)\\\\ &=\\frac{1}{2}\\tan\\theta(13z+14x+15y)\\end{align*} On the other hand, \\begin{align*}[ABC]&=\\sqrt{s(s-a)(s-b)(s-c)}\\\\ &=\\sqrt{21\\cdot6\\cdot7\\cdot8}\\\\ &=84\\end{align*} We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: \\begin{align}x^2+x^2\\tan^2\\theta&=z^2\\tan^2\\theta+(13-z)^2\\\\ z^2+z^2\\tan^2\\theta&=y^2\\tan^2\\theta+(15-y)^2\\\\ y^2+y^2\\tan^2\\theta&=x^2\\tan^2\\theta+(14-x)^2\\end{align} Adding $(1) + (2) + (3)$ gives \\begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\\\ \\Rightarrow13z+14x+15y&=295\\end{align*} Recall that we found that $[ABC]=\\frac{1}{2}\\tan\\theta(13z+14x+15y)=84$. Plugging in $13z+14x+15y=295$, we get $\\tan\\theta=\\frac{168}{295}$, giving us $463.",
"Drop perpendiculars from $P$ to the three sides of $\\triangle ABC$ and let them meet $\\overline{AB}, \\overline{BC},$ and $\\overline{CA}$ at $D, E,$ and $F$ respectively. [asy] import olympiad; real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ pathpen = black +linewidth(0.65); pointpen = black; pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--cycle); /* constructing P, C is there as check */ pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); D(A--MP(\"P\",P,SSW)--B);D(P--C); D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); MP(\"13\",(A+B)/2,S);MP(\"15\",(A+C)/2,NW);MP(\"14\",(C+B)/2,NE); /* constructing D,E,F as foot of perps from P */ pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); D(MP(\"D\",D,NE)--P--MP(\"E\",E,SSW),dashed);D(P--MP(\"F\",F),dashed); D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); [/asy] Let $BE = x, CF = y,$ and $AD = z$. We have that \\begin{align*}DP&=z\\tan\\theta\\\\ EP&=x\\tan\\theta\\\\ FP&=y\\tan\\theta\\end{align*} We can then use the tool of calculating area in two ways \\begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\\\ &=\\frac{1}{2}(13)(z\\tan\\theta)+\\frac{1}{2}(14)(x\\tan\\theta)+\\frac{1}{2}(15)(y\\tan\\theta)\\\\ &=\\frac{1}{2}\\tan\\theta(13z+14x+15y)\\end{align*} On the other hand, \\begin{align*}[ABC]&=\\sqrt{s(s-a)(s-b)(s-c)}\\\\ &=\\sqrt{21\\cdot6\\cdot7\\cdot8}\\\\ &=84\\end{align*} We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: \\begin{align}x^2+x^2\\tan^2\\theta&=z^2\\tan^2\\theta+(13-z)^2\\\\ z^2+z^2\\tan^2\\theta&=y^2\\tan^2\\theta+(15-y)^2\\\\ y^2+y^2\\tan^2\\theta&=x^2\\tan^2\\theta+(14-x)^2\\end{align} Adding $(1) + (2) + (3)$ gives \\begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\\\ \\Rightarrow13z+14x+15y&=295\\end{align*} Recall that we found that $[ABC]=\\frac{1}{2}\\tan\\theta(13z+14x+15y)=84$. Plugging in $13z+14x+15y=295$, we get $\\tan\\theta=\\frac{168}{295}$, giving us $463.",
"Let $AB=c$, $BC=a$, $AC=b$, $PA=x$, $PB=y$, and $PC=z$. So by the Law of Cosines, we have: \\begin{align*}x^2 &= z^2 + b^2 - 2bz\\cos{\\theta}\\\\ y^2 &= x^2 + c^2 - 2cx\\cos{\\theta}\\\\ z^2 &= y^2 + a^2 - 2ay\\cos{\\theta}\\end{align*} Adding these equations and rearranging, we have: \\[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\\cos{\\theta}\\qquad(1)\\] Now $[CAP] + [ABP] + [BCP] = [ABC] = \\sqrt {(21)(8)(7)(6)} = 84$, by Heron's formula. Now the area of a triangle, $[A] = \\frac {mn\\sin{\\beta}}{2}$, where $m$ and $n$ are sides on either side of an angle, $\\beta$. So, \\begin{align*}[CAP] &= \\frac {bz\\sin{\\theta}}{2}\\\\ [ABP] &= \\frac {cx\\sin{\\theta}}{2}\\\\ [BCP] &= \\frac {ay\\sin{\\theta}}{2}\\end{align*} Adding these equations yields: \\begin{align*}[ABC]= 84 &= \\frac {(bz + cx + ay)\\sin{\\theta}}{2}\\\\ \\Rightarrow 168&= (bz + cx + ay)\\sin{\\theta}\\qquad (2)\\end{align*} Dividing $(2)$ by $(1)$, we have: \\begin{align*}\\frac {168}{a^2 + b^2 + c^2} &= \\frac {(bz + cx + ay)\\sin{\\theta}}{(2bz + 2cx + 2ay)\\cos{\\theta}}\\\\ \\Rightarrow \\tan{\\theta} = \\frac {336}{a^2 + b^2 + c^2} &= \\frac {336}{14^2 + 15^2 + 13^2} = \\frac {336}{590} = \\frac {168}{295}\\end{align*} Thus, $m + n = 168 + 295 = 463.",
"Let $\\angle{PAB} = \\angle{PBC} = \\angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have \\begin{align*} b^2 &= a^2 + 169 - 26a \\cos x \\\\ c^2 &= b^2 + 196 - 28b \\cos x \\\\ a^2 &= c^2 + 225 - 30c \\cos x. \\end{align*} Add the three equations up and rearrange to obtain \\[(13a + 14b + 15c) \\cos x = 295.\\] Also, using $[ABC] = \\frac{1}{2}ab \\sin \\angle C$ we have \\[[ABC] = [APB] + [BPC] + [CPA] = \\dfrac{\\sin x}{2}(13a + 14b + 15c) = 84 \\iff (13a + 14b + 15c) \\sin x = 168.\\] Divide the two equations to obtain $\\tan x = \\frac{168}{295} \\iff 463.",
"Firstly, denote angles $ABC$, $BCA$, and $CAB$ as $B$, $A$, and $C$ respectively. Let $\\angle{PAB}=x$. Notice that by angle chasing that $\\angle{BPC}=180-C$ and $\\angle{BPA}=180-B$. Using the nice properties of the 13-14-15 triangle, we have $\\sin B = \\frac{12}{13}$ and $\\sin C = \\frac{4}{5}$. $\\cos C$ is easily computed, so we have $\\cos C=\\frac{3}{5}$. Using Law of Sines, \\begin{align*} \\frac{BP}{\\sin x} &= \\frac{13}{\\sin (180 - B)} \\\\ \\frac{BP}{\\sin (C - x)} &= \\frac{14}{\\sin (180 - C)} \\end{align*} hence, \\begin{align*} \\frac{BP}{\\sin x} &= \\frac{13}{\\sin B} \\\\ \\frac{BP}{\\sin (C - x)} &= \\frac{14}{\\sin C} \\end{align*} Now, computation carries the rest. \\begin{align*} \\frac{13 \\sin x}{\\sin B} &= \\frac{14 \\sin (C-x)}{\\sin C} \\\\ \\frac{169 \\sin x}{12} &= \\frac{210 \\sin (C-x)}{12} \\\\ 169 \\sin x &= 210 (\\sin C \\cos x - \\cos C \\sin x) \\\\ 169 \\sin x &= 210 (\\frac{4}{5} \\cos x - \\frac{3}{5} \\sin x) \\\\ 169 \\sin x &= 168 \\cos x - 126 \\sin x \\\\ 295 \\sin x &= 168 \\cos x \\\\ \\tan x &= \\frac{168}{295} \\end{align*} Extracting yields $168 + 295 = 463.",
"Firstly, denote angles $ABC$, $BCA$, and $CAB$ as $B$, $A$, and $C$ respectively. Let $\\angle{PAB}=x$. Notice that by angle chasing that $\\angle{BPC}=180-C$ and $\\angle{BPA}=180-B$. Using the nice properties of the 13-14-15 triangle, we have $\\sin B = \\frac{12}{13}$ and $\\sin C = \\frac{4}{5}$. $\\cos C$ is easily computed, so we have $\\cos C=\\frac{3}{5}$. Using Law of Sines, \\begin{align*} \\frac{BP}{\\sin x} &= \\frac{13}{\\sin (180 - B)} \\\\ \\frac{BP}{\\sin (C - x)} &= \\frac{14}{\\sin (180 - C)} \\end{align*} hence, \\begin{align*} \\frac{BP}{\\sin x} &= \\frac{13}{\\sin B} \\\\ \\frac{BP}{\\sin (C - x)} &= \\frac{14}{\\sin C} \\end{align*} Now, computation carries the rest. \\begin{align*} \\frac{13 \\sin x}{\\sin B} &= \\frac{14 \\sin (C-x)}{\\sin C} \\\\ \\frac{169 \\sin x}{12} &= \\frac{210 \\sin (C-x)}{12} \\\\ 169 \\sin x &= 210 (\\sin C \\cos x - \\cos C \\sin x) \\\\ 169 \\sin x &= 210 (\\frac{4}{5} \\cos x - \\frac{3}{5} \\sin x) \\\\ 169 \\sin x &= 168 \\cos x - 126 \\sin x \\\\ 295 \\sin x &= 168 \\cos x \\\\ \\tan x &= \\frac{168}{295} \\end{align*} Extracting yields $168 + 295 = 463."
] |
1999-I-15
| 1,999 | 15 |
Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?
| 408 | null |
[
"[asy]defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label(\"\\(A\\)\",A,SW);label(\"\\(B\\)\",B,NW);label(\"\\(C\\)\",C,SE); label(\"\\(D\\)\",foot(A,B,C),NE);label(\"\\(E\\)\",foot(B,A,C),SW);label(\"\\(F\\)\",foot(C,A,B),NW);label(\"\\(P\\)\",P,NW);label(\"\\(Q\\)\",Q,NE);label(\"\\(R\\)\",R,SE);[/asy][asy]import three; defaultpen(linewidth(0.6)); currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); [/asy] As shown in the image above, let $D$, $E$, and $F$ be the midpoints of $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\\triangle ABC$. The crux of this problem is the following lemma. Lemma: The point $O$ is the orthocenter of $\\triangle ABC$. Proof. Observe that \\[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$. Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$. Analogously, $O$ lies on the $B$-altitude and $C$-altitude of $\\triangle ABC$, and so $O$ is, indeed, the orthocenter of $\\triangle ABC$. To find the coordinates of $O$, we need to find the intersection point of altitudes $BE$ and $AD$. The equation of $BE$ is simply $x=16$. $AD$ is perpendicular to line $BC$, so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$. $BC$ has slope $\\frac{24-0}{16-34}=-\\frac{4}{3}$, therefore $y=\\frac{3}{4} x$. These two lines intersect at $(16,12)$, so that's the base of the height of the tetrahedron. Let $S$ be the foot of altitude $BS$ in $\\triangle BPQ$. From the Pythagorean Theorem, $h=\\sqrt{BS^2-SO^2}$. However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$. The area of the base is $102$, so the volume is $\\frac{102*12}{3}=408.~Shen Kislay Kai",
"Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$, $(8, 12, 0)$, and $(25, 12, 0)$. We can compute the area of this triangle as $102$. Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$. Clearly, the height of the pyramid is $z$. The desired volume is thus $\\frac{102z}{3} = 34z$. We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$, $VP = PB$, and $VQ = QC$. We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$. The desired volume is thus $34 \\times 12 = 408.~Shen Kislay Kai",
"The formed tetrahedron has pairwise parallel planar and oppositely equal length ($4\\sqrt{13},15,17$) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras) $p^2+q^2=4^2\\cdot{13}$ $q^2+r^2=15^2$ $r^2+p^2=17^2$ to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\\cdot{17},2^3\\cdot{17},2^3\\cdot{3^2}).$ Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\\tfrac{1}{3}$ and then the volume is $\\tfrac{1}{3}pqr=\\tfrac{1}{3}\\sqrt{2^6\\cdot{3^4}\\cdot{17^2}}=408 Solution by D. Adrian Tanner",
"Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$. By Pythagorean theorem, $EF = \\frac{1}{2} BC = 15, DE = \\frac{1}{2}AB = 4 \\sqrt{13}, DF = \\frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$. Note that $\\triangle DEP \\cong \\triangle EDF$. Create point $F'$ on the plane of $DEF$ such that $\\triangle DEP \\cong \\triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$). Project $F, P$ onto $DE$ as $X, Y$. Note by the definition of $F'$ then $\\angle PYF'$ is the dihedral angle between planes $DEP, DEF$. Now see that by Heron's, \\[[DEP] = [DEF] = \\sqrt{(16 + 2 \\sqrt{13})(16 - 2 \\sqrt{13})(1 + 2 \\sqrt{13})(-1 + 2 \\sqrt{13})} = 102.\\] So $PY$, the hypotenuse $DEP$ has length $\\frac{102 \\cdot 2}{4 \\sqrt{13}} = \\frac{51}{\\sqrt{13}}$. Similarly $F'Y = \\frac{51}{\\sqrt{13}}.$ Further from Pythagoras $DY = \\sqrt{DP^2 - PY^2} = \\frac{18}{\\sqrt{13}}.$ Symmetrically $EX = \\frac{18}{\\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \\frac{16}{\\sqrt{13}}.$ By Law of Cosines on $\\triangle PYF'$, \\begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \\cdot PY \\cdot F'Y \\cos{\\angle PYF'} \\\\ PF^2 - XY^2 &= 2 (\\frac{51}{\\sqrt{13}})^2 \\cos{\\angle PYF'} \\\\ (4\\sqrt{13})^2 - (\\frac{16}{\\sqrt{13}})^2 &= 2 (\\frac{51}{\\sqrt{13}})^2 \\cos{\\angle PYF'} \\\\ \\cos{\\angle PYF'} &= \\frac{9}{17} \\\\ \\sin{\\angle PYF'} &= \\frac{4 \\sqrt{13}}{17}. \\end{align*}. Therefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \\sin{\\angle PYF'} = \\frac{51}{\\sqrt{13}} \\frac{4 \\sqrt{13}}{17} = 12.$ So the area is $\\frac{1}{3}bh = \\frac{1}{3} 12 \\cdot 102 = 408 ~ Aaryabhatta1",
"The Pyramid is a disphenoid, because opposite sides have the same length. The volume of a disphenoid is given by \\[V = \\sqrt{\\frac{(l^{2}+m^{2}-n^{2})(l^{2}-m^{2}+n^{2})(-l^{2}+m^{2}+n^{2})}{72}}.\\] Using the Pythagorean theorem, the side lengths of the smaller triangle are \\(15\\), \\(4\\sqrt{13}\\), and \\(17\\). Plugging in, we get \\[V = \\sqrt{166464} = 408.\\] ~~Disphenoid_lover",
"The Pyramid is a disphenoid, because opposite sides have the same length. The volume of a disphenoid is given by \\[V = \\sqrt{\\frac{(l^{2}+m^{2}-n^{2})(l^{2}-m^{2}+n^{2})(-l^{2}+m^{2}+n^{2})}{72}}.\\] Using the Pythagorean theorem, the side lengths of the smaller triangle are \\(15\\), \\(4\\sqrt{13}\\), and \\(17\\). Plugging in, we get \\[V = \\sqrt{166464} = 408.\\] ~~Disphenoid_lover"
] |
2000-I-1
| 2,000 | 1 |
Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$ .
| 8 |
I
|
[
"If a factor of $10^{n}$ has a $2$ and a $5$ in its prime factorization, then that factor will end in a $0$. Therefore, we have left to consider the case when the two factors have the $2$s and the $5$s separated, so we need to find the first power of 2 or 5 that contains a 0. For $n = 1:$ \\[2^1 = 2 , 5^1 = 5\\] $n = 2:$ \\[2^2 = 4 , 5 ^ 2 =25\\] $n = 3:$ \\[2^3 = 8 , 5 ^3 = 125\\] and so on, until, $n = 8:$ $2^8 = 256$ | $5^8 = 390625$ We see that $5^8$ contains the first zero, so $n = 8."
] |
2000-I-2
| 2,000 | 2 |
Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$ .
| 21 |
I
|
[
"Solution 1 [asy] pointpen = black; pathpen = linewidth(0.7) + black; size(180); pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11); D(D(MP(\"A\\ (u,v)\",A,(1,0)))--D(MP(\"B\",B,N))--D(MP(\"C\",C,N))--D(MP(\"D\",D))--D(MP(\"E\",E))--cycle); D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5)); [/asy] Since $A = (u,v)$, we can find the coordinates of the other points: $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\\frac{1}{2}(2u)(u-v) = u^2 - uv$. Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \\cdot 41$. Since $u,v$ are positive, $u+3v>u$, and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$. Since $v < u$ the latter case is the answer, and $u+v = 021",
"[asy] pointpen = black; pathpen = linewidth(0.7) + black; size(180); pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11); D(D(MP(\"A\\ (u,v)\",A,(1,0)))--D(MP(\"B\",B,N))--D(MP(\"C\",C,N))--D(MP(\"D\",D))--D(MP(\"E\",E))--cycle); D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5)); [/asy] Since $A = (u,v)$, we can find the coordinates of the other points: $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\\frac{1}{2}(2u)(u-v) = u^2 - uv$. Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \\cdot 41$. Since $u,v$ are positive, $u+3v>u$, and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$. Since $v < u$ the latter case is the answer, and $u+v = 021",
"We find the coordinates like in the solution above: $A = (u,v)$, $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. Then we notice pentagon $ABCDE$ fits into a rectangle of side lengths $(u+v)$ and $(2u)$, giving us two triangles, each with hypotenuse $AB$ and $BE$. First, we can solve for the first triangle. Using the coordinates of $A$ and $B$, we discover the side lengths are both $(u-v)$, so the area of the triangle of hypotenuse $AB$ is $\\frac{1}{2}(u-v)^2$. Next, we can solve for the second triangle. Using the coordinates of $A$ and $E$, we discover the side lengths are $(u-v)$ and $(u+v)$, so the area of the triangle of hypotenuse $AE$ is $\\frac{1}{2}(u-v)(u+v) = \\frac{1}{2}(u^2-v^2)$. Now, let’s subtract the area of these 2 triangles from the rectangle giving us $(u+v)(2u)-\\frac{1}{2}(u-v)^2-\\frac{1}{2}(u^2-v^2)=451 —> u^2+3uv=451 -> u(u+3v)=451$. Next, we take note of the fact that $u$ and $u+3v$ are both factors of 451, and since both $u$ and $v$ are positive integers, $u+3v$ must be greater than $u$, thus giving us two cases, where either $u=1$ or $u=11$. After trying both, the only working pair of $(u,v)$ where both $u$ and $v$ are integers are $u=11$ and $v=10$, thus meaning $u + v =$ $021",
"We find the coordinates like in the solution above: $A = (u,v)$, $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. Then we apply the Shoelace Theorem. \\[A = \\frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451\\] \\[\\frac{1}{2}(2u^2 + 6uv) = 451\\] \\[u(u + 3v) = 451\\] This means that $(u,v) = (11, 10)$ or $(1,150)$, but since $v < u$, then the answer is $021"
] |
2000-I-3
| 2,000 | 3 |
In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$ .
| 667 |
I
|
[
"Using the binomial theorem, $\\binom{2000}{1998} b^{1998}a^2 = \\binom{2000}{1997}b^{1997}a^3 \\Longrightarrow b=666a$. Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$, and $a+b=667."
] |
2000-I-4
| 2,000 | 4 |
The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle. [asy]defaultpen(linewidth(0.7)); draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]
| 260 |
I
|
[
"Call the squares' side lengths from smallest to largest $a_1,\\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle. The picture shows that \\begin{align*} a_1+a_2 &= a_3\\\\ a_1 + a_3 &= a_4\\\\ a_3 + a_4 &= a_5\\\\ a_4 + a_5 &= a_6\\\\ a_2 + a_3 + a_5 &= a_7\\\\ a_2 + a_7 &= a_8\\\\ a_1 + a_4 + a_6 &= a_9\\\\ a_6 + a_9 &= a_7 + a_8.\\end{align*} Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$. We can guess that $a_1 = 2$. (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=260.",
"We can just list the equations: \\begin{align*} s_3 &= s_1 + s_2 \\\\ s_4 &= s_3 + s_1 \\\\ s_5 &= s_4 + s_3 \\\\ s_6 &= s_5 + s_4 \\\\ s_7 &= s_5 + s_3 + s_2 \\\\ s_8 &= s_7 + s_2 \\\\ s_9 &= s_8 + s_2 - s_1 \\\\ s_9 + s_8 &= s_7 + s_6 + s_5 \\end{align*}We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \\begin{align*} s_4 &= 2s_1 + s_2 \\\\ s_5 &= 3s_1 +2s_2 \\\\ s_6 &= 5s_1 + 3s_2 \\\\ s_7 &= 4s_1 + 4s_2 \\\\ s_8 &= 4s_1 + 5s_2 \\\\ s_9 &= 3s_1 + 6s_2 \\\\ \\end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \\implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \\[2s_2 = 5s_1 \\implies \\frac{2}{5}s_2 = s_1.\\]Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = 260 ~peelybonehead",
"We can just list the equations: \\begin{align*} s_3 &= s_1 + s_2 \\\\ s_4 &= s_3 + s_1 \\\\ s_5 &= s_4 + s_3 \\\\ s_6 &= s_5 + s_4 \\\\ s_7 &= s_5 + s_3 + s_2 \\\\ s_8 &= s_7 + s_2 \\\\ s_9 &= s_8 + s_2 - s_1 \\\\ s_9 + s_8 &= s_7 + s_6 + s_5 \\end{align*}We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \\begin{align*} s_4 &= 2s_1 + s_2 \\\\ s_5 &= 3s_1 +2s_2 \\\\ s_6 &= 5s_1 + 3s_2 \\\\ s_7 &= 4s_1 + 4s_2 \\\\ s_8 &= 4s_1 + 5s_2 \\\\ s_9 &= 3s_1 + 6s_2 \\\\ \\end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \\implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \\[2s_2 = 5s_1 \\implies \\frac{2}{5}s_2 = s_1.\\]Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = 260 ~peelybonehead",
"We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some \"side length chasing\" and get $4a - 4 = 2a + 5$. Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$. Thus, the perimeter of the rectangle is $2(61 + 69) = 260",
"We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some \"side length chasing\" and get $4a - 4 = 2a + 5$. Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$. Thus, the perimeter of the rectangle is $2(61 + 69) = 260"
] |
2000-I-5
| 2,000 | 5 |
Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $\frac{27}{50},$ and the probability that both marbles are white is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?
| 26 |
I
|
[
"If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$. The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box. Let $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$. Then, $a + b = 25$, and since the $ab$ may be reduced to form $50$ on the denominator of $\\frac{27}{50}$, $50|ab$. It follows that $5|a,b$, so there are 2 pairs of $a$ and $b: (20,5),(15,10)$. Case 1: Then the product of the number of black marbles in each box is $54$, so the only combination that works is $18$ black in first box, and $3$ black in second. Then, $P(\\text{both white}) = \\frac{2}{20} \\cdot \\frac{2}{5} = \\frac{1}{25},$ so $m + n = 26$. Case 2: The only combination that works is 9 black in both. Thus, $P(\\text{both white}) = \\frac{1}{10}\\cdot \\frac{6}{15} = \\frac{1}{25}$. $m + n = 26$. Thus, $m + n = 026.",
"Let $w_1, w_2, b_1,$ and $b_2$ represent the white and black marbles in boxes 1 and 2. Since there are $25$ marbles in the box: $w_1 + w_2 + b_1 + b_2 = 25$ From the fact that there is a $\\frac{27}{50}$ chance of drawing one black marble from each box: $\\frac{b_1 \\cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \\frac{27}{50} = \\frac{54}{100} = \\frac{81}{150}$ Thinking of the numerator and denominator separately, if $\\frac{27}{50}$ was not a reduced fraction when calculating out the probability, then $b_1 \\cdot b_2 = 27$. Since $b_1 < 25$, this forces the variables to be $3$ and $9$ in some permutation. Without loss of generality, let $b_1 = 3$ and $b_2 = 9$. The denominator becomes: $(3 + w_1)(9 + w_2) = 50$ Since there have been $12$ black marbles used, there must be $13$ white marbles. Substituting that in: $(3 + w_1)(9 + (13 - w_1)) = 50$ $(3 + w_1)(22 - w_1) = 50$ Since the factors of $50$ that are greater than $3$ are $5, 10, 25,$ and $50$, the quantity $3 + w_1$ must equal one of those. However, since $w_1 < 13$, testing $2$ and $7$ for $w_1$ does not give a correct product. Thus, $\\frac{27}{50}$ must be a reduced form of the actual fraction. First assume that the fraction was reduced from $\\frac{54}{100}$, yielding the equations $b_1\\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$. Factoring $b_1 \\cdot b_2 = 54$ and saying WLOG that $b_1 < b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$. Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$ Since $w_1 + w_2 = 4$, the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble. Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$, so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$ Thus, the chance of drawing 2 white marbles is $\\frac{w_1 \\cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \\frac{4}{100} = \\frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = 026.",
"We know that $\\frac{27}{50} = \\frac{b_1}{t_1} \\cdot \\frac{b_2}{t_2}$, where $b_1$ and $b_2$ are the number of black marbles in the first and the second box respectively, and $t_1$ and $t_2$ is the total number of marbles in the first and the second boxes respectively. So, $t_1 + t_2 = 25$. Then, we can realize that $\\frac{27}{50} = \\frac{9}{10} \\cdot \\frac{3}{5} = \\frac{9}{10} \\cdot \\frac{9}{15}$, which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes $\\frac{1}{10} \\cdot \\frac{6}{15} = \\frac{1}{25}$. The answer is $1 + 25 = 026 ~Yiyj1"
] |
2000-I-6
| 2,000 | 6 |
For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^{6}$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$ ?
| 997 |
I
|
[
"Solution 1 \\begin{eqnarray*} \\frac{x+y}{2} &=& \\sqrt{xy} + 2\\\\ x+y-4 &=& 2\\sqrt{xy}\\\\ y - 2\\sqrt{xy} + x &=& 4\\\\ \\sqrt{y} - \\sqrt{x} &=& \\pm 2\\end{eqnarray*} Because $y > x$, we only consider $+2$. For simplicity, we can count how many valid pairs of $(\\sqrt{x},\\sqrt{y})$ that satisfy our equation. The maximum that $\\sqrt{y}$ can be is $\\sqrt{10^6} - 1 = 999$ because $\\sqrt{y}$ must be an integer (this is because $\\sqrt{y} - \\sqrt{x} = 2$, an integer). Then $\\sqrt{x} = 997$, and we continue this downward until $\\sqrt{y} = 3$, in which case $\\sqrt{x} = 1$. The number of pairs of $(\\sqrt{x},\\sqrt{y})$, and so $(x,y)$ is then $997.",
"\\begin{eqnarray*} \\frac{x+y}{2} &=& \\sqrt{xy} + 2\\\\ x+y-4 &=& 2\\sqrt{xy}\\\\ y - 2\\sqrt{xy} + x &=& 4\\\\ \\sqrt{y} - \\sqrt{x} &=& \\pm 2\\end{eqnarray*} Because $y > x$, we only consider $+2$. For simplicity, we can count how many valid pairs of $(\\sqrt{x},\\sqrt{y})$ that satisfy our equation. The maximum that $\\sqrt{y}$ can be is $\\sqrt{10^6} - 1 = 999$ because $\\sqrt{y}$ must be an integer (this is because $\\sqrt{y} - \\sqrt{x} = 2$, an integer). Then $\\sqrt{x} = 997$, and we continue this downward until $\\sqrt{y} = 3$, in which case $\\sqrt{x} = 1$. The number of pairs of $(\\sqrt{x},\\sqrt{y})$, and so $(x,y)$ is then $997.",
"Let $a^2$ = $x$ and $b^2$ = $y$, where $a$ and $b$ are positive. Then \\[\\frac{a^2 + b^2}{2} = \\sqrt{{a^2}{b^2}} +2\\] \\[a^2 + b^2 = 2ab + 4\\] \\[(a-b)^2 = 4\\] \\[(a-b) = \\pm 2\\] This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2. Because $\\sqrt{10^6} = 10^3$, then we can use all positive integers less than 1000 for $a$ and $b$. We know that because $x < y$, we get $a < b$. We can count even and odd pairs separately to make things easier*: Odd: \\[(1,3) , (3,5) , (5,7) . . . (997,999)\\] Even: \\[(2,4) , (4,6) , (6,8) . . . (996,998)\\] This makes $499$ odd pairs and $498$ even pairs, for a total of $997.",
"Since the arithmetic mean is 2 more than the geometric mean, $\\frac{x+y}{2} = 2 + \\sqrt{xy}$. We can multiply by 2 to get $x + y = 4 + 2\\sqrt{xy}$. Subtracting 4 and squaring gives \\[((x+y)-4)^2 = 4xy\\] \\[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\\] \\[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\\] Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$, so the problem asks for solutions of \\[(x-y-4)^2 = 16y\\] Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$, $y$ must be from $1^2$ to $999^2$, giving at most 999 options for $y$. However if $y = 1^2$, you get $(x-5)^2 = 16$, which has solutions $x = 9$ and $x = 1$. Both of those solutions are not less than $y$, so $y$ cannot be equal to 1. If $y = 2^2 = 4$, you get $(x - 8)^2 = 64$, which has 2 solutions, $x = 16$, and $x = 0$. 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$, you get exactly 1 solution for $x$, and that gives a total of $999 - 2 = 997.",
"Rearranging our conditions to \\[x^2-2xy+y^2+16-8x-8y=0 \\implies\\] \\[(y-x)^2=8(x+y-2).\\] Thus, $4|y-x.$ Now, let $y = 4k+x.$ Plugging this back into our expression, we get \\[(k-1)^2=x.\\] There, a unique value of $x, y$ is formed for every value of $k$. However, we must have \\[y<10^6 \\implies (k+1)^2< 10^6-1\\] and \\[x=(k-1)^2+1>0.\\] Therefore, there are only $997.",
"Rearranging our conditions to \\[x^2-2xy+y^2+16-8x-8y=0 \\implies\\] \\[(y-x)^2=8(x+y-2).\\] Thus, $4|y-x.$ Now, let $y = 4k+x.$ Plugging this back into our expression, we get \\[(k-1)^2=x.\\] There, a unique value of $x, y$ is formed for every value of $k$. However, we must have \\[y<10^6 \\implies (k+1)^2< 10^6-1\\] and \\[x=(k-1)^2+1>0.\\] Therefore, there are only $997.",
"First we see that our condition is $\\frac{x+y}{2} = 2 + \\sqrt{xy}$. Then we can see that $x+y = 4 + 2\\sqrt{xy}$. From trying a simple example to figure out conditions for $x,y$, we want to find $x-y$ so we can isolate for $x$. From doing the example we can note that we can square both sides and subtract $4xy$: $(x-y)^2 = 16 + 16\\sqrt{xy} \\implies x-y = -2( \\sqrt{1+\\sqrt{xy}})$ (note it is negative because $y > x$. Clearly the square root must be an integer, so now let $\\sqrt{xy} = a^2-1$. Thus $x-y = -2a$. Thus $x = 2 + \\sqrt{xy} - a = 2 + a^2 - 1 -2a$. We can then find $y$, and use the quadratic formula on $x,y$ to ensure they are $>0$ and $<10^6$ respectively. Thus we get that $y$ can go up to 999 and $x$ can go down to $3$, leaving $997$ possibilities for $x,y$."
] |
2000-I-7
| 2,000 | 7 |
Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 5 |
I
|
[
"We can rewrite $xyz=1$ as $\\frac{1}{z}=xy$. Substituting into one of the given equations, we have \\[x+xy=5\\] \\[x(1+y)=5\\] \\[\\frac{1}{x}=\\frac{1+y}{5}.\\] We can substitute back into $y+\\frac{1}{x}=29$ to obtain \\[y+\\frac{1+y}{5}=29\\] \\[5y+1+y=145\\] \\[y=24.\\] We can then substitute once again to get \\[x=\\frac15\\] \\[z=\\frac{5}{24}.\\] Thus, $z+\\frac1y=\\frac{5}{24}+\\frac{1}{24}=\\frac{1}{4}$, so $m+n=005.",
"Let $r = \\frac{m}{n} = z + \\frac {1}{y}$. \\begin{align*} (5)(29)(r)&=\\left(x + \\frac {1}{z}\\right)\\left(y + \\frac {1}{x}\\right)\\left(z + \\frac {1}{y}\\right)\\\\ &=xyz + \\frac{xy}{y} + \\frac{xz}{x} + \\frac{yz}{z} + \\frac{x}{xy} + \\frac{y}{yz} + \\frac{z}{xz} + \\frac{1}{xyz}\\\\ &=1 + x + z + y + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{x} + \\frac{1}{1}\\\\ &=2 + \\left(x + \\frac {1}{z}\\right) + \\left(y + \\frac {1}{x}\\right) + \\left(z + \\frac {1}{y}\\right)\\\\ &=2 + 5 + 29 + r\\\\ &=36 + r \\end{align*} Thus $145r = 36+r \\Rightarrow 144r = 36 \\Rightarrow r = \\frac{36}{144} = \\frac{1}{4}$. So $m + n = 1 + 4 = 5.",
"Since $x+(1/z)=5, 1=z(5-x)=xyz$, so $5-x=xy$. Also, $y=29-(1/x)$ by the second equation. Substitution gives $x=1/5$, $y=24$, and $z=5/24$, so the answer is 4+1 which is equal to $5$.",
"(Hybrid between 1/2) Because $xyz = 1, \\hspace{0.15cm} \\frac{1}{x} = yz, \\hspace{0.15cm} \\frac{1}{y} = xz,$ and $\\hspace{0.05cm}\\frac{1}{z} = xy$. Substituting and factoring, we get $x(y+1) = 5$, $\\hspace{0.15cm}y(z+1) = 29$, and $\\hspace{0.05cm}z(x+1) = k$. Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$, but $xyz$ is $1$, and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$. And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$, and if we make a substitution for $xyz$, and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important. Now, lets add the 3 equations $x(y+1) = 5, \\hspace{0.15cm}y(z+1) = 29$, and $\\hspace{0.05cm}z(x+1) = k$. We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$, so $k = \\frac{1}{4}$ which gives an answer of $1+4 = 005 -AlexLikeMath",
"Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$. Then, since $xyz=1$, we have $\\tfrac 1y-5z=-1$ and $\\tfrac 1z-29x=-1$. Then, since we know that $\\tfrac 1z+x=5$, we can subtract these two equations to get that $30x=6\\implies x=5$. The result follows that $z=\\tfrac 5{24}$ and $y=24$, so $z+\\tfrac 1y=\\tfrac 1{24}+\\tfrac 5{24}=\\tfrac 14$, and the requested answer is $1+4=005",
"Rewrite the equations in terms of x. $x+\\frac{1}{z}=5$ becomes $z=\\frac{1}{x+5}$. $y+\\frac{1}{x}=29$ becomes $y=29-\\frac{1}{x}$ Now express $xyz=1$ in terms of x. $\\frac{1}{5-x}\\cdot(29-\\frac{1}{x})\\cdot x=1$. This evaluates to $29x-1=5-x$, giving us $x=\\frac{1}{5}$. We can now plug x into the other equations to get $y=24$ and $z=\\frac{5}{24}$. Therefore, $z+\\frac{1}{y}=\\frac{5}{24}+\\frac{1}{24}=\\frac{6}{24}=\\frac{1}{4}$. $1+4=5, and we are done. ~MC413551"
] |
2000-I-8
| 2,000 | 8 |
A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$ .
| 52 |
I
|
[
"Solution 1 The scale factor is uniform in all dimensions, so the volume of the liquid is $\\left(\\frac{3}{4}\\right)^{3}$ of the container. The remaining section of the volume is $\\frac{1-\\left(\\frac{3}{4}\\right)^{3}}{1}$ of the volume, and therefore $\\frac{\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}}{1}$ of the height when the vertex is at the top. So, the liquid occupies $\\frac{1-\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}}{1}$ of the height, or $12-12\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}=12-3\\left(37^{1/3}\\right)$. Thus $m+n+p=052. ~MC413551",
"The scale factor is uniform in all dimensions, so the volume of the liquid is $\\left(\\frac{3}{4}\\right)^{3}$ of the container. The remaining section of the volume is $\\frac{1-\\left(\\frac{3}{4}\\right)^{3}}{1}$ of the volume, and therefore $\\frac{\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}}{1}$ of the height when the vertex is at the top. So, the liquid occupies $\\frac{1-\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}}{1}$ of the height, or $12-12\\left(1-\\left(\\frac{3}{4}\\right)^{3}\\right)^{1/3}=12-3\\left(37^{1/3}\\right)$. Thus $m+n+p=052. ~MC413551",
"(Computational) The volume of a cone can be found by $V = \\frac{\\pi}{3}r^2h$. In the second container, if we let $h',r'$ represent the height, radius (respectively) of the air (so $12 -h'$ is the height of the liquid), then the volume of the liquid can be found by $\\frac{\\pi}{3}r^2h - \\frac{\\pi}{3}(r')^2h'$. By similar triangles, we find that the dimensions of the liquid in the first cone to the entire cone is $\\frac{3}{4}$, and that $r' = \\frac{rh'}{h}$; equating, \\begin{align*}\\frac{\\pi}{3}\\left(\\frac{3}{4}r\\right)^2 \\left(\\frac{3}{4}h\\right) &= \\frac{\\pi}{3}\\left(r^2h - \\left(\\frac{rh'}{h}\\right)^2h'\\right)\\\\ \\frac{37}{64}r^2h &= \\frac{r^2}{h^2}(h')^3 \\\\ h' &= \\sqrt[3]{\\frac{37}{64} \\cdot 12^3} = 3\\sqrt[3]{37}\\end{align*} Thus the answer is $12 - h' = 12-3\\sqrt[3]{37}$, and $m+n+p=052. ~MC413551",
"From the formula $V=\\frac{\\pi r^2h}{3}$, we can find that the volume of the container is $100\\pi$. The cone formed by the liquid is similar to the original, but scaled down by $\\frac{3}{4}$ in all directions, so its volume is $100\\pi*\\frac{27}{64}=\\frac{675\\pi}{16}$. The volume of the air in the container is the volume of the container minus the volume of the liquid, which is $\\frac{925\\pi}{16}$, which is $\\frac{37}{64}$ of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have $\\sqrt[3]{\\frac{37}{64}}=\\frac{\\sqrt[3]{37}}{4}$ of the height of the container. This means that the height of the liquid is $12\\left(1-\\frac{\\sqrt[3]{37}}{4}\\right)=12-3\\sqrt[3]{37}$ inches, so our answer is $052. ~MC413551",
"We find that the volume of the cone is $100\\pi$. The volume of the cone with height 9 is $\\frac{675}{16}\\pi$. The difference between the two volumes is $\\frac{925}{16}\\pi$. Note that this is the volume of the cone essentially 'on top of' the frustum described in the problem when the liquid is held with the base horizontal. We can express the volume as $x^2\\pi\\cdot\\frac{12}{5}x\\cdot\\frac{1}{3}$, where x is the radius of the cone. Solving this equation, we get $x=\\frac{5\\sqrt[3]{37}}{4}$. The height of this cone is $\\frac{12}{5}$ of the radius. Then we subtract the value from the height, 12, to get our answer: $12-3\\sqrt[3]{37}$. Therefore, our answer is $12+3+37=52. ~MC413551"
] |
2000-I-9
| 2,000 | 9 |
The system of equations \begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\ \log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\ \end{eqnarray*} has two solutions $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ . Find $y_{1} + y_{2}$ .
| 25 |
I
|
[
"Since $\\log ab = \\log a + \\log b$, we can reduce the equations to a more recognizable form: \\begin{eqnarray*} -\\log x \\log y + \\log x + \\log y - 1 &=& 3 - \\log 2000\\\\ -\\log y \\log z + \\log y + \\log z - 1 &=& - \\log 2\\\\ -\\log x \\log z + \\log x + \\log z - 1 &=& -1\\\\ \\end{eqnarray*} Let $a,b,c$ be $\\log x, \\log y, \\log z$ respectively. Using SFFT, the above equations become (*) \\begin{eqnarray*}(a - 1)(b - 1) &=& \\log 2 \\\\ (b-1)(c-1) &=& \\log 2 \\\\ (a-1)(c-1) &=& 1 \\end{eqnarray*} Small note from different author: $-(3 - \\log 2000) = \\log 2000 - 3 = \\log 2000 - \\log 1000 = \\log 2.$ From here, multiplying the three equations gives \\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\\log 2)^2\\\\ (a-1)(b-1)(c-1) &=& \\pm\\log 2\\end{eqnarray*} Dividing the third equation of (*) from this equation, $b-1 = \\log y - 1 = \\pm\\log 2 \\Longrightarrow \\log y = \\pm \\log 2 + 1$. (Note from different author if you are confused on this step: if $\\pm$ is positive then $\\log y = \\log 2 + 1 = \\log 2 + \\log 10 = \\log 20,$ so $y=20.$ if $\\pm$ is negative then $\\log y = 1 - \\log 2 = \\log 10 - \\log 2 = \\log 5,$ so $y=5.$) This gives $y_1 = 20, y_2 = 5$, and the answer is $y_1 + y_2 = 025.",
"Subtracting the second equation from the first equation yields \\begin{align*} \\log 2000xy-\\log 2yz-((\\log x)(\\log y)-(\\log y)(\\log z)) &= 3 \\\\ \\log\\frac{2000xy}{2yz}-\\log y(\\log x-\\log z) &= 3 \\\\ \\log1000+\\log\\frac{x}{z}-\\log y(\\log\\frac{x}{z}) &= 3 \\\\ 3+\\log\\frac{x}{z}-\\log y(\\log\\frac{x}{z}) &= 3 \\\\ \\log\\frac{x}{z}(1-\\log y) &= 0 \\\\ \\end{align*} If $1-\\log y=0$ then $y=10$. Substituting into the first equation yields $\\log20000=4$ which is not possible. If $\\log\\frac{x}{z}=0$ then $\\frac{x}{z}=1\\Longrightarrow x=z$. Substituting into the third equation gets \\begin{align*} \\log x^2-(\\log x)(\\log x) &= 0 \\\\ \\log x^2-\\log x^x &= 0 \\\\ \\log x^{2-x} &= 0 \\\\ x^{2-x} &= 1 \\\\ \\end{align*} Thus either $x=1$ or $2-x=0\\Longrightarrow x=2$. (Note that here $x\\neq-1$ since logarithm isn't defined for negative number.) Substituting $x=1$ and $x=2$ into the first equation will obtain $y=5$ and $y=20$, respectively. Thus $y_1+y_2=025. ~ Nafer",
"Let $a = \\log x$, $b = \\log y$ and $c = \\log z$. Then the given equations become: \\begin{align*} \\log 2 + a + b - ab = 1 \\\\ \\log 2 + b + c - bc = 1 \\\\ a+c = ac \\\\ \\end{align*} Equating the first and second equations, solving, and factoring, we get $a(1-b) = c(1-b) \\implies{a = c}$. Plugging this result into the third equation, we get $c = 0$ or $2$. Substituting each of these values of $c$ into the second equation, we get $b = 1 - \\log 2$ and $b = 1 + \\log 2$. Substituting backwards from our original substitution, we get $y = 5$ and $y = 20$, respectively, so our answer is $025. ~ anellipticcurveoverq",
"All logs are base 10 by convention. Rearrange the given statements: \\begin{align*} \\log 2000 + \\log x + \\log y - \\log x \\log y = 4, \\quad \\text{which becomes} \\quad \\log x + \\log y = \\log x \\log y + \\log 5. \\\\ \\log 2 + \\log y + \\log z - \\log y \\log z = 1, \\quad \\text{which becomes} \\quad \\log y + \\log z = \\log y \\log z + \\log 5. \\\\ \\log z + \\log x - \\log z \\log x = 0, \\quad \\text{which becomes} \\quad \\log x + \\log z = \\log z \\log x. \\\\ \\end{align*} Subtract the first two equations to obtain $(\\log x - \\log z) = \\log y (\\log x - \\log z).$ This must mean \\(\\log x = \\log z\\) because otherwise \\(\\log y = 1\\) turns the first equation into \\(\\log y = \\log 5\\) which is self-contradictory. With \\(\\log x = \\log z\\) we know each value satisfies \\(2a = a^2\\), so they are both \\(0\\) or both \\(2\\). Finally we arrive at our two solutions, where 0 gives us \\(0 + \\log y = 0 + \\log 5\\), and \\(y = 5\\), and 2 gives us \\(2 + \\log y = 2 \\log y + \\log 5\\), and \\(\\log y = 2 - \\log 5 = \\log 20\\), so \\(y = 20\\). Similar to above we arrive at $025. ~ GrindOlympiads"
] |
2000-I-10
| 2,000 | 10 |
A sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m + n$ .
| 173 |
I
|
[
"Let the sum of all of the terms in the sequence be $\\mathbb{S}$. Then for each integer $k$, $x_k = \\mathbb{S}-x_k-k \\Longrightarrow \\mathbb{S} - 2x_k = k$. Summing this up for all $k$ from $1, 2, \\ldots, 100$, \\begin{align*}100\\mathbb{S}-2(x_1 + x_2 + \\cdots + x_{100}) &= 1 + 2 + \\cdots + 100\\\\ 100\\mathbb{S} - 2\\mathbb{S} &= \\frac{100 \\cdot 101}{2} = 5050\\\\ \\mathbb{S}&=\\frac{2525}{49}\\end{align*} Now, substituting for $x_{50}$, we get $2x_{50}=\\frac{2525}{49}-50=\\frac{75}{49} \\Longrightarrow x_{50}=\\frac{75}{98}$, and the answer is $75+98=173.",
"Consider $x_k$ and $x_{k+1}$. Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\\dfrac{1}{2}.$ In terms of $x_{50},$ the sequence is $x_{50}+\\dfrac{49}{2}, x_{50}+\\dfrac{48}{2},\\cdots,x_{50}+\\dfrac{1}{2}, x_{50}, x_{50}-\\dfrac{1}{2}, \\cdots, x_{50}-\\dfrac{49}{2}, x_{50}-\\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\\dfrac{50}{2}$. Solving, we get $x_{50}=\\dfrac{75}{98}.$ The answer is $75+98=173 - JZ - edited by erinb28lms"
] |
2000-I-11
| 2,000 | 11 |
Let $S$ be the sum of all numbers of the form $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $\frac{S}{10}$ ?
| 248 |
I
|
[
"Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$, it follows that $\\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$, where $-3 \\le x,y \\le 3$. Thus every number in the form of $a/b$ will be expressed one time in the product \\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\\] Using the formula for a geometric series, this reduces to $S = \\frac{2^{-3}(2^7 - 1)}{2-1} \\cdot \\frac{5^{-3}(5^{7} - 1)}{5-1} = \\frac{127 \\cdot 78124}{4000} = 2480 + \\frac{437}{1000}$, and $\\left\\lfloor \\frac{S}{10} \\right\\rfloor = 248.",
"Essentially, the problem asks us to compute \\[\\sum_{a=-3}^3 \\sum_{b=-3}^3 \\frac{2^a}{5^b}\\] which is pretty easy: \\[\\sum_{a=-3}^3 \\sum_{b=-3}^3 \\frac{2^a}{5^b} = \\sum_{a=-3}^3 2^a \\sum_{b=-3}^3 \\frac{1}{5^b} = \\sum_{a=-3}^3 2^a 5^{3}\\bigg( \\frac{1-5^{-7}}{1-\\frac{1}{5}} \\bigg) = 5^{3}\\bigg( \\frac{1-5^{-7}}{1-\\frac{1}{5}} \\bigg) \\sum_{a=-3}^3 2^a = 5^{3}\\bigg( \\frac{1-5^{-7}}{1-\\frac{1}{5}} \\bigg)2^{-3} \\bigg( \\frac{1-2^7}{1-2} \\bigg) = 2480 + \\frac{437}{1000}\\] so our answer is $\\left\\lfloor \\frac{2480 + \\frac{437}{1000}}{10} \\right\\rfloor = 248.",
"The sum is equivalent to $\\sum_{i | 10^6}^{} \\frac{i}{1000}$ Therefore, it's the sum of the factors of $10^6$ divided by $1000$. The sum is $\\frac{127 \\times 19531}{1000}$ by the sum of factors formula. The answer is therefore $248 after some computation. - whatRthose",
"We can organize the fractions and reduce them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that are a combination of a 1 and another divisor. To begin with the first list, list powers of 2 and 5 from 0 to 3. In this specific case I find it easier to augment every denominator to 1000 and then divide by 1000. To do this, write the corresponding divisor under each power. e.g. 2 - 500, 4 - 250, 5 - 200, etc. Call this the \"partner\" of any divisor. Now we now the amount to multiply the numerator if given number is in the denominator. Now we simply combine and reduce these groups. If the powers of 2 are on the denominator, then every power of five will be multiplied by the partner of the power of 2. Essentially, all we have to do is a large scale distributive property application. There is nothing complicated about this except to be careful. Add all powers of 2: 15 Add their partners: 1875 Add all powers of 5: 156 Add their partners: 1248 Then, follow this formula: (sum of powers * sum opposite power's partners)+(sum of powers * sum opposite power's partners) Or, $156*1875+15*1248$ $=311220$ Now, divide by 1000 to compensate for the denominator. $311.22$ Finally, we have to calculate the other list of fractions with 1 and another divisor. e.g. 1 - 250, 1 - 20 etc. (these all count) This time we need to list all divisors of 1000, including 1. Remove all powers of 2 or 5, because we already included those in the other list. Now, notice there are two cases. 1: 1 is in the denominator, making the fraction an integer. 2: 1 is in the numerator. Adding all the integers in the first case gives us 2169. The second case can actually be discarded, but still can be found. Realize that if we include the powers of 2 and 5, then the second case is (sum of all divisors)/1000. Remove all partners of powers of 2 and 5, and we'll get exactly 217/1000, or 0.22. Finally, add all the numbers together: $311.22 + 2169 + 0.22 = 2480.44$ And divide by 10: $248.044$ After an odyssey of bashing: $248 -jackshi2006",
"We can organize the fractions and reduce them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that are a combination of a 1 and another divisor. To begin with the first list, list powers of 2 and 5 from 0 to 3. In this specific case I find it easier to augment every denominator to 1000 and then divide by 1000. To do this, write the corresponding divisor under each power. e.g. 2 - 500, 4 - 250, 5 - 200, etc. Call this the \"partner\" of any divisor. Now we now the amount to multiply the numerator if given number is in the denominator. Now we simply combine and reduce these groups. If the powers of 2 are on the denominator, then every power of five will be multiplied by the partner of the power of 2. Essentially, all we have to do is a large scale distributive property application. There is nothing complicated about this except to be careful. Add all powers of 2: 15 Add their partners: 1875 Add all powers of 5: 156 Add their partners: 1248 Then, follow this formula: (sum of powers * sum opposite power's partners)+(sum of powers * sum opposite power's partners) Or, $156*1875+15*1248$ $=311220$ Now, divide by 1000 to compensate for the denominator. $311.22$ Finally, we have to calculate the other list of fractions with 1 and another divisor. e.g. 1 - 250, 1 - 20 etc. (these all count) This time we need to list all divisors of 1000, including 1. Remove all powers of 2 or 5, because we already included those in the other list. Now, notice there are two cases. 1: 1 is in the denominator, making the fraction an integer. 2: 1 is in the numerator. Adding all the integers in the first case gives us 2169. The second case can actually be discarded, but still can be found. Realize that if we include the powers of 2 and 5, then the second case is (sum of all divisors)/1000. Remove all partners of powers of 2 and 5, and we'll get exactly 217/1000, or 0.22. Finally, add all the numbers together: $311.22 + 2169 + 0.22 = 2480.44$ And divide by 10: $248.044$ After an odyssey of bashing: $248 -jackshi2006",
"Since $1$ is relatively prime to everything, including itself, we start with $b=1$. The numerators will be all factors of 1000, and the sum of the factors of 1000 is $(1+2+4+8)(1+5+25+125) = 15*156=2340$. If the denominator is in the form $2^k$, then the numerator must be in the form $5^j$. The sum of the possible numerators is $1+5+25+125 = 156$, so the sum of all such fractions with denominator $2^k$ is $\\frac{156}{2}+\\frac{156}{4}+\\frac{156}{8} = 136.5$ If the denominator is in the form $5^k$, then the numerator must be in the form $2^j$. The sum of the possible numerators is $1+2+4+8 =15$, so the sum of all such fractions with denominator $5^k$ is $\\frac{15}{5}+\\frac{15}{25}+\\frac{15}{125}$, which is around $3.6$. If the denominator is in the form $2^k5^j$, where $k\\ge 1$ and $j\\ge 1$, the numerator must be $1$. We can get the sum of all such fractions from $(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8})(\\frac{1}{5}+\\frac{1}{25}+\\frac{1}{125}) = \\frac{217}{1000}$. Adding, we get $2480$, with a bit extra, so our answer is $248. ~skibbysiggy",
"Since $1$ is relatively prime to everything, including itself, we start with $b=1$. The numerators will be all factors of 1000, and the sum of the factors of 1000 is $(1+2+4+8)(1+5+25+125) = 15*156=2340$. If the denominator is in the form $2^k$, then the numerator must be in the form $5^j$. The sum of the possible numerators is $1+5+25+125 = 156$, so the sum of all such fractions with denominator $2^k$ is $\\frac{156}{2}+\\frac{156}{4}+\\frac{156}{8} = 136.5$ If the denominator is in the form $5^k$, then the numerator must be in the form $2^j$. The sum of the possible numerators is $1+2+4+8 =15$, so the sum of all such fractions with denominator $5^k$ is $\\frac{15}{5}+\\frac{15}{25}+\\frac{15}{125}$, which is around $3.6$. If the denominator is in the form $2^k5^j$, where $k\\ge 1$ and $j\\ge 1$, the numerator must be $1$. We can get the sum of all such fractions from $(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8})(\\frac{1}{5}+\\frac{1}{25}+\\frac{1}{125}) = \\frac{217}{1000}$. Adding, we get $2480$, with a bit extra, so our answer is $248. ~skibbysiggy"
] |
2000-I-12
| 2,000 | 12 |
Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)$ ?
| 177 |
I
|
[
"\\begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\\end{align*} Since $\\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm) \\[f(x) = f(352 + x)\\] So we need only to consider one period $f(0), f(1), ... f(351)$, which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers. But we also know that $f(x) = f(46 - x) = f(398 - x)$, so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated. This gives a total of \\[352 - (46 - 24 + 1) - (351 - 200 + 1) = 177 (in degrees).",
"This gives the intuition the first solution uses to solve the problem. One can imagine that there must be multiple lines of symmetry for the function $f(x)$, as if a function can be expressed with $f(x)=f(a-x)$ it must be symmetric against line $x=\\frac{a}{2}$. Try this yourself by graphing a polynomial $f(x)$, then graphing $f(n-x)$. If $f(x)=f(n-x)$, their point of intersection at $x=\\frac{n}{2}$ must contain a line of symmetry. For this particular function $f(x)$, it has $3$ lines of symmetry already given: $x=199$, $x=1079$, and $x=1607$. Now imagine these lines of symmetry folding over each other to form new lines of symmetry, because $x=199$ should have another corresponding line of symmetry when being reflected over the lines $x=1079$ and $x=1607$ due to it being, well, symmetric. Doing so, we find that there will be two new lines of symmetry generated by folding $x=199$ over the other 2 lines of symmetry, namely, $x=1079+(1079-199)=1959$ and $x=1607+(1607-199)=3015$. Now, if we fold $x=1079$ over the other 2 lines of symmetry, we will find another 2 lines of symmetry. We can even fold lines of symmetry not originally given by the problem, such as $x=3015$ over $x=1959$ to form even more lines of symmetry. This will result in infinite lines of symmetry, which tells us that this function is periodic. As we have just proven that f(x) has a finite amount of values, now, we need to find what the half period it is, and not a whole period because in periodic functions such as cos(x), the y values will repeat every half period and not a full period. To do this, we go back to the 3 lines of symmetry given by the problem, and we fold $x=1607$ over $x=1079$ to obtain $x=1079-(1607-1079)=551$. We now fold $x=551$ over $x=199$ to find $x=199-(551-199)=-153$. Now we fold $x=-153$ over $x=551$ to get $x=551+(551-(-153))=1255$. Now we fold $x=1255$ over $x=1079$ to find $x=903$, then we fold $x=1079$ over $x=903$, then fold $x=903$ over that line again, until suddenly, we find that after doing a few folds, there will be a line that coincides with $x=199$. Now, if you visualize a graph of all the lines of symmetry, you will find that from $x=199$ to $x=1079$, there will lines of symmetry evenly spaced by $176$ units each. If fold $x=1255$ and $x=1079$ in the other direction, you will find that it will coincide with all other named lines of symmetry and extend to infinity, with one line of symmetry every $176$ units. This tells us that the half period of $f(x)$ is $176$ units, which means that every $176$ units, the y-values will repeat. Now, all we have to do is put in the answer $177 units, it will have counted not the total amount of values, but the total amount of values minus one. That's why we need to add 1 to our answer. -WhatdoHumanitariansEat"
] |
2000-I-13
| 2,000 | 13 |
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $\frac{m}{n}$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 731 |
I
|
[
"Let the intersection of the highways be at the origin $O$, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction. After going $x$ miles, $t=\\frac{d}{r}=\\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\\frac{1}{10}-\\frac{x}{50}$ hours, or $d=rt=14t=1.4-\\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$. All these circles are homothetic with respect to a center at $(5,0)$. [asy] pair truck(pair P){ pair Q = IP(P--P+(7/10,24/10),(35/31,35/31)--(5,0)); D(P--Q,EndArrow(5)); D(CP(P,Q),linewidth(0.5)); return Q; } pointpen = black; pathpen = black+linewidth(0.7); size(250); pair B=(5,0), C=(35/31,35/31); D(D(B)--D(C)--D(B*dir(90))--D(C*dir(90))--D(B*dir(180))--D(C*dir(180))--D(B*dir(270))--D(C*dir(270))--cycle); D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0)); [/asy] [asy] pointpen = black; pathpen = black+linewidth(0.7); size(250); pair O=(0,0), B=(5,0), A=1.4*expi(atan(24/7)), C=1.4*expi(atan(7/24)); D(D(B)--D(A)--D(O)); D(O--D(C)--D(B*dir(90))--D(A*dir(90))--O--D(C*dir(90))--D(B*dir(180))--D(A*dir(180))--O--D(C*dir(180))--D(B*dir(270))--D(A*dir(270))--O--D(C*dir(270))--B,linewidth(0.5)); D(CR(O,1.4)); D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); MP(\"A\",A,N); MP(\"B\",B); MP(\"(5,0)\",B,N); D(MP(\"\\left(\\frac{35}{31},\\frac{35}{31}\\right)\",(35/31,35/31),NE)); D(rightanglemark(O,A,B)); [/asy] Now consider the circle at $(0,0)$. Draw a line tangent to it at $A$ and passing through $B (5,0)$. By the Pythagorean Theorem $AB^2+AO^2=OB^2 \\Longrightarrow AB=\\sqrt{OB^2-AO^2}=\\sqrt{5^2-1.4^2}=\\frac{24}{5}$. Then $\\tan(\\angle ABO)=\\frac{OA}{AB}=\\frac{7}{24}$, so the slope of line $AB$ is $\\frac{-7}{24}$. Since it passes through $(5,0)$ its equation is $y=\\frac{-7}{24}(x-5)$. This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\\left(\\frac{35}{31},\\frac{35}{31}\\right)$. The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$. By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\\frac{700}{31}$ so the answer is $700+31=731."
] |
2000-I-14
| 2,000 | 14 |
In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$ .
| 571 |
I
|
[
"[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label(\"\\(A\\)\",A,(0,1));label(\"\\(B\\)\",B,(-1,-1));label(\"\\(C\\)\",C,(1,-1));label(\"\\(P\\)\",P,(1,1)); label(\"\\(Q\\)\",Q,(-1,1));label(\"\\(R\\)\",R,(1,0)); [/asy] Let point $R$ be in $\\triangle ABC$ such that $QB = BR = RP$. Then $PQBR$ is a rhombus, so $AB \\parallel PR$ and $APRB$ is an isosceles trapezoid. Since $\\overline{PB}$ bisects $\\angle QBR$, it follows by symmetry in trapezoid $APRB$ that $\\overline{RA}$ bisects $\\angle BAC$. Thus $R$ lies on the perpendicular bisector of $\\overline{BC}$, and $BC = BR = RC$. Hence $\\triangle BCR$ is an equilateral triangle. Now $\\angle ABR = \\angle BAC = \\angle ACR$, and the sum of the angles in $\\triangle ABC$ is $\\angle ABR + 60^{\\circ} + \\angle BAC + \\angle ACR + 60^{\\circ} = 3\\angle BAC + 120^{\\circ} = 180^{\\circ} \\Longrightarrow \\angle BAC = 20^{\\circ}$. Then $\\angle APQ = 140^{\\circ}$ and $\\angle ACB = 80^{\\circ}$, so the answer is $\\left\\lfloor 1000 \\cdot \\frac{80}{140} \\right\\rfloor = \\left\\lfloor \\frac{4000}{7} \\right\\rfloor = 571.",
"Let $AP=PQ=QB=BC=x$ and $A$ be the measure of $\\angle BAC$. Since $\\triangle APQ$ and $\\triangle ABC$ are isoceles, $\\angle APQ = 180-2A$ and $\\angle ACB = 90-\\frac{A}{2}$. Because $\\triangle APQ$ and $\\triangle ABC$ both have a side length $x$ opposite $\\angle BAC$, by the law of sines: $\\frac{x}{\\sin A}=\\frac{AQ}{\\sin(180-2A)}=\\frac{AQ+x}{\\sin(90-\\frac{A}{2})}$ Simplifying, this becomes $\\frac{x}{\\sin A}=\\frac{AQ}{\\sin 2A}=\\frac{AQ+x}{\\cos \\frac{A}{2}}$ From the first two fractions, $AQ\\cdot \\sin A = x \\cdot \\sin 2A = x \\cdot (2\\sin A \\cos A) \\Longrightarrow AQ=x\\cdot 2\\cos A$ Substituting, we have from the first and third fractions, $\\frac{x}{\\sin A}=\\frac{x\\cdot 2\\cos A + x}{\\cos \\frac{A}{2}} \\Longrightarrow 2\\cos A\\sin A + \\sin A=\\sin 2A + \\sin A = \\cos \\frac{A}{2}$ By sum-to-product, $\\sin 2A + \\sin A = 2\\sin \\frac{3A}{2} \\cos \\frac{A}{2}$ Thus, $2\\sin \\frac{3A}{2} \\cos \\frac{A}{2} = \\cos \\frac{A}{2} \\Longrightarrow \\sin \\frac{3A}{2} = \\frac{1}{2}$ Because $BC=QB<AB$, $\\angle A$ is acute, so $\\frac{3A}{2}=30 \\Longrightarrow A=20$ $\\angle ACB = \\frac{180-20}{2}=80$, $\\angle APQ = 180-2\\cdot 20 = 140 \\Longrightarrow r=\\frac{4}{7}$ $1000r=\\frac{4000}{7}=571 ~bad_at_mathcounts",
"Let $AP=PQ=QB=BC=x$ and $A$ be the measure of $\\angle BAC$. Since $\\triangle APQ$ and $\\triangle ABC$ are isoceles, $\\angle APQ = 180-2A$ and $\\angle ACB = 90-\\frac{A}{2}$. Because $\\triangle APQ$ and $\\triangle ABC$ both have a side length $x$ opposite $\\angle BAC$, by the law of sines: $\\frac{x}{\\sin A}=\\frac{AQ}{\\sin(180-2A)}=\\frac{AQ+x}{\\sin(90-\\frac{A}{2})}$ Simplifying, this becomes $\\frac{x}{\\sin A}=\\frac{AQ}{\\sin 2A}=\\frac{AQ+x}{\\cos \\frac{A}{2}}$ From the first two fractions, $AQ\\cdot \\sin A = x \\cdot \\sin 2A = x \\cdot (2\\sin A \\cos A) \\Longrightarrow AQ=x\\cdot 2\\cos A$ Substituting, we have from the first and third fractions, $\\frac{x}{\\sin A}=\\frac{x\\cdot 2\\cos A + x}{\\cos \\frac{A}{2}} \\Longrightarrow 2\\cos A\\sin A + \\sin A=\\sin 2A + \\sin A = \\cos \\frac{A}{2}$ By sum-to-product, $\\sin 2A + \\sin A = 2\\sin \\frac{3A}{2} \\cos \\frac{A}{2}$ Thus, $2\\sin \\frac{3A}{2} \\cos \\frac{A}{2} = \\cos \\frac{A}{2} \\Longrightarrow \\sin \\frac{3A}{2} = \\frac{1}{2}$ Because $BC=QB<AB$, $\\angle A$ is acute, so $\\frac{3A}{2}=30 \\Longrightarrow A=20$ $\\angle ACB = \\frac{180-20}{2}=80$, $\\angle APQ = 180-2\\cdot 20 = 140 \\Longrightarrow r=\\frac{4}{7}$ $1000r=\\frac{4000}{7}=571 ~bad_at_mathcounts",
"[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label(\"\\(A\\)\",A,(0,1));label(\"\\(B\\)\",B,(-1,-1));label(\"\\(C\\)\",C,(1,-1));label(\"\\(P\\)\",P,(1,1)); label(\"\\(Q\\)\",Q,(-1,1));label(\"\\(R\\)\",R,(1,0));label(\"\\(S\\)\",S,(-1,0)); [/asy] Again, construct $R$ as above. Let $\\angle BAC = \\angle QBR = \\angle QPR = 2x$ and $\\angle ABC = \\angle ACB = y$, which means $x + y = 90$. $\\triangle QBC$ is isosceles with $QB = BC$, so $\\angle BCQ = 90 - \\frac {y}{2}$. Let $S$ be the intersection of $QC$ and $BP$. Since $\\angle BCQ = \\angle BQC = \\angle BRS$, $BCRS$ is cyclic, which means $\\angle RBS = \\angle RCS = x$. Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\\angle BAC$, $\\angle ABR = \\angle ACR = 2x$. Therefore we have that $\\angle ACB = \\angle ACR + \\angle RCS + \\angle QCB = 2x + x + 90 - \\frac {y}{2} = y$. We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \\frac {y}{2} = y$ to get $x = 10$ and $y = 80$. $\\angle APQ = 180 - 4x = 140$, $\\angle ACB = 80$, so $r = \\frac {80}{140} = \\frac {4}{7}$. $\\left\\lfloor 1000\\left(\\frac {4}{7}\\right)\\right\\rfloor = 571.",
"Let $\\angle BAC= 2\\theta$ and $AP=PQ=QB=BC=x$. $\\triangle APQ$ is isosceles, so $AQ=2x\\cos 2\\theta =2x(1-2\\sin^2\\theta)$ and $AB= AQ+x=x\\left(3-4\\sin^2\\theta\\right)$. $\\triangle{ABC}$ is isosceles too, so $x=BC=2AB\\sin\\theta$. Using the expression for $AB$, we get \\[1=2\\left(3\\sin\\theta-4\\sin^3\\theta\\right)=2\\sin3\\theta\\]by the triple angle formula! Thus $\\theta=10^\\circ$ and $\\angle A = 2\\theta=20^\\circ$. It follows now that $\\angle APQ=140^\\circ$, $\\angle ACB=80^\\circ$, giving $r=\\tfrac{4}{7}$, which implies that $1000r = 571 + \\tfrac 37$. So the answer is $571.",
"Let $\\angle BAC= 2\\theta$ and $AP=PQ=QB=BC=x$. $\\triangle APQ$ is isosceles, so $AQ=2x\\cos 2\\theta =2x(1-2\\sin^2\\theta)$ and $AB= AQ+x=x\\left(3-4\\sin^2\\theta\\right)$. $\\triangle{ABC}$ is isosceles too, so $x=BC=2AB\\sin\\theta$. Using the expression for $AB$, we get \\[1=2\\left(3\\sin\\theta-4\\sin^3\\theta\\right)=2\\sin3\\theta\\]by the triple angle formula! Thus $\\theta=10^\\circ$ and $\\angle A = 2\\theta=20^\\circ$. It follows now that $\\angle APQ=140^\\circ$, $\\angle ACB=80^\\circ$, giving $r=\\tfrac{4}{7}$, which implies that $1000r = 571 + \\tfrac 37$. So the answer is $571.",
"[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(0)+20*dir(80), D=20*dir(0)+20*dir(80)+20*dir(60)+20*dir(100); draw(R--A--B--C--D--A--C);draw(Q--P--R--Q--C); draw(B--P--D); label(\"A\",A,NW); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NE); label(\"P\",P,W); label(\"Q\",Q,W); label(\"R\",R,E);[/asy] Reflect $\\triangle ABC$ over $BC$ and translate it to attach side $AB$ onto $AC$, mapping $\\triangle ABC$ to $\\triangle CAD$. Point $P$ maps to $R$, and $Q$ maps to $P$. Then we have that $BC=BQ=QP=PA=AD=PR=RC$. Notice how $BQ=RC$ and $BQ\\parallel RC$, so $BQRC$ is a parallelogram and $QR=BC$. But $BC=QP=PR$, so $\\triangle QPR$ is actually equilateral. Set $\\angle BAC=\\angle ACD=x$. Then notice that $\\angle QPC=\\angle PQA+\\angle PAQ=2x$, but $\\angle RPC=\\angle PQA=x$. Thus $\\angle QPR=3x=60$, so $x=20$. Thus $\\angle QPA=140^{\\circ}$ and $\\angle BCA=80^{\\circ}$, so $r=\\frac{80}{140}=\\frac{4}{7}$. The answer is $\\left \\lfloor \\frac{4000}{7}\\right \\rfloor =571. ~ethanzhang1001"
] |
2000-I-15
| 2,000 | 15 |
A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$ ?
| 927 |
I
|
[
"We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, $2(1024 - 1000) = 48$, to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \\times 2 = 928$, meaning that there were $927.",
"To simplify matters, we want a power of $2$. Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front. Let the fake cards have positions $1, 3, 5, \\cdots, 95$. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$. From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$, so $927 cards are above it. - Spacesam",
"We work backwards. To reverse the process, we must move the bottom card to the top, and add a new number to the top. Let $d_n$ equal the number of cards below 1999 after $n$ process reversals. Reversing one process, our deck only has $2000$, so we reverse again to obtain $1999, 2000$. So, $d_2 = 1$. When $d_{n-1} > 0$, after a process reversal, the bottom card is moved to the top (it goes above $1999$), so we subtract by one to account for this (the addition of the new card doesn't matter since it goes above $1999$), giving us $d_n = d_{n-1} - 1$. So, $d_3 = 0$. Then, when $d_{n-1} = 0$, after a process reversal, $1999$ moves to the top and one more card is added above $1999$. Since at $d_{n}$, $n$ cards have been added, there must be $n - 2$ cards below $1999$. So, $d_4 = 2$. Then, consider all $d_{n-1} = 0$. Then, $d_n = n-2$ as previously stated. After $n-2$ process reversals, we go back to $d_{2n - 2} = n - 2 - (n - 2) = 0$. Then, $d_{2n-1} = 2n - 3$. Next, we can simply calculate $2 \\cdot 4 - 1 = 7$, $2 \\cdot 7 - 1 = 13$, and so on (which is a very simple bash, as $2000$ is a small number. If you don't want to do this, define sequence $a_n = 2a_{n-1} - 1$, and solve for the closed form, which is very easy). Consequently, we derive $d_{1537} = 1537 - 2 = 1535$. $2000 - 1537 = 463$ steps remain. So, $d_{2000} = d_{1537} - 463 = 1072$. Finally, there are $2000 - 1 - 1072 = \\textbf{927}. ~CrazyVideoGamez",
"We work backwards. To reverse the process, we must move the bottom card to the top, and add a new number to the top. Let $d_n$ equal the number of cards below 1999 after $n$ process reversals. Reversing one process, our deck only has $2000$, so we reverse again to obtain $1999, 2000$. So, $d_2 = 1$. When $d_{n-1} > 0$, after a process reversal, the bottom card is moved to the top (it goes above $1999$), so we subtract by one to account for this (the addition of the new card doesn't matter since it goes above $1999$), giving us $d_n = d_{n-1} - 1$. So, $d_3 = 0$. Then, when $d_{n-1} = 0$, after a process reversal, $1999$ moves to the top and one more card is added above $1999$. Since at $d_{n}$, $n$ cards have been added, there must be $n - 2$ cards below $1999$. So, $d_4 = 2$. Then, consider all $d_{n-1} = 0$. Then, $d_n = n-2$ as previously stated. After $n-2$ process reversals, we go back to $d_{2n - 2} = n - 2 - (n - 2) = 0$. Then, $d_{2n-1} = 2n - 3$. Next, we can simply calculate $2 \\cdot 4 - 1 = 7$, $2 \\cdot 7 - 1 = 13$, and so on (which is a very simple bash, as $2000$ is a small number. If you don't want to do this, define sequence $a_n = 2a_{n-1} - 1$, and solve for the closed form, which is very easy). Consequently, we derive $d_{1537} = 1537 - 2 = 1535$. $2000 - 1537 = 463$ steps remain. So, $d_{2000} = d_{1537} - 463 = 1072$. Finally, there are $2000 - 1 - 1072 = \\textbf{927}. ~CrazyVideoGamez",
"Consider the general problem: with a stack of $n$ cards such that they will be laid out $1, 2, 3, ..., n$ from left to right, how many cards are above the card labeled $n-1$? Let $a_n$ be the answer to the above problem. As a base case, consider $n=2$. Clearly, the stack, from top to bottom, must be $(1, 2)$, so $a_2=0$. Next, let's think about how we can construct a stack of $n+1$ cards from a stack of $n$ cards. First, let us renumber the current stack of $n$ by adding $1$ to the label of each of the cards. Then we must add a card labeled \"$1$\" somewhere in the new deck. Working backwards, we find that we must move the bottom card to the top, then add \"$1$\" to the top of the deck. This is because after one move, the number \"$1$\" will be laid out and the top card will be moved to the bottom, so the deck becomes the same as what we had before, but with everything renumbered correctly such that $2, 3, 4, ...$ will then be laid out in order. Therefore, if $a_n \\ne n-1$ (meaning the card $n-1$ is not at the bottom of the deck and so it won't be moved to the top), then $a_{n+1} = a_n + 2$, since a card from the bottom is moved to be above the $n-1$ card, and the new card \"$1$\" is added to the top. If $a_n = n-1$ (meaning the card $n-1$ is the bottom card), then $a_{n+1}=1$ because the $n-1$ card will move to the top and the card \"$1$\" will be added on top of it. With these recursions and the base case we found earlier, we calculate $a_{2000} = 927. - Frestho",
"Consider the general problem: with a stack of $n$ cards such that they will be laid out $1, 2, 3, ..., n$ from left to right, how many cards are above the card labeled $n-1$? Let $a_n$ be the answer to the above problem. As a base case, consider $n=2$. Clearly, the stack, from top to bottom, must be $(1, 2)$, so $a_2=0$. Next, let's think about how we can construct a stack of $n+1$ cards from a stack of $n$ cards. First, let us renumber the current stack of $n$ by adding $1$ to the label of each of the cards. Then we must add a card labeled \"$1$\" somewhere in the new deck. Working backwards, we find that we must move the bottom card to the top, then add \"$1$\" to the top of the deck. This is because after one move, the number \"$1$\" will be laid out and the top card will be moved to the bottom, so the deck becomes the same as what we had before, but with everything renumbered correctly such that $2, 3, 4, ...$ will then be laid out in order. Therefore, if $a_n \\ne n-1$ (meaning the card $n-1$ is not at the bottom of the deck and so it won't be moved to the top), then $a_{n+1} = a_n + 2$, since a card from the bottom is moved to be above the $n-1$ card, and the new card \"$1$\" is added to the top. If $a_n = n-1$ (meaning the card $n-1$ is the bottom card), then $a_{n+1}=1$ because the $n-1$ card will move to the top and the card \"$1$\" will be added on top of it. With these recursions and the base case we found earlier, we calculate $a_{2000} = 927. - Frestho",
"Let us treat each run through the deck as a separate \"round\". For example, in round one, you would go through all of the $2000$ cards initially in the deck once, in round two, you would go through all $1000$ cards initially in the deck once, so on and so forth. For each round, let us record what the initial and final actions are ($r$ for moving the card to the right, $b$ for moving the card to the bottom), the number of cards moved to the right, the number of cards left, and what the position of the cards moved to the right were in the original $2000$ card deck (as $n = a + ck$ where $n$ is the position, $c$ is the spacing of the cards moved, $a$ is an integer such that the correct first card is moved, and $k$ is an integer greater than or equal to $1$ which represents which card out of all the cards moved to the right you are finding the position of). Then, Round 1: $r$ to $b$, $1000$ to right, $1000$ left in deck, $n = -1 + 2k$, Round 2: $r$ to $b$, $500$ to right, $500$ left in deck, $n = -2 + 4k$, Round 3: $r$ to $b$, $250$ to right, $250$ left in deck, $n = -4 + 8k$, Round 4: $r$ to $b$, $125$ to right, $125$ left in deck, $n = -8 + 16k$. Let us treat the remaining deck of $125$ cards as a totally independent deck, note that the positions of card in this deck are $n = 16k$. Also note that the first action of a new round is never the same action as the last action of the previous round because actions alternate. Also note that for every new round, the spacing between cards moved doubles, and that the cards remaining in the beginning of a new round have position $n = a + c/2 + ck$ for the values $a, c$ of the previous round. Also note that if there are an odd number of cards in an initial deck, the first and last actions are the same. Then, Round 5: $r$ to $r$, $63$ to right, $62$ left in deck, $n = -1 + 2k$, Round 6: $b$ to $r$, $31$ to right, $31$ left in deck, $n = 4k$, because $n = 2k$ positioned cards are left at the beginning of this round and the first card is sent to the bottom, only every second card is sent to the right, and because spacing doubles every round, Round 7: $b$ to $b$, $15$ to right, $16$ left in deck, $n = -2 + 8k$, because $n = 2 + 4k$ positioned cards are left at the beginning and only every second card is sent to the right, and because spacing doubles every round, Round 8: $r$ to $b$, $8$ to right, $8$ left in deck, $n = -14 + 16k$, by similar reasoning, since the cards $n = 2 + 8k$ are left and spacing doubles every round, from here on things get real easy, Round 9: $r$ to $b$, $4$ to right, $4$ left in deck, $n = -22 + 32k$, Round 10: $r$ to $b$, $2$ to right, $2$ left in deck, $n = -38 + 64k$, Round 11: $r$ to $b$, $1$ to right, $1$ left in deck, $n = 58$, since $58 = -38 + 64/2 + 64$. This card must be labeled 1999 since it is second to last. Then, since it is the $58th$ in the deck of $125$, it must be the $58 * 8 = 928th$ card in the original deck, and because we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were $928 - 1 = 927 cards above it in the deck - Romulus and minimaxweii",
"Let us treat each run through the deck as a separate \"round\". For example, in round one, you would go through all of the $2000$ cards initially in the deck once, in round two, you would go through all $1000$ cards initially in the deck once, so on and so forth. For each round, let us record what the initial and final actions are ($r$ for moving the card to the right, $b$ for moving the card to the bottom), the number of cards moved to the right, the number of cards left, and what the position of the cards moved to the right were in the original $2000$ card deck (as $n = a + ck$ where $n$ is the position, $c$ is the spacing of the cards moved, $a$ is an integer such that the correct first card is moved, and $k$ is an integer greater than or equal to $1$ which represents which card out of all the cards moved to the right you are finding the position of). Then, Round 1: $r$ to $b$, $1000$ to right, $1000$ left in deck, $n = -1 + 2k$, Round 2: $r$ to $b$, $500$ to right, $500$ left in deck, $n = -2 + 4k$, Round 3: $r$ to $b$, $250$ to right, $250$ left in deck, $n = -4 + 8k$, Round 4: $r$ to $b$, $125$ to right, $125$ left in deck, $n = -8 + 16k$. Let us treat the remaining deck of $125$ cards as a totally independent deck, note that the positions of card in this deck are $n = 16k$. Also note that the first action of a new round is never the same action as the last action of the previous round because actions alternate. Also note that for every new round, the spacing between cards moved doubles, and that the cards remaining in the beginning of a new round have position $n = a + c/2 + ck$ for the values $a, c$ of the previous round. Also note that if there are an odd number of cards in an initial deck, the first and last actions are the same. Then, Round 5: $r$ to $r$, $63$ to right, $62$ left in deck, $n = -1 + 2k$, Round 6: $b$ to $r$, $31$ to right, $31$ left in deck, $n = 4k$, because $n = 2k$ positioned cards are left at the beginning of this round and the first card is sent to the bottom, only every second card is sent to the right, and because spacing doubles every round, Round 7: $b$ to $b$, $15$ to right, $16$ left in deck, $n = -2 + 8k$, because $n = 2 + 4k$ positioned cards are left at the beginning and only every second card is sent to the right, and because spacing doubles every round, Round 8: $r$ to $b$, $8$ to right, $8$ left in deck, $n = -14 + 16k$, by similar reasoning, since the cards $n = 2 + 8k$ are left and spacing doubles every round, from here on things get real easy, Round 9: $r$ to $b$, $4$ to right, $4$ left in deck, $n = -22 + 32k$, Round 10: $r$ to $b$, $2$ to right, $2$ left in deck, $n = -38 + 64k$, Round 11: $r$ to $b$, $1$ to right, $1$ left in deck, $n = 58$, since $58 = -38 + 64/2 + 64$. This card must be labeled 1999 since it is second to last. Then, since it is the $58th$ in the deck of $125$, it must be the $58 * 8 = 928th$ card in the original deck, and because we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were $928 - 1 = 927 cards above it in the deck - Romulus and minimaxweii",
"Similar to Solution 5, we treat each run-through of the deck from the lowest-indexed card to the highest-indexed card as a separate round. Notice that after each round, approximately half the deck will remain, with the other half having been cast aside to the right in sorted order. Then, we can model each round as if we are running a \"Sieve\" through the deck with powers of 2s, filtering out the cards to put to the right and the cards to be pushed to the bottom. At the beginning, when all $2000$ cards are still in the deck, notice that the first run-through of the deck consists of removing all the cards such that their index $I\\equiv 1\\pmod{2}$, while we leave behind all the cards such that their index $I\\equiv 2\\pmod{2}$. (Since the cards are not $0$-indexed, using $2$ instead of $0\\pmod{2}$ is simpler to keep track of the cards. This will appear again later.) The remaining $1000$ cards, then, will all share the attribute that their index numbers are even. To split this further for round 2, we look more specifically at each index, categorizing them further as $I\\equiv 2\\pmod 4$ or $I\\equiv 4\\pmod 4$ in that order (The card with the smaller remainder will always be the card on the top of the new deck). Then, in Round 2, we will remove all the cards with index values that leave a remainder of $2$ when divided by $4$, while leaving the multiples of $4$ behind. Notice that each time, in each new deck, we are removing all the cards with an odd position in that specific deck, and leaving behind all the cards with an even position in that specific deck. This process will continue until we reach Round 5, where the number of cards that remain ($125$) is no longer a multiple of 2. Therefore, when we remove all the cards with indices $I\\equiv 16\\pmod{32}$ and leave behind all the cards with indices $I\\equiv 32\\pmod{32}$, the last card in the deck will be removed instead of being placed to the bottom, with $125-63=62$ cards remaining. Because of this, when Round 6 begins, we cannot follow our normal procedure, since the top card of the Round 6 deck would have been moved to the bottom! Therefore, when we are using our sieve with $I\\equiv 32\\pmod{64}$ and $I\\equiv 64\\pmod{64}$, the order at which the remainders are presented have flipped, so all the cards with indices that are multiples of 64 will be removed, while all the cards with indices that are 32 more than a multiple of 64 will remain. The $62$ cards from Round 6 will be reduced to $31$, and the last card in the deck will be removed just like in Round 5, which means Round 7 will be affected the same way. Of the choices of $I\\equiv 32\\pmod{128}$ and $I\\equiv 32+64=96\\pmod{128}$, we will remove the latter, leaving behind the cards with indices that are 32 more than a multiple of 128. However, since 31 is an odd number, the last card of the deck will NOT be removed, which means our sieving process will return to normal after Round 7, with $31-15=16$ cards remaining. Since $16$ is a perfect power of $2$, we will not need to worry about this scenario again in the following rounds. Round 8: $I\\equiv 32\\pmod{256}$ removed; $I\\equiv 32+128=160\\pmod{256}$ remains; $16-8=8$ cards left. Round 9: $I\\equiv 160\\pmod{512}$ removed; $I\\equiv 160+256=416\\pmod{512}$ remains; $8-4=4$ cards left. Round 10: $I\\equiv 416\\pmod{1024}$ removed; $I\\equiv 416+512=928\\pmod{1024}$ remains; $4-2=2$ cards left. Round 11: $I\\equiv 928\\pmod{2048}$ removed; $I\\equiv 928+1024=1952\\pmod{2048}$ remains; $2-1=1$ card left. Round 12: The last card in position $1952$ is sorted into place (This is card #$2000$!). It is obvious that the single card put into place in Round 11 is the second-to-last card in the deck, which is our target #$1999$. Then, its position in the original deck leaves a remainder of $928$ when divided by $2056$. It is easy to see that $928$ is the only index that satisfies this, so there will be $928-1=927. ~chz3369",
"Similar to Solution 5, we treat each run-through of the deck from the lowest-indexed card to the highest-indexed card as a separate round. Notice that after each round, approximately half the deck will remain, with the other half having been cast aside to the right in sorted order. Then, we can model each round as if we are running a \"Sieve\" through the deck with powers of 2s, filtering out the cards to put to the right and the cards to be pushed to the bottom. At the beginning, when all $2000$ cards are still in the deck, notice that the first run-through of the deck consists of removing all the cards such that their index $I\\equiv 1\\pmod{2}$, while we leave behind all the cards such that their index $I\\equiv 2\\pmod{2}$. (Since the cards are not $0$-indexed, using $2$ instead of $0\\pmod{2}$ is simpler to keep track of the cards. This will appear again later.) The remaining $1000$ cards, then, will all share the attribute that their index numbers are even. To split this further for round 2, we look more specifically at each index, categorizing them further as $I\\equiv 2\\pmod 4$ or $I\\equiv 4\\pmod 4$ in that order (The card with the smaller remainder will always be the card on the top of the new deck). Then, in Round 2, we will remove all the cards with index values that leave a remainder of $2$ when divided by $4$, while leaving the multiples of $4$ behind. Notice that each time, in each new deck, we are removing all the cards with an odd position in that specific deck, and leaving behind all the cards with an even position in that specific deck. This process will continue until we reach Round 5, where the number of cards that remain ($125$) is no longer a multiple of 2. Therefore, when we remove all the cards with indices $I\\equiv 16\\pmod{32}$ and leave behind all the cards with indices $I\\equiv 32\\pmod{32}$, the last card in the deck will be removed instead of being placed to the bottom, with $125-63=62$ cards remaining. Because of this, when Round 6 begins, we cannot follow our normal procedure, since the top card of the Round 6 deck would have been moved to the bottom! Therefore, when we are using our sieve with $I\\equiv 32\\pmod{64}$ and $I\\equiv 64\\pmod{64}$, the order at which the remainders are presented have flipped, so all the cards with indices that are multiples of 64 will be removed, while all the cards with indices that are 32 more than a multiple of 64 will remain. The $62$ cards from Round 6 will be reduced to $31$, and the last card in the deck will be removed just like in Round 5, which means Round 7 will be affected the same way. Of the choices of $I\\equiv 32\\pmod{128}$ and $I\\equiv 32+64=96\\pmod{128}$, we will remove the latter, leaving behind the cards with indices that are 32 more than a multiple of 128. However, since 31 is an odd number, the last card of the deck will NOT be removed, which means our sieving process will return to normal after Round 7, with $31-15=16$ cards remaining. Since $16$ is a perfect power of $2$, we will not need to worry about this scenario again in the following rounds. Round 8: $I\\equiv 32\\pmod{256}$ removed; $I\\equiv 32+128=160\\pmod{256}$ remains; $16-8=8$ cards left. Round 9: $I\\equiv 160\\pmod{512}$ removed; $I\\equiv 160+256=416\\pmod{512}$ remains; $8-4=4$ cards left. Round 10: $I\\equiv 416\\pmod{1024}$ removed; $I\\equiv 416+512=928\\pmod{1024}$ remains; $4-2=2$ cards left. Round 11: $I\\equiv 928\\pmod{2048}$ removed; $I\\equiv 928+1024=1952\\pmod{2048}$ remains; $2-1=1$ card left. Round 12: The last card in position $1952$ is sorted into place (This is card #$2000$!). It is obvious that the single card put into place in Round 11 is the second-to-last card in the deck, which is our target #$1999$. Then, its position in the original deck leaves a remainder of $928$ when divided by $2056$. It is easy to see that $928$ is the only index that satisfies this, so there will be $928-1=927. ~chz3369"
] |
2000-II-2
| 2,000 | 2 |
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$ ?
| 98 |
II
|
[
"\\[(x-y)(x+y)=2000^2=2^8 \\cdot 5^6\\] Note that $(x-y)$ and $(x+y)$ have the same parities, so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$. We have $2^6 \\cdot 5^6$ left. Since there are $7 \\cdot 7=49$ factors of $2^6 \\cdot 5^6$, and since both $x$ and $y$ can be negative, this gives us $49\\cdot2=098 lattice points.",
"As with solution 1, note that both $x-y$ and $x+y$ must have the same parities, meaning both have to be even. Additionally, we can express both of them in terms of $2^a\\cdot3^b$ and $2^c\\cdot3^d$. Now, $a+c$ must be equal to 6, and both have to be greater than or equal to 1, so there are by stars and bars 7 ways to do this. Similarly, for $b+d$, we have that both only need to be greater than or equal to 0, so this time there are 7 ways to do so. Since both can be negative, we multiply $7\\cdot7\\cdot2$ which gives $098$."
] |
2000-II-3
| 2,000 | 3 |
A deck of forty cards consists of four 1's, four 2's,..., and four 10's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 758 |
II
|
[
"There are ${38 \\choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \\choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\\frac{54+1}{703} = \\frac{55}{703}$, and $m+n = 758.",
"Instead of counting the cases and doing $\\frac{\\text{cases wanted}}{\\text{total amount of cases}}$ we can use probability directly. For sake of simplicity, WLOG, assume that a pair of ones were removed from the deck of forty cards. We can split it into two cases: Case 1: The pair is ones. The probability that a one is chosen is $\\frac{2}{38}.$ The probability that a second one is chosen is $\\frac{1}{37}$ because one card was removed. Therefore, the probability that the pair is ones is $\\frac{2}{38} \\cdot \\frac{1}{37}.$ Case 2: The pair is $2-10.$ The probability that any other number is chosen is $\\frac{36}{38}.$ The probability that a number that is equal to this number is chosen (for example, if two was chosen originally then another two being chosen) is $\\frac{3}{37}.$ Therefore, the probability that the pair is $2-10$ is $\\frac{36}{38} \\cdot \\frac{3}{37}.$ Adding these two probabilities gives $\\frac{2}{38} \\cdot \\frac{1}{37} + \\frac{36}{38} \\cdot \\frac{3}{37} = \\frac{110}{38 \\cdot 37} = \\frac{55}{703}$, and $m+n = 758"
] |
2000-II-4
| 2,000 | 4 |
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
| 180 |
II
|
[
"We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \\cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \\cdots (e_k + 1)$. If a number has $18 = 2 \\cdot 3 \\cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization. Dividing the greatest power of $2$ from $n$, we have an odd integer with six positive divisors, which indicates that it either is ($6 = 2 \\cdot 3$) a prime raised to the $5$th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$, while the smallest example of the latter is $3^2 \\cdot 5 = 45$. Suppose we now divide all of the odd factors from $n$; then we require a power of $2$ with $\\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$. Thus, our answer is $2^2 \\cdot 3^2 \\cdot 5 = 180.",
"Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor. Thus, if $x$ is an odd factor of $n$, then $2x$ and $4x$ must be the two corresponding even factors. So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$ Since there are $18$ factors of $n$, we can write: $(2+1)(a+1)(b+1)(c+1)... = 18$ $(a+1)(b+1)(c+1)... = 6$ Since $6$ only has factors from the set $1, 2, 3, 6$, either $a=5$ and all other variables are $0$, or $a=3$ and $b=2$, with again all other variables equal to $0$. This gives the two numbers $2^2 \\cdot 3^5$ and $2^2 \\cdot 3^2 \\cdot 5$. The latter number is smaller, and is equal to $180.",
"We see that the least number with 6 odd factors is $3^2*5$. Multiplied by $2^2$ (as each factor of 2 doubles the odd factors, as it can be 2n or $2^2n$. Finally, you get $180$ -dragoon"
] |
2000-II-5
| 2,000 | 5 |
Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n$ .
| 376 |
II
|
[
"There are $\\binom{8}{5}$ ways to choose the rings, and there are $5!$ distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. The number of ways to distribute the rings among the fingers is equivalent the number of ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by three dividers. The number of ways to arrange those dividers and balls is just $\\binom {8}{3}$. Multiplying gives the answer: $\\binom{8}{5}\\binom{8}{3}5! = 376320$, and the three leftmost digits are $376."
] |
2000-II-6
| 2,000 | 6 |
One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$ . Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$ .
| 181 |
II
|
[
"Let the shorter base have length $b$ (so the longer has length $b+100$), and let the height be $h$. The length of the midline of the trapezoid is the average of its bases, which is $\\frac{b+b+100}{2} = b+50$. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$. Then, [asy]pathpen = linewidth(0.7); pen d = linetype(\"4 4\") + linewidth(0.7); pair A=(0,0),B=(175,0),C=(105,100),D=(30,100); D(A--B--C--D--cycle); D((A+D)/2 -- (B+C)/2, d); MP(\"b\",(C+D)/2,N);MP(\"b+100\",(A+B)/2); [/asy] \\[\\frac{\\frac 12 (h/2) (b + b+50)}{\\frac 12 (h/2) (b + 50 + b + 100)} = \\frac{2}{3} \\Longrightarrow \\frac{b + 75}{b + 25} = \\frac 32 \\Longrightarrow b = 75\\] Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$. By similar triangles, we easily find that $\\frac{x - 75}{100} = \\frac{x_1+x_2}{100} = \\frac{h_1}{h}$. [asy]pathpen = linewidth(0.7); pen d = linetype(\"4 4\") + linewidth(0.7); pair A=(0,0),B=(175,0),C=(105,100),D=(30,100),E=D*(1.75-(18125)^.5/100),F=IP(B--C,E--(175,E.y)); D(A--B--C--D--cycle); MP(\"75\",(C+D)/2,N);MP(\"175\",(A+B)/2); D(C--(C.x,0),d);D(D--(D.x,0),d); D(E--F,d); D((-20,100)--(-20,0)); MP(\"h\",(-20,50),(-1,0));MP(\"h_1\",(C.x,(C.y+E.y)/2),(-1,0)); MP(\"x_1\",((E.x+D.x)/2,E.y));MP(\"x_2\",((F.x+C.x)/2,E.y)); [/asy] The area of the region including the shorter base must be half of the area of the entire trapezoid, so \\[2 \\cdot \\frac 12 h_1 (75 + x) = \\frac 12 h (75 + 175) \\Longrightarrow x = 125 \\cdot \\frac{h}{h_1} - 75\\] Substituting our expression for $\\frac h{h_1}$ from above, we find that \\[x = \\frac{12500}{x-75} - 75 \\Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \\Longrightarrow x^2 = 18125\\] The answer is $\\left\\lfloor\\frac{x^2}{100}\\right\\rfloor = 181."
] |
2000-II-8
| 2,000 | 8 |
In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$ .
| 110 |
II
|
[
"Solution 1 Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \\perp BD$, it follows that $\\triangle BAC \\sim \\triangle CBD$, so $\\frac{x}{\\sqrt{11}} = \\frac{y}{x} \\Longrightarrow x^2 = y\\sqrt{11}$. Let $E$ be the foot of the altitude from $A$ to $\\overline{CD}$. Then $AE = x$, and $ADE$ is a right triangle. By the Pythagorean Theorem, \\[x^2 + \\left(y-\\sqrt{11}\\right)^2 = 1001 \\Longrightarrow x^4 - 11x^2 - 11^2 \\cdot 9 \\cdot 10 = 0\\] The positive solution to this quadratic equation is $x^2 = 110. ~Nafer ~edits by fermat_sLastAMC",
"Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \\perp BD$, it follows that $\\triangle BAC \\sim \\triangle CBD$, so $\\frac{x}{\\sqrt{11}} = \\frac{y}{x} \\Longrightarrow x^2 = y\\sqrt{11}$. Let $E$ be the foot of the altitude from $A$ to $\\overline{CD}$. Then $AE = x$, and $ADE$ is a right triangle. By the Pythagorean Theorem, \\[x^2 + \\left(y-\\sqrt{11}\\right)^2 = 1001 \\Longrightarrow x^4 - 11x^2 - 11^2 \\cdot 9 \\cdot 10 = 0\\] The positive solution to this quadratic equation is $x^2 = 110. ~Nafer ~edits by fermat_sLastAMC",
"Let $BC=x$. Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\\sqrt{11}+\\sqrt{1001-x^2}$. Therefore, we know that vector $\\vec{BD}=\\langle \\sqrt{11}+\\sqrt{1001-x^2},-x\\rangle$ and vector $\\vec{AC}=\\langle-\\sqrt{11},-x\\rangle$. Now we know that these vectors are perpendicular, so their dot product is 0.\\[\\vec{BD}\\cdot \\vec{AC}=-11-\\sqrt{11(1001-x^2)}+x^2=0\\] \\[(x^2-11)^2=11(1001-x^2)\\] \\[x^4-11x^2-11\\cdot 990=0.\\] As above, we can solve this quadratic to get the positve solution $BC^2=x^2=110. ~Nafer ~edits by fermat_sLastAMC",
"Let $BC=x$ and $CD=y+\\sqrt{11}$. From Pythagoras with $AD$, we obtain $x^2+y^2=1001$. Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$, so we have \\[\\left(y+\\sqrt{11}\\right)^2+11=x^2+1001.\\] Substituting $x^2=1001-y^2$ and simplifying yields \\[y^2+\\sqrt{11}y-990=0,\\] and the quadratic formula gives $y=9\\sqrt{11}$. Then from $x^2+y^2=1001$, we plug in $y$ to find $x^2=110. ~Nafer ~edits by fermat_sLastAMC",
"Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have \\begin{align*} BC^2&=BE^2+CE^2 \\\\ &=(AB^2-AE^2)+(CD^2-DE^2) \\\\ &=CD^2+\\sqrt{11}^2-(AE^2+DE^2) \\\\ &=CD^2+11-AD^2 \\\\ &=CD^2-990 \\end{align*} Followed by dropping the perpendicular like in solution 1, we obtain system of equation \\[BC^2=CD^2-990\\] \\[BC^2+CD^2-2\\sqrt{11}CD=990\\] Rearrange the first equation yields \\[CD^2-BC^2=990\\] Equating it with the second equation we have \\[CD^2-BC^2=BC^2+CD^2-2\\sqrt{11}CD\\] Which gives $CD=\\frac{BC^2}{\\sqrt{11}}$. Substituting into equation 1 obtains the quadratic in terms of $BC^2$ \\[(BC^2)^2-11BC^2-11\\cdot990=0\\] Solving the quadratic to obtain $BC^2=110. ~Nafer ~edits by fermat_sLastAMC"
] |
2000-II-9
| 2,000 | 9 |
Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ .
| 0 |
II
|
[
"Using the quadratic equation on $z^2 - (2 \\cos 3 )z + 1 = 0$, we have $z = \\frac{2\\cos 3 \\pm \\sqrt{4\\cos^2 3 - 4}}{2} = \\cos 3 \\pm i\\sin 3 = \\text{cis}\\,3^{\\circ}$. There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\\theta} = \\cos \\theta + i\\sin \\theta$ and $\\frac 1z= e^{-i\\theta} = \\cos \\theta - i\\sin \\theta$. We have $z+\\frac 1z = 2\\cos \\theta = 2\\cos 3^\\circ$ and $\\theta = 3^\\circ$. Alternatively, we could let $z = a + bi$ and solve to get $z=\\cos 3^\\circ + i\\sin 3^\\circ$. Using De Moivre's Theorem we have $z^{2000} = \\cos 6000^\\circ + i\\sin 6000^\\circ$, $6000 = 16(360) + 240$, so $z^{2000} = \\cos 240^\\circ + i\\sin 240^\\circ$. We want $z^{2000}+\\frac 1{z^{2000}} = 2\\cos 240^\\circ = -1$. Finally, the least integer greater than $-1$ is $000.",
"Let $z=re^{i\\theta}$. Notice that we have $2\\cos(3^{\\circ})=e^{i\\frac{\\pi}{60}}+e^{-i\\frac{\\pi}{60}}=re^{i\\theta}+\\frac{1}{r}e^{-i\\theta}.$ $r$ must be $1$ (or if you take the magnitude would not be the same). Therefore, $z=e^{i\\frac{\\pi}{\\theta}}$ and plugging into the desired expression, we get $e^{i\\frac{100\\pi}{3}}+e^{-i\\frac{100\\pi}{3}}=2\\cos{\\frac{100\\pi}{3}}=-1$. Therefore, the least integer greater is $000 ~solution by williamgolly",
"For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ has the same least integer greater than their solution. We have $z + 1/z = 2\\cos 3$. Since $\\cos 3<1$, $2\\cos 3<2$. If we square the equation $z + 1/z = 2\\cos 3$, we get $z^2 + 2 + 1/(z^2) = 4\\cos^2 3$, or $z^2 + 1/(z^2) = 4\\cos^2 3 - 2$. $4\\cos^2 3 - 2$ is is less than $2$, since $4\\cos^2 3$ is less than $4$. If we square the equation again, we get $z^4 + 1/(z^4) = (4\\cos^2 3 - 2)^2 -2$. Since $4\\cos^2 3 - 2$ is less than 2, $(4\\cos^2 3 - 2)^2$ is less than 4, and $(4\\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\\cos^2 3 - 2)^2 -2$ is also less than $4\\cos^2 3 - 2$. we can see that every time we square the equation, the right-hand side gets smaller and into the negatives. Since the smallest integer that is allowed as an answer is 0, the smallest integer greater is $000 ~ PaperMath ~megaboy6679",
"First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that \\[a+bi+\\frac{a-bi}{a^2+b^2} = 2 \\cos{3^{\\circ}}\\] given the conditions of the problem. Equating imaginary coefficients, we have that \\[b \\left( 1 - \\frac{1}{a^2+b^2}\\right) = 0\\] giving us that either $b=0$ or $|z| = 1$. Let's consider the latter case for now. We now know that $a^2+b^2=1$, so when we equate real coefficients we have that $2a = 2 \\cos{3^{\\circ}}$, therefore $a = \\cos{3^{\\circ}}$. So, $b = \\sin{3^{\\circ}}$ and then we can write $z = \\text{cis}(3)^{\\circ}$. By De Moivre's Theorem, \\[z^{2000} + \\frac{1}{z^{2000}} = \\text{cis} (6000)^{\\circ} + \\text{cis} (-6000)^{\\circ}\\]. The imaginary parts cancel, leaving us with $2 \\cos{6000^{\\circ}}$, which is $240 \\pmod{360}$. Therefore, it is $-1$, and our answer is $000 is not violating our conditions set above."
] |
2000-II-10
| 2,000 | 10 |
A circle is inscribed in quadrilateral $ABCD$ , tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ , $PB=26$ , $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.
| 647 |
II
|
[
"Call the center of the circle $O$. By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed. Thus, $\\angle{AOP}+\\angle{POB}+\\angle{COQ}+\\angle{QOD}=180$, or $(\\arctan(\\tfrac{19}{r})+\\arctan(\\tfrac{26}{r}))+(\\arctan(\\tfrac{37}{r})+\\arctan(\\tfrac{23}{r}))=180$. Take the $\\tan$ of both sides and use the identity for $\\tan(A+B)$ to get \\[\\tan(\\arctan(\\tfrac{19}{r})+\\arctan(\\tfrac{26}{r}))+\\tan(\\arctan(\\tfrac{37}{r})+\\arctan(\\tfrac{23}{r}))=n\\cdot0=0.\\] Use the identity for $\\tan(A+B)$ again to get \\[\\frac{\\tfrac{45}{r}}{1-19\\cdot\\tfrac{26}{r^2}}+\\frac{\\tfrac{60}{r}}{1-37\\cdot\\tfrac{23}{r^2}}=0.\\] Solving gives $r^2=647 quite easily.",
"Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ($a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \\[A = \\sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\\sqrt{647}\\] $r^2=\\frac{A^2}{(a+b+c+d)^2} = 647.",
"Using the formulas established in solution 2, one notices: \\[r^2=\\frac{A^2}{(a+b+c+d)^2}\\] \\[r^2=\\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\\] \\[r^2=\\frac{abc+bcd+acd+abd}{a+b+c+d}\\] \\[r^2=647\\] which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just with less opportunity to cancel coefficients beforehand.",
"Using the formulas established in solution 2, one notices: \\[r^2=\\frac{A^2}{(a+b+c+d)^2}\\] \\[r^2=\\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\\] \\[r^2=\\frac{abc+bcd+acd+abd}{a+b+c+d}\\] \\[r^2=647\\] which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just with less opportunity to cancel coefficients beforehand."
] |
2000-II-11
| 2,000 | 11 |
The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 131 |
II
|
[
"For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$. Suppose $B$ has integer coordinates; then $\\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\\overline{CD}$, and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram, and $\\overrightarrow{AB} = \\overrightarrow{D'C}$. Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.[1] [asy] pathpen = linewidth(0.7); pair A=(0,0), D=(1,7), Da = MP(\"D'\",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--MP(\"D\",D,N)--cycle); D(F--A--Da,linetype(\"4 4\")); [/asy] Let the slope of the perpendicular be $m$. Then the midpoint of $\\overline{DD'}$ lies on the line $y=mx$, so $\\frac{b+7}{2} = m \\cdot \\frac{a+1}{2}$. Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$. Combining these two equations yields \\[a^2 + \\left(7 - (a+1)m\\right)^2 = 50\\] Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \\pm 1, \\pm 7), (\\pm 7, \\pm 1), (\\pm 5, \\pm 5)$. We exclude the cases $(\\pm 1, \\pm 7)$ because they lead to degenerate parallelograms (rectangle, line segment, vertical and horizontal sides). Thus we have \\[7 - 8m = \\pm 1, \\quad 7 + 6m = \\pm 1, \\quad 7 - 6m = \\pm 5, 7 + 4m = \\pm 5\\] These yield $m = 1, \\frac 34, -1, -\\frac 43, 2, \\frac 13, -3, - \\frac 12$. Therefore, the corresponding slopes of $\\overline{AB}$ are $-1, -\\frac 43, 1, \\frac 34, -\\frac 12, -3, \\frac 13$, and $2$. The sum of their absolute values is $\\frac{119}{12}$. The answer is $m+n= 131. ~inaccessibles",
"A very natural solution: Shift $A$ to the origin. Suppose point $B$ is $(x, kx)$. Note $k$ is the slope we're looking for. Note that point $C$ must be of the form: $(x \\pm 1, kx \\pm 7)$ or $(x \\pm 7, kx \\pm 1)$ or $(x \\pm 5, kx \\pm 5)$. Note that we want the slope of the line connecting $D$ and $C$ so also be $k$, since $AB$ and $CD$ are parallel. Instead of dealing with the 12 cases, we consider point $C$ of the form $(x \\pm Y, kx \\pm Z)$ where we plug in the necessary values for $Y$ and $Z$ after simplifying. Since the slopes of $AB$ and $CD$ must both be $k$, $\\frac{7 - kx \\pm Z}{1 - x \\pm Y} = k \\implies k = \\frac{7 \\pm Z}{1 \\pm Y}$. Plugging in the possible values of $\\pm 7, \\pm 1, \\pm 5$ in their respective pairs and ruling out degenerate cases, we find the sum is $\\frac{119}{12} \\implies m + n = 131 due to degeneracy.)"
] |
2000-II-12
| 2,000 | 12 |
The points $A$ , $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ , $BC=14$ , $CA=15$ , and that the distance from $O$ to triangle $ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ , $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+k$ .
| 118 |
II
|
[
"Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$. By the Pythagorean Theorem on triangles $\\triangle OAD$, $\\triangle OBD$ and $\\triangle OCD$ we get: \\[DA^2=DB^2=DC^2=20^2-OD^2\\] It follows that $DA=DB=DC$, so $D$ is the circumcenter of $\\triangle ABC$. By Heron's Formula the area of $\\triangle ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles): \\[K = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{21(21-15)(21-14)(21-13)} = 84\\] From $R = \\frac{abc}{4K}$, we know that the circumradius of $\\triangle ABC$ is: \\[R = \\frac{abc}{4K} = \\frac{(13)(14)(15)}{4(84)} = \\frac{65}{8}\\] Thus by the Pythagorean Theorem again, \\[OD = \\sqrt{20^2-R^2} = \\sqrt{20^2-\\frac{65^2}{8^2}} = \\frac{15\\sqrt{95}}{8}.\\] So the final answer is $15+95+8=118.",
"We know the radii to $A$,$B$, and $C$ form a triangular pyramid $OABC$. We know the lengths of the edges $OA = OB = OC = 20$. First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$. Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$, and $z$ occupy the other dimension, with the origin as $C$. We look at vectors $OA$ and $OC$. Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$, hence it is $7$ units along the $x$ axis. Hence for vector $OC$, in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$, since $A$ is $9$ along $x$, and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$. We know both vector magnitudes are $20$. Solving for $h$ yields $\\frac{15\\sqrt{95} }{8}$, so Answer = $118. ~fermat_sLastAMC",
"We know the radii to $A$,$B$, and $C$ form a triangular pyramid $OABC$. We know the lengths of the edges $OA = OB = OC = 20$. First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$. Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$, and $z$ occupy the other dimension, with the origin as $C$. We look at vectors $OA$ and $OC$. Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$, hence it is $7$ units along the $x$ axis. Hence for vector $OC$, in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$, since $A$ is $9$ along $x$, and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$. We know both vector magnitudes are $20$. Solving for $h$ yields $\\frac{15\\sqrt{95} }{8}$, so Answer = $118. ~fermat_sLastAMC",
"The distance from $O$ to $ABC$ forms a right triangle with the circumradius of the triangle and the radius of the sphere. The hypotenuse has length $20$, since it is the radius of the sphere. The circumradius of a $13$, $14$, $15$ triangle is $\\frac{65}{8}$, so the distance from $O$ to $ABC$ is given by $\\sqrt{20^2-(\\frac{65}{8})^2} = \\frac{15\\sqrt{95}}{8}$, and $15+95+8 = 118. -skibbysiggy"
] |
2000-II-13
| 2,000 | 13 |
The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ , $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$ .
| 200 |
II
|
[
"We may factor the equation as:[1] \\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\\\ \\end{align*} Now $100x^4+10x^2+1\\ge 1>0$ for real $x$. Thus the real roots must be the roots of the equation $20x^2+x-2=0$. By the quadratic formula the roots of this are: \\[x=\\frac{-1\\pm\\sqrt{1^2-4(-2)(20)}}{40} = \\frac{-1\\pm\\sqrt{1+160}}{40} = \\frac{-1\\pm\\sqrt{161}}{40}.\\] Thus $r=\\frac{-1+\\sqrt{161}}{40}$, and so the final answer is $-1+161+40 = 200.",
"It would be really nice if the coefficients were symmetrical. What if we make the substitution $x = -\\frac{i}{\\sqrt{10}}y$? Then the equation becomes $-2y^6 - \\left(\\frac{i}{\\sqrt{10}}\\right)y^5 + \\left(\\frac{i}{\\sqrt{10}}\\right)y^3 - \\left(\\frac{i}{\\sqrt{10}}\\right)y - 2 = 0$. It's symmetric! Dividing by $y^3$ and rearranging, we get $-2\\left(y^3 + \\frac{1}{y^3}\\right) - \\left(\\frac{i}{\\sqrt{10}}\\right)\\left(y^2 + \\frac{1}{y^2}\\right) + \\left(\\frac{i}{\\sqrt{10}}\\right)$ Now, if we let $z = y + \\frac{1}{y}$, we can get the equations $z = y + \\frac{1}{y}$ $z^2 - 2 = y^2 + \\frac{1}{y^2}$ and $z^3 - 3z = y^3 + \\frac{1}{y^3}$ (These come from squaring $z$ and subtracting $2$ then multiplying that result by $z$ and subtracting $z$.) Plugging this into our polynomial, expanding, and rearranging, we get $-2z^3 - \\left(\\frac{i}{\\sqrt{10}}\\right)z^2 + 6z + \\left(\\frac{3i}{\\sqrt{10}}\\right)$ Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$, so what squared equals 3? Plugging in $z = \\sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\\sqrt{3}$, we see that it also works! Great, we use long division on the polynomial by $\\left(z - \\sqrt{3}\\right)\\left(z + \\sqrt{3}\\right) = z^2 - 3$ and we get $2z - \\left(\\frac{i}{\\sqrt{10}}\\right) = 0$. We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant and then multiply by $-\\left(\\frac{i}{\\sqrt{10}}\\right)$. We get that $z = \\left(\\frac{i}{-2\\sqrt{10}}\\right)$. Solving for $y$ (using $y + \\frac{1}{y} = z$) we get that $y = \\frac{-i \\pm \\sqrt{161}i}{4\\sqrt{10}}$, and multiplying this by $-\\left(\\frac{i}{\\sqrt{10}}\\right)$ (because $x = -\\left(\\frac{i}{\\sqrt{10}}\\right)y$) we get that $x = \\frac{-1 \\pm \\sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = 200 ~Grizzy ~formatting edits by fermat_sLastAMC",
"It would be really nice if the coefficients were symmetrical. What if we make the substitution $x = -\\frac{i}{\\sqrt{10}}y$? Then the equation becomes $-2y^6 - \\left(\\frac{i}{\\sqrt{10}}\\right)y^5 + \\left(\\frac{i}{\\sqrt{10}}\\right)y^3 - \\left(\\frac{i}{\\sqrt{10}}\\right)y - 2 = 0$. It's symmetric! Dividing by $y^3$ and rearranging, we get $-2\\left(y^3 + \\frac{1}{y^3}\\right) - \\left(\\frac{i}{\\sqrt{10}}\\right)\\left(y^2 + \\frac{1}{y^2}\\right) + \\left(\\frac{i}{\\sqrt{10}}\\right)$ Now, if we let $z = y + \\frac{1}{y}$, we can get the equations $z = y + \\frac{1}{y}$ $z^2 - 2 = y^2 + \\frac{1}{y^2}$ and $z^3 - 3z = y^3 + \\frac{1}{y^3}$ (These come from squaring $z$ and subtracting $2$ then multiplying that result by $z$ and subtracting $z$.) Plugging this into our polynomial, expanding, and rearranging, we get $-2z^3 - \\left(\\frac{i}{\\sqrt{10}}\\right)z^2 + 6z + \\left(\\frac{3i}{\\sqrt{10}}\\right)$ Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$, so what squared equals 3? Plugging in $z = \\sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\\sqrt{3}$, we see that it also works! Great, we use long division on the polynomial by $\\left(z - \\sqrt{3}\\right)\\left(z + \\sqrt{3}\\right) = z^2 - 3$ and we get $2z - \\left(\\frac{i}{\\sqrt{10}}\\right) = 0$. We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant and then multiply by $-\\left(\\frac{i}{\\sqrt{10}}\\right)$. We get that $z = \\left(\\frac{i}{-2\\sqrt{10}}\\right)$. Solving for $y$ (using $y + \\frac{1}{y} = z$) we get that $y = \\frac{-i \\pm \\sqrt{161}i}{4\\sqrt{10}}$, and multiplying this by $-\\left(\\frac{i}{\\sqrt{10}}\\right)$ (because $x = -\\left(\\frac{i}{\\sqrt{10}}\\right)y$) we get that $x = \\frac{-1 \\pm \\sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = 200 ~Grizzy ~formatting edits by fermat_sLastAMC",
"Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$. The expression in parentheses is a geometric series with common factor $10x^2$. Using the geometric sum formula, we rewrite as $2000x^6-2+\\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\\neq0$. Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\\frac{x}{10x^2-1}=0, 10x^2-1\\neq0 \\implies$ $(1000x^6-1)(2+\\frac{x}{10x^2-1})=0, 10x^2-1\\neq0$. Note that setting $1000x^6-1=0$ gives $10x^2-1=0$, which is clearly extraneous. Hence, we set $2+\\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\\frac{-1+\\sqrt{161}}{40} \\implies -1+161+40=200 ~keeper1098",
"Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$. The expression in parentheses is a geometric series with common factor $10x^2$. Using the geometric sum formula, we rewrite as $2000x^6-2+\\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\\neq0$. Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\\frac{x}{10x^2-1}=0, 10x^2-1\\neq0 \\implies$ $(1000x^6-1)(2+\\frac{x}{10x^2-1})=0, 10x^2-1\\neq0$. Note that setting $1000x^6-1=0$ gives $10x^2-1=0$, which is clearly extraneous. Hence, we set $2+\\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\\frac{-1+\\sqrt{161}}{40} \\implies -1+161+40=200 ~keeper1098",
"If we look at the polynomial's terms, we can see that the number of zeros in a term more or less correlates to the power of $x^2$. Thus, we let $y=10x^2$. The equation then becomes $2y^3+xy^2+xy+x-2=0$, or $x(y^2+y+1)=2(1-y^3)$. By the difference of cubes formula, $2(1-y^3)=2(1-y)(1+y+y^2)$, so we have two cases: either $y^2+y+1=0$, or $x=2(1-y)$. We start with the second formula as it is simpler. Solving with the quadratic formula after re-substitution, we see that $x=\\frac{-1\\pm\\sqrt{161}}{40}$, so the answer is $-1+161+40=200. Thus our solution is correct. ~eevee9406"
] |
2000-II-14
| 2,000 | 14 |
Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$ .
| 495 |
II
|
[
"Solution 1 Note that $1+\\sum_{k=1}^{n-1} {k\\cdot k!} = 1+\\sum_{k=1}^{n-1} {((k+1)\\cdot k!- k!)} = 1+\\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \\cdots + (n! - (n-1)!)) = n!$. Thus for all $m\\in\\mathbb{N}$, $(32m+16)!-(32m)! = \\left(1+\\sum_{k=1}^{32m+15} {k\\cdot k!}\\right)-\\left(1+\\sum_{k=1}^{32m-1} {k\\cdot k!}\\right) = \\sum_{k=32m}^{32m+15}k\\cdot k!.$ So now, \\begin{align*} 16!-32!+48!-64!+\\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\\cdots+(2000!-1984!)\\\\ &=16! +\\sum_{m=1}^{62}\\left((32m+16)!-(32m)!\\right)\\\\ &=16! +\\sum_{m=1}^{62}\\sum_{k=32m}^{32m+15}k\\cdot k! \\end{align*} Therefore we have $f_{16} = 1$, $f_k=k$ if $32m\\le k \\le 32m+15$ for some $m=1,2,\\ldots,62$, and $f_k = 0$ for all other $k$. Therefore we have: \\begin{align*} f_1-f_2+f_3-f_4+\\cdots+(-1)^{j+1}f_j &= (-1)^{17}\\cdot 1 + \\sum_{m=1}^{62}\\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\\\ &= -1 + \\sum_{m=1}^{62}\\left[\\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\\right]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}1\\\\ &= -1 + \\sum_{m=1}^{62}8\\\\ &= -1 + 8\\cdot 62\\\\ &= 495.",
"Note that $1+\\sum_{k=1}^{n-1} {k\\cdot k!} = 1+\\sum_{k=1}^{n-1} {((k+1)\\cdot k!- k!)} = 1+\\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \\cdots + (n! - (n-1)!)) = n!$. Thus for all $m\\in\\mathbb{N}$, $(32m+16)!-(32m)! = \\left(1+\\sum_{k=1}^{32m+15} {k\\cdot k!}\\right)-\\left(1+\\sum_{k=1}^{32m-1} {k\\cdot k!}\\right) = \\sum_{k=32m}^{32m+15}k\\cdot k!.$ So now, \\begin{align*} 16!-32!+48!-64!+\\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\\cdots+(2000!-1984!)\\\\ &=16! +\\sum_{m=1}^{62}\\left((32m+16)!-(32m)!\\right)\\\\ &=16! +\\sum_{m=1}^{62}\\sum_{k=32m}^{32m+15}k\\cdot k! \\end{align*} Therefore we have $f_{16} = 1$, $f_k=k$ if $32m\\le k \\le 32m+15$ for some $m=1,2,\\ldots,62$, and $f_k = 0$ for all other $k$. Therefore we have: \\begin{align*} f_1-f_2+f_3-f_4+\\cdots+(-1)^{j+1}f_j &= (-1)^{17}\\cdot 1 + \\sum_{m=1}^{62}\\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\\\ &= -1 + \\sum_{m=1}^{62}\\left[\\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\\right]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}1\\\\ &= -1 + \\sum_{m=1}^{62}8\\\\ &= -1 + 8\\cdot 62\\\\ &= 495.",
"This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition. Let us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$, (such as $10000_{10} - 100_{10}$) the result will be an integer in base $n$ composed only of the digits $n - 1$ and $0$ (in this example, $9900$). More specifically, the difference $(n^k)_n - (n^j)_n$, $j<k$ , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$’s followed by $j$ $0$’s. It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$, the largest digit value is $n - 1$, in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$. However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$. $31_!$ should be represented as $101_!$, and is $7_{10}$.) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$, but rather is $650000_!$. Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$’s everywhere else, and then using a standard carry/borrow system accounting for the place value. With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \\cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\\cdots(1984)$ followed by $1983$ $0$’s in the factorial base. $1968! - 1952! = (1967)(1966)\\cdots(1952)$ followed by $1951$ $0$’s, and so on for the rest of the terms, except $16!$, which will merely have a $1$ in the $16!$ place followed by $0$’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \\leq k \\leq 32m+15$ if $1\\leq m \\in\\mathbb{Z} \\leq 62$, $f_{16} = 1$, and $f_k = 0$ for all other $k$. Therefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$, and this will continue. Therefore, $f_{1999} - f_{1998} + \\cdots - f_{1984} = 8$. We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$, so our answer is almost $8 \\cdot 62$. However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \\cdot 62 - 1 = 495.",
"This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition. Let us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$, (such as $10000_{10} - 100_{10}$) the result will be an integer in base $n$ composed only of the digits $n - 1$ and $0$ (in this example, $9900$). More specifically, the difference $(n^k)_n - (n^j)_n$, $j<k$ , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$’s followed by $j$ $0$’s. It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$, the largest digit value is $n - 1$, in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$. However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$. $31_!$ should be represented as $101_!$, and is $7_{10}$.) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$, but rather is $650000_!$. Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$’s everywhere else, and then using a standard carry/borrow system accounting for the place value. With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \\cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\\cdots(1984)$ followed by $1983$ $0$’s in the factorial base. $1968! - 1952! = (1967)(1966)\\cdots(1952)$ followed by $1951$ $0$’s, and so on for the rest of the terms, except $16!$, which will merely have a $1$ in the $16!$ place followed by $0$’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \\leq k \\leq 32m+15$ if $1\\leq m \\in\\mathbb{Z} \\leq 62$, $f_{16} = 1$, and $f_k = 0$ for all other $k$. Therefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$, and this will continue. Therefore, $f_{1999} - f_{1998} + \\cdots - f_{1984} = 8$. We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$, so our answer is almost $8 \\cdot 62$. However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \\cdot 62 - 1 = 495.",
"Let $S = 16!-32!+\\cdots-1984!+2000!$. Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \\cdots$ is significantly smaller than $2000!$), it follows that $1999! < S < 2000!$. Hence $f_{2000} = 0$. Then $2000! = 2000 \\cdot 1999! = 1999 \\cdot 1999! + 1999!$, and as $S - 2000! << 1999!$, it follows that $1999 \\cdot 1999! < S < 2000 \\cdot 1999!$. Hence $f_{1999} = 1999$, and we now need to find the factorial base expansion of \\[S_2 = S - 1999 \\cdot 1999! = 1999! - 1984! + 1962! - 1946! + \\cdots + 16!\\] Since $|S_2 - 1999!| << 1999!$, we can repeat the above argument recursively to yield $f_{1998} = 1998$, and so forth down to $f_{1985} = 1985$. Now $S_{16} = 1985! - 1984! + 1962! + \\cdots = 1984 \\cdot 1984! + 1962! + \\cdots$, so $f_{1984} = 1984$. The remaining sum is now just $1962! - 1946! + \\cdots + 16!$. We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$, and $f_k=k$ if $32m\\le k \\le 32m+15$ for some $m=1,2,\\ldots,62$, and $f_k = 0$ for all other $k$. Now for each $m$, we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \\cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \\cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \\cdots + 1 + 1$ $= 8$. Thus, our answer is $-f_{16} + 8 \\cdot 62 = 495.",
"Let $S = 16!-32!+\\cdots-1984!+2000!$. Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \\cdots$ is significantly smaller than $2000!$), it follows that $1999! < S < 2000!$. Hence $f_{2000} = 0$. Then $2000! = 2000 \\cdot 1999! = 1999 \\cdot 1999! + 1999!$, and as $S - 2000! << 1999!$, it follows that $1999 \\cdot 1999! < S < 2000 \\cdot 1999!$. Hence $f_{1999} = 1999$, and we now need to find the factorial base expansion of \\[S_2 = S - 1999 \\cdot 1999! = 1999! - 1984! + 1962! - 1946! + \\cdots + 16!\\] Since $|S_2 - 1999!| << 1999!$, we can repeat the above argument recursively to yield $f_{1998} = 1998$, and so forth down to $f_{1985} = 1985$. Now $S_{16} = 1985! - 1984! + 1962! + \\cdots = 1984 \\cdot 1984! + 1962! + \\cdots$, so $f_{1984} = 1984$. The remaining sum is now just $1962! - 1946! + \\cdots + 16!$. We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$, and $f_k=k$ if $32m\\le k \\le 32m+15$ for some $m=1,2,\\ldots,62$, and $f_k = 0$ for all other $k$. Now for each $m$, we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \\cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \\cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \\cdots + 1 + 1$ $= 8$. Thus, our answer is $-f_{16} + 8 \\cdot 62 = 495.",
"First consider how we would express $48!-32!$. We can't use any multiples of $48!$, since we'll never be able to make the $-32!$. Instead, we have to start with using $47!$s. Having $47*47!$ is the closest we can get to $48!$, since $48! = 48*47!$. Thus, $f_{47}=47$. Now consider what we have left to express. We wanted to express $48!-32!$, and now we have $47*47!$. This gives us $47!-32!$ left. Clearly, we can't use anymore $47!$s, so we have to use $46!$s. The most $46!$s we could use is $46$ of them. Thus, $f_{46}=46$. Since $47!=47*46!$, we are left with $46!-32!$ to express. Clearly, each time, we have $f_n = n$, leaving us with $n!-32!$ to express. Eventually, we will get down to needing to express $34!-32!$. We will need $33$ $33!$s, leaving $33!-32!$ to be expressed. Since $33!=33*32!$, $33!-32! = 32*32!$. Thus, $48!-32!$ can be expressed as $\\sum_{i=32}^{47} i*(i!)$. Returning to the problem, notice we can break the expression into groups of $2$, having $16!+(48!-32!)+(80!-64!)+...(2000!-1984!)$. Each pair has an alternating sum of $8$, and since there are $62$ pairs, the sum is $496$. Including the $16!$ at the start, which accounts for $-1$, we get $495. -skibbysiggy",
"First consider how we would express $48!-32!$. We can't use any multiples of $48!$, since we'll never be able to make the $-32!$. Instead, we have to start with using $47!$s. Having $47*47!$ is the closest we can get to $48!$, since $48! = 48*47!$. Thus, $f_{47}=47$. Now consider what we have left to express. We wanted to express $48!-32!$, and now we have $47*47!$. This gives us $47!-32!$ left. Clearly, we can't use anymore $47!$s, so we have to use $46!$s. The most $46!$s we could use is $46$ of them. Thus, $f_{46}=46$. Since $47!=47*46!$, we are left with $46!-32!$ to express. Clearly, each time, we have $f_n = n$, leaving us with $n!-32!$ to express. Eventually, we will get down to needing to express $34!-32!$. We will need $33$ $33!$s, leaving $33!-32!$ to be expressed. Since $33!=33*32!$, $33!-32! = 32*32!$. Thus, $48!-32!$ can be expressed as $\\sum_{i=32}^{47} i*(i!)$. Returning to the problem, notice we can break the expression into groups of $2$, having $16!+(48!-32!)+(80!-64!)+...(2000!-1984!)$. Each pair has an alternating sum of $8$, and since there are $62$ pairs, the sum is $496$. Including the $16!$ at the start, which accounts for $-1$, we get $495. -skibbysiggy"
] |
2001-I-1
| 2,001 | 1 |
Find the sum of all positive two-digit integers that are divisible by each of their digits.
| 630 |
I
|
[
"Let our number be $10a + b$, $a,b \\neq 0$. Then we have two conditions: $10a + b \\equiv 10a \\equiv 0 \\pmod{b}$ and $10a + b \\equiv b \\pmod{a}$, or $a$ divides into $b$ and $b$ divides into $10a$. Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$, then $b$ would not be a digit). For $b = a$, we have $n = 11a$ for nine possibilities, giving us a sum of $11 \\cdot \\frac {9(10)}{2} = 495$. For $b = 2a$, we have $n = 12a$ for four possibilities (the higher ones give $b > 9$), giving us a sum of $12 \\cdot \\frac {4(5)}{2} = 120$. For $b = 5a$, we have $n = 15a$ for one possibility (again, higher ones give $b > 9$), giving us a sum of $15$. If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = 630.",
"By testing every 2-digit number, we can list out all of the possibilities: \\[11+12+15+22+24+33+36+44+48+55+66+77+88+99=630.\\]",
"To further expand on solution 2, it would be tedious to test all $90$ two-digit numbers. We can reduce the amount to look at by focusing on the tens digit. First, we cannot have any number that is a multiple of $10$. We also note that any number with the same digits is a number that satisfies this problem. This gives \\[11, 22, 33, ... 99.\\] We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers $11, 12, 13, ... 19$ and numbers $22, 24, 26, 28$. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of $5$ or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.",
"In this solution, we will do casework on the ones digit. Before we start, let's make some variables. Let $a$ be the ones digit, and $b$ be the tens digit. Let $n$ equal our number. Our number can be expressed as $10b+a$. We can easily see that $b|a$, since $b|n$, and $b|10b$. Therefore, $b|(n-10b)$. Now, let's start with the casework. Case 1: $a=1$ Since $b|a$, $b=1$. From this, we get that $n=11$ satisfies the condition. Case 2: $a=2$ We either have $b=1$, or $b=2$. From this, we get that $n=12$ and $n=22$ satisfy the condition. Case 3: $a=3$ We have $b=3$. From this, we get that $n=33$ satisfies the condition. Note that $b=1$ was not included because $3$ does not divide $13$. Case 4: $a=4$ We either have $b=2$ or $b=4$. From this, we get that $n=24$ and $n=44$ satisfy the condition. $b=1$ was not included for similar reasons as last time. Case 5: $a=5$ We either have $b=1$ or $b=5$. From this, we get that $n=15$ and $n=55$ satisfy the condition. Continuing with this process up to $a=9$, we get that $n$ could be $11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99$. Summing, we get that the answer is $630. -bronzetruck2016"
] |
2001-I-2
| 2,001 | 2 |
A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$ .
| 651 |
I
|
[
"Let $x$ be the mean of $\\mathcal{S}$. Let $a$ be the number of elements in $\\mathcal{S}$. Then, the given tells us that $\\frac{ax+1}{a+1}=x-13$ and $\\frac{ax+2001}{a+1}=x+27$. Subtracting, we have \\begin{align*}\\frac{ax+2001}{a+1}-40=\\frac{ax+1}{a+1} \\Longrightarrow \\frac{2000}{a+1}=40 \\Longrightarrow a=49\\end{align*} We plug that into our very first formula, and get: \\begin{align*}\\frac{49x+1}{50}&=x-13 \\\\ 49x+1&=50x-650 \\\\ x&=651.\\end{align*}",
"Since this is a weighted average problem, the mean of $S$ is $\\frac{13}{27}$ as far from $1$ as it is from $2001$. Thus, the mean of $S$ is $1 + \\frac{13}{13 + 27}(2001 - 1) = 651."
] |
2001-I-3
| 2,001 | 3 |
Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.
| 500 |
I
|
[
"From Vieta's formulas, in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \\cdots + a_0 = 0$, then the sum of the roots is $\\frac{-a_{n-1}}{a_n}$. From the Binomial Theorem, the first term of $\\left(\\frac 12-x\\right)^{2001}$ is $-x^{2001}$, but $x^{2001}+-x^{2001}=0$, so the term with the largest degree is $x^{2000}$. So we need the coefficient of that term, as well as the coefficient of $x^{1999}$. \\begin{align*}\\binom{2001}{1} \\cdot (-x)^{2000} \\cdot \\left(\\frac{1}{2}\\right)^1&=\\frac{2001x^{2000}}{2}\\\\ \\binom{2001}{2} \\cdot (-x)^{1999} \\cdot \\left(\\frac{1}{2}\\right)^2 &=\\frac{-x^{1999}*2001*2000}{8}=-2001 \\cdot 250x^{1999} \\end{align*} Applying Vieta's formulas, we find that the sum of the roots is $-\\frac{-2001 \\cdot 250}{\\frac{2001}{2}}=250 \\cdot 2=500.",
"We find that the given equation has a $2000^{\\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\\frac{1}{2}$ to get a sum of $500.",
"Note that if $r$ is a root, then $\\frac{1}{2}-r$ is a root and they sum up to $\\frac{1}{2}.$ We make the substitution $y=x-\\frac{1}{4}$ so \\[(\\frac{1}{4}+y)^{2001}+(\\frac{1}{4}-y)^{2001}=0.\\] Expanding gives \\[2\\cdot\\frac{1}{4}\\cdot\\binom{2001}{1}y^{2000}-0y^{1999}+\\cdots\\] so by Vieta, the sum of the roots of $y$ is 0. Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is \\[2000(\\sum_{n=1}^{2000} y+\\frac{1}{4})=2000(0+\\frac{1}{4})=500.\\]"
] |
2001-I-4
| 2,001 | 4 |
In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ , $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$ .
| 291 |
I
|
[
"After chasing angles, $\\angle ATC=75^{\\circ}$ and $\\angle TCA=75^{\\circ}$, meaning $\\triangle TAC$ is an isosceles triangle and $AC=24$. Using law of sines on $\\triangle ABC$, we can create the following equation: $\\frac{24}{\\sin(\\angle ABC)}$ $=$ $\\frac{BC}{\\sin(\\angle BAC)}$ $\\angle ABC=45^{\\circ}$ and $\\angle BAC=60^{\\circ}$, so $BC = 12\\sqrt{6}$. We can then use the Law of Sines area formula $\\frac{1}{2} \\cdot BC \\cdot AC \\cdot \\sin(\\angle BCA)$ to find the area of the triangle. $\\sin(75)$ can be found through the sin addition formula. $\\sin(75)$ $=$ $\\frac{\\sqrt{6} + \\sqrt{2}}{4}$ Therefore, the area of the triangle is $\\frac{\\sqrt{6} + \\sqrt{2}}{4}$ $\\cdot$ $24$ $\\cdot$ $12\\sqrt{6}$ $\\cdot$ $\\frac{1}{2}$ $72\\sqrt{3} + 216$ $72 + 3 + 216 =$ $291",
"First, draw a good diagram. We realize that $\\angle C = 75^\\circ$, and $\\angle CAT = 30^\\circ$. Therefore, $\\angle CTA = 75^\\circ$ as well, making $\\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$. [asy] size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*\"$24$\", A--T, N); label(rotate(degrees(C-A))*\"$24$\", A--C, 2*NW); label(\"$12\\sqrt 3$\", C--D, E); label(\"$12\\sqrt 3$\", D--B, S); label(\"$12$\", A--D, S); pen p = fontsize(8)+red; MA(\"45^\\circ\", C,B,A,2); MA(\"30^\\circ\", B,A,T,2.5); MA(\"30^\\circ\", T,A,C,3.5); dot(\"$A$\", A, SW); dot(\"$B$\", B, SE); dot(\"$C$\", C, N); dot(\"$T$\", T, NE); dot(\"$D$\", D, S); [/asy] By 30-60-90 triangles, $AD=12$ and $CD=12\\sqrt{3}$. We also notice that $\\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$, which makes $BD=12\\sqrt{3}$. The base $AB$ is $12+12\\sqrt{3}$, and the altitude $CD=12\\sqrt{3}$. We can easily find that the area of triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=291. -youyanli",
"First, draw a good diagram. We realize that $\\angle C = 75^\\circ$, and $\\angle CAT = 30^\\circ$. Therefore, $\\angle CTA = 75^\\circ$ as well, making $\\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$. We now drop an altitude from $C$, and call the foot this altitude point $D$. [asy] size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*\"$24$\", A--T, N); label(rotate(degrees(C-A))*\"$24$\", A--C, 2*NW); label(\"$12\\sqrt 3$\", C--D, E); label(\"$12\\sqrt 3$\", D--B, S); label(\"$12$\", A--D, S); pen p = fontsize(8)+red; MA(\"45^\\circ\", C,B,A,2); MA(\"30^\\circ\", B,A,T,2.5); MA(\"30^\\circ\", T,A,C,3.5); dot(\"$A$\", A, SW); dot(\"$B$\", B, SE); dot(\"$C$\", C, N); dot(\"$T$\", T, NE); dot(\"$D$\", D, S); [/asy] By 30-60-90 triangles, $AD=12$ and $CD=12\\sqrt{3}$. We also notice that $\\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$, which makes $BD=12\\sqrt{3}$. The base $AB$ is $12+12\\sqrt{3}$, and the altitude $CD=12\\sqrt{3}$. We can easily find that the area of triangle $ABC$ is $216+72\\sqrt{3}$, so $a+b+c=291. -youyanli",
"After drawing line $AT$, we see that we have two triangles: $\\triangle ABT,$ with $45$, $30$, and $105$ degrees, and $\\triangle ATC$, with $30$, $75$, $75$ degrees. If we can sum these two triangles' areas, we have our answer. Let's take care of $\\triangle ATC$ first. We see that $\\triangle ATC$ is a isosceles triangle, with $AT = AC = 24$. Because the area of a triangle is $\\frac{1}{2}ab\\sin C$, we have $\\frac{1}{2}\\cdot 24^2\\cdot\\frac{1}{2}$, which is equal to $144.$ Now, on to $\\triangle ABT$. Draw the altitude from angle $\\angle T$ to $AB$, and call the point of intersection $D$. This splits $\\triangle ABT$ into $2$ triangles, one with $30-60-90$ ($\\triangle ADT$), and another with $45-45-90$ ($\\triangle BDT$). Now, because we know that $AT$ is $24$, we have by special right triangle ratios. The area of $\\triangle ADT$ is $\\frac{12\\sqrt{3}\\cdot 12}{2}$, and the area of $\\triangle BDT$ is $\\frac{12\\cdot 12}{2}$, which adds to $72\\sqrt{3} + 72$. Adding this to $\\triangle ATC$ we get a total sum of $216 + 72\\sqrt{3}.$ Thus, $a + b + c$ would be $216 + 72 + 3 = 291 ~MathCosine",
"After drawing line $AT$, we see that we have two triangles: $\\triangle ABT,$ with $45$, $30$, and $105$ degrees, and $\\triangle ATC$, with $30$, $75$, $75$ degrees. If we can sum these two triangles' areas, we have our answer. Let's take care of $\\triangle ATC$ first. We see that $\\triangle ATC$ is a isosceles triangle, with $AT = AC = 24$. Because the area of a triangle is $\\frac{1}{2}ab\\sin C$, we have $\\frac{1}{2}\\cdot 24^2\\cdot\\frac{1}{2}$, which is equal to $144.$ Now, on to $\\triangle ABT$. Draw the altitude from angle $\\angle T$ to $AB$, and call the point of intersection $D$. This splits $\\triangle ABT$ into $2$ triangles, one with $30-60-90$ ($\\triangle ADT$), and another with $45-45-90$ ($\\triangle BDT$). Now, because we know that $AT$ is $24$, we have by special right triangle ratios. The area of $\\triangle ADT$ is $\\frac{12\\sqrt{3}\\cdot 12}{2}$, and the area of $\\triangle BDT$ is $\\frac{12\\cdot 12}{2}$, which adds to $72\\sqrt{3} + 72$. Adding this to $\\triangle ATC$ we get a total sum of $216 + 72\\sqrt{3}.$ Thus, $a + b + c$ would be $216 + 72 + 3 = 291 ~MathCosine"
] |
2001-I-5
| 2,001 | 5 |
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the length of each side is $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 937 |
I
|
[
"[asy] pointpen = black; pathpen = black + linewidth(0.7); path e = xscale(2)*unitcircle; real x = -8/13*3^.5; D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */ D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle); [/asy] Solution 1 Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$ Note that the slope of $\\overline{AC}$ is $\\tan 60^\\circ = \\sqrt {3}.$ Hence, the equation of the line containing $\\overline{AC}$ is \\[y = x\\sqrt {3} + 1.\\] This will intersect the ellipse when \\begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\\sqrt {3} + 1)^{2} \\\\ & = & x^{2} + 4(3x^{2} + 2x\\sqrt {3} + 1) \\implies x(13x+8\\sqrt 3)=0\\implies x = \\frac { - 8\\sqrt {3}}{13}. \\end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3. Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\\left(\\frac {8\\sqrt {3}}{13},y_{0}\\right)$ and $\\left(\\frac { - 8\\sqrt {3}}{13},y_{0}\\right),$ respectively, for some value of $y_{0}.$ It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$. Our answer is \\[BC = 2*\\frac {8\\sqrt {3}}{13}=\\sqrt {4\\left(\\frac {8\\sqrt {3}}{13}\\right)^{2}} = \\sqrt {\\frac {768}{169}}\\implies m + n = 937",
"Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$ Note that the slope of $\\overline{AC}$ is $\\tan 60^\\circ = \\sqrt {3}.$ Hence, the equation of the line containing $\\overline{AC}$ is \\[y = x\\sqrt {3} + 1.\\] This will intersect the ellipse when \\begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\\sqrt {3} + 1)^{2} \\\\ & = & x^{2} + 4(3x^{2} + 2x\\sqrt {3} + 1) \\implies x(13x+8\\sqrt 3)=0\\implies x = \\frac { - 8\\sqrt {3}}{13}. \\end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3. Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\\left(\\frac {8\\sqrt {3}}{13},y_{0}\\right)$ and $\\left(\\frac { - 8\\sqrt {3}}{13},y_{0}\\right),$ respectively, for some value of $y_{0}.$ It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$. Our answer is \\[BC = 2*\\frac {8\\sqrt {3}}{13}=\\sqrt {4\\left(\\frac {8\\sqrt {3}}{13}\\right)^{2}} = \\sqrt {\\frac {768}{169}}\\implies m + n = 937",
"Solving for $y$ in terms of $x$ gives $y=\\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\\sqrt{4-x^2}/2)$ and $(-x,\\sqrt{4-x^2}/2)$, which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\\sqrt{4-x^2}/2)$ and $(0,1)$, so by the distance formula we have \\[2x=\\sqrt{x^2+(1-\\sqrt{4-x^2}/2)^2}.\\] Squaring both sides and simplifying through algebra yields $x^2=192/169$, so $2x=\\sqrt{768/169}$ and the answer is $937",
"Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$, and WLOG, we can say that $x$ is positive. Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base): $\\sqrt{x^2 + (y-1)^2} = 2x$ Square both sides, $x^2 + (y-1)^2 = 4x^2\\implies (y-1)^2 = 3x^2$ Now, with the equation of the ellipse: $x^2 + 4y^2 = 4$ $x^2 = 4-4y^2$ $3x^2 = 12-12y^2$ Substituting, $12-12y^2 = y^2 - 2y +1$ Moving stuff around and solving: $y = \\frac{-11}{13}, 1$ The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be: $\\sqrt{\\frac{768}{169}}$ and so we get the desired answer of $937",
"Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$-axis and $C$ as the vertex to the right of the $y$-axis. Let $D$ be the intersection of $BC$ and the $y$-axis. Let $x_0$ be the $x$-coordinate of $C.$ This implies \\[C=\\left(x_0 , \\sqrt{\\frac{4-x_0^2}{4}}\\right)\\] and \\[B=\\left(-x_0 , \\sqrt{\\frac{4-x_0^2}{4}}\\right).\\] Note that $BC=2x_0$ and \\[\\frac{BC}{\\sqrt3}=AD=1-\\sqrt{\\frac{4-x_0^2}{4}}.\\] This yields \\[\\frac{2x_0}{\\sqrt3}=1-\\sqrt{\\frac{4-x_0^2}{4}}.\\] Re-arranging and squaring, we have \\[\\frac{4-x_0^2}{4}=\\frac{4x_0^2}{3}-\\frac{4x_0}{\\sqrt3} +1.\\] Simplifying and solving for $x_0$, we have \\[x_0=\\frac{48}{13\\sqrt 3}.\\] As the length of each side is $2x_0,$ our desired length is \\[4x_0^2=\\frac{768}{169}\\] which means our desired answer is \\[768+169=937",
"Notice that $x^2+4y^2=4$ can be rewritten as $(x)^2+(2y)^2=2^2$. The points of the triangle are $(0, 1)$, $(-x, 1-x\\sqrt{3})$, and $(x, 1-x\\sqrt{3})$. When plugging the second coordinate into the equation, we get $x^2+4-8x\\sqrt{3}+12x^2=4$, which equals $13x^2-8x\\sqrt{3}=0$. This yields $x(13x-8\\sqrt{3})=0$. Obviously x can't be 0, so $x=\\frac{8\\sqrt{3}}{13}$. The side length of the equilateral triangle is twice of this, so $\\frac{16\\sqrt{3}}{13}$. This can be rewritten as $\\sqrt{\\frac{256\\cdot3}{169}}=\\sqrt{\\frac{768}{169}}$. $768+169=937",
"Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\\sqrt{13}, n\\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\\sqrt{3}.$ Using power of a point on X, \\[n^2=2n\\sqrt{3}(2-2n\\sqrt{3})\\] Simplifying gets us to $13n^2=4n\\sqrt{3}.$ Then, $n=\\dfrac{4\\sqrt{3}}{13}$ which means the side length is $\\dfrac{16\\sqrt{3}}{13}=\\sqrt{\\dfrac{768}{169}}.$ Thus, the answer is $768+169=937"
] |
2001-I-6
| 2,001 | 6 |
A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 79 |
I
|
[
"Solution 1 Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$. In the diagram below, the lowest $y$-coordinate at each of $a$, $b$, $c$, and $d$ corresponds to the value of the roll. The red path corresponds to the sequence of rolls $2, 3, 5, 5$. This establishes a bijection between valid dice roll sequences and block walking paths. The solution to this problem is therefore $\\dfrac{\\binom{9}{4}}{6^4} = {\\dfrac{7}{72}}$. So the answer is $079. ~MC413551",
"Recast the problem entirely as a block-walking problem. Call the respective dice $a, b, c, d$. In the diagram below, the lowest $y$-coordinate at each of $a$, $b$, $c$, and $d$ corresponds to the value of the roll. The red path corresponds to the sequence of rolls $2, 3, 5, 5$. This establishes a bijection between valid dice roll sequences and block walking paths. The solution to this problem is therefore $\\dfrac{\\binom{9}{4}}{6^4} = {\\dfrac{7}{72}}$. So the answer is $079. ~MC413551",
"Let $a, b, c,$ and $d$ be the results of rolling the four dice respectively. We have the range $1\\leq a\\leq b\\leq c\\leq d\\leq 6$, and there are $6^4=1296$ total outcomes from rolling the dice. To transfer the inequality into a strictly increasing inequality, we can transform it into $1\\leq a<b+1<c+2<d+3\\leq 9$. Now, lets suppose $a'=a, b'=b+1, c'=c+2,$ and $d'=d+3$. Then, $1\\leq a'<b'<c'<d' \\leq 9$. Clearly, there are $\\binom{9}{4}=126$ values that satisfy this equation, so our answer is $\\dfrac{\\binom{9}{4}}{6^4} = {\\dfrac{7}{72}}\\implies 079. ~MC413551",
"If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\\{1,2,3,4,5,6\\}$. We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are ${9\\choose4} = 126$ outcomes of four dice. The solution is therefore $\\frac{126}{6^4} = \\frac{7}{72}$, and $7 + 72 = 079. ~MC413551",
"Call the dice rolls $a, b, c, d$. The difference between the $a$ and $d$ distinguishes the number of possible rolls there are. If $a - d = 0$, then the values of $b,\\ c$ are set, and so there are $6$ values for $a,\\ d$. If $a - d = 1$, then there are ${3\\choose2} = 3$ ways to arrange for values of $b,\\ c$, but only $5$ values for $a,\\ d$. If $a - d = 2$, then there are ${4\\choose2} = 6$ ways to arrange $b, c$, and there are only $6 - 2 = 4$ values for $a, d$. Continuing, we see that the sum is equal to $\\sum_{i = 0}^{5}{i+2\\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126$. The requested probability is $\\frac{126}{6^4} = \\frac{7}{72}$ and our answer is $79. ~MC413551",
"The dice rolls can be in the form ABCD AABC AABB AAAB AAAA where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.) Now, for the first case, there are ${6\\choose4} = 15$ ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation. Second case: ${6\\choose3} = 20$ ways to pick 3 numbers, ${3\\choose1}$ ways to pick 1 of those 3 to duplicate. A total of 60 for this case. Third case: ${6\\choose2}=15$ ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters. Fourth case: ${6\\choose2} = 15$ ways to pick 2 numbers. We pick one to duplicate with ${2\\choose1} = 2$, so there are a total of 30 in this case. Fifth case: ${6\\choose1} = 6$; all get duplicated so nothing else matters. There are a total of $6^4$ possible dice rolls. Thus, $\\frac{m}{n} = \\frac{15 + 60 + 15 + 30 + 6}{6^4} = \\frac{126}{6^4} = \\frac{7}{72}$ Therefore, our answer is $079. ~MC413551",
"Consider the number of possible dice roll combinations which work after $1$ roll, after $2$ rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on. Suppose we generalize this as a function, say $f(l,r)$ return the number of possible combinations after $l$ rolls and $r$ being the beginning value of the first roll. It becomes clear that from above, $f(1,r) = 1$; every value of $l$ after that is equal to the sum of the number of combinations of $l - 1$ rolls that have a starting value of at least $r$. If we slowly count through and add up all the possible combinations we get $\\frac{7}{72}$ possibilities. Solution 7 In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is $\\le$ the minimum value of the second subset. If the first subset ends in a 1, there is $1$ such subset and there are $6 + 5 + 4 + 3 + 2 + 1 = \\frac{6}{2}(6 + 1) = 21$ ways of making the second subset. If the first subset ends in a 2, there is $2$ such subsets and there are $5 + \\ldots + 1 = \\frac{5}{2}(5 + 1) = 15$ ways of making the second subset. Thus, the number of combinations is $\\sum_{i=1}^6 i \\cdot \\left(\\frac{(7 - i)(8 - i)}{2}\\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126$, and the probability again is $\\frac7{72}$, giving $m+n=079. ~MC413551",
"In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is $\\le$ the minimum value of the second subset. If the first subset ends in a 1, there is $1$ such subset and there are $6 + 5 + 4 + 3 + 2 + 1 = \\frac{6}{2}(6 + 1) = 21$ ways of making the second subset. If the first subset ends in a 2, there is $2$ such subsets and there are $5 + \\ldots + 1 = \\frac{5}{2}(5 + 1) = 15$ ways of making the second subset. Thus, the number of combinations is $\\sum_{i=1}^6 i \\cdot \\left(\\frac{(7 - i)(8 - i)}{2}\\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126$, and the probability again is $\\frac7{72}$, giving $m+n=079. ~MC413551",
"If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer. We just need to work out the valid cases for each roll. Denote by $N_{k}(n)$ the number of valid cases in the $k+1$-th roll when the number $n$ is rolled, for $k=1,2,3$ and $1 <= n <= 6$. Then we have the following recursion formula: \\[N_{1}(n) = n\\] \\[N_{2}(n) = N_{1}(1) + N_{1}(2) + ... + N_{1}(n)=N_{2}(n-1)+N_{1}(n)\\] \\[N_{3}(n) = N_{2}(1) + N_{2}(2) + ... + N_{2}(n)=N_{3}(n-1)+N_{2}(n)\\] The logic is that, if $n$ is rolled, then the number of valid cases is the subtotal of valid cases in the preceding roll for the outcome of 1 to $n$. The recursion can be easily calculated by hand when $N_k(n)$ are put in columns side by side, given the fact that te numbers are smaller than 100. Finally, the total number of cases is \\[N = \\sum_{n=1}^{6} N_{3}(n)\\] and \\[P = \\frac{N}{6^4}\\] Solution 9: Observation of each case Lets try casework and observe the cases. Notice that if the last roll is a $1$, then the only dice rolls may be $1-1-1-1$, which is only $1$ possibility. Observe that if the last roll is $2$, then there are $4 = 1 + 3$ possibilities. When the last roll is a $3$, there are $10 = 1 + 3 + 6$ possibilities. Notice when the last roll is $n$, the number of cases is the sum of the first $n$ positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls. So there are a total of $1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126$ possibilities. So the probability is $\\frac{126}{6^4} = \\frac{7}{72}$ and $7 + 72 = 079. ~MC413551",
"Lets try casework and observe the cases. Notice that if the last roll is a $1$, then the only dice rolls may be $1-1-1-1$, which is only $1$ possibility. Observe that if the last roll is $2$, then there are $4 = 1 + 3$ possibilities. When the last roll is a $3$, there are $10 = 1 + 3 + 6$ possibilities. Notice when the last roll is $n$, the number of cases is the sum of the first $n$ positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls. So there are a total of $1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126$ possibilities. So the probability is $\\frac{126}{6^4} = \\frac{7}{72}$ and $7 + 72 = 079. ~MC413551",
"This is equivalent to picking a four-element sequence of $\\{1, 2, 3, 4, 5, 6\\}$ with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions. Now count the number of such four-element sequences, let $a$ be the number of $1$s in the sequence, $b$ be the number of $2$s, $c$ $3$s, $d$ $4$s, $e$ $5$s, and $f$ $6$s. Now we see that we must have \\[a + b + c + d + e + f = 4\\] with $a, b, c, d, e, f$ being nonnegative integers since there are a total of $4$ numbers picked. The number of solutions to this is $\\dbinom{9}{4},$ so our total number is equal to $\\dfrac{\\binom{9}{4}}{6^4} = \\dfrac{7}{72},$ making our answer $079. ~MC413551",
"Let the rolls be $a,b,c$ and $d\\newline$ let $z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\\newline$ $z+e+f+g+h=5\\newline$ This equation has $C(5+5-1, 5-1)=126$ integer solutions$\\newline$ $126/1296=7/72\\newline$ $7+72=79. ~MC413551",
"Let the rolls be $a,b,c$ and $d\\newline$ let $z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\\newline$ $z+e+f+g+h=5\\newline$ This equation has $C(5+5-1, 5-1)=126$ integer solutions$\\newline$ $126/1296=7/72\\newline$ $7+72=79. ~MC413551",
"Let's say the four dice values are all different. These can only be arranged in one way to satisfy our conditions, so there are $\\binom{6}{4}=15$ ways. If there are three different values, there are $\\binom{6}{3}$ to choose the numbers and $\\binom{3}{1}$ to choose which number will have the repeat, so $20\\cdot3=60$ ways. If there are two different values, there are $\\binom{6}{2}$ ways to choose the numbers. If there are two of each, then there is one way. If there are three of one and one of another, then there are $\\binom{2}{1}$ ways. Therefore $15\\cdot(1+2)=45$. Now if all the numbers are the same, there are $\\binom{6}{1}=6$ ways. Altogether we have $15+60+45+6=126$ ways. $\\frac{126}{6^4}=\\frac{21}{216}=\\frac{7}{72}$. $7+72=79. ~MC413551"
] |
2001-I-7
| 2,001 | 7 |
Triangle $ABC$ has $AB=21$ , $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 923 |
I
|
[
"[asy] pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--cycle); D(MP(\"I\",I,NE)); D(MP(\"E\",E,NE)--MP(\"D\",D,NW)); // D((A.x,0)--A,linetype(\"4 4\")+linewidth(0.7)); D((I.x,0)--I,linetype(\"4 4\")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); D(B--I--C); MP(\"20\",(B+C)/2); MP(\"21\",(A+B)/2,NW); MP(\"22\",(A+C)/2,NE); [/asy] Let $I$ be the incenter of $\\triangle ABC$, so that $BI$ and $CI$ are angle bisectors of $\\angle ABC$ and $\\angle ACB$ respectively. Then, $\\angle BID = \\angle CBI = \\angle DBI,$ so $\\triangle BDI$ is isosceles, and similarly $\\triangle CEI$ is isosceles. It follows that $DE = DB + EC$, so the perimeter of $\\triangle ADE$ is $AD + AE + DE = AB + AC = 43$. Hence, the ratio of the perimeters of $\\triangle ADE$ and $\\triangle ABC$ is $\\frac{43}{63}$, which is the scale factor between the two similar triangles, and thus $DE = \\frac{43}{63} \\times 20 = \\frac{860}{63}$. Thus, $m + n = 923.",
"[asy] pointpen = black; pathpen = black+linewidth(0.7); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--cycle); D(incircle(A,B,C)); D(MP(\"I\",I,NE)); D(MP(\"E\",E,NE)--MP(\"D\",D,NW)); D((A.x,0)--A,linetype(\"4 4\")+linewidth(0.7)); D((I.x,0)--I,linetype(\"4 4\")+linewidth(0.7)); D(rightanglemark(B,(A.x,0),A,30)); MP(\"20\",(B+C)/2); MP(\"21\",(A+B)/2,NW); MP(\"22\",(A+C)/2,NE); [/asy] The semiperimeter of $ABC$ is $s = \\frac{20 + 21 + 22}{2} = \\frac{63}{2}$. By Heron's formula, the area of the whole triangle is $A = \\sqrt{s(s-a)(s-b)(s-c)} = \\frac{21\\sqrt{1311}}{4}$. Using the formula $A = rs$, we find that the inradius is $r = \\frac{A}{s} = \\frac{\\sqrt{1311}}6$. Since $\\triangle ADE \\sim \\triangle ABC$, the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$. From $A=\\frac{1}{2}bh$, we find $h_{ABC} = \\frac{21\\sqrt{1311}}{40}$. Thus, we have $\\frac{h_{ADE}}{h_{ABC}} = \\frac{h_{ABC}-r}{h_{ABC}} = \\frac{21\\sqrt{1311}/40-\\sqrt{1311}/6}{21\\sqrt{1311}/40}=\\frac{DE}{20}.$ Solving for $DE$ gives $DE=\\frac{860}{63},$ so the answer is $m+n=923.",
"[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype(\"4 4\"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--cycle); D(incircle(A,B,C)); D(MP(\"P\",I,(1,2))); D(MP(\"E\",E,NE)--MP(\"D\",D,NW)); MP(\"20\",(B+C)/2); MP(\"21\",(A+B)/2,NW); MP(\"22\",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(B,C,A,A--B),d); D(anglebisector(C,B,A,A--C),d); [/asy] Let $P$ be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$, and name the intersection $F$. Using the angle bisector theorem, we know the ratio $BF:CF$ is $21:22$, thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$, giving $F$ a weight of $43$. In the same manner, using another bisector, we find that $A$ has a weight of $20$. So, now we know $P$ has a weight of $63$, and the ratio of $FP:PA$ is $20:43$. Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$. So, $DE$ is $43/63$ the size of $BC$. Multiplying this ratio by the length of $BC$, we find $DE$ is $860/63 = m/n$. Therefore, $m+n=923.",
"[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype(\"4 4\"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--cycle); D(incircle(A,B,C)); D(MP(\"P\",I,(1,2))); D(MP(\"E\",E,NE)--MP(\"D\",D,NW)); MP(\"20\",(B+C)/2); MP(\"21\",(A+B)/2,NW); MP(\"22\",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(B,C,A,A--B),d); D(anglebisector(C,B,A,A--C),d); [/asy] Let $P$ be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$, and name the intersection $F$. Using the angle bisector theorem, we know the ratio $BF:CF$ is $21:22$, thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$, giving $F$ a weight of $43$. In the same manner, using another bisector, we find that $A$ has a weight of $20$. So, now we know $P$ has a weight of $63$, and the ratio of $FP:PA$ is $20:43$. Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$. So, $DE$ is $43/63$ the size of $BC$. Multiplying this ratio by the length of $BC$, we find $DE$ is $860/63 = m/n$. Therefore, $m+n=923.",
"More directly than Solution 2, we have \\[DE=BC\\left(\\frac{h_a-r}{h_a}\\right)=20\\left(1-\\frac{r}{\\frac{[ABC]}{\\frac{BC}{2}}}\\right)=20\\left(1-\\frac{10r}{sr}\\right)=20\\left(1-\\frac{10}{\\frac{63}{2}}\\right)=\\frac{860}{63}\\implies 923.\\]",
"More directly than Solution 2, we have \\[DE=BC\\left(\\frac{h_a-r}{h_a}\\right)=20\\left(1-\\frac{r}{\\frac{[ABC]}{\\frac{BC}{2}}}\\right)=20\\left(1-\\frac{10r}{sr}\\right)=20\\left(1-\\frac{10}{\\frac{63}{2}}\\right)=\\frac{860}{63}\\implies 923.\\]",
"Diagram borrowed from Solution 3. [asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype(\"4 4\"); pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--cycle); D(incircle(A,B,C)); D(MP(\"P\",I,(1,2))); D(MP(\"E\",E,NE)--MP(\"D\",D,NW)); MP(\"20\",(B+C)/2); MP(\"21\",(A+B)/2,NW); MP(\"22\",(A+C)/2,NE); /* construct angle bisectors */ path anglebisector (pair X, pair Y, pair Z, path K) { return Y -- IP(Y -- Y + 30 * (bisectorpoint(X,Y,Z)-Y) , K); } D(anglebisector(C,A,B,B--C), d); D(anglebisector(C,B,A,A--C),d); [/asy] Let the angle bisector of $\\angle{A}$ intersects $BC$ at $F$. Applying the Angle Bisector Theorem on $\\angle{A}$ we have \\[\\frac{AB}{BF}=\\frac{AC}{CF}\\] \\[BF=BC\\cdot(\\frac{AB}{AB+AC})\\] \\[BF=\\frac{420}{43}\\] Since $BP$ is the angle bisector of $\\angle{B}$, we can once again apply the Angle Bisector Theorem on $\\angle{B}$ which gives \\[\\frac{BA}{AP}=\\frac{BF}{FP}\\] \\[\\frac{AP}{PF}=\\frac{AB}{BF}=\\frac{41}{20}\\] Since $\\bigtriangleup ADE\\sim\\bigtriangleup ABC$ we have \\[\\frac{DE}{BC}=\\frac{AP}{AF}\\] \\[DE=BC\\cdot(\\frac{AP}{(\\frac{61}{41})\\cdot AP})\\] Solving gets $DE=\\frac{860}{63}$. Thus $m+n=860+63=923. ~ Nafer",
"Let $A'$ be the foot of the altitude from $A$ to $\\overline {BC}$ and $K$ be the foot of the altitude from $A$ to $\\overline{DE}$. Evidently, \\[\\frac{AK}{AA'} = 1- \\frac{r}{AA'} = 1 - \\frac{T/s}{T/BC}\\] where $r$ is the inradius, $T = [ABC]$, and $s$ is the semiperimeter. So, \\[\\frac{AK}{AA'} = 1 - \\frac{BC}{s} = 1 - \\frac{20}{63}= \\frac{43}{63}\\] Therefore, by similar triangles, we have $DE = BC * \\frac{AK}{AA'} = 20 * \\frac{AK}{AA'}= \\frac{860}{63}.",
"Label $P$ the point the angle bisector of $A$ intersects ${BC}$. First we find ${BP}$ and ${PC}$. By the Angle Bisector Theorem, $\\frac{BP}{PC} = \\frac{21}{22}$ and solving for each using the fact that ${BC} = 20$, we see that ${BP} = \\frac{420}{43}$ and $PC = \\frac{440}{43}$. Because ${AP}$ is the angle bisector of $<A$, we can simply calculate it using Stewarts, \\[{AP} = 21*22 - \\frac{440}{43}\\cdot\\frac{420}{43}\\] \\[{AP} = 21*22 - \\frac{440\\cdot420}{43^2}\\] Now we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \\frac{43}{63} \\cdot[21\\cdot22 - \\frac{420*440}{43^2}] = \\frac{43^2\\cdot22 - 20\\cdot440}{43\\cdot3}$. Now because $\\triangle{APE} ~ \\triangle{ABC}$, we can find ${DE}$ by $\\frac{AO}{AP} \\cdot 20$. Dividing and simplifying, we see that $\\frac{1}{21}\\cdot\\frac{43}{3}\\cdot20 = \\frac{860}{63}$. So the answer is $923 ~YBSuburbanTea",
"To solve this problem, we can use the fact that, in $\\triangle ABC$, the vector representation of the incenter is $\\overrightarrow I = \\frac{a\\overrightarrow A + b\\overrightarrow B + c\\overrightarrow C}{a+b+c}$ and that that the vector of the foot of the bisector of $\\angle BAC$ on $\\overline{BC}$ is $\\overrightarrow P = \\frac{b\\overrightarrow B + c\\overrightarrow C}{b+c}$, where $a=BC,$ $b=AC,$ and $c=AB$. Let point $A$ be the origin of the coordinate plane. Then, $\\overrightarrow A$ is the zero vector, so we can simplify our expression for $\\overrightarrow I$ to $\\frac{b\\overrightarrow B + c\\overrightarrow C}{a+b+c}$. Now, note that the vector components of $\\overrightarrow I$ and $\\overrightarrow P$ are the same, but they are multiplied by different scalars. Thus, the ratio of these scalars is the ratio of these vectors' magnitudes. Thus, we have $\\frac{|\\overrightarrow I|}{|\\overrightarrow P|}=\\frac{\\tfrac1{a+b+c}}{\\tfrac1{b+c}}=\\frac{b+c}{a+b+c}=\\frac{43}{63}$. Let $D \\in \\overline{AB}$ and $E \\in \\overline{AC}$. Because $\\triangle AIE \\sim \\triangle APC$, we have $\\frac{AI}{AP}=\\frac{AE}{AC}$. Further, because $\\triangle ADE \\sim \\triangle ABC$, we have $\\frac{AE}{AC}=\\frac{DE}{BC}$. Thus, by transitivity, $\\frac{AI}{AP}=\\frac{DE}{BC}$. We know that $\\frac{AI}{AP}=\\frac{43}{63}$, so $DE=\\frac{AI}{AD}\\cdot BC = \\frac{43}{63}\\cdot 20 = \\frac{860}{63}$. Thus, our answer is $860+63=923.",
"To solve this problem, we can use the fact that, in $\\triangle ABC$, the vector representation of the incenter is $\\overrightarrow I = \\frac{a\\overrightarrow A + b\\overrightarrow B + c\\overrightarrow C}{a+b+c}$ and that that the vector of the foot of the bisector of $\\angle BAC$ on $\\overline{BC}$ is $\\overrightarrow P = \\frac{b\\overrightarrow B + c\\overrightarrow C}{b+c}$, where $a=BC,$ $b=AC,$ and $c=AB$. Let point $A$ be the origin of the coordinate plane. Then, $\\overrightarrow A$ is the zero vector, so we can simplify our expression for $\\overrightarrow I$ to $\\frac{b\\overrightarrow B + c\\overrightarrow C}{a+b+c}$. Now, note that the vector components of $\\overrightarrow I$ and $\\overrightarrow P$ are the same, but they are multiplied by different scalars. Thus, the ratio of these scalars is the ratio of these vectors' magnitudes. Thus, we have $\\frac{|\\overrightarrow I|}{|\\overrightarrow P|}=\\frac{\\tfrac1{a+b+c}}{\\tfrac1{b+c}}=\\frac{b+c}{a+b+c}=\\frac{43}{63}$. Let $D \\in \\overline{AB}$ and $E \\in \\overline{AC}$. Because $\\triangle AIE \\sim \\triangle APC$, we have $\\frac{AI}{AP}=\\frac{AE}{AC}$. Further, because $\\triangle ADE \\sim \\triangle ABC$, we have $\\frac{AE}{AC}=\\frac{DE}{BC}$. Thus, by transitivity, $\\frac{AI}{AP}=\\frac{DE}{BC}$. We know that $\\frac{AI}{AP}=\\frac{43}{63}$, so $DE=\\frac{AI}{AD}\\cdot BC = \\frac{43}{63}\\cdot 20 = \\frac{860}{63}$. Thus, our answer is $860+63=923."
] |
2001-I-8
| 2,001 | 8 |
Call a positive integer $N$ a $\textit{7-10 double}$ if the digits of the base-7 representation of $N$ form a base-10 number that is twice $N$ . For example, $51$ is a 7-10 double because its base-7 representation is $102$ . What is the largest 7-10 double?
| 315 |
I
|
[
"We let $N_7 = \\overline{a_na_{n-1}\\cdots a_0}_7$; we are given that \\[2(a_na_{n-1}\\cdots a_0)_7 = (a_na_{n-1}\\cdots a_0)_{10}\\] (This is because the digits in $N$ ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2) Expanding, we find that \\[2 \\cdot 7^n a_n + 2 \\cdot 7^{n-1} a_{n-1} + \\cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \\cdots + a_0\\] or re-arranging, \\[a_0 + 4a_1 = 2a_2 + 314a_3 + \\cdots + (10^n - 2 \\cdot 7^n)a_n\\] Since the $a_i$s are base-$7$ digits, it follows that $a_i < 7$, and the LHS is less than or equal to $30$. Hence our number can have at most $3$ digits in base-$7$. Letting $a_2 = 6$, we find that $630_7 = 315 is our largest 7-10 double.",
"Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation. Given this is an AIME problem, $A<1000$. If we look at $B$ in base $10$, it must be equal to $2A$, so $B<2000$ when $B$ is looked at in base $10.$ If $B$ in base $10$ is less than $2000$, then $B$ as a number in base $7$ must be less than $2*7^3=686$. $686$ is non-existent in base $7$, so we're gonna have to bump that down to $666_7$. This suggests that $A$ is less than $\\frac{666}{2}=333$. Guess and check shows that $A<320$, and checking values in that range produces $315.",
"Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation. Given this is an AIME problem, $A<1000$. If we look at $B$ in base $10$, it must be equal to $2A$, so $B<2000$ when $B$ is looked at in base $10.$ If $B$ in base $10$ is less than $2000$, then $B$ as a number in base $7$ must be less than $2*7^3=686$. $686$ is non-existent in base $7$, so we're gonna have to bump that down to $666_7$. This suggests that $A$ is less than $\\frac{666}{2}=333$. Guess and check shows that $A<320$, and checking values in that range produces $315.",
"Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be \\[abc\\] in base 7. Then the number in expanded form is \\[49a+7b+c\\] in base 7 and \\[100a+10b+c\\] in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. \\[98a+14b+2c=100a+10b+c\\] which simplifies to \\[2a=4b+c.\\] The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, $b$ is $3$ and $c$ is $0$. Therefore, the largest 7-10 double is 630 in base 7, or $315 in base 10."
] |
2001-I-9
| 2,001 | 9 |
In triangle $ABC$ , $AB=13$ , $BC=15$ and $CA=17$ . Point $D$ is on $\overline{AB}$ , $E$ is on $\overline{BC}$ , and $F$ is on $\overline{CA}$ . Let $AD=p\cdot AB$ , $BE=q\cdot BC$ , and $CF=r\cdot CA$ , where $p$ , $q$ , and $r$ are positive and satisfy $p+q+r=2/3$ and $p^2+q^2+r^2=2/5$ . The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 61 |
I
|
[
"Solution 1 [asy] /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); D(D(MP(\"A\",A))--D(MP(\"B\",B))--D(MP(\"C\",C,N))--cycle); D(D(MP(\"D\",D))--D(MP(\"E\",E,NE))--D(MP(\"F\",F,NW))--cycle); [/asy] We let $[\\ldots]$ denote area; then the desired value is $\\frac mn = \\frac{[DEF]}{[ABC]} = \\frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$ Using the formula for the area of a triangle $\\frac{1}{2}ab\\sin C$, we find that $\\frac{[ADF]}{[ABC]} = \\frac{\\frac 12 \\cdot p \\cdot AB \\cdot (1-r) \\cdot AC \\cdot \\sin \\angle CAB}{\\frac 12 \\cdot AB \\cdot AC \\cdot \\sin \\angle CAB} = p(1-r)$ and similarly that $\\frac{[BDE]}{[ABC]} = q(1-p)$ and $\\frac{[CEF]}{[ABC]} = r(1-q)$. Thus, we wish to find \\begin{align*}\\frac{[DEF]}{[ABC]} &= 1 - \\frac{[ADF]}{[ABC]} - \\frac{[BDE]}{[ABC]} - \\frac{[CEF]}{[ABC]} \\\\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\\\ &= (pq + qr + rp) - (p + q + r) + 1 \\end{align*} We know that $p + q + r = \\frac 23$, and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \\Longleftrightarrow pq + qr + rp = \\frac{\\left(\\frac 23\\right)^2 - \\frac 25}{2} = \\frac{1}{45}$. Substituting, the answer is $\\frac 1{45} - \\frac 23 + 1 = \\frac{16}{45}$, and $m+n = 061, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.",
"[asy] /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ real p = 0.5, q = 0.1, r = 0.05; /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ pointpen = black; pathpen = linewidth(0.7) + black; pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); D(D(MP(\"A\",A))--D(MP(\"B\",B))--D(MP(\"C\",C,N))--cycle); D(D(MP(\"D\",D))--D(MP(\"E\",E,NE))--D(MP(\"F\",F,NW))--cycle); [/asy] We let $[\\ldots]$ denote area; then the desired value is $\\frac mn = \\frac{[DEF]}{[ABC]} = \\frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$ Using the formula for the area of a triangle $\\frac{1}{2}ab\\sin C$, we find that $\\frac{[ADF]}{[ABC]} = \\frac{\\frac 12 \\cdot p \\cdot AB \\cdot (1-r) \\cdot AC \\cdot \\sin \\angle CAB}{\\frac 12 \\cdot AB \\cdot AC \\cdot \\sin \\angle CAB} = p(1-r)$ and similarly that $\\frac{[BDE]}{[ABC]} = q(1-p)$ and $\\frac{[CEF]}{[ABC]} = r(1-q)$. Thus, we wish to find \\begin{align*}\\frac{[DEF]}{[ABC]} &= 1 - \\frac{[ADF]}{[ABC]} - \\frac{[BDE]}{[ABC]} - \\frac{[CEF]}{[ABC]} \\\\ &= 1 - p(1-r) - q(1-p) - r(1-q)\\\\ &= (pq + qr + rp) - (p + q + r) + 1 \\end{align*} We know that $p + q + r = \\frac 23$, and also that $(p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \\Longleftrightarrow pq + qr + rp = \\frac{\\left(\\frac 23\\right)^2 - \\frac 25}{2} = \\frac{1}{45}$. Substituting, the answer is $\\frac 1{45} - \\frac 23 + 1 = \\frac{16}{45}$, and $m+n = 061, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.",
"By the barycentric area formula, our desired ratio is equal to \\begin{align*} \\begin{vmatrix} 1-p & p & 0 \\\\ 0 & 1-q & q \\\\ r & 0 & 1-r \\notag \\end{vmatrix} &=1-p-q-r+pq+qr+pr\\\\ &=1-(p+q+r)+\\frac{(p+q+r)^2-(p^2+q^2+r^2)}{2}\\\\ &=1-\\frac{2}{3}+\\frac{\\frac{4}{9}-\\frac{2}{5}}{2}\\\\ &=\\frac{16}{45}, \\end{align*} so the answer is $061, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.",
"Since the only conditions are that $p + q + r = \\frac{2}{3}$ and $p^2 + q^2 + r^2 = \\frac{2}{5}$, we can simply let one of the variables be equal to 0. In this case, let $p = 0$. Then, $q + r = \\frac{2}{3}$ and $q^2 + r^2$ = $\\frac{2}{5}$. Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$. Solving this system of equations, we get $q = \\frac{1}{3} \\pm \\sqrt{\\frac{4}{45}}$, and $r = \\frac{1}{3} \\mp \\sqrt{\\frac{4}{45}}$. Plugging back into $(1-q)(1-r)$, we get $\\frac{16}{45}$, so the answer is $061, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all.",
"Since the only conditions are that $p + q + r = \\frac{2}{3}$ and $p^2 + q^2 + r^2 = \\frac{2}{5}$, we can simply let one of the variables be equal to 0. In this case, let $p = 0$. Then, $q + r = \\frac{2}{3}$ and $q^2 + r^2$ = $\\frac{2}{5}$. Note that the ratio between the area of $DEF$ and $ABC$ is equivalent to $(1-q)(1-r)$. Solving this system of equations, we get $q = \\frac{1}{3} \\pm \\sqrt{\\frac{4}{45}}$, and $r = \\frac{1}{3} \\mp \\sqrt{\\frac{4}{45}}$. Plugging back into $(1-q)(1-r)$, we get $\\frac{16}{45}$, so the answer is $061, are actually not necessary to solve the problem. This is clearly demonstrated in all of the above solutions, as the side lengths are not used at all."
] |
2001-I-10
| 2,001 | 10 |
Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 200 |
I
|
[
"Solution 1 The distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore, For $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities. For $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(0,2)$, $(2,0)$, $(1,3)$, and $(3,1)$, $8$ possibilities. For $z$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(0,2)$, $(0,4)$, $(2,0)$, $(4,0)$, $(2,4)$, $(4,2)$, $(1,3)$, and $(3,1)$, $13$ possibilities. However, we have $3\\cdot 4\\cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\\frac {5\\cdot 8\\cdot 13 - 60}{60\\cdot 59} = \\frac {23}{177}\\Longrightarrow m+n = 200. Solution by jackshi2006",
"The distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore, For $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities. For $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(0,2)$, $(2,0)$, $(1,3)$, and $(3,1)$, $8$ possibilities. For $z$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(0,2)$, $(0,4)$, $(2,0)$, $(4,0)$, $(2,4)$, $(4,2)$, $(1,3)$, and $(3,1)$, $13$ possibilities. However, we have $3\\cdot 4\\cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\\frac {5\\cdot 8\\cdot 13 - 60}{60\\cdot 59} = \\frac {23}{177}\\Longrightarrow m+n = 200. Solution by jackshi2006",
"There are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. We group the points by parity of each individual coordinate -- that is, if $x$ is even or odd, $y$ is even or odd, and $z$ is even or odd. Note that to have something that works, the two points must have this same type of classification (otherwise, if one doesn't match, the resulting sum for the coordinates will be odd at that particular spot). There are $12$ EEEs (the first position denotes the parity of $x,$ the second $y,$ and the third $z.$), $8$ EEOs, $12$ EOEs, $6$ OEEs, $8$ EOOs, $4$ OEOs, $6$ OOEs, and $4$ OOOs. Doing a sanity check, $12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,$ which is the total number of points. Now, we can see that there are $12 \\cdot 11$ ways to choose two EEEs (respective to order), $8 \\cdot 7$ ways to choose two EEOs, and so on. Therefore, we get \\[12\\cdot11 + 8\\cdot7 + 12\\cdot11 + 6\\cdot5 + 8\\cdot7 + 4\\cdot3 + 6\\cdot5 + 4\\cdot3 = 460\\] ways to choose two points where order matters. There are $60 \\cdot 59$ total ways to do this, so we get a final answer of \\[\\dfrac{460}{60 \\cdot 59} = \\dfrac{23}{3 \\cdot 59} = \\dfrac{23}{177},\\] for our answer of $23 + 177 = 200. Solution by jackshi2006",
"Similarly to Solution 2, we note that there are $60$ points and that the parities of the two points' coordinates must be the same in order for the midpoint to be in $S$. Ignore the distinct points condition. The probability that the midpoint is in $S$ is then \\[\\left(\\left(\\frac 23\\right)^2+\\left(\\frac 13\\right)^2\\right)\\left(\\left(\\frac 24\\right)^2+\\left(\\frac 24\\right)^2\\right)\\left(\\left(\\frac 35\\right)^2+\\left(\\frac 25\\right)^2\\right)=\\frac{13}{90}.\\] Note that $\\frac{13}{90}=\\frac{520}{3600}$. Since there are $3600$ total ways to choose $2$ points from $S$, there must be $520$ pairs of points that have their midpoint in $S$. Of these pairs, $60$ of them contain identical points (not distinct). Subtracting these cases, our answer is $\\frac{520-60}{3600-60}=\\frac{23}{177}\\implies200. Solution by jackshi2006",
"There are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. Note that in order for the midpoint of the line segment to be a lattice point, the lengths on the x, y, and z axis must be even numbers. We will define all segments by denoting the amount they extend in each dimension: $(x, y, z)$. For example, the longest diagonal possible will be $(2,3,4)$, the space diagonal of the box. Thus, any line segment must have dimensions that are even. For $x$ the segment may have a value of $0$ for $x$, (in which case the segment would be two dimensional) or a value of $2$. The same applies for $y$, because although it is three units long the longest even integer is two. For $z$ the value may be $0$, $2$, or $4$. Notice that if a value is zero, then the segment will pertain to only two dimensions. If two values are zero then the line segment becomes one dimensional. Then the total number of possibilities will be $2 \\cdot 2 \\cdot 3$. Listing them out appears as follows: $2,2,4$ $2,2,2$ $2,2,0$ $2,0,4$ $2,0,2$ $2,0,0$ $0,2,4$ $0,2,2$ $0,2,0$ $0,0,4$ $0,0,2$ $0,0,0$ * this value is a single point Now, picture every line segment to be the space diagonal of a box. Allow this box to define the space the segment occupies. The question now transforms into \"how many ways can we arrange this smaller box in the two by three by four?\". Notice that the amount an edge can shift inside the larger box is the length of an edge of the larger box (2, 3, or 4) minus the edge of the smaller box (also known as the edge), plus one. For example, (0, 2, 2) would be $3 \\cdot 2 \\cdot 3$. Repeat this process. $2,2,4$ 2 $2,2,2$ 6 $2,2,0$ 10 $2,0,4$ 4 $2,0,2$ 12 $2,0,0$ 20 $0,2,4$ 6 $0,2,2$ 18 $0,2,0$ 30 $0,0,4$ 12 $0,0,2$ 36 $0,0,0$ 60 * this won't be included, but notice that sixty the number of lattice points Finally, we remember that there are four distinct space diagonals in a box, so we should multiply every value by four, right? Unfortunately we forgot to consider that some values have only one or two dimensions. They should be multiplied by one or two, respectively. This is because segments with two dimensions are the diagonals of a rectangle and thus have two orientations. Then any value on our list without any zeroes will be multiplied by four, and any value on our list with only one zero will be multiplied by two, and finally any value on our list with two zeroes will be multiplied by one: $2,2,4$ 2 8 $2,2,2$ 6 24 $2,2,0$ 10 20 $2,0,4$ 4 8 $2,0,2$ 12 24 $2,0,0$ 20 20 $0,2,4$ 6 12 $0,2,2$ 18 36 $0,2,0$ 30 30 $0,0,4$ 12 12 $0,0,2$ 36 36 $0,0,0$ 60 * it's nice to point out that this value will be multiplied by zero add every value on the rightmost side of each term and we will receive $230$. Multiply by two because each segment can be flipped, to receive $460$. There are $60 \\cdot 59$ ways to choose two distinct points, so we get \\[\\dfrac{460}{60 \\cdot 59} = \\dfrac{23}{3 \\cdot 59} = \\dfrac{23}{177},\\] for our answer of $23 + 177 = 200. Solution by jackshi2006"
] |
2001-I-11
| 2,001 | 11 |
In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$
| 149 |
I
|
[
"Let each point $P_i$ be in column $c_i$. The numberings for $P_i$ can now be defined as follows. \\begin{align*}x_i &= (i - 1)N + c_i\\\\ y_i &= (c_i - 1)5 + i \\end{align*} We can now convert the five given equalities. \\begin{align}x_1&=y_2 & \\Longrightarrow & & c_1 &= 5 c_2-3\\\\ x_2&=y_1 & \\Longrightarrow & & N+c_2 &= 5 c_1-4\\\\ x_3&=y_4 & \\Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\\\ x_4&=y_5 & \\Longrightarrow & & 3 N+c_4 &= 5 c_5\\\\ x_5&=y_3 & \\Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \\end{align} Equations $(1)$ and $(2)$ combine to form \\[N = 24c_2 - 19\\] Similarly equations $(3)$, $(4)$, and $(5)$ combine to form \\[117N +51 = 124c_3\\] Take this equation modulo 31 \\[24N+20\\equiv 0 \\pmod{31}\\] And substitute for N \\[24 \\cdot 24 c_2 - 24 \\cdot 19 +20\\equiv 0 \\pmod{31}\\] \\[18 c_2 \\equiv 2 \\pmod{31}\\] Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \\cdot 7 - 19 = 149$ The column values can also easily be found by substitution \\begin{align*}c_1&=32\\\\ c_2&=7\\\\ c_3&=141\\\\ c_4&=88\\\\ c_5&=107 \\end{align*} As these are all positive and less than $N$, $149 is the solution."
] |
2001-I-12
| 2,001 | 12 |
A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 5 |
I
|
[
"[asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label(\"$I$\",I,W); label(\"$A$\",A,S); label(\"$B$\",B,S); label(\"$C$\",C,W*-1); label(\"$D$\",D,W*-1); [/asy] The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$ The signed distance between a plane and a point $I$ can be calculated as $\\frac{(I-G) \\cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC. A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\\times(B-C)=\\langle 8, 12, 24 \\rangle$ Thus $\\frac{(I-C) \\cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$ \\begin{align*}\\frac{(I-C) \\cdot P}{|P|}&=-r\\\\ \\frac{(\\langle r, r, r \\rangle-\\langle 0, 0, 2 \\rangle) \\cdot P}{|P|}&=-r\\\\ \\frac{\\langle r, r, r-2 \\rangle \\cdot \\langle 8, 12, 24 \\rangle}{\\langle 8, 12, 24 \\rangle}&=-r\\\\ \\frac{44r -48}{28}&=-r\\\\ 44r-48&=-28r\\\\ 72r&=48\\\\ r&=\\frac{2}{3} \\end{align*} Finally $2+3=005",
"Notice that we can split the tetrahedron into $4$ smaller tetrahedrons such that the height of each tetrahedron is $r$ and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the tetrahedrons are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be $V$ and surface area be $F$, using the volume formula for each pyramid(base times height divided by 3) we have $\\dfrac{rF}{3}=V$. The surface area of the pyramid is $\\dfrac{6\\cdot{4}+6\\cdot{2}+4\\cdot{2}}{2}+[ABC]=22+[ABC]$. We know triangle ABC's side lengths, $\\sqrt{2^{2}+4^{2}}, \\sqrt{2^{2}+6^{2}},$ and $\\sqrt{4^{2}+6^{2}}$, so using the expanded form of heron's formula, \\begin{align*}[ABC]&=\\sqrt{\\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}\\\\ &=\\sqrt{2(5\\cdot{13}+10\\cdot{5}+13\\cdot{10})-5^{2}-10^{2}-13^{2}}\\\\ &=\\sqrt{196}\\\\ &=14\\end{align*} Therefore, the surface area is $14+22=36$, and the volume is $\\dfrac{[BCD]\\cdot{6}}{3}=\\dfrac{4\\cdot{2}\\cdot{6}}{3\\cdot{2}}=8$, and using the formula above that $\\dfrac{rF}{3}=V$, we have $12r=8$ and thus $r=\\dfrac{2}{3}$, so the desired answer is $2+3=005. (Solution by Shaddoll)",
"The intercept form equation of the plane $ABC$ is $\\frac{x}{6}+\\dfrac{y}{4}+\\dfrac{z}{2}=1.$ Its normal form is $\\dfrac{2}{7}x+\\dfrac{3}{7}y+\\dfrac{6}{7}z-\\dfrac{12}{7}=0$ (square sum of the coefficients equals 1). The distance from $(r,r,r)$ to the plane is $\\left |\\dfrac{2}{7}r+\\dfrac{3}{7}r+\\dfrac{6}{7}r-\\dfrac{12}{7}\\right |$. Since $(r,r,r)$ and $(0,0,0)$ are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in $(0,0,0)$). Therefore we have $-\\left (\\dfrac{2}{7}r+\\dfrac{3}{7}r+\\dfrac{6}{7}r-\\dfrac{12}{7}\\right )=r.$ So $r=\\dfrac{2}{3},$ which solves the problem. Additionally, if $(r,r,r)$ is on the other side of $ABC$, we have $\\left (\\dfrac{2}{7}r+\\dfrac{3}{7}r+\\dfrac{6}{7}r-\\dfrac{12}{7}\\right )=r$, which yields $r=\\dfrac{12}{5},$ corresponding an \"ex-sphere\" that is tangent to face $ABC$ as well as the extensions of the other 3 faces. -JZ",
"First let us find the equation of the plane passing through $(6,0,0), (0,0,2), (0,4,0)$. The \"point-slope form\" is $A(6-x1)+B(0-y1)+C(0-z1)=0.$ Plugging in $(0,0,2)$ gives $A(6)+B(0)+C(-2)=0.$ Plugging in $(0,4,0)$ gives $A(6)+B(-4)+C(0)=0.$ We can then use Cramer's rule/cross multiplication to get $A/(0-8)=-B/(0+12)=C/(-24)=k.$ Solve for A, B, C to get $2k, 3k, 6k$ respectively. We can then get $2k(x-x1)+3k(y-y1)+6k(z-z1)=0.$ Cancel out k on both sides. Next, let us substitute $(0,0,2)$. We can then get $2x+3y+6z=12$as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get $2x/7+2y/7+6z/7=12/7$ to be the normal form. Note that the point is going to be at $(r,r,r).$ We find the distance from $(r,r,r)$ to the plane as $2/7r+3/7r+6/7r-12/7/(\\sqrt{(4/49+9/49+36/49)})$, which is $+/-(11r/7-12/7)$. We take the negative value of this because if we plug in $(0,0,0)$ to the equation of the plane we get a negative value. We equate that value to r and we get the equation $-(11r/7-12/7)=r$ to solve $r={2/3}$, so the answer is $005.",
"Clearly, if the radius of the sphere is $r$, the center of the sphere lies on $(r, r, r)$. We find the equation of plane $ABC$ to be $\\frac16 x+\\frac14 y+\\frac12 z=1$. From the definition of the insphere, it must be true that the distance from the center of the sphere to plane $ABC$ is equal to the length of the radius of the sphere. By point-to-plane, we have \\[r=\\frac{|\\frac16 r+\\frac14 r+\\frac12 r-1|}{\\sqrt{\\left(\\frac16\\right)^2+\\left(\\frac14\\right)^2+\\left(\\frac12\\right)^2}} \\implies r=\\frac23,\\] so the answer is $005. -pqr.",
"The radius of the insphere in a tetrahedron can be calculated using the formula $r = \\frac{3V}{A}$, where $r$ is the radius, $V$ is the volume of the tetrahedron, and $A$ is the total surface area of the tetrahedron. We calculate the volume of the tetrahedron as: \\[V = \\frac{2 \\times 4 \\times 6}{6} = 8\\] Next, we find the total surface area as the sum of the areas of the four triangular faces: \\[\\text{Surface Area} = [ABD] + [BDC] + [ACD] + [ABC] = 12 + 4 + 6 + [ABC]\\] The side lengths of triangle $ABC$ are $\\sqrt{52}$, $\\sqrt{40}$, and $\\sqrt{20}$. Constructing an altitude from $A$ and using the system of equations $x^2 + h^2 = 40$ and $(2\\sqrt{5} - x)^2 + h^2 = 20$, we solve for $h$ and get: \\[h = \\frac{14\\sqrt{5}}{5}\\] Thus, the area of triangle $ABC$ is: \\[\\text{Area of } [ABC] = 14\\] Now, the total surface area is: \\[A = 12 + 4 + 6 + 14 = 36\\] Finally, using the formula for the radius, we have: \\[r = \\frac{3 \\times 8}{36} = \\frac{2}{3}\\] $005",
"The radius of the insphere in a tetrahedron can be calculated using the formula $r = \\frac{3V}{A}$, where $r$ is the radius, $V$ is the volume of the tetrahedron, and $A$ is the total surface area of the tetrahedron. We calculate the volume of the tetrahedron as: \\[V = \\frac{2 \\times 4 \\times 6}{6} = 8\\] Next, we find the total surface area as the sum of the areas of the four triangular faces: \\[\\text{Surface Area} = [ABD] + [BDC] + [ACD] + [ABC] = 12 + 4 + 6 + [ABC]\\] The side lengths of triangle $ABC$ are $\\sqrt{52}$, $\\sqrt{40}$, and $\\sqrt{20}$. Constructing an altitude from $A$ and using the system of equations $x^2 + h^2 = 40$ and $(2\\sqrt{5} - x)^2 + h^2 = 20$, we solve for $h$ and get: \\[h = \\frac{14\\sqrt{5}}{5}\\] Thus, the area of triangle $ABC$ is: \\[\\text{Area of } [ABC] = 14\\] Now, the total surface area is: \\[A = 12 + 4 + 6 + 14 = 36\\] Finally, using the formula for the radius, we have: \\[r = \\frac{3 \\times 8}{36} = \\frac{2}{3}\\] $005"
] |
2001-I-13
| 2,001 | 13 |
In a certain circle, the chord of a $d$ -degree arc is 22 centimeters long, and the chord of a $2d$ -degree arc is 20 centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $m + n.$
| 174 |
I
|
[
"Solution 1 Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$-degree arcs and one chord of one $3d$-degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$-degree arcs. Let $AB$, $AC$, and $BD$ be the chords of the $d$-degree arcs, and let $CD$ be the chord of the $3d$-degree arc. Also let $x$ be equal to the chord length of the $3d$-degree arc. Hence, the length of the chords, $AD$ and $BC$, of the $2d$-degree arcs can be represented as $x + 20$, as given in the problem. Using Ptolemy's theorem, \\[AB(CD) + AC(BD) = AD(BC)\\] \\[22x + 22(22) = (x + 20)^2\\] \\[22x + 484 = x^2 + 40x + 400\\] \\[0 = x^2 + 18x - 84\\] We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. \\[x = \\frac{-18 + \\sqrt{18^2 + 4(84)}}{2}\\] \\[x = \\frac{-18 + \\sqrt{660}}{2}\\] $x$ simplifies to $\\frac{-18 + 2\\sqrt{165}}{2},$ which equals $-9 + \\sqrt{165}.$ Thus, the answer is $9 + 165 = 174.",
"Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$-degree arcs and one chord of one $3d$-degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$-degree arcs. Let $AB$, $AC$, and $BD$ be the chords of the $d$-degree arcs, and let $CD$ be the chord of the $3d$-degree arc. Also let $x$ be equal to the chord length of the $3d$-degree arc. Hence, the length of the chords, $AD$ and $BC$, of the $2d$-degree arcs can be represented as $x + 20$, as given in the problem. Using Ptolemy's theorem, \\[AB(CD) + AC(BD) = AD(BC)\\] \\[22x + 22(22) = (x + 20)^2\\] \\[22x + 484 = x^2 + 40x + 400\\] \\[0 = x^2 + 18x - 84\\] We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. \\[x = \\frac{-18 + \\sqrt{18^2 + 4(84)}}{2}\\] \\[x = \\frac{-18 + \\sqrt{660}}{2}\\] $x$ simplifies to $\\frac{-18 + 2\\sqrt{165}}{2},$ which equals $-9 + \\sqrt{165}.$ Thus, the answer is $9 + 165 = 174.",
"Let $z=\\frac{d}{2},$ and $R$ be the circumradius. From the given information, \\[2R\\sin z=22\\] \\[2R(\\sin 2z-\\sin 3z)=20\\] Dividing the latter by the former, \\[\\frac{2\\sin z\\cos z-(3\\cos^2z\\sin z-\\sin^3 z)}{\\sin z}=2\\cos z-(3\\cos^2z-\\sin^2z)=1+2\\cos z-4\\cos^2z=\\frac{10}{11}\\] \\[4\\cos^2z-2\\cos z-\\frac{1}{11}=0 (1)\\] We want to find \\[\\frac{22\\sin (3z)}{\\sin z}=22(3-4\\sin^2z)=22(4\\cos^2z-1).\\] From $(1),$ this is equivalent to $44\\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\\sqrt{165}-9,$ so our answer is $174.",
"Let $z=\\frac{d}{2}$, $R$ be the circumradius, and $a$ be the length of 3d degree chord. Using the extended sine law, we obtain: \\[22=2R\\sin(z)\\] \\[20+a=2R\\sin(2z)\\] \\[a=2R\\sin(3z)\\] Dividing the second from the first we get $\\cos(z)=\\frac{20+a}{44}$ By the triple angle formula we can manipulate the third equation as follows: $a=2R\\times \\sin(3z)=\\frac{22}{\\sin(z)} \\times (3\\sin(z)-4\\sin^3(z)) = 22(3-4\\sin^2(z))=22(4\\cos^2(z)-1)=\\frac{(20+a)^2}{22}-22$ Solving the quadratic equation gives the answer to be $174."
] |
2001-I-14
| 2,001 | 14 |
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?
| 351 |
I
|
[
"",
"Let $0$ represent a house that does not receive mail and $1$ represent a house that does receive mail. This problem is now asking for the number of $19$-digit strings of $0$'s and $1$'s such that there are no two consecutive $1$'s and no three consecutive $0$'s. The last two digits of any $n$-digit string can't be $11$, so the only possibilities are $00$, $01$, and $10$. Let $a_n$ be the number of $n$-digit strings ending in $00$, $b_n$ be the number of $n$-digit strings ending in $01$, and $c_n$ be the number of $n$-digit strings ending in $10$. If an $n$-digit string ends in $00$, then the previous digit must be a $1$, and the last two digits of the $n-1$ digits substring will be $10$. So \\[a_{n} = c_{n-1}.\\] If an $n$-digit string ends in $01$, then the previous digit can be either a $0$ or a $1$, and the last two digits of the $n-1$ digits substring can be either $00$ or $10$. So \\[b_{n} = a_{n-1} + c_{n-1}.\\] If an $n$-digit string ends in $10$, then the previous digit must be a $0$, and the last two digits of the $n-1$ digits substring will be $01$. So \\[c_{n} = b_{n-1}.\\] Clearly, $a_2=b_2=c_2=1$. Using the recursive equations and initial values: \\[\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \\multicolumn{19}{c}{}\\\\\\hline n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\\\\hline b_n&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114&151\\\\\\hline c_n&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114\\\\\\hline \\end{array}\\] As a result $a_{19}+b_{19}+c_{19}=351.",
"Let $M_n$ represent the number of mail delivery patterns that end with the last house receiving mail. This is $b_n$ in Solution 1. Similarly define $A_n$ to be the number of mail delivery patterns that end with last house not receiving mail. This is just $a_n$ and $c_n$ in solution 1. Let $T_n$ be the total number of mail delivery patterns. Here are the possible ending cases: the string ends in $1, 10,$ or $100$. The first case is just $M_n$. The second case is $M_{n-1}$. The third case is $M_{n-2}$. So we have $T_n = M_n + M_{n-1} + M_{n-2}$. Since we want $T_{19}$, it is just $M_{18} + M_{17} + M_{16}$. Now using the same logic as above we can find $M_n = M_{n-2} + M_{n-3}$ ( the cases are 01 and 001). We can refer back to solution 1's table and only keep track of $b_n$, ignoring both $a_n$ and $c_n$. - MathLegend27",
"Let $M_n$ represent the number of mail delivery patterns that end with the last house receiving mail. This is $b_n$ in Solution 1. Similarly define $A_n$ to be the number of mail delivery patterns that end with last house not receiving mail. This is just $a_n$ and $c_n$ in solution 1. Let $T_n$ be the total number of mail delivery patterns. Here are the possible ending cases: the string ends in $1, 10,$ or $100$. The first case is just $M_n$. The second case is $M_{n-1}$. The third case is $M_{n-2}$. So we have $T_n = M_n + M_{n-1} + M_{n-2}$. Since we want $T_{19}$, it is just $M_{18} + M_{17} + M_{16}$. Now using the same logic as above we can find $M_n = M_{n-2} + M_{n-3}$ ( the cases are 01 and 001). We can refer back to solution 1's table and only keep track of $b_n$, ignoring both $a_n$ and $c_n$. - MathLegend27",
"We split the problem into cases using the number of houses that get mail. Let \"|\" represent a house that gets mail, and \"o\" represent a house that doesn't. With a fixed number of |, an o can be inserted between 2 |'s or on the very left or right. There cannot be more than one o that is free to arrange to be placed between two |'s because no three o's can be adjacent, but there can be a maximum of two o's placed on the very left or right. Note that according to the Pigeonhole Principle, no more than 10 houses can get mail on the same day. Case 1: 10 houses get mail. No 2 adjacent houses can get mail on the same day, so there must be an o between every two |. $10-1=9$ o's are fixed so we count the number of ways to insert $19 - 10 - 9 = 0$ o's to $10+1 = 11$ spots, or $\\binom{11}{0} = 1$. Case 2: 9 houses get mail. In this case, $9-1 = 8$ o's are fixed so we count the number of ways to insert $19 - 9 - 8 = 2$ o's to $9+1=10$ spots. However, there is also the case where two o's are both on the very left / right. When both o's that are free to arrange are put on a side, there are $10-1=9$ spots left to insert $2-2=0$ o's. Hence the total number of ways in this case is $\\binom{10}{2} + 2\\binom{9}{0} = 47$. Case 3: 8 houses get mail. In this case, $8-1=7$ o's are fixed so we count the number of ways to insert $19-8-7=4$ o's to $8+1=9$ spots. When two o's are put to the very left / right, there are $9-1=8$ spots left to insert $4-2=2$ o's. We also need to take care of the case where two o's are on the very left and two o's are on the very right: we have $9-1-1=7$ spots to insert $4-2-2=0$ o's. Hence the total number of ways in this case is $\\binom{9}{4} + 2\\binom{8}{2} + \\binom{7}{0} = 183$. Case 4: 7 houses get mail. In this case, $7-1=6$ o's are fixed so we count the number of ways to insert $19-7-6=6$ o's to $7+1=8$ spots. When two o's are put to the very left / right, there are $8-1=7$ spots left to insert $6-2=4$ o's. When two o's are on the very left and two o's are on the very right, we have $8-1-1=6$ spots to insert $6-2-2=2$ o's. Hence the total number of ways in this case is $\\binom{8}{6} + 2\\binom{7}{4} + \\binom{6}{2} = 113$. Case 5: 6 houses get mail. We have to be careful in this case: $6-1=5$ o's are fixed so we are inserting $19-6-5=8$ o's to $6+1=7$ spots, which means that at least 1 of the 2 sides must have two o's. When 1 of the 2 sides have two o's, there are $7-1=6$ spots to insert $8-2=6$ o's. When both sides have two o's, there are $7-1-1=5$ spots to insert $8-2-2=4$ o's. Hence the total number of ways in this case is $2\\binom{6}{6} + \\binom{5}{4} = 7$. When less than 6 houses get(s) mail, it's again not possible since at least three o's must be together (again, according to the Pigeonhole Principle). Therefore, the desired answer is $1+47+183+113+7=351. - MP8148",
"We split the problem into cases using the number of houses that get mail. Let \"|\" represent a house that gets mail, and \"o\" represent a house that doesn't. With a fixed number of |, an o can be inserted between 2 |'s or on the very left or right. There cannot be more than one o that is free to arrange to be placed between two |'s because no three o's can be adjacent, but there can be a maximum of two o's placed on the very left or right. Note that according to the Pigeonhole Principle, no more than 10 houses can get mail on the same day. Case 1: 10 houses get mail. No 2 adjacent houses can get mail on the same day, so there must be an o between every two |. $10-1=9$ o's are fixed so we count the number of ways to insert $19 - 10 - 9 = 0$ o's to $10+1 = 11$ spots, or $\\binom{11}{0} = 1$. Case 2: 9 houses get mail. In this case, $9-1 = 8$ o's are fixed so we count the number of ways to insert $19 - 9 - 8 = 2$ o's to $9+1=10$ spots. However, there is also the case where two o's are both on the very left / right. When both o's that are free to arrange are put on a side, there are $10-1=9$ spots left to insert $2-2=0$ o's. Hence the total number of ways in this case is $\\binom{10}{2} + 2\\binom{9}{0} = 47$. Case 3: 8 houses get mail. In this case, $8-1=7$ o's are fixed so we count the number of ways to insert $19-8-7=4$ o's to $8+1=9$ spots. When two o's are put to the very left / right, there are $9-1=8$ spots left to insert $4-2=2$ o's. We also need to take care of the case where two o's are on the very left and two o's are on the very right: we have $9-1-1=7$ spots to insert $4-2-2=0$ o's. Hence the total number of ways in this case is $\\binom{9}{4} + 2\\binom{8}{2} + \\binom{7}{0} = 183$. Case 4: 7 houses get mail. In this case, $7-1=6$ o's are fixed so we count the number of ways to insert $19-7-6=6$ o's to $7+1=8$ spots. When two o's are put to the very left / right, there are $8-1=7$ spots left to insert $6-2=4$ o's. When two o's are on the very left and two o's are on the very right, we have $8-1-1=6$ spots to insert $6-2-2=2$ o's. Hence the total number of ways in this case is $\\binom{8}{6} + 2\\binom{7}{4} + \\binom{6}{2} = 113$. Case 5: 6 houses get mail. We have to be careful in this case: $6-1=5$ o's are fixed so we are inserting $19-6-5=8$ o's to $6+1=7$ spots, which means that at least 1 of the 2 sides must have two o's. When 1 of the 2 sides have two o's, there are $7-1=6$ spots to insert $8-2=6$ o's. When both sides have two o's, there are $7-1-1=5$ spots to insert $8-2-2=4$ o's. Hence the total number of ways in this case is $2\\binom{6}{6} + \\binom{5}{4} = 7$. When less than 6 houses get(s) mail, it's again not possible since at least three o's must be together (again, according to the Pigeonhole Principle). Therefore, the desired answer is $1+47+183+113+7=351. - MP8148",
"There doesn't seem to be anything especially noticeable about the number nineteen in this problem, meaning that we can replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candidate for recursion. At first, it's not immediately clear how to relate the state of $n$ houses in general to that of $n - 1, n - 2,$ or $n - 3.$ We thus break it up into cases, based on whether the first house gets mail or not. Let $p_n$ be the number of ways to distribute the mail to $n$ houses. Assume that the first house gets mail. Therefore, since no two adjacent houses get mail on the same day, the second house must not get mail. Starting from the third house, however, things start to look messy, and it looks like we have to break our recurrence down into even smaller cases, which is something that we don't like -- we want to keep our relations as simple as possible. Therefore, seeing that we can't work forwards anymore, we try to work backwards. Once the mail carrier delivers the mail to the first and (lack of mail) to the second houses, have him deliver mail to the remaining $n - 3$ houses at the end of the row, skipping the third house. There are $p_{n - 3}$ ways to do this. Now, we see that the availability of mail at the third house is fixed -- if the fourth house doesn't receive mail, the third one must, and if the fourth house receives mail, the third one can't. Therefore, there are simply $p_{n-3}$ ways to deliver the mail if the first house gets mail. If the first house doesn't get mail, then we use the same logic -- have the mail carrier skip the second house and deliver the remaining mail to the $n - 2$ houses in $p_{n-2}$ ways. Then, the availability of mail for the second house is fixed, so there are $p_{n - 2}$ ways to deliver the mail in this case. We thus have established a recurrence relation -- since the first house either gets mail or it doesn't, and cannot achieve both at the same time, we are confident about the validity of our relation: \\[p_n = p_{n-2} + p_{n-3}.\\]Now, we simply calculate $p_1, p_2,$ and $p_3.$ Then, it's off to the races for computation! $p_1 = 2,$ because the first house can either gets mail or it doesn't -- there are no restrictions. $p_2 = 3,$ because all of the possible deliveries are valid (of which there are $2 \\cdot 2 = 4$) except the one where both houses receive mail. $p_3 = 4,$ as there are $4$ possible ways (here, M represents that that house gets mail and N represents no mail): MNM, MNN, NNM, NMN. Using our recurrence relation, we eventually get that $p_{19} = 351 and we're done. Solution by Ilikeapos",
"There doesn't seem to be anything especially noticeable about the number nineteen in this problem, meaning that we can replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candidate for recursion. At first, it's not immediately clear how to relate the state of $n$ houses in general to that of $n - 1, n - 2,$ or $n - 3.$ We thus break it up into cases, based on whether the first house gets mail or not. Let $p_n$ be the number of ways to distribute the mail to $n$ houses. Assume that the first house gets mail. Therefore, since no two adjacent houses get mail on the same day, the second house must not get mail. Starting from the third house, however, things start to look messy, and it looks like we have to break our recurrence down into even smaller cases, which is something that we don't like -- we want to keep our relations as simple as possible. Therefore, seeing that we can't work forwards anymore, we try to work backwards. Once the mail carrier delivers the mail to the first and (lack of mail) to the second houses, have him deliver mail to the remaining $n - 3$ houses at the end of the row, skipping the third house. There are $p_{n - 3}$ ways to do this. Now, we see that the availability of mail at the third house is fixed -- if the fourth house doesn't receive mail, the third one must, and if the fourth house receives mail, the third one can't. Therefore, there are simply $p_{n-3}$ ways to deliver the mail if the first house gets mail. If the first house doesn't get mail, then we use the same logic -- have the mail carrier skip the second house and deliver the remaining mail to the $n - 2$ houses in $p_{n-2}$ ways. Then, the availability of mail for the second house is fixed, so there are $p_{n - 2}$ ways to deliver the mail in this case. We thus have established a recurrence relation -- since the first house either gets mail or it doesn't, and cannot achieve both at the same time, we are confident about the validity of our relation: \\[p_n = p_{n-2} + p_{n-3}.\\]Now, we simply calculate $p_1, p_2,$ and $p_3.$ Then, it's off to the races for computation! $p_1 = 2,$ because the first house can either gets mail or it doesn't -- there are no restrictions. $p_2 = 3,$ because all of the possible deliveries are valid (of which there are $2 \\cdot 2 = 4$) except the one where both houses receive mail. $p_3 = 4,$ as there are $4$ possible ways (here, M represents that that house gets mail and N represents no mail): MNM, MNN, NNM, NMN. Using our recurrence relation, we eventually get that $p_{19} = 351 and we're done. Solution by Ilikeapos",
"We use similar wording as in solution 1. In this problem, we divide into 3 cases: Case 1: The first house gets mail. In other words, the sequence starts with a $1.$ We first introduce two variables. Let $x$ be the number of 01's, and let $y$ be the number of 001's in the sequence. For each case, we will divide further into three sub-cases, based on the pattern of mail delivery for the last three houses. Sub-case 1: The last house that gets mail is the 17th house. Thus, we have the equation $2x+3y=16,$ which has solutions $(2,4),(5,2),$ and $(8,0).$ There are $\\binom{2+4}{2}+\\binom{5+2}{2}+\\binom{8+0}{8}=37$ total sequences. Sub-case 2: The last house that gets mail is the 18th house. We have $2x+3y=17,$ and finding all solutions yields 49 total sequences. Sub-case 3: The last house that gets mail is the 19th house. We have $2x+3y=18,$ which gives us 65 total sequences. There are 151 sequences for this case. Case 2: The first house that gets mail is the 2nd house. The sequence begins with $01.$ Our sub-cases are still the same. However, our equations become $2x+3y=15,16,17.$ Computing yields $28+37+49=114$ sequences. Case 3: The first house that gets mail is the 3rd house. The sequence starts with $001.$ We have the equations $2x+3y=14,15,16$ so we get $21+28+37=86$ sequences. Totaling up gives us $151+114+86=351$ different sequences. -math129",
"Let $w_n$ be the number of possible ways if the last house has mail, and $b_n$ be the number of possible ways if the last house does not have mail. If the last house has mail, then, the next house can't have mail, meaning that $b_n = w_{n - 1}$. If the last house doesn't have mail, then the next house can either have mail or not have mail. If the next house has mail, then we simply count the number of ways that the row ends in a house with mail, so that means so far, our recursive rule is $w_n = b_{n - 1} + \\text{something}$. If the next house does not have mail, then the next house after that must have mail, meaning that $w_n = b_{n - 1} + b_{n - 2}$. Recursing all the way up to $b_{19}$ and $w_{19}$, we get $100 + 251 = 351",
"Let $w_n$ be the number of possible ways if the last house has mail, and $b_n$ be the number of possible ways if the last house does not have mail. If the last house has mail, then, the next house can't have mail, meaning that $b_n = w_{n - 1}$. If the last house doesn't have mail, then the next house can either have mail or not have mail. If the next house has mail, then we simply count the number of ways that the row ends in a house with mail, so that means so far, our recursive rule is $w_n = b_{n - 1} + \\text{something}$. If the next house does not have mail, then the next house after that must have mail, meaning that $w_n = b_{n - 1} + b_{n - 2}$. Recursing all the way up to $b_{19}$ and $w_{19}$, we get $100 + 251 = 351",
"Let $a_n$ be the number of ways if the first house has mail, and let $b_n$ be the number of ways if the first house does not get mail. $a_n=a_{n-2}+a_{n-3}$ because if the first house gets mail, the next house that gets mail must either be the third or fourth house. $b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house. Note that we only need list out values of $a_n$ as $b$ depends on $a$. $a_1=1, a_2=1, a_3=2, a_4=2, \\ldots$. $a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\\cdot a_{17}+a_{18}=65+2\\cdot 86+114=351.",
"Let $a_n$ be the number of ways if the first house has mail, and let $b_n$ be the number of ways if the first house does not get mail. $a_n=a_{n-2}+a_{n-3}$ because if the first house gets mail, the next house that gets mail must either be the third or fourth house. $b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house. Note that we only need list out values of $a_n$ as $b$ depends on $a$. $a_1=1, a_2=1, a_3=2, a_4=2, \\ldots$. $a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\\cdot a_{17}+a_{18}=65+2\\cdot 86+114=351.",
"Like the previous solutions we use $1$s and $0$s to indicate whether a house get mails or not. Call a sequence of $1$s and $0$s valid if it has no two consecutive $1$s or three consecutive $0$s. We claim there is a $1$-to-$1$ correspondence between (A) a valid sequence of length $19$; and (B) a valid sequence of length $25$ that starts and ends with $1$. For each valid sequence in (B), we can remove the first three numbers as well as the last three numbers to get a valid sequence in (A). On the other hand, for each valid sequence in (A), we can append three numbers to each end by the following rules to get a valid sequence in (B): 1) If the sequence starts/ends with $1$, we append $100$/$001$ to the start/end; 2) If the sequence starts/ends with $0$, we append $101$ to the start/end. Now, since all valid sequences in (B) start with either $1010$ or $1001$, and end with either $0101$ or $1001$, it's easy to see this is a $1$-$1$ correspondence. To compute the number of valid sequences in (B), we start with $n$ of $1$s with $n-1$ of $0$s in between, then distribute the rest $25-(2n-1)=26-2n$ of $0$s to the $n-1$ slots. The total number of valid sequences is then \\[\\sum_n \\binom{n-1}{26-2n} = \\binom{12}{0} + \\binom{11}{2} + \\binom{10}{4} + \\binom{9}{6} + \\binom{8}{8} = 1 + 55 + 210 + 84 + 1 = 351.\\] ~asops",
"Like the previous solutions we use $1$s and $0$s to indicate whether a house get mails or not. Call a sequence of $1$s and $0$s valid if it has no two consecutive $1$s or three consecutive $0$s. We claim there is a $1$-to-$1$ correspondence between (A) a valid sequence of length $19$; and (B) a valid sequence of length $25$ that starts and ends with $1$. For each valid sequence in (B), we can remove the first three numbers as well as the last three numbers to get a valid sequence in (A). On the other hand, for each valid sequence in (A), we can append three numbers to each end by the following rules to get a valid sequence in (B): 1) If the sequence starts/ends with $1$, we append $100$/$001$ to the start/end; 2) If the sequence starts/ends with $0$, we append $101$ to the start/end. Now, since all valid sequences in (B) start with either $1010$ or $1001$, and end with either $0101$ or $1001$, it's easy to see this is a $1$-$1$ correspondence. To compute the number of valid sequences in (B), we start with $n$ of $1$s with $n-1$ of $0$s in between, then distribute the rest $25-(2n-1)=26-2n$ of $0$s to the $n-1$ slots. The total number of valid sequences is then \\[\\sum_n \\binom{n-1}{26-2n} = \\binom{12}{0} + \\binom{11}{2} + \\binom{10}{4} + \\binom{9}{6} + \\binom{8}{8} = 1 + 55 + 210 + 84 + 1 = 351.\\] ~asops",
"When we see this problem, we immediately think of recursion. Let $a_k$ be the point such that the rightmost house of k houses receives mail. Then $a_1=1$, $a_2=1$, $a_3=2$. Now, we find the recurrence relation \\[a_n=a_{n-3}+a_{n-2}\\] because we can either have 1 space in between the mailed rightmost house of $a_n$, or two spaces, as in our previous a’s we already have the a_n mailed rightmost. Solving, we get $a_17=86, a_18=114, a_19=151$. Now, we realize that the number of ways for the 19 houses to get mail is the sum of $a_17,a_18,a_19$ because we can either have 2 empty no mail houses, 1 house, or the rightmost house is lit. Summing, we get 351. ~MathCosine"
] |
2001-I-15
| 2,001 | 15 |
The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where 8 and 1 are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 85 |
I
|
[
"Choose one face of the octahedron randomly and label it with $1$. There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face. Clearly, the labels for the A-faces must come from the set $\\{3,4,5,6,7\\}$, since these faces are all adjacent to $1$. There are thus $5 \\cdot 4 \\cdot 3 = 60$ ways to assign the labels for the A-faces. The labels for the B-faces and C-face are the two remaining numbers from the above set, plus $2$ and $8$. The number on the C-face must not be consecutive to any of the numbers on the B-faces. From here it is easiest to brute force the $10$ possibilities for the $4$ numbers on the B and C faces: 2348 (2678): 8(2) is the only one not adjacent to any of the others, so it goes on the C-face. 4(6) has only one B-face it can go to, while 2 and 3 (7 and 8) can be assigned randomly to the last two. 2 possibilities here. 2358 (2578): 5 cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so just 1 possibility here. 2368 (2478): 6(4) cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so 1 possibility here. 2458 (2568): All of the numbers have only one B-face they could go to. 2 and 4 (6 and 8) can go on the same, so one must go to the C-face. Only 2(8) is not consecutive with any of the others, so it goes on the C-face. 1 possibility. 2378: None of the numbers can go on the C-face because they will be consecutive with one of the B-face numbers. So this possibility is impossible. 2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, so this possibility is impossible. There is a total of $10$ possibilities. There are $3!=6$ permutations (rotations/reflections) of each, so $60$ acceptable ways to fill in the rest of the octahedron given the $1$. There are $7!=5040$ ways to randomly fill in the rest of the octahedron. So the probability is $\\frac {60}{5040} = \\frac {1}{84}$. The answer is $085.",
"Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits ($B$ for black and $W$ for white). Type I: $BB-WWWW-BB$. There are $4!$ ways to arrange the black vertices and consequently two of the white vertices and $2!$ ways to arrange the other two white vertices. Since the template has a period of $8$, there are $4!\\cdot 2!\\cdot 8 = 384$ circuits of type I. Type II: $B-WW-BB-WW-B$. There are $4!$ ways to arrange the black vertices and consequently the white vertices. Since the template has a period of $4$, there are $4! \\cdot 4 = 96$ circuits of type II. Thus, there are $384+96=480$ circuits satisfying the given condition, out of the $8!$ possible circuits. Therefore, the desired probability is $\\frac{480}{8!} = \\frac{1}{84}$. The answer is $085.",
"As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices. The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in $5\\cdot4\\cdot3=60$ ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices. The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus $\\frac{8\\cdot60}{8!}=\\frac{8\\cdot5\\cdot4\\cdot3}{8!}=\\frac{1}{84}$, from which the answer is $085. , the average is one per choice of 3 to border the 1, ex 5,6,7 border 2 solutions, 3,5,7 border 0 solution",
"[asy] import three; draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1)); draw((0,0,0)--(1,0,0)); draw((0,0,1)--(1,0,1)); for(int i = 0; i < 2; ++i) { for(int j = 0; j < 2; ++j) { for(int k = 0; k < 2; ++k) { dot((i,j,k)); } } } // dot((0,0,1),blue); // dot((0,1,0),green); // dot((1,0,0),red); draw((0,0,0)--(1,1,0)); draw((0,1,0)--(1,0,0)); draw((0,0,1)--(1,1,1)); draw((0,1,1)--(1,0,1)); [/asy] The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $AR,BR,CR,DR,AX,BX,CX,DX$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$ Case 1: Four of the vertical edges are used. $6\\cdot 2=12.$ Case 2: Two of the vertical edges are used. $4\\cdot 3 \\cdot 2\\cdot 2=48.$ So, the probability is $\\frac{60}{5040}=\\frac{1}{84}.$ Therefore, our answer is $085",
"As with some of the previous solutions, consider the cube formed by connecting the centroids of the faces on the octahedron. We choose a random vertex(hence fixing the diagram), giving us $7!$ ways as our denominator. WLOG, we color this start vertex red, and we color all $3$ vertices adjacent to it blue. We repeat this for the other vertices. Note that there is a 1-1 correspondence between the number of valid face numberings with rotational symmetry and the number of ways to move a particle to every vertex of the cube and returning to the start vertex with only diagonal moves. We can move along either short diagonals and long diagonals. Next, note that we can only move from the red vertices to the blue vertices and back with long diagonals(since short diagonals keep us on the color we are currently on). Thus, it is easy to check that the only possible sequences of long and short diagonals are cyclic permutations of $SSSLSSSL$ and $SLSLSLSL$. Case 1: $SSSLSSSL$. There are $4$ cyclic permutations, and we can traverse the entirety of the red tetrahedron in $4$ steps in $3!$ ways. Then, after the move to the blue tetrahedron from one of the non-starting red vertices, we realize that our first step cannot be to the blue vertex opposite the starting red vertex. Hence, there are $2$ possibilities for our first step. However, once we make our first move, the path is fixed. This leaves us with $4! \\cdot 2$ ways. Case 2: $SLSLSLSL$ There are $2$ cyclic permutations in this case. After we choose one of the $3$ red vertices to go to from the starting vertex and move to the blue tetrahedron, we are once again left with $2$ choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our choices are fixed. This leaves us with $2 \\cdot 3 \\cdot 2$ ways. Finally, summing and dividing by $7!$ gives us $\\frac{60}{7!} = \\frac{1}{84} \\rightarrow 085, as desired. - Spacesam"
] |
2001-II-1
| 2,001 | 1 |
Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$ ?
| 816 |
II
|
[
"The two-digit perfect squares are $16, 25, 36, 49, 64, 81$. We try making a sequence starting with each one: $16 - 64 - 49$. This terminates since none of them end in a $9$, giving us $1649$. $25$. $36 - 64 - 49$, $3649$. $49$. $64 - 49$, $649$. $81 - 16 - 64 - 49$, $81649$. The largest is $81649$, so our answer is $816."
] |
2001-II-2
| 2,001 | 2 |
Each of the 2001 students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between 80 percent and 85 percent of the school population, and the number who study French is between 30 percent and 40 percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M-m$ .
| 298 |
II
|
[
"Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \\cap F$ be the number of students who study both. Then $\\left\\lceil 80\\% \\cdot 2001 \\right\\rceil = 1601 \\le S \\le \\left\\lfloor 85\\% \\cdot 2001 \\right\\rfloor = 1700$, and $\\left\\lceil 30\\% \\cdot 2001 \\right\\rceil = 601 \\le F \\le \\left\\lfloor 40\\% \\cdot 2001 \\right\\rfloor = 800$. By the Principle of Inclusion-Exclusion, \\[S+F- S \\cap F = S \\cup F = 2001\\] For $m = S \\cap F$ to be smallest, $S$ and $F$ must be minimized. \\[1601 + 601 - m = 2001 \\Longrightarrow m = 201\\] For $M = S \\cap F$ to be largest, $S$ and $F$ must be maximized. \\[1700 + 800 - M = 2001 \\Longrightarrow M = 499\\] Therefore, the answer is $M - m = 499 - 201 = 298."
] |
2001-II-3
| 2,001 | 3 |
Given that \begin{align*}x_{1}&=211,\\ x_{2}&=375,\\ x_{3}&=420,\\ x_{4}&=523,\ \text{and}\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*} find the value of $x_{531}+x_{753}+x_{975}$ .
| 898 |
II
|
[
"We find that $x_5 = 267$ by the recursive formula. Summing the recursions \\begin{align*} x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\\\ x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5} \\end{align*} yields $x_{n} = -x_{n-5}$. Thus $x_n = (-1)^k x_{n-5k}$. Since $531 = 106 \\cdot 5 + 1,\\ 753 = 150 \\cdot 5 + 3,\\ 975 = 194 \\cdot 5 + 5$, it follows that \\[x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = 898.\\]",
"The recursive formula suggests telescoping. Indeed, if we add $x_n$ and $x_{n-1}$, we have $x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}$. Subtracting $x_{n-1}$ yields $x_n = -x_{n-5} \\implies x_n = -(-(x_{n-10})) = x_{n-10}$. Thus, \\[x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = 898 at all."
] |
2001-II-4
| 2,001 | 4 |
Let $R = (8,6)$ . The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$ , respectively, such that $R$ is the midpoint of $\overline{PQ}$ . The length of $PQ$ equals $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 67 |
II
|
[
"[asy] pointpen = black; pathpen = black+linewidth(0.7); pair R = (8,6), P = (32,60)/7, Q= (80,24)/7; D((0,0)--MP(\"x\",(13,0),E),EndArrow(6)); D((0,0)--MP(\"y\",(0,10),N),EndArrow(6)); D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6)); D(D(MP(\"P\",P,NW))--D(MP(\"Q\",Q),SE),linetype(\"4 4\")); D(MP(\"R\",R,NE)); [/asy] The coordinates of $P$ can be written as $\\left(a, \\frac{15a}8\\right)$ and the coordinates of point $Q$ can be written as $\\left(b,\\frac{3b}{10}\\right)$. By the midpoint formula, we have $\\frac{a+b}2=8$ and $\\frac{15a}{16}+\\frac{3b}{20}=6$. Solving for $b$ gives $b= \\frac{80}{7}$, so the point $Q$ is $\\left(\\frac{80}7, \\frac{24}7\\right)$. The answer is twice the distance from $Q$ to $(8,6)$, which by the distance formula is $\\frac{60}{7}$. Thus, the answer is $067."
] |
2001-II-5
| 2,001 | 5 |
A set of positive numbers has the $triangle~property$ if it has three distinct elements that are the lengths of the sides of a triangle whose area is positive. Consider sets $\{4, 5, 6, \ldots, n\}$ of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of $n$ ?
| 253 |
II
|
[
"Out of all ten-element subsets with distinct elements that do not possess the triangle property, we want to find the one with the smallest maximum element. Call this subset $\\mathcal{S}$. Without loss of generality, consider any $a, b, c \\,\\in \\mathcal{S}$ with $a < b < c$. $\\,\\mathcal{S}$ does not possess the triangle property, so $c \\geq a + b$. We use this property to build up $\\mathcal{S}$ from the smallest possible $a$ and $b$: \\[\\mathcal{S} = \\{\\, 4,\\, 5,\\, 4+5, \\,5+(4+5),\\, \\ldots\\,\\} = \\{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\\}\\] $\\mathcal{S}$ is the \"smallest\" ten-element subset without the triangle property, and since the set $\\{4, 5, 6, \\ldots, 253\\}$ is the largest set of consecutive integers that does not contain this subset, it is also the largest set of consecutive integers in which all ten-element subsets possess the triangle property. Thus, our answer is $n = 253 term. (The latter part is generally pretty annoying). ~Dhillonr25 ~Minor edit by Yiyj1"
] |
2001-II-6
| 2,001 | 6 |
Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. The ratio of the area of square $EFGH$ to the area of square $ABCD$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$ .
| 251 |
II
|
[
"Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$, $2b$ be the side length of $EFGH$. By the Pythagorean Theorem, the radius of $\\odot O = OC = a\\sqrt{2}$. [asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype(\"4 4\") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C,N)--MP(\"D\",D)--cycle); D(MP(\"E\",E,SW)--MP(\"F\",F,NW)--MP(\"G\",G,NE)--MP(\"H\",H,SE)--cycle); D(CP(D(MP(\"O\",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); [/asy] Now consider right triangle $OGI$, where $I$ is the midpoint of $\\overline{GH}$. Then, by the Pythagorean Theorem, \\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\\\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \\end{align*} Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\\frac{[EFGH]}{[ABCD]} = \\left(\\frac 15\\right)^2 = \\frac{1}{25}$, and the answer is $10n + m = 251 to 1.",
"Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$, $2b$ be the side length of $EFGH$. By the Pythagorean Theorem, the radius of $\\odot O = OC = a\\sqrt{2}$. [asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype(\"4 4\") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP(\"A\",A)--MP(\"B\",B,N)--MP(\"C\",C,N)--MP(\"D\",D)--cycle); D(MP(\"E\",E,SW)--MP(\"F\",F,NW)--MP(\"G\",G,NE)--MP(\"H\",H,SE)--cycle); D(CP(D(MP(\"O\",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); [/asy] Now consider right triangle $OGI$, where $I$ is the midpoint of $\\overline{GH}$. Then, by the Pythagorean Theorem, \\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\\\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \\end{align*} Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\\frac{[EFGH]}{[ABCD]} = \\left(\\frac 15\\right)^2 = \\frac{1}{25}$, and the answer is $10n + m = 251 to 1.",
"Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$. Let $DF$ = $b$ and diameter $HI$ go through $J$ the midpoint of $EF$. Since a diameter always bisects a chord perpendicular to it, $DJ$ = $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$. $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$$\\sqrt{2} / 2$, half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$. Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$, and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$, meaning the ratio of areas $((a-2b)/a)^2$ = $(1/5)^2$ = $1/25$ = $m/n.$ Then $10n + m = 25 * 10 + 1 = 251.",
"Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$. Let $DF$ = $b$ and diameter $HI$ go through $J$ the midpoint of $EF$. Since a diameter always bisects a chord perpendicular to it, $DJ$ = $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$. $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$$\\sqrt{2} / 2$, half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$. Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$, and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$, meaning the ratio of areas $((a-2b)/a)^2$ = $(1/5)^2$ = $1/25$ = $m/n.$ Then $10n + m = 25 * 10 + 1 = 251."
] |
2001-II-7
| 2,001 | 7 |
Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ , $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$ ?
| 725 |
II
|
[
"Solution 1 (analytic) [asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP(\"P\",P)--MP(\"Q\",Q)--MP(\"R\",R,W)--cycle); D(MP(\"S\",S,W) -- MP(\"T\",T,NE)); D(MP(\"U\",U) -- MP(\"V\",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype(\"4 4\")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); [/asy] Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\\triangle PQR$, where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$), $s = \\frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \\frac 12 bh = 5400$ is the area, we find $r_{1} = \\frac As = 30$. Or, the inradius could be directly by using the formula $\\frac{a+b-c}{2}$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles.) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$, and so $RS = 60, UQ = 30$. Note that $\\triangle PQR \\sim \\triangle STR \\sim \\triangle UQV$. Since the ratio of corresponding lengths of similar figures are the same, we have \\[\\frac{r_{1}}{PR} = \\frac{r_{2}}{RS} \\Longrightarrow r_{2} = 15\\ \\text{and} \\ \\frac{r_{1}}{PQ} = \\frac{r_{3}}{UQ} \\Longrightarrow r_{3} = 10.\\] Let the centers of $\\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$, respectively; then by the distance formula we have $O_2O_3 = \\sqrt{55^2 + 65^2} = \\sqrt{10 \\cdot 725}$. Therefore, the answer is $n = 725.",
"[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP(\"P\",P)--MP(\"Q\",Q)--MP(\"R\",R,W)--cycle); D(MP(\"S\",S,W) -- MP(\"T\",T,NE)); D(MP(\"U\",U) -- MP(\"V\",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype(\"4 4\")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); [/asy] Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\\triangle PQR$, where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$), $s = \\frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \\frac 12 bh = 5400$ is the area, we find $r_{1} = \\frac As = 30$. Or, the inradius could be directly by using the formula $\\frac{a+b-c}{2}$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles.) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$, and so $RS = 60, UQ = 30$. Note that $\\triangle PQR \\sim \\triangle STR \\sim \\triangle UQV$. Since the ratio of corresponding lengths of similar figures are the same, we have \\[\\frac{r_{1}}{PR} = \\frac{r_{2}}{RS} \\Longrightarrow r_{2} = 15\\ \\text{and} \\ \\frac{r_{1}}{PQ} = \\frac{r_{3}}{UQ} \\Longrightarrow r_{3} = 10.\\] Let the centers of $\\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$, respectively; then by the distance formula we have $O_2O_3 = \\sqrt{55^2 + 65^2} = \\sqrt{10 \\cdot 725}$. Therefore, the answer is $n = 725.",
"[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP(\"P\",P)--MP(\"Q\",Q)--MP(\"R\",R,W)--cycle); D(MP(\"S\",S,W) -- MP(\"T\",T,NE)); D(MP(\"U\",U) -- MP(\"V\",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype(\"4 4\")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); [/asy] Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\\triangle PQR$, where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$), $s = \\frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \\frac 12 bh = 5400$ is the area, we find $r_{1} = \\frac As = 30$. Or, the inradius could be directly by using the formula $\\frac{a+b-c}{2}$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles.) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$, and so $RS = 60, UQ = 30$. Note that $\\triangle PQR \\sim \\triangle STR \\sim \\triangle UQV$. Since the ratio of corresponding lengths of similar figures are the same, we have \\[\\frac{r_{1}}{PR} = \\frac{r_{2}}{RS} \\Longrightarrow r_{2} = 15\\ \\text{and} \\ \\frac{r_{1}}{PQ} = \\frac{r_{3}}{UQ} \\Longrightarrow r_{3} = 10.\\] Let the centers of $\\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$, respectively; then by the distance formula we have $O_2O_3 = \\sqrt{55^2 + 65^2} = \\sqrt{10 \\cdot 725}$. Therefore, the answer is $n = 725.",
"[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP(\"P\",P)--MP(\"Q\",Q)--MP(\"R\",R,W)--cycle); D(MP(\"S\",S,W) -- MP(\"T\",T,NE)); D(MP(\"U\",U) -- MP(\"V\",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype(\"4 4\")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); pair A2 = IP(incircle(R,S,T), Q--R), A3 = IP(incircle(Q,U,V), Q--R); D(D(MP(\"A_2\",A2,NE)) -- O2, linetype(\"4 4\")+linewidth(0.6)); D(D(MP(\"A_3\",A3,NE)) -- O3 -- foot(O3, A2, O2), linetype(\"4 4\")+linewidth(0.6)); [/asy] We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$. By the Two Tangent Theorem, we find that $A_{1}Q = 60$, $A_{1}R = 90$. Using the similar triangles, $RA_{2} = 45$, $QA_{3} = 20$, so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$. Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\\implies n=725.",
"[asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP(\"P\",P)--MP(\"Q\",Q)--MP(\"R\",R,W)--cycle); D(MP(\"S\",S,W) -- MP(\"T\",T,NE)); D(MP(\"U\",U) -- MP(\"V\",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype(\"4 4\")); D(CR(D(O1), 30)); D(CR(D(O2), 15)); D(CR(D(O3), 10)); pair A2 = IP(incircle(R,S,T), Q--R), A3 = IP(incircle(Q,U,V), Q--R); D(D(MP(\"A_2\",A2,NE)) -- O2, linetype(\"4 4\")+linewidth(0.6)); D(D(MP(\"A_3\",A3,NE)) -- O3 -- foot(O3, A2, O2), linetype(\"4 4\")+linewidth(0.6)); [/asy] We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$. By the Two Tangent Theorem, we find that $A_{1}Q = 60$, $A_{1}R = 90$. Using the similar triangles, $RA_{2} = 45$, $QA_{3} = 20$, so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$. Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\\implies n=725.",
"The radius of an incircle is $r=A_t/\\text{semiperimeter}$. The area of the triangle is equal to $\\frac{90\\times120}{2} = 5400$ and the semiperimeter is equal to $\\frac{90+120+150}{2} = 180$. The radius, therefore, is equal to $\\frac{5400}{180} = 30$. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$, $\\frac{1}{2}(90) = 45$, and $\\frac{1}{2}\\times150 = 75$. The radius of the circle inscribed in this triangle with dimensions $45\\times60\\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$. Defining $P$ as $(0,0)$, $C_2$ is equal to $(60+15,15)$ or $(75,15)$. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$, $\\frac{1}{3}\\times120$, $\\frac{1}{3}\\times150$ or $30,40,50$. The radius of $C_3$ by using the formula mentioned at the beginning is $10$. Using $P$ as $(0,0)$, $C_3$ is equal to $(10, 60+10)$ or $(10,70)$. Using the distance formula, the distance between $C_2$ and $C_3$: $\\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\\sqrt{7250}$ or $\\sqrt{725\\times10}$, thus $n$ is $725.",
"We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$, as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$, which is $15$ less than the square plus the inradius of $C_3$. Our final answer is $\\sqrt{65^2 + 55^2} = \\sqrt{7250} = \\sqrt{(10)(725)} \\rightarrow 725.",
"We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$, as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$, which is $15$ less than the square plus the inradius of $C_3$. Our final answer is $\\sqrt{65^2 + 55^2} = \\sqrt{7250} = \\sqrt{(10)(725)} \\rightarrow 725."
] |
2001-II-8
| 2,001 | 8 |
A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1 - |x - 2|$ for $1\leq x \leq 3$ . Find the smallest $x$ for which $f(x) = f(2001)$ .
| 429 |
II
|
[
"Iterating the condition $f(3x) = 3f(x)$, we find that $f(x) = 3^kf\\left(\\frac{x}{3^k}\\right)$ for positive integers $k$. We know the definition of $f(x)$ from $1 \\le x \\le 3$, so we would like to express $f(2001) = 3^kf\\left(\\frac{2001}{3^k}\\right),\\ 1 \\le \\frac{2001}{3^k} \\le 3 \\Longrightarrow k = 6$. Indeed, \\[f(2001) = 729\\left[1 - \\left| \\frac{2001}{729} - 2\\right|\\right] = 186.\\] We now need the smallest $x$ such that $f(x) = 3^kf\\left(\\frac{x}{3^k}\\right) = 186$. The range of $f(x),\\ 1 \\le x \\le 3$, is $0 \\le f(x) \\le 1$. So when $1 \\le \\frac{x}{3^k} \\le 3$, we have $0 \\le f\\left(\\frac{x}{3^k}\\right) = \\frac{186}{3^k} \\le 1$. Multiplying by $3^k$: $0 \\le 186 \\le 3^k$, so the smallest value of $k$ is $k = 5$. Then, \\[186 = {3^5}f\\left(\\frac{x}{3^5}\\right).\\] Because we forced $1 \\le \\frac{x}{3^5} \\le 3$, so \\[186 = {3^5}f\\left(\\frac{x}{3^5}\\right) = 243\\left[1 - \\left| \\frac{x}{243} - 2\\right|\\right] \\Longrightarrow x = \\pm 57 + 2 \\cdot 243.\\] We want the smaller value of $x = 429 at each iteration.",
"First, we start by graphing the function when $1\\leq{x}\\leq3$, which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$. Similarly, using $f(3x)=3f(x)$, we get a dilation of our initial figure by a factor of 3 for the next interval and so on. Observe that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$. First, we compute $f(2001)$. The nearest intersection point is $(1458,729)$ when $a=7$. Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$, we compute $f(2001)=729-543=186$. However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$, namely $(486,243)$. Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$, we get $x=486-57=429. ~Magnetoninja",
"First, we start by graphing the function when $1\\leq{x}\\leq3$, which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$. Similarly, using $f(3x)=3f(x)$, we get a dilation of our initial figure by a factor of 3 for the next interval and so on. Observe that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$. First, we compute $f(2001)$. The nearest intersection point is $(1458,729)$ when $a=7$. Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$, we compute $f(2001)=729-543=186$. However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$, namely $(486,243)$. Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$, we get $x=486-57=429. ~Magnetoninja",
"We evaluate the first few terms of f(x) to try to find a pattern. F(1)=0 F(2)=1 F(3)=0 F(4)=1 F(5) = 3(F($\\frac{5}{3}$)) = 2 That doesn‘t seem to be getting us anywhere. We notice what we did with f(5) will probably work with f(2001). $F(2001) = 3f(667)=9f(\\frac{667}{3}) = 27f(\\frac{667}{9}) = 81f(\\frac{667}{27})=243f(\\frac{667}{81})=729f(\\frac{667}{243})$ From here, we can evaluate f(2001) = $186$ when we plug in $\\frac{667}{243}$ into $1 - |x - 2|$. So all we need to find is the least number, let‘s call it, say y such that f(y)=186. Repeating the same process we did before with f(2001), $186 = F(y)= 3f(\\frac{y}{3}) = 9f(\\frac{y}{9}) = 27f(\\frac{y}{27})=81f(\\frac{y}{81}) = 243f(\\frac{y}{243})$ Notice that we stopped at $243f(\\frac{y}{243})$ because $\\frac{186}{243}$ is inside the range of $1-|x-2|$, which is [0,1]. Now, f(y/243) = 186/243. Setting $186/243 = 1-|x-2|$, we get 2 solutions for x: $\\frac{543}{243}$ and $\\frac{429}{243}$. Now, the problem asks for the smallest solution, so we obviously choose 429/243 as the solution for 243f(y/243) because it is smaller. We found that $\\frac{y}{243}=\\frac{429}{243}$, and solving this equation gives our answer $429 ~MathCosine",
"We evaluate the first few terms of f(x) to try to find a pattern. F(1)=0 F(2)=1 F(3)=0 F(4)=1 F(5) = 3(F($\\frac{5}{3}$)) = 2 That doesn‘t seem to be getting us anywhere. We notice what we did with f(5) will probably work with f(2001). $F(2001) = 3f(667)=9f(\\frac{667}{3}) = 27f(\\frac{667}{9}) = 81f(\\frac{667}{27})=243f(\\frac{667}{81})=729f(\\frac{667}{243})$ From here, we can evaluate f(2001) = $186$ when we plug in $\\frac{667}{243}$ into $1 - |x - 2|$. So all we need to find is the least number, let‘s call it, say y such that f(y)=186. Repeating the same process we did before with f(2001), $186 = F(y)= 3f(\\frac{y}{3}) = 9f(\\frac{y}{9}) = 27f(\\frac{y}{27})=81f(\\frac{y}{81}) = 243f(\\frac{y}{243})$ Notice that we stopped at $243f(\\frac{y}{243})$ because $\\frac{186}{243}$ is inside the range of $1-|x-2|$, which is [0,1]. Now, f(y/243) = 186/243. Setting $186/243 = 1-|x-2|$, we get 2 solutions for x: $\\frac{543}{243}$ and $\\frac{429}{243}$. Now, the problem asks for the smallest solution, so we obviously choose 429/243 as the solution for 243f(y/243) because it is smaller. We found that $\\frac{y}{243}=\\frac{429}{243}$, and solving this equation gives our answer $429 ~MathCosine"
] |
2001-II-9
| 2,001 | 9 |
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 929 |
II
|
[
"We can use complementary counting, counting all of the colorings that have at least one red $2\\times 2$ square. For at least one red $2 \\times 2$ square: There are four $2 \\times 2$ squares to choose which one will be red. Then there are $2^5$ ways to color the rest of the squares. $4*32=128$ For at least two $2 \\times 2$ squares: There are two cases: those with two $2x2$ red squares on one side and those without red squares on one side but at two corners. The first case is easy: 4 ways to choose which the side the squares will be on, and $2^3$ ways to color the rest of the squares, so 32 ways to do that. For the second case, there will be only two ways to pick two squares, and $2^2$ ways to color the other squares. $32+8=40$ For at least three $2 \\times 2$ squares: Choosing three such squares leaves only one square left, with four places to place it. This is $2 \\cdot 4 = 8$ ways. For at least four $2 \\times 2$ squares, we clearly only have one way. By the Principle of Inclusion-Exclusion, there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \\times 2$ square. There are $2^9=512$ ways to paint the $3 \\times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \\times 2$ red square is $\\frac{417}{512}$, and $417+512=929.",
"We consider how many ways we can have 2*2 grid $(1)$: All the grids are red--$1$ case $(2)$: One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases $(3)$: Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases. Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\\binom 9 2-8-2=26$ cases $(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\\cdot (\\binom 5 3)-4=36$ $(5)$ Four unit squares are blue, no overcounted case will be considered. there are $4\\cdot \\binom 5 4=20$ $(6)$ Five unit squares are blue, $4$ cases in all Sum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed. In all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\\frac{95}{512}=\\frac{417}{512}$ leads to $929 ~bluesoul",
"\\[\\begin{array}{|c|c|c|} \\hline C_{11} & C_{12} & C_{13}\\\\ \\hline C_{21} & C_{22} & C_{23}\\\\ \\hline C_{31} & C_{32} & C_{33}\\\\ \\hline \\end{array}\\] Case 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square Assume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$. To make sure there are no more 2-by-2 red squares, $C_{31} \\text{and} C_{32}$ can't both be red and $C_{13} \\text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \\text{and} C_{32}$ and $C_{13} \\text{and} C_{23}$. $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$, there are $3 \\cdot 3 \\cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \\cdot 4 = 68$ coloring methods. Case 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares Case 2.1: $2$ 2-by-2 red squares take up $6$ unit grids Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$. To make sure there are no more 2-by-2 red squares, $C_{13} \\text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \\text{and} C_{23}$. $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$, there are $3 \\cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \\cdot 4 = 20$ coloring methods. Case 2.2: $2$ 2-by-2 red squares take up $7$ unit grids Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$. To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$, there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \\cdot 2 = 2$ coloring methods. Hence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods. Case 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares Assume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$. To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$. For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$, there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \\cdot 4 = 4$ coloring methods. Case 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares If the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method. In total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \\[\\frac{m}{n}=1-\\frac{95}{2^9}=\\frac{417}{512}\\] \\[m+n=417+512=\\textbf{929}\\] ~isabelchen",
"\\[\\begin{array}{|c|c|c|} \\hline C_{11} & C_{12} & C_{13}\\\\ \\hline C_{21} & C_{22} & C_{23}\\\\ \\hline C_{31} & C_{32} & C_{33}\\\\ \\hline \\end{array}\\] Case 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square Assume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$. To make sure there are no more 2-by-2 red squares, $C_{31} \\text{and} C_{32}$ can't both be red and $C_{13} \\text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \\text{and} C_{32}$ and $C_{13} \\text{and} C_{23}$. $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$, there are $3 \\cdot 3 \\cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \\cdot 4 = 68$ coloring methods. Case 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares Case 2.1: $2$ 2-by-2 red squares take up $6$ unit grids Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$. To make sure there are no more 2-by-2 red squares, $C_{13} \\text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \\text{and} C_{23}$. $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$, there are $3 \\cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \\cdot 4 = 20$ coloring methods. Case 2.2: $2$ 2-by-2 red squares take up $7$ unit grids Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$. To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$, there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \\cdot 2 = 2$ coloring methods. Hence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods. Case 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares Assume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$. To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$. For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$, there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \\cdot 4 = 4$ coloring methods. Case 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares If the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method. In total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \\[\\frac{m}{n}=1-\\frac{95}{2^9}=\\frac{417}{512}\\] \\[m+n=417+512=\\textbf{929}\\] ~isabelchen"
] |
2001-II-10
| 2,001 | 10 |
How many positive integer multiples of 1001 can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$ ?
| 784 |
II
|
[
"The prime factorization of $1001 = 7\\times 11\\times 13$. We have $7\\times 11\\times 13\\times k = 10^j - 10^i = 10^i(10^{j - i} - 1)$. Since $\\text{gcd}\\,(10^i = 2^i \\times 5^i, 7 \\times 11 \\times 13) = 1$, we require that $1001 = 10^3 + 1 | 10^{j-i} - 1$. From the factorization $10^6 - 1 = (10^3 + 1)(10^{3} - 1)$, we see that $j-i = 6$ works; also, $a-b | a^n - b^n$ implies that $10^{6} - 1 | 10^{6k} - 1$, and so any $j-i \\equiv 0 \\pmod{6}.",
"Observation: We see that there is a pattern with $10^k \\pmod{1001}$. \\[10^0 \\equiv 1 \\pmod{1001}\\] \\[10^1 \\equiv 10 \\pmod{1001}\\] \\[10^2 \\equiv 100 \\pmod{1001}\\] \\[10^3 \\equiv -1 \\pmod{1001}\\] \\[10^4 \\equiv -10 \\pmod{1001}\\] \\[10^5 \\equiv -100 \\pmod{1001}\\] \\[10^6 \\equiv 1 \\pmod{1001}\\] \\[10^7 \\equiv 10 \\pmod{1001}\\] \\[10^8 \\equiv 100 \\pmod{1001}\\] So, this pattern repeats every 6. Also, $10^j-10^i \\equiv 0 \\pmod{1001}$, so $10^j \\equiv 10^i \\pmod{1001}$, and thus, \\[j \\equiv i \\pmod{6}\\]. Continue with the 2nd paragraph of solution 1, and we get the answer of $784. -AlexLikeMath",
"Note that $1001=7\\cdot 11\\cdot 13,$ and note that $10^3 \\equiv \\pmod{p}$ for prime $p | 1001$; therefore, the order of 10 modulo $7,11$, and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\\cdot 16 + 6 \\cdot \\frac{15 \\cdot 16}{2} = 784. ~Dhillonr25",
"$10^j - 10^i \\equiv 0 \\pmod{1001} \\iff 10^{j - i} - 1 \\equiv 0 \\pmod{1001} \\iff 10^{j - i} \\equiv 1 \\pmod{1001} \\iff j \\equiv i \\pmod 6$. If $j \\equiv i \\equiv n \\pmod 6$ for $n = 0, 1, 2, 3$, there are $17$ choices for each value of $n$, yielding $4 \\cdot \\dbinom{17}{2} = 544$. However, if $n = 4, 5$, there are only $16$ choices, giving us $2 \\cdot \\dbinom{16}{2} = 240$. So, our final answer is $544 + 240 = 784. ~Puck_0"
] |
2001-II-11
| 2,001 | 11 |
Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 341 |
II
|
[
"Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle, the desired probability is half the probability that Club Truncator does not have the same number of wins and losses. The possible ways to achieve the same number of wins and losses are $0$ ties, $3$ wins and $3$ losses; $2$ ties, $2$ wins, and $2$ losses; $4$ ties, $1$ win, and $1$ loss; or $6$ ties. Since there are $6$ games, there are $\\frac{6!}{3!3!}$ ways for the first, and $\\frac{6!}{2!2!2!}$, $\\frac{6!}{4!}$, and $1$ ways for the rest, respectively, out of a total of $3^6$. This gives a probability of $141/729$. Then the desired answer is $\\frac{1 - \\frac{141}{729}}{2} = \\frac{98}{243}$, so the answer is $m+n = 341.",
"At first, it wins $6$ games, only one way Secondly, it wins $5$ games, the other game can be either win or loss, there are $\\binom{6}{5}\\cdot 2=12$ ways Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\\binom{6}{4}\\cdot 2^2=60$ ways Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\\binom{6}{3}\\cdot (2^3-1)=140$ ways Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\\binom{6}{2}(1+\\binom{4}{1})=75$ cases Last case, it only wins $1$ game so the rest games must be all draw, $1$ game The answer is $\\frac{1+12+60+140+75+6}{3^6}=\\frac{98}{243}$ leads to $341 ~bluesoul"
] |
2001-II-12
| 2,001 | 12 |
Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i + 1}$ , replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $P_{3}$ is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 101 |
II
|
[
"On the first construction, $P_1$, four new tetrahedra will be constructed with side lengths $\\frac 12$ of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume $\\left(\\frac 12\\right)^3 = \\frac 18$. The total volume added here is then $\\Delta P_1 = 4 \\cdot \\frac 18 = \\frac 12$. We now note that for each midpoint triangle we construct in step $P_{i}$, there are now $6$ places to construct new midpoint triangles for step $P_{i+1}$. The outward tetrahedron for the midpoint triangle provides $3$ of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other $3$. This is because if you read this question carefully, it asks to add new tetrahedra to each face of $P_{i}$ which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of $\\frac 18$. Thus we have the recursion $\\Delta P_{i+1} = \\frac{6}{8} \\Delta P_i$, and so $\\Delta P_i = \\frac 12 \\cdot \\left(\\frac{3}{4}\\right)^{i-1} P_1$. The volume of $P_3 = P_0 + \\Delta P_1 + \\Delta P_2 + \\Delta P_3 = 1 + \\frac 12 + \\frac 38 + \\frac 9{32} = \\frac{69}{32}$, and $m+n=101. Note that the summation was in fact a geometric series."
] |
2001-II-13
| 2,001 | 13 |
In quadrilateral $ABCD$ , $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ , $AB = 8$ , $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 69 |
II
|
[
"Extend $\\overline{AD}$ and $\\overline{BC}$ to meet at $E$. Then, since $\\angle BAD = \\angle ADC$ and $\\angle ABD = \\angle DCE$, we know that $\\triangle ABD \\sim \\triangle DCE$. Hence $\\angle ADB = \\angle DEC$, and $\\triangle BDE$ is isosceles. Then $BD = BE = 10$. [asy] /* We arbitrarily set AD = x */ real x = 60^.5, anglesize = 28; pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype(\"6 6\")+linewidth(0.7); pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5)); D(MP(\"A\",A)--MP(\"B\",B,NW)--MP(\"C\",C,NW)--MP(\"D\",D)--cycle); D(B--D); D(A--MP(\"E\",E)--B,d); D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize)); MP(\"10\",(B+D)/2,SW);MP(\"8\",(A+B)/2,W);MP(\"6\",(B+C)/2,NW); [/asy] Using the similarity, we have: \\[\\frac{AB}{BD} = \\frac 8{10} = \\frac{CD}{CE} = \\frac{CD}{16} \\Longrightarrow CD = \\frac{64}5\\] The answer is $m+n = 069",
"Draw a line from $B$, parallel to $\\overline{AD}$, and let it meet $\\overline{CD}$ at $M$. Note that $\\triangle{DAB}$ is similar to $\\triangle{BMC}$ by AA similarity, since $\\angle{ABD}=\\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\\angle{BMC}=\\angle{ADM}=\\angle{DAB}$. Now since $ADMB$ is an isosceles trapezoid, $MD=8$. By the similarity, we have $MC=AB\\cdot \\frac{BC}{BD}=8\\cdot \\frac{6}{10}=\\frac{24}{5}$, hence $CD=MC+MD=\\frac{24}{5}+8=\\frac{64}{5}\\implies 64+5=069.",
"Since $\\angle{BAD}=\\angle{ADM}$, if we extend AB and DC, they must meet at one point to form a isosceles triangle $\\triangle{ADM}$.Now, since the problem told that $\\angle{ABD}=\\angle{BCD}$, we can imply that $\\angle{DBM}=\\angle{BCM}$ Since $\\angle{M}=\\angle{M}$, so $\\triangle{CBM}\\sim\\triangle{BDM}$. Assume the length of $BM=x$;Since $\\frac{BC}{MB}=\\frac{DB}{MD}$ we can get $\\frac{6}{x}=\\frac{10}{8+x}$, we get that $x=12$.So $AM=DM=20$ similarly, we use the same pair of similar triangle we get $\\frac{CM}{BM}=\\frac{BM}{DM}$, we get that $CM=\\frac{36}{5}$. Finally, $CD=MD-MC=\\frac{64}{5}\\implies 64+5=69=069 ~bluesoul",
"Denote $\\angle{BAD}=\\angle{CDA}=x$, and $\\angle{ABD}=\\angle{BCD}=y$. Note that $\\angle{ADB}=180^\\circ-x-y$, and $\\angle{DBC}=360^\\circ-2x-2y$. This motivates us to draw the angle bisector of $\\angle{DBC}$ because $\\angle{DBC} = 2 \\angle{ADB}$, so we do so and consider the intersection with $CD$ as $E$. By the angle bisector theorem, we have $\\frac{CE}{DE} = \\frac{BC}{BD} = \\frac{3}{5}$, so we write $CE=3z$ and $DE=5z$. We also know that $\\angle{EBC}=\\angle{ADB}$ and $\\angle{BCE}=\\angle{DBA}$, so $\\triangle{ADB} \\sim \\triangle{EBC}$. Hence, $\\frac{CE}{BC}=\\frac{AB}{BD}$, so we have $3z=\\frac{24}{5}$. As $CD=8z$, it must be that $CD=\\frac{64}{5}$, so the final answer is $069."
] |
2001-II-14
| 2,001 | 14 |
There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $|z| = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ .
| 840 |
II
|
[
"$z$ can be written in the form $\\text{cis\\,}\\theta$. Rearranging, we find that $\\text{cis\\,}{28}\\theta = \\text{cis\\,}{8}\\theta+1$ Since the real part of $\\text{cis\\,}{28}\\theta$ is one more than the real part of $\\text{cis\\,} {8}\\theta$ and their imaginary parts are equal, it is clear that either $\\text{cis\\,}{28}\\theta = \\frac{1}{2}+\\frac {\\sqrt{3}}{2}i$ and $\\text{cis\\,} {8}\\theta = -\\frac{1}{2}+\\frac {\\sqrt{3}}{2}i$, or $\\text{cis\\,}{28}\\theta = \\frac{1}{2} - \\frac{\\sqrt{3}}{2}i$ and $\\text{cis\\,} {8}\\theta = -\\frac{1}{2}- \\frac{\\sqrt{3}}{2}i$ Case 1 : $\\text{cis\\,}{28}\\theta = \\frac{1}{2}+ \\frac{\\sqrt{3}}{2}i$ and $\\text{cis\\,} {8}\\theta = -\\frac{1}{2}+\\frac{\\sqrt{3}}{2}i$ Setting up and solving equations, $Z^{28}= \\text{cis\\,}{60^\\circ}$ and $Z^8= \\text{cis\\,}{120^\\circ}$, we see that the solutions common to both equations have arguments $15^\\circ , 105^\\circ, 195^\\circ,$ and $\\ 285^\\circ$. We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values. Case 2 : $\\text{cis\\,}{28}\\theta = \\frac{1}{2} -\\frac {\\sqrt{3}}{2}i$ and $\\text{cis\\,} {8}\\theta = -\\frac {1}{2} -\\frac{\\sqrt{3}}{2}i$ Again setting up equations ($Z^{28}= \\text{cis\\,}{300^\\circ}$ and $Z^{8} = \\text{cis\\,}{240^\\circ}$) we see that the common solutions have arguments of $75^\\circ, 165^\\circ, 255^\\circ,$ and $345^\\circ$ Listing all of these values, we find that $\\theta_{2} + \\theta_{4} + \\ldots + \\theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\\circ$ which is equal to $840 degrees. We only want the sum of a certain number of theta, not all of it."
] |
2001-II-15
| 2,001 | 15 |
Let $EFGH$ , $EFDC$ , and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$ , and let $A$ be the eighth vertex of the cube. Let $I$ , $J$ , and $K$ , be the points on $\overline{EF}$ , $\overline{EH}$ , and $\overline{EC}$ , respectively, so that $EI = EJ = EK = 2$ . A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$ , and containing the edges, $\overline{IJ}$ , $\overline{JK}$ , and $\overline{KI}$ . The surface area of $S$ , including the walls of the tunnel, is $m + n\sqrt {p}$ , where $m$ , $n$ , and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$ .
| 417 |
II
|
[
"[asy] import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(\"10 2\"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy] [asy] import three; currentprojection = orthographic(camera=(1/2,1/3,-1/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(\"10 2\"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy] Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little visualization (involving some similar triangles, because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$, and $(0,1,0)$ to $(2,2,0)$, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$, and the other two faces of the tunnel are congruent to this shape. Observe that this shape is made up of two congruent trapezoids each with height $\\sqrt {2}$ and bases $7\\sqrt {3}$ and $6\\sqrt {3}$. Together they make up an area of $\\sqrt {2}(7\\sqrt {3} + 6\\sqrt {3}) = 13\\sqrt {6}$. The total area of the tunnel is then $3\\cdot13\\sqrt {6} = 39\\sqrt {6}$. Around the corner $E$ we're missing an area of $6$, the same goes for the corner opposite $E$ . So the outside area is $6\\cdot 64 - 2\\cdot 6 = 372$. Thus the the total surface area is $372 + 39\\sqrt {6}$, and the answer is $372 + 39 + 6 = 417."
] |
2002-I-1
| 2,002 | 1 |
Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 59 |
I
|
[
"Consider the three-digit arrangement, $\\overline{aba}$. There are $10$ choices for $a$ and $10$ choices for $b$ (since it is possible for $a=b$), and so the probability of picking the palindrome is $\\frac{10 \\times 10}{10^3} = \\frac 1{10}$. Similarly, there is a $\\frac 1{26}$ probability of picking the three-letter palindrome. By the Principle of Inclusion-Exclusion, the total probability is $\\frac{1}{26}+\\frac{1}{10}-\\frac{1}{260}=\\frac{35}{260}=\\frac{7}{52}\\quad\\Longrightarrow\\quad7+52=59",
"Using complementary counting, we count all of the license plates that do not have the desired property. To not be a palindrome, the first and third characters of each string must be different. Therefore, there are $10\\cdot 10\\cdot 9$ three-digit non-palindromes, and there are $26\\cdot 26\\cdot 25$ three-letter non-palindromes. As there are $10^3\\cdot 26^3$ total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is $\\frac{10\\cdot 10\\cdot 9\\cdot 26\\cdot 26\\cdot 25}{10^3\\cdot 26^3}=\\frac{45}{52}$. We subtract this from 1 to get $1-\\frac{45}{52}=\\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=59. ~minor edit by Yiyj1",
"Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is \\[\\frac{25}{26}\\cdot \\frac{9}{10}=\\frac{45}{52}\\] thus we have $1-\\frac{45}{52}=\\frac{7}{52}$ so our answer is $7+52 = 59 ~Dhillonr25",
"We can find the probability of getting a letter and number palindrome through Solution One, which gives us $\\frac{1}{26},$ and $\\frac{1}{10},$ respectively. Then, we can use casework to solve the question. We begin by creating the cases: \\begin{align*} \\bullet\\ \\text{Case 1: The license plate includes only a letter palindrome, and no number palindrome} \\\\ \\bullet\\ \\text{Case 2: The license plate includes only a number palindrome, and no letter palindrome} \\\\ \\bullet\\ \\text{Case 3: The license plate includes both a number palindrome, and a letter palindrome} \\end{align*} We know that the complement of these probabilities gives us the probability that the numbers and letters are NOT palindromes, so we can use that in our cases to get: \\begin{align} \\frac{1}{26} \\cdot \\frac{9}{10} &= \\frac{9}{260} & \\text{Case 1}\\\\ \\frac{25}{26} \\cdot \\frac{1}{10} &= \\frac{25}{260} & \\text{Case 2}\\\\ \\frac{1}{26} \\cdot \\frac{1}{10} &= \\frac{1}{260} & \\text{Case 3} \\end{align} Finally, we can add them all together to get: $\\frac{9 + 25 + 1}{260} = \\frac{35}{260} = \\frac{7}{52} = \\frac{m}{n}.$ Thus, we have $m + n = 059 ~ Cheetahboy93"
] |
2002-I-2
| 2,002 | 2 |
The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\frac{1}{2}\left(\sqrt{p}-q\right)$ , where $p$ and $q$ are positive integers. Find $p+q$ . [asy] size(250);real x=sqrt(3); int i; draw(origin--(14,0)--(14,2+2x)--(0,2+2x)--cycle); for(i=0; i<7; i=i+1) { draw(Circle((2*i+1,1), 1)^^Circle((2*i+1,1+2x), 1)); } for(i=0; i<6; i=i+1) { draw(Circle((2*i+2,1+x), 1)); } [/asy]
| 154 |
I
|
[
"Let the radius of the circles be $r$. The longer dimension of the rectangle can be written as $14r$, and by the Pythagorean Theorem, we find that the shorter dimension is $2r\\left(\\sqrt{3}+1\\right)$. Therefore, $\\frac{14r}{2r\\left(\\sqrt{3}+1\\right)}= \\frac{7}{\\sqrt{3} + 1} \\cdot \\left[\\frac{\\sqrt{3}-1}{\\sqrt{3}-1}\\right] = \\frac{1}{2}\\left(7\\sqrt{3} - 7\\right) = \\frac{1}{2}\\left(\\sqrt{p}-q\\right)$. Thus we have $p=147$ and $q=7$, so $p+q=154.",
"Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$. So, this means that each of the radii is $1$. Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\\sqrt{3}$, and the shorter side of the triangle is therefore $2\\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\\frac{1}{2} \\cdot (\\sqrt{147}-7)$ and the final answer is $147+7 = 154."
] |
2002-I-3
| 2,002 | 3 |
Jane is 25 years old. Dick is older than Jane. In $n$ years, where $n$ is a positive integer, Dick's age and Jane's age will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let $d$ be Dick's present age. How many ordered pairs of positive integers $(d,n)$ are possible?
| 25 |
I
|
[
"Let Jane's age $n$ years from now be $10a+b$, and let Dick's age be $10b+a$. If $10b+a>10a+b$, then $b>a$. The possible pairs of $a,b$ are: $(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \\dots , (8,9)$ That makes 36. But $10a+b>25$, so we subtract all the extraneous pairs: $(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),$ and $(1,9)$. $36-11=025",
"Start by assuming that $n < 5$ (essentially, Jane is in the 20s when their ages are 'reverses' of each other). Then we get the pairs \\[(61,1),(70,2),(79,3),(88,4).\\] Repeating this for the 30s gives \\[(34,9),(43,10),(52,11),(61,12),(70,13),(79,14).\\] From here, it's pretty clear that every decade we go up we get $(d,n+11)$ as a pair. Since both ages must always be two-digit numbers, we can show that each decade after the 30s, we get 1 fewer option. Therefore, our answer is $4+6+5+\\dots+2+1=4+21=025 ~Dhillonr25"
] |
2002-I-4
| 2,002 | 4 |
Consider the sequence defined by $a_k=\frac 1{k^2+k}$ for $k\ge 1$ . Given that $a_m+a_{m+1}+\cdots+a_{n-1}=1/29$ , for positive integers $m$ and $n$ with $m<n$ , find $m+n$ .
| 840 |
I
|
[
"Using partial fraction decomposition yields $\\dfrac{1}{k^2+k}=\\dfrac{1}{k(k+1)}=\\dfrac{1}{k}-\\dfrac{1}{k+1}$. Thus, $a_m+a_{m+1}+\\cdots +a_{n-1}=\\dfrac{1}{m}-\\dfrac{1}{m+1}+\\dfrac{1}{m+1}-\\dfrac{1}{m+2}+\\cdots +\\dfrac{1}{n-1}-\\dfrac{1}{n}=\\dfrac{1}{m}-\\dfrac{1}{n}$ Which means that $\\dfrac{n-m}{mn}=\\dfrac{1}{29}$ Since we need a factor of 29 in the denominator, we let $n=29t$.* Substituting, we get $29t-m=mt$ so $\\frac{29t}{t+1} = m$ Since $m$ is an integer, $t+1 = 29$, so $t=28$. It quickly follows that $n=29(28)$ and $m=28$, so $m+n = 30(28) = 840.",
"Note that $a_1 + a_2 + \\cdots + a_i = \\dfrac{i}{i+1}$. This can be proven by induction. Thus, $\\sum\\limits_{i=m}^{n-1} a_i = \\sum\\limits_{i=1}^{n-1} a_i - \\sum\\limits_{i=1}^{m-1} a_i = \\dfrac{n-1}{n} - \\dfrac{m-1}{m} = \\dfrac{n-m}{mn} = 1/29$. Cross-multiplying yields $29n - 29m - mn = 0$, and adding $29^2$ to both sides gives $(29-m)(29+n) = 29^2$. Clearly, $m < n \\implies 29 - m = 1$ and $29 + n = 29^2$. Hence, $m = 28$, $n = 812$, and $m+n = 840. ~ keeper1098",
"To solve this problem, I need to find two positive integers $m$ and $n$ where $m < n$ and the sum of sequence terms equals $\\frac{1}{29}$. First, let me simplify $a_k = \\frac{1}{k^2 + k}$ using partial fractions. $a_k = \\frac{1}{k(k+1)} = \\frac{1}{k} - \\frac{1}{k+1}$ Express the sum $a_m + a_{m+1} + \\cdots + a_{n-1}$ using this simplification. $a_m + a_{m+1} + \\cdots + a_{n-1} = \\sum_{k=m}^{n-1} \\left(\\frac{1}{k} - \\frac{1}{k+1}\\right)$ This is a telescoping series where intermediate terms cancel: $a_m + a_{m+1} + \\cdots + a_{n-1} = \\frac{1}{m} - \\frac{1}{n}$ Use the given condition that this sum equals $\\frac{1}{29}$. $\\frac{1}{m} - \\frac{1}{n} = \\frac{1}{29}$ Multiplying both sides by $mn$: $n - m = \\frac{mn}{29}$ Rearranging: $29(n - m) = mn$ $29n - 29m = mn$ $29n - mn = 29m$ $n(29 - m) = 29m$ Solve for $n$ in terms of $m$. $n = \\frac{29m}{29-m}$ Since $n$ must be a positive integer, $29-m$ must divide $29m$ evenly. Since $29$ is prime, for $29-m$ to divide $29m$ (when $m < 29$), we need $29-m$ to divide $m$. This means $m = k(29-m)$ for some positive integer $k$. $m = k(29-m)$ $m = 29k - km$ $m(1+k) = 29k$ $m = \\frac{29k}{1+k}$ For $m$ to be an integer, $1+k$ must divide $29k$. When $k = 28$, we get $m = \\frac{29(28)}{29} = 28$ Calculate $n$ using our value of $m$. $n = \\frac{29(28)}{29-28} = \\frac{812}{1} = 812$ Therefore, $m + n = 28 + 812 = 840 ~ brandonyee"
] |
2002-I-5
| 2,002 | 5 |
Let $A_1, A_2, A_3, \ldots, A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\ldots,A_{12}\}$ ?
| 183 |
I
|
[
"There are 66 ways of picking two vertices. Note with any two vertices one can draw three squares (two with the vertices forming a side, another with the vertices forming the diagonal). So so far we have $66(3)=198$ squares, but we have overcounted since some squares have their other two vertices in the dodecagon as well. All 12 combinations of two distinct vertices that form a square side only form 3 squares, and all 12 combinations of two vertices that form a square diagonal only form 6 squares. So in total, we have overcounted by $9+6=15$, and $198-15=183.",
"Proceed as above to initially get 198 squares (with overcounting). Then note that any square with all four vertices on the dodecagon has to have three sides \"between\" each vertex, giving us a total of three squares. However, we counted these squares with all four of their sides plus both of their diagonals, meaning we counted them 6 times. Therefore, our answer is $198-3(6-1)=198-15=183 ~Dhillonr25"
] |
2002-I-6
| 2,002 | 6 |
The solutions to the system of equations \begin{align*} \log_{225}{x}+\log_{64}{y} = 4\\ \log_{x}{225}- \log_{y}{64} = 1 \end{align*} are $(x_1,y_1)$ and $(x_2, y_2)$ . Find $\log_{30}{(x_1y_1x_2y_2)}$ .
| 12 |
I
|
[
"Let $A=\\log_{225}x$ and let $B=\\log_{64}y$. From the first equation: $A+B=4 \\Rightarrow B = 4-A$. Plugging this into the second equation yields $\\frac{1}{A}-\\frac{1}{B}=\\frac{1}{A}-\\frac{1}{4-A}=1 \\Rightarrow A = 3\\pm\\sqrt{5}$ and thus, $B=1\\pm\\sqrt{5}$. So, $\\log_{225}(x_1x_2)=\\log_{225}(x_1)+\\log_{225}(x_2)=(3+\\sqrt{5})+(3-\\sqrt{5})=6$ $\\Rightarrow x_1x_2=225^6=15^{12}$. And $\\log_{64}(y_1y_2)=\\log_{64}(y_1)+\\log_{64}(y_2)=(1+\\sqrt{5})+(1-\\sqrt{5})=2$ $\\Rightarrow y_1y_2=64^2=2^{12}$. Thus, $\\log_{30}\\left(x_1y_1x_2y_2\\right) = \\log_{30}\\left(15^{12}\\cdot2^{12} \\right) = \\log_{30}\\left(30^{12} \\right) = 012."
] |
2002-I-8
| 2,002 | 8 |
Find the smallest integer $k$ for which the conditions (1) $a_1, a_2, a_3, \ldots$ is a nondecreasing sequence of positive integers (2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$ (3) $a_9=k$ are satisfied by more than one sequence.
| 748 |
I
|
[
"From $(2)$, $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$ Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$. Since $\\gcd(13,21)=1$, the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$. Then, $a_2=y_0-13$. By $(1)$, $a_2 \\ge a_1 \\Rightarrow y_0-13 \\ge x_0+21 \\Rightarrow y_0 \\ge x_0+34 \\ge 35$. So the smallest value of $k$ is attained when $(x_0,y_0)=(1,35)$ which yields $(a_1,a_2)=(1,35)$ or $(22,22)$. Thus, $k=13(1)+21(35)=748."
] |
2002-I-9
| 2,002 | 9 |
Harold, Tanya, and Ulysses paint a very long picket fence. Harold starts with the first picket and paints every $h$ th picket; Tanya starts with the second picket and paints every $t$ th picket; and Ulysses starts with the third picket and paints every $u$ th picket. Call the positive integer $100h+10t+u$ $\textit{paintable}$ when the triple $(h,t,u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
| 757 |
I
|
[
"Solution 1 Note that it is impossible for any of $h,t,u$ to be $1$, since then each picket will have been painted one time, and then some will be painted more than once. $h$ cannot be $2$, or that will result in painting the third picket twice. If $h=3$, then $t$ may not equal anything not divisible by $3$, and the same for $u$. Now for fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable. If $h$ is $4$, then $t$ must be even. The same for $u$, except that it can't be $2 \\mod 4$. Thus $u$ is $0 \\mod 4$ and $t$ is $2 \\mod 4$. Since this is all $\\mod 4$, $t$ must be $2$ and $u$ must be $4$, in order for $5,6$ to be paint-able. Thus $424$ is paintable. $h$ cannot be greater than $4$, since if that were the case then the answer would be greater than $999$, which would be impossible for the AIME. Thus the sum of all paintable numbers is $757. Solution 4 The wording of this problem is a bit dissatisfying and could have been improved by stating that there are at least h + t + u pickets instead of \"very long\". For example, a very long fence could have 5 pickets, h = 78, t = 47, u = 1. To compensate for this we should lean on the fact that the answer cannot exceed 999. Clearly we have to accept that there are 4 pickets or more, and the 4th picket is not reached from setting u = 1. If the 4th picket is reached from h = 3, then the 5th picket will not be reached via u = 2, or else we'd be unable to have 7 pickets. So h = 3 => t = 3 => u = 3, and 333 is a solution. We now have h > 3 for any remaining cases, and thus we only have room for 1 more case. Since the 4th picket must henceforth be reached via t = 2, we either have the 5th picket reached via u = 2 (requiring h to be any number exceeding the number of pickets, and we're stipulating that this is not allowed), or we have the 5th picket reached via h = 4, from which it follows that u = 4. So we have accumulated the solutions 333 and 424 and we have exhausted both the space of reasonable possibilities as well as the available space before we would exceed 999.",
"Note that it is impossible for any of $h,t,u$ to be $1$, since then each picket will have been painted one time, and then some will be painted more than once. $h$ cannot be $2$, or that will result in painting the third picket twice. If $h=3$, then $t$ may not equal anything not divisible by $3$, and the same for $u$. Now for fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable. If $h$ is $4$, then $t$ must be even. The same for $u$, except that it can't be $2 \\mod 4$. Thus $u$ is $0 \\mod 4$ and $t$ is $2 \\mod 4$. Since this is all $\\mod 4$, $t$ must be $2$ and $u$ must be $4$, in order for $5,6$ to be paint-able. Thus $424$ is paintable. $h$ cannot be greater than $4$, since if that were the case then the answer would be greater than $999$, which would be impossible for the AIME. Thus the sum of all paintable numbers is $757. Solution 4 The wording of this problem is a bit dissatisfying and could have been improved by stating that there are at least h + t + u pickets instead of \"very long\". For example, a very long fence could have 5 pickets, h = 78, t = 47, u = 1. To compensate for this we should lean on the fact that the answer cannot exceed 999. Clearly we have to accept that there are 4 pickets or more, and the 4th picket is not reached from setting u = 1. If the 4th picket is reached from h = 3, then the 5th picket will not be reached via u = 2, or else we'd be unable to have 7 pickets. So h = 3 => t = 3 => u = 3, and 333 is a solution. We now have h > 3 for any remaining cases, and thus we only have room for 1 more case. Since the 4th picket must henceforth be reached via t = 2, we either have the 5th picket reached via u = 2 (requiring h to be any number exceeding the number of pickets, and we're stipulating that this is not allowed), or we have the 5th picket reached via h = 4, from which it follows that u = 4. So we have accumulated the solutions 333 and 424 and we have exhausted both the space of reasonable possibilities as well as the available space before we would exceed 999.",
"Again, note that $h,t,u \\neq 1$. The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \\equiv 1 \\pmod{h},\\ n \\equiv 2 \\pmod{t},\\ n \\equiv 3 \\pmod{u}$. By the Chinese Remainder Theorem, the greatest common divisor of any pair of the three numbers $\\{h,t,u\\}$ cannot be $1$ (since otherwise without loss of generality consider $\\text{gcd}\\,(h,t) = 1$; then there will be a common solution $\\pmod{h \\times t}$). Now for $4$ to be paint-able, we require either $h = 3$ or $t=2$, but not both. In the former condition, since $\\text{gcd}\\,(h,t),\\ \\text{gcd}\\,(h,u) \\neq 1$, it follows that $3|t,u$. For $5$ and $6$ to be paint-able, we require $t = u = 3$, and it is easy to see that $333$ works. In the latter condition, similarly we require that $2|h,u$. All even numbers are painted. We now renumber the remaining odd pickets to become the set of all positive integers ($1,3,5, \\ldots \\rightarrow 1',2',3', \\ldots$, where $n' = \\frac{n+1}{2}$), which requires the transformation $h' = h/2,\\ u' = u/2$, and with the painting starting respectively at $1',2'$. Our new number system retains the same conditions as above, except without $t$. We thus need $\\text{gcd}\\, (h',u') \\neq 1, h',u' \\neq 1$. Then for $3',4'$ to be painted, we require $h' = u' = 2$. This translates to $424$, which we see works. Thus the answer is $333+424 = 757. Solution 4 The wording of this problem is a bit dissatisfying and could have been improved by stating that there are at least h + t + u pickets instead of \"very long\". For example, a very long fence could have 5 pickets, h = 78, t = 47, u = 1. To compensate for this we should lean on the fact that the answer cannot exceed 999. Clearly we have to accept that there are 4 pickets or more, and the 4th picket is not reached from setting u = 1. If the 4th picket is reached from h = 3, then the 5th picket will not be reached via u = 2, or else we'd be unable to have 7 pickets. So h = 3 => t = 3 => u = 3, and 333 is a solution. We now have h > 3 for any remaining cases, and thus we only have room for 1 more case. Since the 4th picket must henceforth be reached via t = 2, we either have the 5th picket reached via u = 2 (requiring h to be any number exceeding the number of pickets, and we're stipulating that this is not allowed), or we have the 5th picket reached via h = 4, from which it follows that u = 4. So we have accumulated the solutions 333 and 424 and we have exhausted both the space of reasonable possibilities as well as the available space before we would exceed 999.",
"The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \\equiv 1 \\pmod{h},\\ n \\equiv 2 \\pmod{t},\\ n \\equiv 3 \\pmod{u}$. Note that the smallest number, $min \\{ h,t,u \\},$ divides the other $2$, and the next smallest divide the largest number, otherwise there is a common solution by the Chinese Remainder Theorem. It is also a necessary condition so that it paints exactly once. Note that the smallest number can't be at least $5$, otherwise not all picket will be painted. We are left with few cases (we could also exclude $1$ as the possibility) which could be done quickly. Thus the answer is $333+424 = 757. Solution 4 The wording of this problem is a bit dissatisfying and could have been improved by stating that there are at least h + t + u pickets instead of \"very long\". For example, a very long fence could have 5 pickets, h = 78, t = 47, u = 1. To compensate for this we should lean on the fact that the answer cannot exceed 999. Clearly we have to accept that there are 4 pickets or more, and the 4th picket is not reached from setting u = 1. If the 4th picket is reached from h = 3, then the 5th picket will not be reached via u = 2, or else we'd be unable to have 7 pickets. So h = 3 => t = 3 => u = 3, and 333 is a solution. We now have h > 3 for any remaining cases, and thus we only have room for 1 more case. Since the 4th picket must henceforth be reached via t = 2, we either have the 5th picket reached via u = 2 (requiring h to be any number exceeding the number of pickets, and we're stipulating that this is not allowed), or we have the 5th picket reached via h = 4, from which it follows that u = 4. So we have accumulated the solutions 333 and 424 and we have exhausted both the space of reasonable possibilities as well as the available space before we would exceed 999.",
"The wording of this problem is a bit dissatisfying and could have been improved by stating that there are at least h + t + u pickets instead of \"very long\". For example, a very long fence could have 5 pickets, h = 78, t = 47, u = 1. To compensate for this we should lean on the fact that the answer cannot exceed 999. Clearly we have to accept that there are 4 pickets or more, and the 4th picket is not reached from setting u = 1. If the 4th picket is reached from h = 3, then the 5th picket will not be reached via u = 2, or else we'd be unable to have 7 pickets. So h = 3 => t = 3 => u = 3, and 333 is a solution. We now have h > 3 for any remaining cases, and thus we only have room for 1 more case. Since the 4th picket must henceforth be reached via t = 2, we either have the 5th picket reached via u = 2 (requiring h to be any number exceeding the number of pickets, and we're stipulating that this is not allowed), or we have the 5th picket reached via h = 4, from which it follows that u = 4. So we have accumulated the solutions 333 and 424 and we have exhausted both the space of reasonable possibilities as well as the available space before we would exceed 999."
] |
2002-I-10
| 2,002 | 10 |
In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the integer closest to the area of quadrilateral $DCFG$ . [asy] size(250); pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D); draw(A--B--C--A--D^^E--F); pair point=G+1*dir(250); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); markscalefactor=0.1; draw(rightanglemark(A,B,C)); label("10", A--F, dir(90)*dir(A--F)); label("27", F--C, dir(90)*dir(F--C)); label("3", (0,10), W); label("9", (0,4), W); [/asy]
| 148 |
I
|
[
"By the Pythagorean Theorem, $BC=35$. Letting $BD=x$ we can use the Angle Bisector Theorem on triangle $ABC$ to get $x/12=(35-x)/37$, and solving gives $BD=60/7$ and $DC=185/7$. The area of triangle $AGF$ is $10/3$ that of triangle $AEG$, since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$. Since the area of a triangle is $\\frac{ab\\sin{C}}2$, the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$. The area of triangle $ABD$ is $360/7$, and the area of the entire triangle $ABC$ is $210$. Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $148 as the answer.",
"By the Pythagorean Theorem, $BC=35$. From now on, every number we get will be rounded to two decimal points. Using law of cosines on AEF, EF is around 9.46. Using the angle bisector theorem, GF and DC are 7.28 and (185/7), respectively. We also know the lengths of EG and BD, so a quick Stewart's Theorem for triangles ABC and AEF gives lengths 3.76 and 14.75 for AG and AD, respectively. Now we have all segments of triangles AGF and ADC. Joy! It's time for some Heron's Formula. This gives area 10.95 for triangle AGF and 158.68 for triangle ADC. Subtracting gives us 147.73. Round to the nearest whole number and we get $148. -jackshi2006",
"By the Pythagorean Theorem, $BC=35$. By the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$. We can find the coordinates of F, and use that the find the equation of line EF. Then, we can find the coordinates of G. Triangle $ADC = 1110/7$ and we can find the area triangle $AGF$ with the shoelace theorem, so subtracting that from $ADC$ gives us $148 as the closest integer. -jackshi2006",
"By the Pythagorean Theorem, $BC = 35$, and by the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$. Draw a perpendicular from $F$ to $\\overline{AE}$. Let the intersection of $F$ and $\\overline{AE}$ be $H$. triangle $AHF$ is similar to $ABC$ by $AA$ similarity. thus, $AF/AC = HF/BC$. We find that $HF = 350/37$, so the area of $AEF = 525/37$. The area of triangle $AGF$ is $10/3$ that of triangle $AEG$, since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$, so the area of $AGF = 5250/481$. The area of triangle $ADC$ is $1110/7$, since the base is $185/7$ and the height is $12$. Thus, the area of $DCFG$ equals the area of $ADC - AGF$, or rounded to the nearest integer, $148 ~ PaperMath"
] |
2002-I-11
| 2,002 | 11 |
Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$ . A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$ , which is $7$ units from $\overline{BG}$ and $5$ units from $\overline{BC}$ . The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$ , where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$ .
| 230 |
I
|
[
"When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane. In each reflection, the distance traveled by the light is $\\sqrt{ (12^2) + (5^2) + (7^2) }$ = $\\sqrt{218}$. This happens 12 times, so the total distance is $12\\sqrt{218}$. $m=12$ and $n=218$, so therefore, the answer is $m+n=230.",
"When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane. In each reflection, the distance traveled by the light is $\\sqrt{ (12^2) + (5^2) + (7^2) }$ = $\\sqrt{218}$. This happens 12 times, so the total distance is $12\\sqrt{218}$. $m=12$ and $n=218$, so therefore, the answer is $m+n=230.",
"We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided into cubes identical to the one we have. Now let's follow two photons of light that start in $A$ at the same time: one of them will reflect as given in the problem statement, the second will simply fly straight through all cubes. It can easily be seen that at any moment in time the photons are in exactly the same position relative to the cubes they are inside at the moment. In other words, we can take the cube with the first photon, translate it and flip if necessary, to get the cube with the other photon. It follows that both photons will hit a vertex at the same time, and at this moment they will have travelled the same distance. Now, the path of the second photon is simply a half-line given by the vector $(12,7,5)$. That is, the points visited by the photon are of the form $(12t,7t,5t)$ for $t\\geq 0$. We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube). Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$. At this moment the second photon will be at the coordinates $(12\\cdot 12,7\\cdot 12,5\\cdot 12)$. Then the distance it travelled is $\\sqrt{ (12\\cdot 12)^2 + (7\\cdot 12)^2 + (5\\cdot 12)^2 } = 12\\sqrt{12^2 + 7^2 + 5^2}=12\\sqrt{218}$. And as the factorization of $218$ is $218=2\\cdot 109$, we have $m=12$ and $n=218$, hence $m+n=230.",
"We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided into cubes identical to the one we have. Now let's follow two photons of light that start in $A$ at the same time: one of them will reflect as given in the problem statement, the second will simply fly straight through all cubes. It can easily be seen that at any moment in time the photons are in exactly the same position relative to the cubes they are inside at the moment. In other words, we can take the cube with the first photon, translate it and flip if necessary, to get the cube with the other photon. It follows that both photons will hit a vertex at the same time, and at this moment they will have travelled the same distance. Now, the path of the second photon is simply a half-line given by the vector $(12,7,5)$. That is, the points visited by the photon are of the form $(12t,7t,5t)$ for $t\\geq 0$. We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube). Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$. At this moment the second photon will be at the coordinates $(12\\cdot 12,7\\cdot 12,5\\cdot 12)$. Then the distance it travelled is $\\sqrt{ (12\\cdot 12)^2 + (7\\cdot 12)^2 + (5\\cdot 12)^2 } = 12\\sqrt{12^2 + 7^2 + 5^2}=12\\sqrt{218}$. And as the factorization of $218$ is $218=2\\cdot 109$, we have $m=12$ and $n=218$, hence $m+n=230."
] |
2002-I-12
| 2,002 | 12 |
Let $F(z)=\frac{z+i}{z-i}$ for all complex numbers $z\not= i$ , and let $z_n=F(z_{n-1})$ for all positive integers $n$ . Given that $z_0=\frac 1{137}+i$ and $z_{2002}=a+bi$ , where $a$ and $b$ are real numbers, find $a+b$ .
| 275 |
I
|
[
"Iterating $F$ we get: \\begin{align*} F(z) &= \\frac{z+i}{z-i}\\\\ F(F(z)) &= \\frac{\\frac{z+i}{z-i}+i}{\\frac{z+i}{z-i}-i} = \\frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \\frac{z+i+zi+1}{z+i-zi-1}= \\frac{(z+1)(i+1)}{(z-1)(1-i)}\\\\ &= \\frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \\frac{(z+1)(2i)}{(z-1)(2)}= \\frac{z+1}{z-1}i\\\\ F(F(F(z))) &= \\frac{\\frac{z+1}{z-1}i+i}{\\frac{z+1}{z-1}i-i} = \\frac{\\frac{z+1}{z-1}+1}{\\frac{z+1}{z-1}-1} = \\frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \\frac{2z}{2} = z. \\end{align*} From this, it follows that $z_{k+3} = z_k$, for all $k$. Thus $z_{2002} = z_{3\\cdot 667+1} = z_1 = \\frac{z_0+i}{z_0-i} = \\frac{(\\frac{1}{137}+i)+i}{(\\frac{1}{137}+i)-i}= \\frac{\\frac{1}{137}+2i}{\\frac{1}{137}}= 1+274i.$ Thus $a+b = 1+274 = 275."
] |
2002-I-13
| 2,002 | 13 |
In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths 18 and 27, respectively, and $AB = 24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ .
| 63 |
I
|
[
"[asy] size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C)); D(D(MP(\"A\",A))--D(MP(\"B\",B))--D(MP(\"C\",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP(\"F\",F)); D(A--D); D(C--F); D(A--F--B); D(MP(\"E\",E,NE)); D(MP(\"D\",D,NE)); MP(\"12\",(A+E)/2,SE,f);MP(\"12\",(B+E)/2,f); MP(\"27\",(C+E)/2,SW,f); MP(\"18\",(A+D)/2,SE,f); [/asy] Applying Stewart's Theorem to medians $AD, CE$, we have: \\begin{align*} BC^2 + 4 \\cdot 18^2 &= 2\\left(24^2 + AC^2\\right) \\\\ 24^2 + 4 \\cdot 27^2 &= 2\\left(AC^2 + BC^2\\right) \\end{align*} Substituting the first equation into the second and simplification yields $24^2 = 2\\left(3AC^2 + 2 \\cdot 24^2 - 4 \\cdot 18^2\\right)- 4 \\cdot 27^2$ $\\Longrightarrow AC = \\sqrt{2^5 \\cdot 3 + 2 \\cdot 3^5 + 2^4 \\cdot 3^3 - 2^7 \\cdot 3} = 3\\sqrt{70}$. By the Power of a Point Theorem on $E$, we get $EF = \\frac{12^2}{27} = \\frac{16}{3}$. The Law of Cosines on $\\triangle ACE$ gives \\begin{align*} \\cos \\angle AEC = \\left(\\frac{12^2 + 27^2 - 9 \\cdot 70}{2 \\cdot 12 \\cdot 27}\\right) = \\frac{3}{8} \\end{align*} Hence $\\sin \\angle AEC = \\sqrt{1 - \\cos^2 \\angle AEC} = \\frac{\\sqrt{55}}{8}$. Because $\\triangle AEF, BEF$ have the same height and equal bases, they have the same area, and $[ABF] = 2[AEF] = 2 \\cdot \\frac 12 \\cdot AE \\cdot EF \\sin \\angle AEF = 12 \\cdot \\frac{16}{3} \\cdot \\frac{\\sqrt{55}}{8} = 8\\sqrt{55}$, and the answer is $8 + 55 = 063.",
"Let $AD$ and $CE$ intersect at $P$. Since medians split one another in a 2:1 ratio, we have \\begin{align*} AP = 12, PE = 9 \\end{align*} This gives isosceles $APE$ and thus an easy area calculation. After extending the altitude to $PE$ and using the fact that it is also a median, we find \\begin{align*} [APE] = \\frac{27\\sqrt{55}}{4} \\end{align*} Using Power of a Point, we have \\begin{align*} EF=\\frac{16}{3} \\end{align*} By Same Height Different Base, \\begin{align*} \\frac{[AFE]}{[APE]}=\\frac{[AFE]}{(\\frac{27\\sqrt{55}}{4})}=\\frac{EF}{PE}=\\frac{(\\frac{16}{3})}{9}=\\frac{16}{27} \\end{align*} Solving gives \\begin{align*} [AFE] = 4\\sqrt{55} \\end{align*} and \\begin{align*} [AFB]=2[AFE]=8\\sqrt{55} \\end{align*} Thus, our answer is $8+55=063.",
"Note that, as above, it is quite easy to get that $\\sin \\angle AEP = \\frac{\\sqrt{55}}{8}$ (equate Heron's and $\\frac{1}{2}ab\\sin C$ to find this). Now note that $\\angle FEA = \\angle BEC$ because they are vertical angles, $\\angle FAE = \\angle ECB$, and $\\angle EFA = \\angle ABC$ (the latter two are derived from the inscribed angle theorem). Therefore $\\Delta AEF$ ~ $\\Delta CEB$ and so $FE = \\frac{144}{27}$ and $\\sin \\angle FEA = \\frac{\\sqrt{55}}{8}$ so the area of $\\Delta BFA$ is $8\\sqrt{55}$ giving us $063 as our answer. (One may just get the area via triangle similarity too--this is if you are tired by the end of test and just want to bash some stuff out--it may also serve as a useful check). ~Dhillonr25",
"Note that, as above, it is quite easy to get that $\\sin \\angle AEP = \\frac{\\sqrt{55}}{8}$ (equate Heron's and $\\frac{1}{2}ab\\sin C$ to find this). Now note that $\\angle FEA = \\angle BEC$ because they are vertical angles, $\\angle FAE = \\angle ECB$, and $\\angle EFA = \\angle ABC$ (the latter two are derived from the inscribed angle theorem). Therefore $\\Delta AEF$ ~ $\\Delta CEB$ and so $FE = \\frac{144}{27}$ and $\\sin \\angle FEA = \\frac{\\sqrt{55}}{8}$ so the area of $\\Delta BFA$ is $8\\sqrt{55}$ giving us $063 as our answer. (One may just get the area via triangle similarity too--this is if you are tired by the end of test and just want to bash some stuff out--it may also serve as a useful check). ~Dhillonr25",
"Apply barycentric coordinates on $\\triangle ABC$. We know that $D=\\left(0, \\frac{1}{2}, \\frac{1}{2}\\right), E=\\left(\\frac{1}{2}, \\frac{1}{2}, 0\\right)$. We can now get the displacement vectors $\\overrightarrow{AD} = \\left(1, -\\frac{1}{2}, -\\frac{1}{2}\\right)$ and $\\overrightarrow{CE}=\\left(-\\frac{1}{2}, -\\frac{1}{2}, 1\\right)$. Now, applying the distance formula and simplifying gives us the two equations \\begin{align*} 2b^2+2c^2-a^2&=1296 \\\\ 2a^2+2b^2-c^2&=2916. \\\\ \\end{align*} Substituting $c=24$ and solving with algebra now gives $a=6\\sqrt{31}, b=3\\sqrt{70}$. Now we can find $F$. Note that $CE$ can be parameterized as $(1:1:t)$, so plugging into the circumcircle equation and solving for $t$ gives $t=\\frac{-c^2}{a^2+b^2}$ so $F=(a^2+b^2:a^2+b^2:-c^2)$. Plugging in for $a,b$ gives us $F=(1746:1746:-576)$. Thus, by the area formula, we have\\[\\frac{[AFB]}{[ABC]}= \\left|\\begin{matrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ \\frac{97}{162} & \\frac{97}{162} & -\\frac{16}{81} \\end{matrix}\\right|=\\frac{16}{81}.\\]By Heron's Formula, we have $[ABC]=\\frac{81\\sqrt{55}}{2}$ which immediately gives $[AFB]=8\\sqrt{55}$ from our ratio, extracting $063. -Taco12",
"Apply barycentric coordinates on $\\triangle ABC$. We know that $D=\\left(0, \\frac{1}{2}, \\frac{1}{2}\\right), E=\\left(\\frac{1}{2}, \\frac{1}{2}, 0\\right)$. We can now get the displacement vectors $\\overrightarrow{AD} = \\left(1, -\\frac{1}{2}, -\\frac{1}{2}\\right)$ and $\\overrightarrow{CE}=\\left(-\\frac{1}{2}, -\\frac{1}{2}, 1\\right)$. Now, applying the distance formula and simplifying gives us the two equations \\begin{align*} 2b^2+2c^2-a^2&=1296 \\\\ 2a^2+2b^2-c^2&=2916. \\\\ \\end{align*} Substituting $c=24$ and solving with algebra now gives $a=6\\sqrt{31}, b=3\\sqrt{70}$. Now we can find $F$. Note that $CE$ can be parameterized as $(1:1:t)$, so plugging into the circumcircle equation and solving for $t$ gives $t=\\frac{-c^2}{a^2+b^2}$ so $F=(a^2+b^2:a^2+b^2:-c^2)$. Plugging in for $a,b$ gives us $F=(1746:1746:-576)$. Thus, by the area formula, we have\\[\\frac{[AFB]}{[ABC]}= \\left|\\begin{matrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ \\frac{97}{162} & \\frac{97}{162} & -\\frac{16}{81} \\end{matrix}\\right|=\\frac{16}{81}.\\]By Heron's Formula, we have $[ABC]=\\frac{81\\sqrt{55}}{2}$ which immediately gives $[AFB]=8\\sqrt{55}$ from our ratio, extracting $063. -Taco12",
"Since $AD$ is the median, let $BD=BC=x$. Since $CE$ is a median, $AE=BE=12$. Applying Power of a Point with respect to point $E$, we see that $EF=\\frac{16}{3}$. Applying Stewart's Theorem on triangles $\\triangle ADC$ and $\\triangle ABC$, we get that $x=3\\sqrt{31}$ and $y=3\\sqrt{70}$. The area of $\\triangle AFB$ is simply $\\frac{1}{2} \\cdot \\sin{\\angle FBA}\\cdot FB\\cdot AB$. We know $AB=24$. Also, we know that $\\angle FBA = \\angle FCA$. Then, applying Law of Cosines on triangle $EAC$, we get that $\\cos{\\angle FCA}=\\frac{3\\sqrt{70}}{28}$ which means that $\\sin{\\angle FCA=\\angle FBA}=\\frac{\\sqrt{154}}{28}$. Then, applying Stewart's Theorem on triangle $FBC$ with cevian $BE$ allows us to receive that $FB=\\frac{4\\sqrt{70}}{3}$. Now, plugging into our earlier area formula, we receive $\\frac{1}{2} \\cdot \\frac{\\sqrt{154}}{28} \\cdot \\frac{4\\sqrt{70}}{3} \\cdot 24 = 8\\sqrt{55}.$ Therefore, the desired answer is $8+55=063. ~SirAppel",
"Since $AD$ is the median, let $BD=BC=x$. Since $CE$ is a median, $AE=BE=12$. Applying Power of a Point with respect to point $E$, we see that $EF=\\frac{16}{3}$. Applying Stewart's Theorem on triangles $\\triangle ADC$ and $\\triangle ABC$, we get that $x=3\\sqrt{31}$ and $y=3\\sqrt{70}$. The area of $\\triangle AFB$ is simply $\\frac{1}{2} \\cdot \\sin{\\angle FBA}\\cdot FB\\cdot AB$. We know $AB=24$. Also, we know that $\\angle FBA = \\angle FCA$. Then, applying Law of Cosines on triangle $EAC$, we get that $\\cos{\\angle FCA}=\\frac{3\\sqrt{70}}{28}$ which means that $\\sin{\\angle FCA=\\angle FBA}=\\frac{\\sqrt{154}}{28}$. Then, applying Stewart's Theorem on triangle $FBC$ with cevian $BE$ allows us to receive that $FB=\\frac{4\\sqrt{70}}{3}$. Now, plugging into our earlier area formula, we receive $\\frac{1}{2} \\cdot \\frac{\\sqrt{154}}{28} \\cdot \\frac{4\\sqrt{70}}{3} \\cdot 24 = 8\\sqrt{55}.$ Therefore, the desired answer is $8+55=063. ~SirAppel"
] |
2002-I-14
| 2,002 | 14 |
A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer. Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is the greatest number of elements that $\mathcal{S}$ can have?
| 30 |
I
|
[
"Let the sum of the integers in $\\mathcal{S}$ be $N$, and let the size of $|\\mathcal{S}|$ be $n+1$. After any element $x$ is removed, we are given that $n|N-x$, so $x\\equiv N\\pmod{n}$. Since $1\\in\\mathcal{S}$, $N\\equiv1\\pmod{n}$, and all elements are congruent to 1 mod $n$. Since they are positive integers, the largest element is at least $n^2+1$, the $(n+1)$th positive integer congruent to 1 mod $n$. We are also given that this largest member is 2002, so $2002\\equiv1\\pmod{n}$, and $n|2001=3\\cdot23\\cdot29$. Also, we have $n^2+1\\le2002$, so $n \\leq 44$. The largest factor of 2001 less than 45 is 29, so $n=29$ and $n+1$ $\\Rightarrow{30, for instance."
] |
2002-I-15
| 2,002 | 15 |
Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given that $EG^2 = p - q\sqrt {r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$
| 163 |
I
|
[
"[asy] size(200); import three; import graph; defaultpen(linewidth(0.7)+fontsize(8)); currentprojection=orthographic(-30,50,40); triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12); F = 1/2*H+1/2*B; G = 1/2*H+1/2*A; draw((A--B--C--D--A)^^(D--E--C)^^(A--G--F--B)^^(G--E--F));draw((G--H--F)^^(H--E1),gray(0.6)); dot(H1^^H,linewidth(2)); label(\"$A$\",A,( 0,-1, 0)); label(\"$B$\",B,( 0, 1, 0)); label(\"$C$\",C,( 0, 1, 0)); label(\"$D$\",D,( 0,-1, 0)); label(\"$E$\",E,(-1,-1, 1)); label(\"$F$\",F,( 0, 1, 0)); label(\"$G$\",G,(-1,-1, 1)); label(\"$H$\",H,( 1,-1, 1)); label(\"$H'$\",H1,(-1,-1, 1)); [/asy] Let's put the polyhedron onto a coordinate plane. For simplicity, let the origin be the center of the square: $A(-6,6,0)$, $B(-6,-6,0)$, $C(6,-6,0)$ and $D(6,6,0)$. Since $ABFG$ is an isosceles trapezoid and $CDE$ is an isosceles triangle, we have symmetry about the $xz$-plane. Therefore, the $y$-component of $E$ is 0. We are given that the $z$ component is 12, and it lies over the square, so we must have $E(2,0,12)$ so $CE=\\sqrt{4^2+6^2+12^2}=\\sqrt{196}=14$ (the other solution, $E(10,0,12)$ does not lie over the square). Now let $F(a,-3,b)$ and $G(a,3,b)$, so $FG=6$ is parallel to $\\overline{AB}$. We must have $BF=8$, so $(a+6)^2+b^2=8^2-3^2=55$. The last piece of information we have is that $ADEG$ (and its reflection, $BCEF$) are faces of the polyhedron, so they must all lie in the same plane. Since we have $A$, $D$, and $E$, we can derive this plane.* Let $H$ be the extension of the intersection of the lines containing $\\overline{AG}, \\overline{BF}$. It follows that the projection of $\\triangle AHB$ onto the plane $x = 6$ must coincide with the $\\triangle CDE'$, where $E'$ is the projection of $E$ onto the plane $x = 6$. $\\triangle GHF \\sim \\triangle AHB$ by a ratio of $1/2$, so the distance from $H$ to the plane $x = -6$ is \\[\\sqrt{\\left(\\sqrt{(2 \\times 8)^2 - 6^2}\\right)^2 - 12^2} = 2\\sqrt{19};\\] and by the similarity, the distance from $G$ to the plane $x = -6$ is $\\sqrt{19}$. The altitude from $G$ to $ABCD$ has height $12/2 = 6$. By similarity, the x-coordinate of $G$ is $-6/2 = -3$. Then $G = (-6 \\pm \\sqrt{19}, -3, 6)$. Now that we have located $G$, we can calculate $EG^2$: \\[EG^2=(8\\pm\\sqrt{19})^2+3^2+6^2=64\\pm16\\sqrt{19}+19+9+36=128\\pm16\\sqrt{19}.\\] Taking the negative root because the answer form asks for it, we get $128-16\\sqrt{19}$, and $128+16+19=163.",
"We let $A$ be the origin, or $(0,0,0)$, $B = (0,0,12)$, and $D = (12,0,0)$. Draw the perpendiculars from F and G to AB, and let their intersections be X and Y, respectively. By symmetry, $FX = GY = \\frac{12-6}2 = 3$, so $G = (a,b,3)$, where a and b are variables. We can now calculate the coordinates of E. Drawing the perpendicular from E to CD and letting the intersection be Z, we have $CZ = DZ = 6$ and $EZ = 4\\sqrt{10}$. Therefore, the x coordinate of $E$ is $12-\\sqrt{(4\\sqrt{10})^2-12^2}=12-\\sqrt{16}=12-4=8$, so $E = (8,12,6)$. We also know that $A,D,E,$ and $G$ are coplanar, so they all lie on the plane $z = Ax+By+C$. Since $(0,0,0)$ is on it, then $C = 0$. Also, since $(12,0,0)$ is contained, then $A = 0$. Finally, since $(8,12,6)$ is on the plane, then $B = \\frac 12$. Therefore, $b = 6$. Since $GA = 8$, then $a^2+6^2+3^2=8^2$, or $a = \\pm \\sqrt{19}$. Therefore, the two permissible values of $EG^2$ are $(8 \\pm \\sqrt{19})^2+6^2+3^2 = 128 \\pm 16\\sqrt{19}$. The only one that satisfies the conditions of the problem is $128 - 16\\sqrt{19}$, from which the answer is $128+16+19=163.",
"Denote the foot of the altitude from $E$ to $ABCD$ be $X$. Let the projection of $X$ onto $AD$ be $Y$. We seek $YD=a$. Let $E=(0, 0, 0)$. Then we get $X=(0, 0, -12)$. Because the diagram is symmetrical, $Y=(a, -6, -12)$. So, $a^2+6^2+12^2=14^2 \\rightarrow a=4$. We find $EA=2\\sqrt{61}$. Extend $EG$ and $FE$ to meet the plane $z=0$. Since $EGAD$ is a quadrilateral and all on a plane, then the extension of $EG$ and $FE$ will meet the lines $AD$ and $BC$, respectively. Call these intersections $A'$ and $B'$. Let $EA'=a, AA'=b$. Using the Law of Cosines on $\\triangle EAD$ gives $\\cos(\\angle EAD)=\\frac{4}{\\sqrt{61}}$. Using Law of Cosines on $\\triangle EA'D$ gives the equation $a^2=b^2+244+16b$. Now, using Apollonius' Theorem on the same triangle gives $a^2=2b^2+232$. Equating the two gives $b^2-16b-12=0$. Solving gives us $b=8-2\\sqrt{19}, b^2=140-32\\sqrt{19}$. Finally, plugging into either expression for $a$ gives $a^2=512-64\\sqrt{19}$. Since $FG=\\frac{1}{2}A'B'$ and is parallel to $A'B'$, by the midpoint theorem, \\[EG=\\frac{1}{2} A'B' \\rightarrow EG^2=\\frac{1}{4}A'B'^2 \\rightarrow = 128-16\\sqrt{19}\\]. Then $128+16+19=163$. -RyanZhu817",
"Denote the foot of the altitude from $E$ to $ABCD$ be $X$. Let the projection of $X$ onto $AD$ be $Y$. We seek $YD=a$. Let $E=(0, 0, 0)$. Then we get $X=(0, 0, -12)$. Because the diagram is symmetrical, $Y=(a, -6, -12)$. So, $a^2+6^2+12^2=14^2 \\rightarrow a=4$. We find $EA=2\\sqrt{61}$. Extend $EG$ and $FE$ to meet the plane $z=0$. Since $EGAD$ is a quadrilateral and all on a plane, then the extension of $EG$ and $FE$ will meet the lines $AD$ and $BC$, respectively. Call these intersections $A'$ and $B'$. Let $EA'=a, AA'=b$. Using the Law of Cosines on $\\triangle EAD$ gives $\\cos(\\angle EAD)=\\frac{4}{\\sqrt{61}}$. Using Law of Cosines on $\\triangle EA'D$ gives the equation $a^2=b^2+244+16b$. Now, using Apollonius' Theorem on the same triangle gives $a^2=2b^2+232$. Equating the two gives $b^2-16b-12=0$. Solving gives us $b=8-2\\sqrt{19}, b^2=140-32\\sqrt{19}$. Finally, plugging into either expression for $a$ gives $a^2=512-64\\sqrt{19}$. Since $FG=\\frac{1}{2}A'B'$ and is parallel to $A'B'$, by the midpoint theorem, \\[EG=\\frac{1}{2} A'B' \\rightarrow EG^2=\\frac{1}{4}A'B'^2 \\rightarrow = 128-16\\sqrt{19}\\]. Then $128+16+19=163$. -RyanZhu817"
] |
2002-II-1
| 2,002 | 1 |
Given that $x$ and $y$ are both integers between $100$ and $999$ , inclusive; $y$ is the number formed by reversing the digits of $x$ ; and $z=|x-y|$ . How many distinct values of $z$ are possible?
| 9 |
II
|
[
"We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$. From this, we have \\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\\\&=&|99a-99c|\\\\&=&99|a-c|\\\\ \\end{eqnarray*} Because $a$ and $c$ are digits, and $a$ and $c$ are both between 1 and 9 (from condition 1), there are $009 can be expressed this way)."
] |
2002-II-2
| 2,002 | 2 |
Three vertices of a cube are $P=(7,12,10)$ , $Q=(8,8,1)$ , and $R=(11,3,9)$ . What is the surface area of the cube?
| 294 |
II
|
[
"$PQ=\\sqrt{(8-7)^2+(8-12)^2+(1-10)^2}=\\sqrt{98}$ $PR=\\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\\sqrt{98}$ $QR=\\sqrt{(11-8)^2+(3-8)^2+(9-1)^2}=\\sqrt{98}$ So, $PQR$ is an equilateral triangle. Let the side of the cube be $a$. $a\\sqrt{2}=\\sqrt{98}$ So, $a=7$, and hence the surface area is $6a^2=294."
] |
2002-II-3
| 2,002 | 3 |
It is given that $\log_{6}a + \log_{6}b + \log_{6}c = 6$ , where $a$ , $b$ , and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c$ .
| 111 |
II
|
[
"$abc=6^6$. Since they form an increasing geometric sequence, $b$ is the geometric mean of the product $abc$. $b=\\sqrt[3]{abc}=6^2=36$. Since $b-a$ is the square of an integer, we can find a few values of $a$ that work: $11, 20, 27, 32,$ and $35$. Out of these, the only value of $a$ that works is $a=27$, from which we can deduce that $c=\\dfrac{36}{27}\\cdot 36=\\dfrac{4}{3}\\cdot 36=48$. Thus, $a+b+c=27+36+48=111",
"Let $r$ be the common ratio of the geometric sequence. Since it is increasing, that means that $b = ar$, and $c = ar^2$. Simplifying the logarithm, we get $\\log_6(a^3 \\cdot r^3) = 6$. Therefore, $a^3 \\cdot r^3 = 6^6$. Taking the cube root of both sides, we see that $ar = 6^2 = 36$. Now since $ar = b$, that means $b = 36$. Using the trial and error shown in solution 1, we get $a = 27$, and $r = \\frac{4}{3}$. Now, $27 \\cdot r^2= c = 48$. Therefore, the answer is $27+36+48 = 111 ~idk12345678",
"Let $r$ be the common ratio of the geometric sequence. Since it is increasing, that means that $b = ar$, and $c = ar^2$. Simplifying the logarithm, we get $\\log_6(a^3 \\cdot r^3) = 6$. Therefore, $a^3 \\cdot r^3 = 6^6$. Taking the cube root of both sides, we see that $ar = 6^2 = 36$. Now since $ar = b$, that means $b = 36$. Using the trial and error shown in solution 1, we get $a = 27$, and $r = \\frac{4}{3}$. Now, $27 \\cdot r^2= c = 48$. Therefore, the answer is $27+36+48 = 111 ~idk12345678"
] |
2002-II-4
| 2,002 | 4 |
Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$ . AIME 2002 II Problem 4.gif If $n=202$ , then the area of the garden enclosed by the path, not including the path itself, is $m\left(\sqrt3/2\right)$ square units, where $m$ is a positive integer. Find the remainder when $m$ is divided by $1000$ .
| 803 |
II
|
[
"When $n>1$, the path of blocks has $6(n-1)$ blocks total in it. When $n=1$, there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is \\[(1+6+12+18+\\cdots +1200)A=(1+6(1+2+3...+200))A\\], where $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of the first $n$ integers: \\[(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A\\]. Since $A=\\dfrac{3\\sqrt{3}}{2}$, the area of the garden is \\[120601\\cdot \\dfrac{3\\sqrt{3}}{2}=\\dfrac{361803\\sqrt{3}}{2}\\]. $m=361803$, $\\dfrac{m}{1000}=361$ Remainder $803.",
"Note that this is just the definition for a centered hexagonal number, and the formula for $(n-1)^{th}$ term is $3n(n+1)+1$. Applying this for $200$ as we want the inner area gives $120601$. Then continue as above."
] |
2002-II-5
| 2,002 | 5 |
Find the sum of all positive integers $a=2^n3^m$ where $n$ and $m$ are non-negative integers, for which $a^6$ is not a divisor of $6^a$ .
| 42 |
II
|
[
"Substitute $a=2^n3^m$ into $a^6$ and $6^a$, and find all pairs of non-negative integers (n,m) for which $(2^n3^m)^{6}$ is not a divisor of $6^{2^n3^m}$ Simplifying both expressions: $2^{6n} \\cdot 3^{6m}$ is not a divisor of $2^{2^n3^m} \\cdot 3^{2^n3^m}$ Comparing both exponents (noting that there must be either extra powers of 2 or extra powers of 3 in the left expression): $6n > 2^n3^m$ OR $6m > 2^n3^m$ Using the first inequality $6n > 2^n3^m$ and going case by case starting with n $\\in$ {0, 1, 2, 3...}: n=0: $0>1 \\cdot 3^m$ which has no solution for non-negative integers m n=1: $6 > 2 \\cdot 3^m$ which is true for m=0 but fails for higher integers $\\Rightarrow (1,0)$ n=2: $12 > 4 \\cdot 3^m$ which is true for m=0 but fails for higher integers $\\Rightarrow (2,0)$ n=3: $18 > 8 \\cdot 3^m$ which is true for m=0 but fails for higher integers $\\Rightarrow (3,0)$ n=4: $24 > 16 \\cdot 3^m$ which is true for m=0 but fails for higher integers $\\Rightarrow (4,0)$ n=5: $30 > 32 \\cdot 3^m$ which has no solution for non-negative integers m There are no more solutions for higher $n$, as polynomials like $6n$ grow slower than exponentials like $2^n$. Using the second inequality $6m > 2^n3^m$ and going case by case starting with m $\\in$ {0, 1, 2, 3...}: m=0: $0>2^n \\cdot 1$ which has no solution for non-negative integers n m=1: $6>2^n \\cdot 3$ which is true for n=0 but fails for higher integers $\\Rightarrow (0,1)$ m=2: $12>2^n \\cdot 9$ which is true for n=0 but fails for higher integers $\\Rightarrow (0,2)$ m=3: $18>2^n \\cdot 27$ which has no solution for non-negative integers n There are no more solutions for higher $m$, as polynomials like $6m$ grow slower than exponentials like $3^m$. Thus there are six numbers corresponding to (1,0), (2,0), (3,0), (4,0), (0,1), and (0,2). Plugging them back into the original expression, these numbers are 2, 4, 8, 16, 3, and 9, respectively. Their sum is $042.",
"Notice that the condition is equivalent to saying \\[v_2(a^6) \\geq v_2(6^a) \\implies 6n \\geq a\\] \\[v_3(a^6) \\geq v_3(6^a) \\implies 6m \\geq a.\\] Notice that we cannot have both expressions to be equality state, as that would result in $a^6 = 6^a.$ Testing, we see the possible pairs $(n, m)$ are $(1, 0), (2, 0), (3, 0), (4, 0), (0, 1), (0, 2).$ Since the $a$ grows much faster than the left-hand side of the above inequalities, these are all possible solutions. Adding, we get $042. ~Solution by Williamgolly",
"Notice that the condition is equivalent to saying \\[v_2(a^6) \\geq v_2(6^a) \\implies 6n \\geq a\\] \\[v_3(a^6) \\geq v_3(6^a) \\implies 6m \\geq a.\\] Notice that we cannot have both expressions to be equality state, as that would result in $a^6 = 6^a.$ Testing, we see the possible pairs $(n, m)$ are $(1, 0), (2, 0), (3, 0), (4, 0), (0, 1), (0, 2).$ Since the $a$ grows much faster than the left-hand side of the above inequalities, these are all possible solutions. Adding, we get $042. ~Solution by Williamgolly"
] |
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