ID stringlengths 8 10 | Year int64 1.98k 2.02k | Problem Number int64 1 15 | Question stringlengths 37 2.66k | Answer int64 0 997 | Part stringclasses 2 values | Solution listlengths 1 25 |
|---|---|---|---|---|---|---|
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\\] ... |
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}\... |
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 $|M_1-M_2| \\leq m$. A... |
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 rel... |
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); ... |
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 angle... |
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 en... |
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... |
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 ... |
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 $02... |
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 w... |
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 $(2x+q, 2x-q)=(35, 1)$... |
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 tri... |
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... |
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. Look... |
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 $... |
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 ... |
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\\... |
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 -... |
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 th... |
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,... |
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... |
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 chec... |
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),N... |
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... |
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)--(0,1... |
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_... |
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 le... |
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... |
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... |
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... |
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,... |
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\\mat... |
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^{... |
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), ...... |
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}... |
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));labe... |
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 w... |
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 ... |
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 ha... |
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 a... |
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... |
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... |
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 $... |
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} =... |
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}))+(\\ar... |
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' ... |
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 ... |
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... |
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+\\... |
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 $n = 11a$... |
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=... |
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... |
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: $\\frac{24}{\\sin(\\angle ABC)}$ $=$ $\\frac{BC}{\\sin(\\angle BAC)}$ $\\angle ABC=45^{\\circ}$... |
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... |
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 biject... |
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(\... |
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 \\cd... |
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(... |
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 $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, ... |
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 & ... |
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); dra... |
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 $d$-degre... |
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 ca... |
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 ... |
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 i... |
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\\% ... |
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 ... |
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\... |
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 t... |
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), ... |
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) ... |
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 $f(2001) = 3^kf\\left(\\frac{2001}{3^k}\\right),\\ 1 \\le \\frac{2001}{3^k} \\le 3 \\Longrightarrow k = 6$... |
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 $... |
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^{... |
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... |
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\\ri... |
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 */ ... |
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 $\... |
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,... |
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-... |
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... |
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), (... |
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}=... |
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 a... |
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_... |
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 y... |
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 th... |
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 $1... |
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 ... |
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\\... |
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)... |
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,... |
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-... |
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 whic... |
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 th... |
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 ... |
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