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 |
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2006-I-6 | 2,006 | 6 | Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits. Find the sum of the elements of $\mathcal{S}.$ | 360 | I | [
"Numbers of the form $0.\\overline{abc}$ can be written as $\\frac{abc}{999}$. There are $10\\times9\\times8=720$ such numbers. Each digit will appear in each place value $\\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\\frac{45\\times72\\times111}{999}= ... |
2006-I-7 | 2,006 | 7 | An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is $\frac{11}{5}$ . Find the ratio of shaded region $D$ to the area of shaded region $A$ . [asy] size(6cm); defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S); [/asy] | 408 | I | [
"Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof. Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at $0, 1, 2 \\ldots$. The base of region $\\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\\mathcal{D}$ is on the line $x =... |
2006-I-8 | 2,006 | 8 | Hexagon $ABCDEF$ is divided into five rhombuses, $P, Q, R, S,$ and $T$ , as shown. Rhombuses $P, Q, R,$ and $S$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $T$ . Given that $K$ is a positive integer, find the number of possible values for $K.$ [asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy] | 89 | I | [
"Let $x$ denote the common side length of the rhombi, and let $y$ denote one of the smaller interior angles of rhombus $\\mathcal{P}$. Then $x^2\\sin(y)=\\sqrt{2006}$. We also see that $K=x^2\\sin(2y) \\implies K=2x^2\\sin y \\cdot \\cos y \\implies K = 2\\sqrt{2006}\\cdot \\cos y$. Thus $K$ can be any positive int... |
2006-I-9 | 2,006 | 9 | The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$ | 46 | I | [
"\\[\\log_8 a_1+\\log_8 a_2+\\ldots+\\log_8 a_{12}= \\log_8 a+\\log_8 (ar)+\\ldots+\\log_8 (ar^{11}) \\\\ = \\log_8(a\\cdot ar\\cdot ar^2\\cdot \\cdots \\cdot ar^{11}) = \\log_8 (a^{12}r^{66})\\] So our question is equivalent to solving $\\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers. $a^{12}r^{66}=8^{20... |
2006-I-10 | 2,006 | 10 | Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$ 's equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ [asy] unitsize(0.50cm); draw((0,-1)--(0,6)); draw((-1,0)--(6,0)); draw(shift(1,1)*unitcircle); draw(shift(1,3)*unitcircle); draw(shift(1,5)*unitcircle); draw(shift(3,1)*unitcircle); draw(shift(3,3)*unitcircle); draw(shift(3,5)*unitcircle); draw(shift(5,1)*unitcircle); draw(shift(5,3)*unitcircle); [/asy] | 65 | I | [
"[asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j<array[i]; j=j+1) { draw(Circle((1+2*i,1+2*j),1)); }} label(\"x\", (7,0)); label(\"y\", (0,7)); draw((5/3,0)--(23/6,6.5),blue);[/asy] Solution 1 The line passing thr... |
2006-I-12 | 2,006 | 12 | Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x,$ where $x$ is measured in degrees and $100< x< 200.$ | 906 | I | [
"Observe that $2\\cos 4x\\cos x = \\cos 5x + \\cos 3x$ by the sum-to-product formulas. Defining $a = \\cos 3x$ and $b = \\cos 5x$, we have $a^3 + b^3 = (a+b)^3 \\rightarrow ab(a+b) = 0$. But $a+b = 2\\cos 4x\\cos x$, so we require $\\cos x = 0$, $\\cos 3x = 0$, $\\cos 4x = 0$, or $\\cos 5x = 0$. Hence we see by car... |
2006-I-13 | 2,006 | 13 | For each even positive integer $x,$ let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square. | 899 | I | [
"Given $g : x \\mapsto \\max_{j : 2^j | x} 2^j$, consider $S_n = g(2) + \\cdots + g(2^n)$. Define $S = \\{2, 4, \\ldots, 2^n\\}$. There are $2^0$ elements of $S$ that are divisible by $2^n$, $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \\ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ ... |
2006-I-14 | 2,006 | 14 | A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$ ) | 183 | I | [
"[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,dd... |
2006-I-15 | 2,006 | 15 | Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$ | 27 | I | [
"Solution 1 Suppose $b_{i} = \\frac {x_{i}}3$. We have \\[\\sum_{i = 1}^{2006}b_{i}^{2} = \\sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \\sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)\\] So \\[\\sum_{i = 0}^{2005}b_{i} = \\frac {b_{2006}^{2} - 2006}2\\] Now \\[\\sum_{i = 1}^{2006}b_{i} = \\frac {b_{2006}^{2} + 2b_{2006} - 200... |
2006-II-1 | 2,006 | 1 | In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$ . | 46 | II | [
"Let the side length be called $x$, so $x=AB=BC=CD=DE=EF=AF$. The diagonal $BF=\\sqrt{AB^2+AF^2}=\\sqrt{x^2+x^2}=x\\sqrt{2}$. Then the areas of the triangles AFB and CDE in total are $\\frac{x^2}{2}\\cdot 2$, and the area of the rectangle BCEF equals $x\\cdot x\\sqrt{2}=x^2\\sqrt{2}$ Then we have to solve the equat... |
2006-II-2 | 2,006 | 2 | The lengths of the sides of a triangle with positive area are $\log_{10} 12$ , $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$ . | 893 | II | [
"By the Triangle Inequality and applying the well-known logarithmic property $\\log_{c} a + \\log_{c} b = \\log_{c} ab$, we have that $\\log_{10} 12 + \\log_{10} n > \\log_{10} 75$ $\\log_{10} 12n > \\log_{10} 75$ $12n > 75$ $n > \\frac{75}{12} = \\frac{25}{4} = 6.25$ Also, $\\log_{10} 12 + \\log_{10} 75 > \\log_{1... |
2006-II-3 | 2,006 | 3 | Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$ . | 49 | II | [
"Note that the product of the first $100$ positive odd integers can be written as $1\\cdot 3\\cdot 5\\cdot 7\\cdots 195\\cdot 197\\cdot 199=\\frac{1\\cdot 2\\cdots200}{2\\cdot4\\cdots200} = \\frac{200!}{2^{100}\\cdot 100!}$ Hence, we seek the number of threes in $200!$ decreased by the number of threes in $100!.$ T... |
2006-II-5 | 2,006 | 5 | When rolling a certain unfair six-sided die with faces numbered $1, 2, 3, 4, 5$ , and $6$ , the probability of obtaining face $F$ is greater than $\frac{1}{6}$ , the probability of obtaining the face opposite is less than $\frac{1}{6}$ , the probability of obtaining any one of the other four faces is $\frac{1}{6}$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $\frac{47}{288}$ . Given that the probability of obtaining face $F$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | 29 | II | [
"Without loss of generality, assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a proba... |
2006-II-6 | 2,006 | 6 | Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$ | 12 | II | [
"[asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C,... |
2006-II-7 | 2,006 | 7 | Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | 738 | II | [
"Solution 1 There are $\\left\\lfloor\\frac{999}{10}\\right\\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multip... |
2006-II-8 | 2,006 | 8 | There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color. Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed? [asy] size(50); pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy] | 336 | II | [
"If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle ... |
2006-II-9 | 2,006 | 9 | Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope, and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y),$ and that $x=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$ | 27 | II | [
"Call the centers $O_1, O_2, O_3$, the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$, and $s_2$ on $\\mathcal{C}_2$), and the intersection of each common internal tangent to the X-axis $r, s$. $\\triangle O_1r_1r \\sim \\triangle O_2r_2r$ since both triangles have a right angle and hav... |
2006-II-10 | 2,006 | 10 | Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 831 | II | [
"Solution 1 The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$. We let this probability be $p$; then the probability that $A$ and $B$ end with the s... |
2006-II-11 | 2,006 | 11 | A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | 834 | II | [
"Solution 1 Define the sum as $s$. Since $a_n\\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, the sum will be: $s = a_{28} + \\sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\\\ s = a_{28} + \\left(\\sum^{30}_{k=4} a_{k} - \\sum^{29}_{k=3} a_{k}\\right) - \\left(\\sum^{28}_{k=2} a_{k}\\right)\\\\ s = a_{28} + (a_{30} - a_{3}) -... |
2006-II-12 | 2,006 | 12 | Equilateral $\triangle ABC$ is inscribed in a circle of radius 2. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$ | 865 | II | [
"[asy] size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8); pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D; path O = CP((0,-2),A); pair G = OP(A--F,O); D(MP(\"A\",A,N,s)--MP(\"B\",B,W,s)--MP(\"C\",C,E,s)--cycle);D(O); D(B--MP(\"D\",D,W,s)--MP(\"F\",... |
2006-II-13 | 2,006 | 13 | How many integers $N$ less than 1000 can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$ ? | 15 | II | [
"Let the first odd integer be $2n+1$, $n\\geq 0$. Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$. The odd integers form an arithmetic sequence with sum $N = j\\left(\\frac{(2n+1) + (2(n+j)-1)}{2}\\right) = j(2n+j)$. Thus, $j$ is a factor of $N$. Since $n\\geq 0$, it follows that $2n+j \\geq j$ and $j\\l... |
2006-II-14 | 2,006 | 14 | Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from $1$ to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer. | 63 | II | [
"Let $K = \\sum_{i=1}^{9}{\\frac{1}{i}}$. Examining the terms in $S_1$, we see that $S_1 = K + 1$ since each digit $n$ appears once and 1 appears an extra time. Now consider writing out $S_2$. Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so ... |
2006-II-15 | 2,006 | 15 | Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$ | 9 | II | [
"Let $\\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all altitudes are in the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the Pythagorean Theore... |
2007-I-1 | 2,007 | 1 | How many positive perfect squares less than $10^6$ are multiples of 24? | 83 | I | [
"The prime factorization of $24$ is $2^3\\cdot3$. Thus, each square must have at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$. This means that each square is in the form $(12c)^2$, where $12 c$ is a positive integer less than $\\sqrt{10^6}$. T... |
2007-I-2 | 2,007 | 2 | A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person. | 52 | I | [
"Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there i... |
2007-I-3 | 2,007 | 3 | The complex number $z$ is equal to $9+bi$ , where $b$ is a positive real number and $i^{2}=-1$ . Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to? | 15 | I | [
"Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$. Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \\ldots + (9\\cdot 18 + 81)bi - b^3i$. Setting these two equal, we get that $18bi = 243bi - b^3i$, so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$. Since $b > 0$, the soluti... |
2007-I-4 | 2,007 | 4 | Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are $60$ , $84$ , and $140$ years. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again? | 105 | I | [
"Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \\frac{t \\pi}{30}$, $b(t) = \\frac{t \\pi}{42}$, $c(t) = \\frac{t \\pi}{70}$. In order for the planets and the central... |
2007-I-5 | 2,007 | 5 | The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer. For how many integer Fahrenheit temperatures between $32$ and $1000$ inclusive does the original temperature equal the final temperature? | 539 | I | [
"Solution 1 Examine $F - 32$ modulo 9. If $F - 32 \\equiv 0 \\pmod{9}$, then we can define $9x = F - 32$. This shows that $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(F-32)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x) + 32\\right] \\Longrightarrow F = 9x + 32$. This case works. If $F - 32 \\equi... |
2007-I-6 | 2,007 | 6 | A frog is placed at the origin on the number line , and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of $3$ , or to the closest point with a greater integer coordinate that is a multiple of $13$ . A move sequence is a sequence of coordinates which correspond to valid moves, beginning with $0$ , and ending with $39$ . For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog? | 169 | I | [
"Solution 1 Let us keep a careful tree of the possible number of paths around every multiple of $13$. From $0 \\Rightarrow 13$, we can end at either $12$ (mult. of 3) or $13$ (mult. of 13). Only $1$ path leads to $12$ Continuing from $12$, there is $1 \\cdot 1 = 1$ way to continue to $24$ There are $1 \\cdot \\left... |
2007-I-7 | 2,007 | 7 | Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )$ Find the remainder when $N$ is divided by 1000. ( $\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$ , and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$ .) | 477 | I | [
"The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer); otherwise, it is equal to 1. Thus, we need to find when or not $\\log_{\\sqrt{2}} k$ is an integer. The change of base formula shows that $\\frac{\\log k}{\\log \\sqrt{2}} = \\frac{2 \\log k}{\\log 2}$. For t... |
2007-I-8 | 2,007 | 8 | The polynomial $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$ ? | 30 | I | [
"Solution 1 We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic. Let this root be $a$. We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \\frac{-k}{5}$. We then know that $\\frac{-k}{5}$ is a root ... |
2007-I-9 | 2,007 | 9 | In right triangle $ABC$ with right angle $C$ , $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ . Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius of either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 737 | I | [
"Solution 1 Label the points as in the diagram above. If we draw $\\overline{O_1A}$ and $\\overline{O_2B}$, we form two right triangles. As $\\overline{AF}$ and $\\overline{AD}$ are both tangents to the circle, we see that $\\overline{O_1A}$ is an angle bisector. Thus, $\\triangle AFO_1 \\cong \\triangle ADO_1$. Ca... |
2007-I-10 | 2,007 | 10 | In a $6 \times 4$ grid ( $6$ rows, $4$ columns), $12$ of the $24$ squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by $1000$ . AIME I 2007-10.png | 860 | I | [
"Solution 1 Consider the first column. There are ${6\\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer. Now consider the 3x3 that... |
2007-I-11 | 2,007 | 11 | For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | 955 | I | [
"$\\left(k- \\frac 12\\right)^2=k^2-k+\\frac 14$ and $\\left(k+ \\frac 12\\right)^2=k^2+k+ \\frac 14$. Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\\leq k^2+k$. There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$. $44<\\sqrt{2007}<45$, so all numbers $1$... |
2007-I-12 | 2,007 | 12 | In isosceles triangle $ABC$ , $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $(p-q+r-s)/2$ . | 875 | I | [
"[asy] defaultpen(fontsize(12)+0.6); size(300); var theta=15; pair A=origin, B=(20,0), C=extension(A,dir(75),B/2,bisectorpoint(A,B)), Cp=rotate(theta,A)*C, Bp=rotate(theta,A)*B, X=extension(A,Bp,B,C), Y=extension(B,C,Bp,Cp); draw(A--B--C--A); draw(A--Bp--Cp--A, royalblue); markscalefactor=0.1; draw(rightanglemark(Y... |
2007-I-13 | 2,007 | 13 | A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length 4. A plane passes through the midpoints of $AE$ , $BC$ , and $CD$ . The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$ . Find $p$ . | 80 | I | [
"Solution 1 Note first that the intersection is a pentagon. Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\\ B(2,2,0),\\ C(2,-2,0),\\ D(-2,-2,0),\\ E(0,0,2\\sqrt{2})$. Using the coordinates of the three points of intersec... |
2007-I-14 | 2,007 | 14 | A sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ , $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$ | 224 | I | [
"Solution 1 Define a function $f(n)$ on the non-negative integers, as \\[f(n) = \\frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \\frac{a_n}{a_{n+1}}+\\frac{a_{n+1}}{a_{n}}\\] We want $\\left\\lfloor f(2006) \\right\\rfloor$. Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$. Dividing through by $a_{n}a_{n-1}$, we get \... |
2007-I-15 | 2,007 | 15 | Let $ABC$ be an equilateral triangle, and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ . Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ . The area of triangle $DEF$ is $14\sqrt{3}$ . The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime. Find $r$ . | 989 | I | [
"Denote the length of a side of the triangle $x$, and of $\\overline{AE}$ as $y$. The area of the entire equilateral triangle is $\\frac{x^2\\sqrt{3}}{4}$. Add up the areas of the triangles using the $\\frac{1}{2}ab\\sin C$ formula (notice that for the three outside triangles, $\\sin 60 = \\frac{\\sqrt{3}}{2}$): $\... |
2007-II-1 | 2,007 | 1 | A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in $2007$ . No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in $2007$ . A set of plates in which each possible sequence appears exactly once contains N license plates. Find N/10. | 372 | II | [
"There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice. If $0$ appears 0 or 1 times amongst the sequence, there are $\\frac{7!}{(7-5)!} = 2520$ sequences possible. If $0$ appears twice in the sequence, there are ${5\\choose2} = 10$ places to place the $0$s. The... |
2007-II-2 | 2,007 | 2 | Find the number of ordered triples $(a,b,c)$ where $a$ , $b$ , and $c$ are positive integers , $a$ is a factor of $b$ , $a$ is a factor of $c$ , and $a+b+c=100$ . | 200 | II | [
"Denote $x = \\frac{b}{a}$ and $y = \\frac{c}{a}$. The last condition reduces to $a(1 + x + y) = 100$. Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$. Subtracting the one, we see that $x + y = \\{0,1,3,4,9,19,24,49,99\\}$. There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y ... |
2007-II-4 | 2,007 | 4 | The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$ . | 450 | II | [
"Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit. Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on): $100 = 300x + 200y$ $2(60) = 240x + 300y$ $3(50) = 150x + my$ Solve the system of ... |
2007-II-5 | 2,007 | 5 | The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant ? | 888 | II | [
"Solution 1 There are $223 \\cdot 9 = 2007$ squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are $(223,0),\\ (0,9)$. Count the number of squares that the diagonal of the rectangle passes through. Since the ... |
2007-II-6 | 2,007 | 6 | An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd , and $a_{i}>a_{i+1}$ if $a_{i}$ is even . How many four-digit parity-monotonic integers are there? | 640 | II | [
"Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of $a_1,\\ a_2,\\ a_3$ because of the given conditions. (Also note that 0 cannot appear as 0 cannot be the first digit of an integer.) A clear pattern emerges. For example, for $3$ in the second column, we note that $3$ is less than $4,6... |
2007-II-7 | 2,007 | 7 | Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$ Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$ | 553 | II | [
"Solution 1 For $x = 1$, we see that $\\sqrt[3]{1} \\ldots \\sqrt[3]{7}$ all work, giving 7 integers. For $x=2$, we see that in $\\sqrt[3]{8} \\ldots \\sqrt[3]{26}$, all of the even numbers work, giving 10 integers. For $x = 3$, we get 13, and so on. We can predict that at $x = 22$ we get 70. To prove this, note th... |
2007-II-9 | 2,007 | 9 | Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$ | 259 | II | [
"Solution 1 Several Pythagorean triples exist amongst the numbers given. $BE = DF = \\sqrt{63^2 + 84^2} = 21\\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \\sqrt{63^2 + (448 - 2\\cdot84)^2} = 7\\sqrt{9^2 + 40^2} = 287$. Use the Two Tangent Theorem on $\\triangle BEF$. Since both circles are inscribed in congru... |
2007-II-10 | 2,007 | 10 | Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ , $n$ , and $r$ are positive integers , $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$ ) | 710 | II | [
"Use casework: $B$ has 6 elements: Probability: $\\frac{1}{2^6} = \\frac{1}{64}$ $A$ must have either 0 or 6 elements, probability: $\\frac{2}{2^6} = \\frac{2}{64}$. $B$ has 5 elements: Probability: ${6\\choose5}/64 = \\frac{6}{64}$ $A$ must have either 0, 6, or 1, 5 elements. The total probability is $\\frac{2}{64... |
2007-II-11 | 2,007 | 11 | Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface . The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$ . It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime . Find $a+b+c.$ | 179 | II | [
"If it weren’t for the small tube, the larger tube would travel $144\\pi$. Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube. Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in th... |
2007-II-12 | 2,007 | 12 | The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that $\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$ find $\log_{3}(x_{14}).$ | 91 | II | [
"Suppose that $x_0 = a$, and that the common ratio between the terms is $r$. The first conditions tells us that $\\log_3 a + \\log_3 ar + \\ldots + \\log_3 ar^7 = 308$. Using the rules of logarithms, we can simplify that to $\\log_3 a^8r^{1 + 2 + \\ldots + 7} = 308$. Thus, $a^8r^{28} = 3^{308}$. Since all of the te... |
2007-II-13 | 2,007 | 13 | A triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3$ ? [asy] for (int i=0; i<12; ++i){ for (int j=0; j<i; ++j){ //dot((-j+i/2,-i)); draw((-j+i/2,-i)--(-j+i/2+1,-i)--(-j+i/2+1,-i+1)--(-j+i/2,-i+1)--cycle); } } [/asy] | 640 | II | [
"Label each of the bottom squares as $x_0, x_1 \\ldots x_9, x_{10}$. Through induction, we can find that the top square is equal to ${10\\choose0}x_0 + {10\\choose1}x_1 + {10\\choose2}x_2 + \\ldots {10\\choose10}x_{10}$. (This also makes sense based on a combinatorial argument: the number of ways a number can \"tra... |
2007-II-14 | 2,007 | 14 | Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ , $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | 676 | II | [
"Let $r$ be a root of $f(x)$. Then we have $f(r)f(2r^2)=f(2r^3+r)$; since $r$ is a root, we have $f(r)=0$; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$, so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has n... |
2007-II-15 | 2,007 | 15 | Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ , $\omega_{B}$ to $BC$ and $BA$ , $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 389 | II | [
"[asy] defaultpen(fontsize(12)+0.8); size(350); pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcente... |
2008-I-1 | 2,008 | 1 | Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance? | 252 | I | [
"Solution 1 Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance. Thus, $\\dfrac{3k}{5k + 20} = \\dfrac{29}{50}$, solving gives $k = 116$. Thus, the number of people that like to dance is $2k + 20 = 252",... |
2008-I-2 | 2,008 | 2 | Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$ , and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$ . | 25 | I | [
"Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\\cdot10=50$. This implies that vertex G must be located outside of square $AIME$. [asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw... |
2008-I-3 | 2,008 | 3 | Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates. | 314 | I | [
"Let the biking rate be $b$, swimming rate be $s$, jogging rate be $j$, all in km/h. We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$. Subtracting the second from twice the first gives $4j + 5s = 57$. Mod 4, we need $s\\equiv1\\pmod{4}$. Thus, $(j,s) = (13,1),(8,5),(3,9)$. $(13,1)$ and $(3,9)$ give non-integral $b$, b... |
2008-I-4 | 2,008 | 4 | There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$ . | 80 | I | [
"Completing the square, $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$. Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares. Since $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \\cdot 61$, the factors must... |
2008-I-5 | 2,008 | 5 | A right circular cone has base radius $r$ and height $h$ . The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making $17$ complete rotations. The value of $h/r$ can be written in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ . | 14 | I | [
"The path is a circle with radius equal to the slant height of the cone, which is $\\sqrt {r^{2} + h^{2}}$. Thus, the length of the path is $2\\pi\\sqrt {r^{2} + h^{2}}$. Also, the length of the path is 17 times the circumference of the base, which is $34r\\pi$. Setting these equal gives $\\sqrt {r^{2} + h^{2}} = 1... |
2008-I-6 | 2,008 | 6 | A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$ ? | 17 | I | [
"Let the $k$th number in the $n$th row be $a(n,k)$. Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$.[1] We wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$. Since $2^{n-1}$ and $67$ are relatively prime, it follows that $67|n+2k-2$. Since every row has one less element than the p... |
2008-I-7 | 2,008 | 7 | Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$ . For example, $S_4$ is the set ${400,401,402,\ldots,499}$ . How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square? | 708 | I | [
"The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$, which means that all squares above $50^2 = 2500$ are more than $100$ apart. Then the first $26$ sets ($S_0,\\cdots S_{25}$) each have at least one perfect square because the differences between consecutive squares in them are all less than $... |
2008-I-8 | 2,008 | 8 | Find the positive integer $n$ such that \[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\] | 47 | I | [
"Since we are dealing with acute angles, $\\tan(\\arctan{a}) = a$. Note that $\\tan(\\arctan{a} + \\arctan{b}) = \\dfrac{a + b}{1 - ab}$, by tangent addition. Thus, $\\arctan{a} + \\arctan{b} = \\arctan{\\dfrac{a + b}{1 - ab}}$. Applying this to the first two terms, we get $\\arctan{\\dfrac{1}{3}} + \\arctan{\\dfra... |
2008-I-9 | 2,008 | 9 | Ten identical crates each of dimensions $3$ ft $\times$ $4$ ft $\times$ $6$ ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41$ ft tall, where $m$ and $n$ are relatively prime positive integers. Find $m$ . | 190 | I | [
"Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following: \\begin{align*}3a + 4b + 6c &= 41\\\\ a + b + c &= 10\\end{align*} Subtracting 3 times the second from the first gives $b + 3c = 11$, or $(b,c) = (2,3),(5,2),(8,1),(11,0)$. The last doesn't work, o... |
2008-I-10 | 2,008 | 10 | Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$ . The diagonals have length $10\sqrt {21}$ , and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$ , respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$ . The distance $EF$ can be expressed in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ . | 32 | I | [
"[asy] size(300); defaultpen(1); pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0); draw(F--C--B--A); draw(E--A--D--C); draw(A--C,dashed); draw(circle(A,abs(C-A)),dotted); label(\"\\(A\\)\",A,S); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NE); label(\"\\(D\\)\",D,SE); label(\"\\(... |
2008-I-11 | 2,008 | 11 | Consider sequences that consist entirely of $A$ 's and $B$ 's and that have the property that every run of consecutive $A$ 's has even length, and every run of consecutive $B$ 's has odd length. Examples of such sequences are $AA$ , $B$ , and $AABAA$ , while $BBAB$ is not such a sequence. How many such sequences have length 14? | 172 | I | [
"Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$. If a sequence ends in an $A$, then it must have been formed by appending two $A$s to the end of a string of length $n-2$. If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a str... |
2008-I-12 | 2,008 | 12 | On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by 10. | 375 | I | [
"Let $n$ be the number of car lengths that separates each car (it is easy to see that this should be the same between each pair of consecutive cars.) Then their speed is at most $15n$. Let a unit be the distance between the cars (front to front). Then the length of each unit is $4(n + 1)$. To maximize, in a unit, t... |
2008-I-13 | 2,008 | 13 | Let $p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3$ . Suppose that $p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1) = p(1,1) = p(1, - 1) = p(2,2) = 0$ . There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$ , $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$ . Find $a + b + c$ . | 40 | I | [
"Solution 1 \\begin{align*} p(0,0) &= a_0 \\\\ &= 0 \\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\\\ &= a_1 + a_3 + a_6 \\\\ &= 0 \\\\ p(-1,0) &= -a_1 + a_3 - a_6 \\\\ &= 0 \\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ give... |
2008-I-14 | 2,008 | 14 | Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of segment $BP$ . Find $m^{2}$ . | 432 | I | [
"Solution 1 [asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rig... |
2008-II-1 | 2,008 | 1 | Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$ . | 100 | II | [
"Rewriting this sequence with more terms, we have \\begin{align*} N &= 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + 95^2 - 94^2 - 93^2 + 92^2 + 91^2 + \\ldots - 10^2 - 9^2 + 8^2 + 7^2 - 6^2 - 5^2 + 4^2 + 3^2 - 2^2 - 1^2 \\mbox{, and reordering, we get}\\\\ N &= (100^2 - 98^2) + (99^2 - 97^2) + (96^2 - 94^2) + (95^2 - 93^2) ... |
2008-II-2 | 2,008 | 2 | Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -mile mark at exactly the same time. How many minutes has it taken them? | 620 | II | [
"Let Rudolf bike at a rate $r$, so Jennifer bikes at the rate $\\dfrac 34r$. Let the time both take be $t$. Then Rudolf stops $49$ times (because the rest after he reaches the finish does not count), losing a total of $49 \\cdot 5 = 245$ minutes, while Jennifer stops $24$ times, losing a total of $24 \\cdot 5 = 120... |
2008-II-3 | 2,008 | 3 | A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off? | 729 | II | [
"Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$, so that the desired volume is $abc$. Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$. By AM-GM, $\\frac{a+b+c}{3} = 9 \\ge \\sqrt[3]{abc} \\Longrightarrow abc ... |
2008-II-4 | 2,008 | 4 | There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ integers $a_k$ ( $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$ . | 21 | II | [
"In base $3$, we find that $\\overline{2008}_{10} = \\overline{2202101}_{3}$. In other words, $2008 = 2 \\cdot 3^{6} + 2 \\cdot 3^{5} + 2 \\cdot 3^3 + 1 \\cdot 3^2 + 1 \\cdot 3^0$ In order to rewrite as a sum of perfect powers of $3$, we can use the fact that $2 \\cdot 3^k = 3^{k+1} - 3^k$: $2008 = (3^7 - 3^6) + (3... |
2008-II-5 | 2,008 | 5 | In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ , $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ . | 504 | II | [
"Solution 1 Extend $\\overline{AB}$ and $\\overline{CD}$ to meet at a point $E$. Then $\\angle AED = 180 - 53 - 37 = 90^{\\circ}$. [asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M ... |
2008-II-6 | 2,008 | 6 | The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$ . | 561 | II | [
"Rearranging the definitions, we have \\[\\frac{a_n}{a_{n-1}} = \\frac{a_{n-1}}{a_{n-2}} + 1,\\quad \\frac{b_n}{b_{n-1}} = \\frac{b_{n-1}}{b_{n-2}} + 1\\] from which it follows that $\\frac{a_n}{a_{n-1}} = 1+ \\frac{a_{n-1}}{a_{n-2}} = \\cdots = (n-1) + \\frac{a_{1}}{a_0} = n$ and $\\frac{b_n}{b_{n-1}} = (n-1) + \\... |
2008-II-7 | 2,008 | 7 | Let $r$ , $s$ , and $t$ be the three roots of the equation \[8x^3 + 1001x + 2008 = 0.\] Find $(r + s)^3 + (s + t)^3 + (t + r)^3$ . | 753 | II | [
"By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is \\[(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).\\] Also by Vieta's formulas we have \\[rst = -rs(r + s) = -\\dfrac{2008}{8} = -251\\] so negating both sides and multiplying thr... |
2008-II-8 | 2,008 | 8 | Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer. | 251 | II | [
"Solution 1 By the product-to-sum identities, we have that $2\\cos a \\sin b = \\sin (a+b) - \\sin (a-b)$. Therefore, this reduces to a telescope series: \\begin{align*} \\sum_{k=1}^{n} 2\\cos(k^2a)\\sin(ka) &= \\sum_{k=1}^{n} [\\sin(k(k+1)a) - \\sin((k-1)ka)]\\\\ &= -\\sin(0) + \\sin(2a)- \\sin(2a) + \\sin(6a) - \... |
2008-II-9 | 2,008 | 9 | A particle is located on the coordinate plane at $(5,0)$ . Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle's position after $150$ moves is $(p,q)$ , find the greatest integer less than or equal to $|p| + |q|$ . | 19 | II | [
"Solution 1 Let $P(x, y)$ be the position of the particle on the $xy$-plane, $r$ be the length $OP$ where $O$ is the origin, and $\\theta$ be the inclination of OP to the x-axis. If $(x', y')$ is the position of the particle after a move from $P$, then we have two equations for $x'$ and $y'$: \\[x'=r\\cos(\\pi/4+\\... |
2008-II-11 | 2,008 | 11 | In triangle $ABC$ , $AB = AC = 100$ , and $BC = 56$ . Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to circle $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\bigtriangleup\overline{ABC}$ . The radius of circle $Q$ can be expressed in the form $m - n\sqrt{k}$ ,where $m$ , $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m +nk$ . | 254 | II | [
"[asy] size(200); pathpen=black;pointpen=black;pen f=fontsize(9); real r=44-6*35^.5; pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); path PC=CR(P,16),QC=CR(Q,r); D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); D(PC); D(QC); MP(\"A\",A,N,f);MP(\"B\",B,f);MP(... |
2008-II-12 | 2,008 | 12 | There are two distinguishable flagpoles, and there are $19$ flags, of which $10$ are identical blue flags, and $9$ are identical green flags. Let $N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when $N$ is divided by $1000$ . | 310 | II | [
"Solution 1 The well known problem of ordering $x$ elements of a string of $y$ elements such that none of the $x$ elements are next to each other has ${y-x+1\\choose x}$ solutions. (1) We generalize for $a$ blues and $b$ greens. Consider a string of $a+b$ elements such that we want to choose the greens such that no... |
2008-II-13 | 2,008 | 13 | A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ . | 29 | II | [
"If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate ... |
2008-II-14 | 2,008 | 14 | Let $a$ and $b$ be positive real numbers with $a \ge b$ . Let $\rho$ be the maximum possible value of $\dfrac{a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a-x)^2 + (b-y)^2\] has a solution $(x,y)$ satisfying $0 \le x < a$ and $0 \le y < b$ . Then $\rho^2$ can be expressed as a fraction $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 7 | II | [
"Solution 1 Notice that the given equation implies $a^2 + y^2 = b^2 + x^2 = 2(ax + by)$ We have $2by \\ge y^2$, so $2ax \\le a^2 \\implies x \\le \\frac {a}{2}$. Then, notice $b^2 + x^2 = a^2 + y^2 \\ge a^2$, so $b^2 \\ge \\frac {3}{4}a^2 \\implies \\rho^2 \\le \\frac {4}{3}$. The solution $(a, b, x, y) = \\left(1,... |
2009-I-1 | 2,009 | 1 | Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | 840 | I | [
"Assume that the largest geometric number starts with a $9$. We know that the common ratio must be a rational of the form $k/3$ for some integer $k$, because a whole number should be attained for the 3rd term as well. When $k = 1$, the number is $931$. When $k = 2$, the number is $964$. When $k = 3$, we get $999$, ... |
2009-I-2 | 2,009 | 2 | There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that \[\frac {z}{z + n} = 4i.\] Find $n$ . | 697 | I | [
"Let $z = a + 164i$. Then \\[\\frac {a + 164i}{a + 164i + n} = 4i\\] and \\[a + 164i = \\left (4i \\right ) \\left (a + n + 164i \\right ) = 4i \\left (a + n \\right ) - 656.\\] By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation, we conclude that \\[a = -656.\\] By... |
2009-I-3 | 2,009 | 3 | A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 11 | I | [
"The probability of three heads and five tails is $\\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\\binom {8}{3}p^5(1-p)^3$. \\begin{align*} 25\\binom {8}{3}p^3(1-p)^5&=\\binom {8}{3}p^5(1-p)^3 \\\\ 25(1-p)^2&=p^2 \\\\ 25p^2-50p+25&=p^2 \\\\ 24p^2-50p+25&=0 \\\\ p&=\\frac {5}{6}\\en... |
2009-I-4 | 2,009 | 4 | In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ . | 177 | I | [
"One of the ways to solve this problem is to make this parallelogram a straight line. So the whole length of the line is $APC$($AMC$ or $ANC$), and $ABC$ is $1000x+2009x=3009x.$ $AP$($AM$ or $AN$) is $17x.$ So the answer is $3009x/17x = 177",
"Draw a diagram with all the given points and lines involved. Construct... |
2009-I-5 | 2,009 | 5 | Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$ . | 72 | I | [
"[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,m... |
2009-I-6 | 2,009 | 6 | How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ ? | 412 | I | [
"First, $x$ must be less than $5$, since otherwise $x^{\\lfloor x\\rfloor}$ would be at least $3125$ which is greater than $1000$. Because ${\\lfloor x\\rfloor}$ must be an integer, let’s do case work based on ${\\lfloor x\\rfloor}$: For ${\\lfloor x\\rfloor}=0$, $N=1$ as long as $x \\neq 0$. This gives us $1$ valu... |
2009-I-7 | 2,009 | 7 | The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$ . | 41 | I | [
"The best way to solve this problem is to get the iterated part out of the exponent: \\[5^{(a_{n + 1} - a_n)} = \\frac {1}{n + \\frac {2}{3}} + 1\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {n + \\frac {5}{3}}{n + \\frac {2}{3}}\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {3n + 5}{3n + 2}\\] \\[a_{n + 1} - a_n = \\log_5{\\left(\... |
2009-I-8 | 2,009 | 8 | Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ . | 398 | I | [
"When computing $N$, the number $2^x$ will be added $x$ times (for terms $2^x-2^0$, $2^x-2^1$, ..., $2^x - 2^{x-1}$), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\\cdot 2^{10} + 8\\cdot 2^9 + 6\\cdot 2^8 + \\cdots - 8\\cdot 2^1 - 10\\cdot 2^0$. Evaluating $N \\bmod {1000}$ yields: \\begin{align*}... |
2009-I-9 | 2,009 | 9 | A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $\text{\textdollar}1$ to $\text{\textdollar}9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$ . Find the total number of possible guesses for all three prizes consistent with the hint. | 420 | I | [
"[Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$, that could have been on that day.] Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$, then the string is \\[... |
2009-I-10 | 2,009 | 10 | The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Earthling must occupy chair $15$ . Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $N \cdot (5!)^3$ . Find $N$ . | 346 | I | [
"Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es. Pretend the table only seats $3$ \"people\", with $1$ \"person\" from each planet. Counting ... |
2009-I-11 | 2,009 | 11 | Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer. | 600 | I | [
"Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). $\\det \\left(\\begin{array}{c} P \\\\ Q\\end{array}\\right)=\\det \\left(\\begin{arr... |
2009-I-12 | 2,009 | 12 | In right $\triangle ABC$ with hypotenuse $\overline{AB}$ , $AC = 12$ , $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 11 | I | [
"First, note that $AB=37$; let the tangents from $I$ to $\\omega$ have length $x$. Then the perimeter of $\\triangle ABI$ is equal to \\[2(x+AD+DB)=2(x+37).\\] It remains to compute $\\dfrac{2(x+37)}{37}=2+\\dfrac{2}{37}x$. Observe $CD=\\dfrac{12\\cdot 35}{37}=\\dfrac{420}{37}$, so the radius of $\\omega$ is $\\dfr... |
2009-I-13 | 2,009 | 13 | The terms of the sequence $\{a_i\}$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ . | 90 | I | [
"Our expression is \\[a_{n + 2} = \\frac {a_n + 2009} {1 + a_{n + 1}}.\\] Manipulate this to obtain: \\[a_{n + 2}a_{n + 1}+a_{n+2}= a_n + 2009.\\] Our goal is to cancel terms. If we substitute in $n+1$ for $n,$ we get: \\[a_{n+3}a_{n+2}+a_{n+3}=a_{n+1}+2009.\\] Subtracting these two equations and manipulating the e... |
2009-I-14 | 2,009 | 14 | For $t = 1, 2, 3, 4$ , define $S_t = \sum_{i = 1}^{350}a_i^t$ , where $a_i \in \{1,2,3,4\}$ . If $S_1 = 513$ and $S_4 = 4745$ , find the minimum possible value for $S_2$ . | 905 | I | [
"Because the order of the $a$'s doesn't matter, we simply need to find the number of $1$s $2$s $3$s and $4$s that minimize $S_2$. So let $w, x, y,$ and $z$ represent the number of $1$s, $2$s, $3$s, and $4$s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $a... |
2009-I-15 | 2,009 | 15 | In triangle $ABC$ , $AB = 10$ , $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed 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 $a + b + c$ . | 150 | I | [
"First, by the Law of Cosines, we have \\[\\cos BAC = \\frac {16^2 + 10^2 - 14^2}{2\\cdot 10 \\cdot 16} = \\frac {256+100-196}{320} = \\frac {1}{2},\\] so $\\angle BAC = 60^\\circ$. Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$, respectively. We first compute \\[\\angle BO_1D = \\angle B... |
2009-II-1 | 2,009 | 1 | Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left. | 114 | II | [
"Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$. After painting the wall, the total amount of paint is $482-4x$. Because the total amount in each color is the same after finishing, we have \\[\\frac{482-4x}{3} = 130-x\\] \\[482-4x=390-3x\\] \\[x=92\\] Our answer is $482-4\... |
2009-II-2 | 2,009 | 2 | Suppose that $a$ , $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ , $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\] | 469 | II | [
"First, we have: \\[x^{(\\log_y z)^2} = x^{\\left( (\\log_y z)^2 \\right) } = x^{(\\log_y z) \\cdot (\\log_y z) } = \\left( x^{\\log_y z} \\right)^{\\log_y z}\\] Now, let $x=y^w$, then we have: \\[x^{\\log_y z} = \\left( y^w \\right)^{\\log_y z} = y^{w\\log_y z} = y^{\\log_y (z^w)} = z^w\\] This is all we need to e... |
2009-II-3 | 2,009 | 3 | In rectangle $ABCD$ , $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ . | 141 | II | [
"Solution 1 [asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label(\"\\(E\\)\",E,N); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,SW); label(\"\\(C\\)\",C,SE); label(\"\\(D\\)\",D,NE); label(\"\\(F\\)\",F,W); label(\"\\(... |
2009-II-4 | 2,009 | 4 | A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ . | 89 | II | [
"Resolving the ambiguity The problem statement is confusing, as it can be interpreted in two ways: Either as \"there is a $k>1$ such that the child in $k$-th place had eaten $n+2-2k$ grapes\", or \"for all $k$, the child in $k$-th place had eaten $n+2-2k$ grapes\". The second meaning was apparently the intended one... |
2009-II-5 | 2,009 | 5 | Equilateral triangle $T$ is inscribed in circle $A$ , which has radius $10$ . Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$ . Circles $C$ and $D$ , both with radius $2$ , are internally tangent to circle $A$ at the other two vertices of $T$ . Circles $B$ , $C$ , and $D$ are all externally tangent to circle $E$ , which has radius $\dfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep}; draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5)); dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy] | 32 | II | [
"Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX =4$. Assume $AE$ = $p$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$. $AC = 8$, and angle $CAE = 60$ degrees because we a... |
2009-II-6 | 2,009 | 6 | Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ . | 750 | II | [
"We can use complementary counting. We can choose a five-element subset in ${14\\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$s and 5 $B$s, thereby showing that there are ${10\\choose 5}$... |
2009-II-7 | 2,009 | 7 | Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ . | 401 | II | [
"First, note that $(2n)!! = 2^n \\cdot n!$, and that $(2n)!! \\cdot (2n-1)!! = (2n)!$. We can now take the fraction $\\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$. We get that this fraction is equal to $\\dfrac{(2i)!}{(2i)!!^2} = \\dfrac{(2i)!}{2^{2i}(i!)^2}$. Now we can... |
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