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 |
|---|---|---|---|---|---|---|
2009-II-8 | 2,009 | 8 | Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m+n$ . | 41 | II | [
"Solution 1 There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at ... |
2009-II-9 | 2,009 | 9 | Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$ , and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$ . Find the remainder when $m-n$ is divided by $1000$ . | 0 | II | [
"Solution 1 It is actually reasonably easy to compute $m$ and $n$ exactly. First, note that if $4x+3y+2z=2009$, then $y$ must be odd. Let $y=2y'-1$. We get $4x + 6y' - 3 + 2z = 2009$, which simplifies to $2x + 3y' + z = 1006$. For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly on... |
2009-II-10 | 2,009 | 10 | Four lighthouses are located at points $A$ , $B$ , $C$ , and $D$ . The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$ , the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$ , and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$ . To an observer at $A$ , the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$ , the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac{p\sqrt{r}}{q}$ , where $p$ , $q$ , and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p+q+r$ . | 96 | II | [
"Let $O$ be the intersection of $BC$ and $AD$. By the Angle Bisector Theorem, $\\frac {5}{BO}$ = $\\frac {13}{CO}$, so $BO$ = $5x$ and $CO$ = $13x$, and $BO$ + $OC$ = $BC$ = $12$, so $x$ = $\\frac {2}{3}$, and $OC$ = $\\frac {26}{3}$. Let $P$ be the foot of the altitude from $D$ to $OC$. It can be seen that triangl... |
2009-II-11 | 2,009 | 11 | For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$ . Find the sum of all possible values of the product $m \cdot n$ . | 125 | II | [
"We have $\\log m - \\log k = \\log \\left( \\frac mk \\right)$, hence we can rewrite the inequality as follows: \\[- \\log n < \\log \\left( \\frac mk \\right) < \\log n\\] We can now get rid of the logarithms, obtaining: \\[\\frac 1n < \\frac mk < n\\] And this can be rewritten in terms of $k$ as \\[\\frac mn < k... |
2009-II-12 | 2,009 | 12 | From the set of integers $\{1,2,3,\dots,2009\}$ , choose $k$ pairs $\{a_i,b_i\}$ with $a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$ . Find the maximum possible value of $k$ . | 803 | II | [
"Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\\cdots+2k=k(2k+1)$. On the other hand, as the sum of each pair is distinct and at most equal to $2009$, the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \\cdots + (2009-(k-1)) = \\fr... |
2009-II-13 | 2,009 | 13 | Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$ . The arc is divided into seven congruent arcs by six equally spaced points $C_1,C_2,\dots,C_6$ . All chords of the form $\overline{AC_i}$ or $\overline{BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the remainder when $n$ is divided by $1000$ . | 672 | II | [
"Solution 1 Let the radius be 1 instead. All lengths will be halved so we will multiply by $2^{12}$ at the end. Place the semicircle on the complex plane, with the center of the circle being 0 and the diameter being the real axis. Then $C_1,\\ldots, C_6$ are 6 of the 14th roots of unity. Let $\\omega=\\text{cis}\\f... |
2009-II-14 | 2,009 | 14 | The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}$ for $n \geq 0$ . Find the greatest integer less than or equal to $a_{10}$ . | 983 | II | [
"The \"obvious\" substitution An obvious way how to get the $4^n$ from under the square root is to use the substitution $a_n = 2^n b_n$. Then the square root simplifies as follows: $\\sqrt{4^n - a_n^2} = \\sqrt{4^n - (2^n b_n)^2} = \\sqrt{4^n - 4^n b_n^2} = 2^n \\sqrt{1 - b_n^2}$. The new recurrence then becomes $b... |
2009-II-15 | 2,009 | 15 | Let $\overline{MN}$ be a diameter of a circle with diameter $1$ . Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\dfrac 35$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with the chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt t$ , where $r$ , $s$ , and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$ . | 14 | II | [
"Solution 1 (Quick Calculus) Let $V = \\overline{NM} \\cap \\overline{AC}$ and $W = \\overline{NM} \\cap \\overline{BC}$. Further more let $\\angle NMC = \\alpha$ and $\\angle MNC = 90^\\circ - \\alpha$. Angle chasing reveals $\\angle NBC = \\angle NAC = \\alpha$ and $\\angle MBC = \\angle MAC = 90^\\circ - \\alpha... |
2010-I-1 | 2,010 | 1 | Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 107 | I | [
"$2010^2 = 2^2\\cdot3^2\\cdot5^2\\cdot67^2$. Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$). Therefore the probability is \\[\\frac {2\\cdot2^4\\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \\frac {26}{81} \\Longrightarrow 26+ 81 = 107.\\]",
"Th... |
2010-I-2 | 2,010 | 2 | Find the remainder when $9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}$ is divided by $1000$ . | 109 | I | [
"Note that $999\\equiv 9999\\equiv\\dots \\equiv\\underbrace{99\\cdots9}_{\\text{999 9's}}\\equiv -1 \\pmod{1000}$ (see modular arithmetic). That is a total of $999 - 3 + 1 = 997$ integers, so all those integers multiplied out are congruent to $- 1\\pmod{1000}$. Thus, the entire expression is congruent to $- 1\\tim... |
2010-I-3 | 2,010 | 3 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ . | 529 | I | [
"Substitute $y = \\frac34x$ into $x^y = y^x$ and solve. \\[x^{\\frac34x} = \\left(\\frac34x\\right)^x\\] \\[x^{\\frac34x} = \\left(\\frac34\\right)^x \\cdot x^x\\] \\[x^{-\\frac14x} = \\left(\\frac34\\right)^x\\] \\[x^{-\\frac14} = \\frac34\\] \\[x = \\frac{256}{81}\\] \\[y = \\frac34x = \\frac{192}{81}\\] \\[x + y... |
2010-I-4 | 2,010 | 4 | Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 515 | I | [
"This can be solved quickly and easily with generating functions. Let $x^n$ represent flipping $n$ heads. The generating functions for these coins are $(1+x)$,$(1+x)$,and $(3+4x)$ in order. The product is $3+10x+11x^2+4x^3$. ($ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, ... |
2010-I-5 | 2,010 | 5 | Positive integers $a$ , $b$ , $c$ , and $d$ satisfy $a > b > c > d$ , $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ . | 501 | I | [
"Using the difference of squares, $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \\ge a + b + c + d = 2010$, where equality must hold so $b = a - 1$ and $d = c - 1$. Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $501",
"Since $a+b$ must be greater than $1005$, i... |
2010-I-6 | 2,010 | 6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ . | 406 | I | [
"Solution 1 [asy] import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype(\"6 2\")+linewidth(0.7)); draw(graph(R,min,max),linety... |
2010-I-7 | 2,010 | 7 | Define an ordered triple $(A, B, C)$ of sets to be $\textit{minimally intersecting}$ if $|A \cap B| = |B \cap C| = |C \cap A| = 1$ and $A \cap B \cap C = \emptyset$ . For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $\{1,2,3,4,5,6,7\}$ . Find the remainder when $N$ is divided by $1000$ . Note : $|S|$ represents the number of elements in the set $S$ . | 760 | I | [
"Let each pair of two sets have one element in common. Label the common elements as $x$, $y$, $z$. Set $A$ will have elements $x$ and $y$, set $B$ will have $y$ and $z$, and set $C$ will have $x$ and $z$. There are $7 \\cdot 6 \\cdot 5 = 210$ ways to choose values of $x$, $y$ and $z$. There are $4$ unpicked numbers... |
2010-I-8 | 2,010 | 8 | For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$ , where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$ . | 132 | I | [
"The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\\pm5,0), (0,\\pm5), (\\pm3,\\pm4), (\\pm4,\\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\\left(\\frac12,\\frac12\\right)$. The distance from $\\... |
2010-I-9 | 2,010 | 9 | Let $(a,b,c)$ be a real solution of the system of equations $x^3 - xyz = 2$ , $y^3 - xyz = 6$ , $z^3 - xyz = 20$ . The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 158 | I | [
"Solution 1 Add the three equations to get $a^3 + b^3 + c^3 = 28 + 3abc$. Now, let $abc = p$. $a = \\sqrt [3]{p + 2}$, $b = \\sqrt [3]{p + 6}$ and $c = \\sqrt [3]{p + 20}$, so $p = abc = (\\sqrt [3]{p + 2})(\\sqrt [3]{p + 6})(\\sqrt [3]{p + 20})$. Now cube both sides; the $p^3$ terms cancel out. Solve the remaining... |
2010-I-10 | 2,010 | 10 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ . | 202 | I | [
"If we choose $a_3$ and $a_1$ such that $(10^3)(a_3) + (10)(a_1) \\leq 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3 = 0$ or $a_3 = 1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3 = 2$ then ... |
2010-I-11 | 2,010 | 11 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$ . When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$ , the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$ , where $m$ , $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$ . | 365 | I | [
"[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle... |
2010-I-12 | 2,010 | 12 | Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$ . Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$ , $b$ , and $c$ (not necessarily distinct) such that $ab = c$ . Note : a partition of $S$ is a pair of sets $A$ , $B$ such that $A \cap B = \emptyset$ , $A \cup B = S$ . | 243 | I | [
"We claim that $243$ is the minimal value of $m$. Let the two partitioned sets be $A$ and $B$; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality, we place $3$ in $A$. Then $9$ must be placed in $B$, so $81$ must be placed in $A$, and $27$ ... |
2010-I-13 | 2,010 | 13 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ , $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ , $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ . | 69 | I | [
"Diagram [asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),z... |
2010-I-14 | 2,010 | 14 | For each positive integer $n,$ let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$ . Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ . | 109 | I | [
"Observe that $f$ is strictly increasing in $n$. We realize that we need $100$ terms to add up to around $300$, so we need some sequence of $2$s, $3$s, and then $4$s. It follows that $n \\approx 100$ (alternatively, use binary search to get to this, with $n\\le 1000$). Manually checking shows that $f(109) = 300$ an... |
2010-I-15 | 2,010 | 15 | In $\triangle{ABC}$ with $AB = 12$ , $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ . | 45 | I | [
"[asy] /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200); /* segments and figures */ draw((0,0)--(15,0)); draw((15,0)--(6.66667,9.97775)); draw((6.66667,9.97775)--(0,0)); draw((7.33333,0)--(6.66667,9.97775)); draw(circle((4.66667,2.49... |
2010-II-1 | 2,010 | 1 | Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000$ . | 640 | II | [
"If an integer is divisible by $36$, it must also be divisible by $9$ since $9$ is a factor of $36$. It is a well-known fact that, if $N$ is divisible by $9$, the sum of the digits of $N$ is a multiple of $9$. Hence, if $N$ contains all the even digits, the sum of the digits would be $0 + 2 + 4 + 6 + 8 = 20$, which... |
2010-II-2 | 2,010 | 2 | A point $P$ is chosen at random in the interior of a unit square $S$ . Let $d(P)$ denote the distance from $P$ to the closest side of $S$ . The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 281 | II | [
"Any point outside the square with side length $\\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\\frac{3}{5}$ that has the same center and orientation as the unit square has $\\frac{1}{5}\\le d(P)\\le\\frac{1}{3}$. [asy] unitsize(1mm); defaultpen(lin... |
2010-II-3 | 2,010 | 3 | Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$ . Find the greatest positive integer $n$ such that $2^n$ divides $K$ . | 150 | II | [
"In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\\cdots(19^1)$ (or alternatively, $19! \\cdot 18! \\cdots 1!$.) When we count the number of factors of $2$, we have 4... |
2010-II-4 | 2,010 | 4 | Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 52 | II | [
"Solution 1 There are $12 \\cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart. If we number the gates $1$ through $12$, then gates $1$ and... |
2010-II-5 | 2,010 | 5 | Positive numbers $x$ , $y$ , and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$ . Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$ . | 75 | II | [
"Using the properties of logarithms, $\\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\\log_{10}x)(\\log_{10}y) + (\\log_{10}x)(\\log_{10}z) + (\\log_{10}y)(\\log_{10}z)= 468$ by using the fact that $\\log_{10}ab = \\log_{10}a + \\log_{10}b$. Through further simplification, we find that $\\log_{1... |
2010-II-6 | 2,010 | 6 | Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients. | 8 | II | [
"You can factor the polynomial into two quadratic factors or a linear and a cubic factor. For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that \\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\\] Therefore, again setting coefficients equal, $... |
2010-II-7 | 2,010 | 7 | Let $P(z)=z^3+az^2+bz+c$ , where $a$ , $b$ , and $c$ are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ , $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ . | 136 | II | [
"Set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$. Since $a,b,c\\in{R}$, the imaginary part of $a,b,c$ must be $0$. Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$, and therefore: $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$. Now, do the part where the imaginary part of c is 0 ... |
2010-II-9 | 2,010 | 9 | Let $ABCDEF$ be a regular hexagon. Let $G$ , $H$ , $I$ , $J$ , $K$ , and $L$ be the midpoints of sides $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , and $AF$ , respectively. The segments $\overline{AH}$ , $\overline{BI}$ , $\overline{CJ}$ , $\overline{DK}$ , $\overline{EL}$ , and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 11 | II | [
"[asy] defaultpen(0.8pt+fontsize(12pt)); pair A,B,C,D,E,F; pair G,H,I,J,K,L; A=dir(0); B=dir(60); C=dir(120); D=dir(180); E=dir(240); F=dir(300); draw(A--B--C--D--E--F--cycle,blue); G=(A+B)/2; H=(B+C)/2; I=(C+D)/2; J=(D+E)/2; K=(E+F)/2; L=(F+A)/2; int i; for (i=0; i<6; i+=1) { draw(rotate(60*i)*(A--H),dotted); } pa... |
2010-II-10 | 2,010 | 10 | Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ . | 163 | II | [
"Solution 1 Let $f(x) = a(x-r)(x-s)$. Then $ars=2010=2\\cdot3\\cdot5\\cdot67$. First consider the case where $r$ and $s$ (and thus $a$) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$, $r$, and $s$. However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$, we... |
2010-II-12 | 2,010 | 12 | Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$ . Find the minimum possible value of their common perimeter. | 676 | II | [
"Let $s$ be the semiperimeter of the two triangles. Also, let the base of the longer triangle be $16x$ and the base of the shorter triangle be $14x$ for some arbitrary factor $x$. Then, the dimensions of the two triangles must be $s-8x,s-8x,16x$ and $s-7x,s-7x,14x$. By Heron's Formula, we have \\[\\sqrt{s(8x)(8x)(s... |
2010-II-13 | 2,010 | 13 | The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each pick a card from the deck randomly and without replacement. The two people with lower numbered cards form a team, and the two people with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$ , and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 263 | II | [
"Once the two cards are drawn, there are $\\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$, which occurs in $\\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above ... |
2010-II-14 | 2,010 | 14 | Triangle $ABC$ with right angle at $C$ , $\angle BAC < 45^\circ$ and $AB = 4$ . Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$ . The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$ , where $p$ , $q$ , $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$ . | 7 | II | [
"Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\\angle{DOA} = 2\\angle ACP = \\angle{APC} = \\angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$. Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean The... |
2010-II-15 | 2,010 | 15 | In triangle $ABC$ , $AC=13$ , $BC=14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$ . | 218 | II | [
"Define the function $f:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}$ by \\[f(X)=\\text{Pow}_{(AMN)}(X)-\\text{Pow}_{(ADE)}(X)\\] for points $X$ in the plane. Then $f$ is linear, so $\\frac{BQ}{CQ}=\\frac{f(B)-f(Q)}{f(Q)-f(C)}$. But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$, $(ADE)$ thus \\[\\frac{BQ}{CQ}=-\... |
2011-I-1 | 2,011 | 1 | Jar $A$ contains four liters of a solution that is $45\%$ acid. Jar $B$ contains five liters of a solution that is $48\%$ acid. Jar $C$ contains one liter of a solution that is $k\%$ acid. From jar $C$ , $\frac{m}{n}$ liters of the solution is added to jar $A$ , and the remainder of the solution in jar $C$ is added to jar B. At the end both jar $A$ and jar $B$ contain solutions that are $50\%$ acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ . | 85 | I | [
"Jar A contains $\\frac{11}{5}$ liters of water, and $\\frac{9}{5}$ liters of acid; jar B contains $\\frac{13}{5}$ liters of water and $\\frac{12}{5}$ liters of acid. The gap between the amount of water and acid in the first jar, $\\frac{2}{5}$, is double that of the gap in the second jar, $\\frac{1}{5}$. Therefore... |
2011-I-2 | 2,011 | 2 | In rectangle $ABCD$ , $AB = 12$ and $BC = 10$ . Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9$ , $DF = 8$ , $\overline{BE} \parallel \overline{DF}$ , $\overline{EF} \parallel \overline{AB}$ , and line $BE$ intersects segment $\overline{AD}$ . The length $EF$ can be expressed in the form $m \sqrt{n} - p$ , where $m$ , $n$ , and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p$ . | 36 | I | [
"Let us call the point where $\\overline{EF}$ intersects $\\overline{AD}$ point $G$, and the point where $\\overline{EF}$ intersects $\\overline{BC}$ point $H$. Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are simila... |
2011-I-3 | 2,011 | 3 | Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A = (24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B = (5,6)$ . The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha + \beta$ . | 31 | I | [
"Given that $L$ has slope $\\frac{5}{12}$ and contains the point $A=(24,-1)$, we may write the point-slope equation for $L$ as $y+1=\\frac{5}{12}(x-24)$. Since $M$ is perpendicular to $L$ and contains the point $B=(5,6)$, we have that the slope of $M$ is $-\\frac{12}{5}$, and consequently that the point-slope equat... |
2011-I-4 | 2,011 | 4 | In triangle $ABC$ , $AB = 125$ , $AC = 117$ , and $BC = 120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$ . | 56 | I | [
"Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively. [asy] defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bise... |
2011-I-5 | 2,011 | 5 | The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements. | 144 | I | [
"First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$. It is simplest to do this by looking at each of the digits $\\bmod{3}$. We see that the numbers $1, 4,$ and $7$ are congruent to $1 \\pmod{3}$, that the numbers $2, 5,$ and $8$ are congruent to $2 \... |
2011-I-6 | 2,011 | 6 | Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ . | 11 | I | [
"If the vertex is at $\\left(\\frac{1}{4}, -\\frac{9}{8}\\right)$, the equation of the parabola can be expressed in the form \\[y=a\\left(x-\\frac{1}{4}\\right)^2-\\frac{9}{8}.\\] Expanding, we find that \\[y=a\\left(x^2-\\frac{x}{2}+\\frac{1}{16}\\right)-\\frac{9}{8},\\] and \\[y=ax^2-\\frac{ax}{2}+\\frac{a}{16}-\... |
2011-I-7 | 2,011 | 7 | Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$ , $x_1$ , $\dots$ , $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\] | 16 | I | [
"$m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}$. Now, divide by $m^{x_0}$ to get $1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}$. Notice that since we can choose all nonnegative $x_0,...,x_{2011}$, we can make $x_n-x_0$ whatever we desire. WLOG, let $x_0\\geq...\\geq x_{2011}$ and let $a_n=x_n-x_0$. Noti... |
2011-I-9 | 2,011 | 9 | Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$ . Find $24 \cot^2 x$ . | 192 | I | [
"We can rewrite the given expression as \\[\\sqrt{24^3\\sin^3 x}=24\\cos x\\] Square both sides and divide by $24^2$ to get \\[24\\sin ^3 x=\\cos ^2 x\\] Rewrite $\\cos ^2 x$ as $1-\\sin ^2 x$ \\[24\\sin ^3 x=1-\\sin ^2 x\\] \\[24\\sin ^3 x+\\sin ^2 x - 1=0\\] Testing values using the rational root theorem gives $\... |
2011-I-10 | 2,011 | 10 | The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ . | 503 | I | [
"Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater). Break up the problem into two cas... |
2011-I-11 | 2,011 | 11 | Let $R$ be the set of all possible remainders when a number of the form $2^n$ , $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$ . Find the remainder when $S$ is divided by 1000. | 7 | I | [
"Note that $x \\equiv y \\pmod{1000} \\Leftrightarrow x \\equiv y \\pmod{125}$ and $x \\equiv y \\pmod{8}$. So we must find the first two integers $i$ and $j$ such that $2^i \\equiv 2^j \\pmod{125}$ and $2^i \\equiv 2^j \\pmod{8}$ and $i \\neq j$. Note that $i$ and $j$ will be greater than 2 since remainders of $1,... |
2011-I-12 | 2,011 | 12 | Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent. | 594 | I | [
"Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men: _(2)_(2)_(2)_ _(3... |
2011-I-13 | 2,011 | 13 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r - \sqrt{s}}{t}$ , where $r$ , $s$ , and $t$ are positive integers. Find $r + s + t$ . | 330 | I | [
"Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \\[\\frac{AX+BY+CZ+D}{\\sqrt{A^2+B^2+C^2}},\\] so the (directed) distance from any point $(x,y,z)$ to the pl... |
2011-I-14 | 2,011 | 14 | Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ , $M_3$ , $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ , $\overline{A_3 A_4}$ , $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ , $R_3 \perp R_5$ , $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ , $R_3$ and $R_5$ , $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ , $B_3$ , $B_5$ , $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ . | 37 | I | [
"We use coordinates. Let the octagon have side length $2$ and center $(0, 0)$. Then all of its vertices have the form $(\\pm 1, \\pm\\left(1+\\sqrt{2}\\right))$ or $(\\pm\\left(1+\\sqrt{2}\\right), \\pm 1)$. By symmetry, $B_{1}B_{3}B_{5}B_{7}$ is a square. Thus lines $\\overleftrightarrow{B_{1}B_{3}}$ and $\\overle... |
2011-I-15 | 2,011 | 15 | For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ , $b$ , and $c$ . Find $|a| + |b| + |c|$ . | 98 | I | [
"From Vieta's formulas, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$. Thus $a = -(b+c)$. All three of $a$, $b$, and $c$ are non-zero: say, if $a=0$, then $b=-c=\\pm\\sqrt{2011}$ (which is not an integer). $\\textsc{wlog}$, let $|a| \\ge |b| \\ge |c|$. If $a > 0$, then $b,c < 0$ and if $a < 0$, then $b,c > 0,$ f... |
2011-II-1 | 2,011 | 1 | Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$ . | 37 | II | [
"Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $\\frac{1}{2} - 2x =\\frac{2}{9} (1-x)$, or $9 - 36x = 4 - 4x$, or $32x = 5$ or $x = 5/32$. Therefore, $m + n = 5 + 32 = 037.",
"WLOG, Gary purchased \\( n \\) liters and consumed \\( m \\) liters. After this, he purchased \\( \\frac{... |
2011-II-2 | 2,011 | 2 | On square $ABCD$ , point $E$ lies on side $AD$ and point $F$ lies on side $BC$ , so that $BE=EF=FD=30$ . Find the area of the square $ABCD$ . | 810 | II | [
"Drawing the square and examining the given lengths, [asy] size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label(\"$A$\", A, plain.SW); pair B = (3, 0); dot(B); label(\"$B$\", B, plain.SE); pair C = (3, 3); dot(C); label(\"$C$\", C, plain.NE); pair D = (0, 3); dot(D); label(\"$D$\", D, plai... |
2011-II-3 | 2,011 | 3 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle. | 143 | II | [
"Solution 1 The average angle in an 18-gon is $160^\\circ$. In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to $160^\\circ$. Thus for some positive (the sequence is increasing and thus non-constant) integer $d$, the middle two terms are $(160-d)^\\cir... |
2011-II-4 | 2,011 | 4 | In triangle $ABC$ , $AB=20$ and $AC=11$ . The angle bisector of angle $A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of intersection of $AC$ and the line $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 51 | II | [
"[asy] pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP(\"A\",D(A))--MP(\"B\",D(B),N)--MP(\"C\",D(C))--cycle); D(A--MP(\"D\",D(D),NE)--MP(\"D'\",D(D2))); D(B--MP(\"P\"... |
2011-II-5 | 2,011 | 5 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | 542 | II | [
"Since the sum of the first $2011$ terms is $200$, and the sum of the first $4022$ terms is $380$, the sum of the second $2011$ terms is $180$. This is decreasing from the first 2011, so the common ratio is less than one. Because it is a geometric sequence and the sum of the first 2011 terms is $200$, second $2011$... |
2011-II-6 | 2,011 | 6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | 80 | II | [
"Rearranging the inequality we get $d-c > b-a$. Let $e = 11$, then $(a, b-a, c-b, d-c, e-d)$ is a partition of 11 into 5 positive integers or equivalently: $(a-1, b-a-1, c-b-1, d-c-1, e-d-1)$ is a partition of 6 into 5 non-negative integer parts. Via a standard stars and bars argument, the number of ways to partiti... |
2011-II-7 | 2,011 | 7 | Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let $m$ be the maximum number of red marbles for which such an arrangement is possible, and let $N$ be the number of ways he can arrange the $m+5$ marbles to satisfy the requirement. Find the remainder when $N$ is divided by $1000$ . | 3 | II | [
"We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 \"not the same colors\" and 0 \"same colors.\" Now, for every red marble we add, we will add one \"same color\" pair and keep all 10 \"n... |
2011-II-8 | 2,011 | 8 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ . | 784 | II | [
"The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$. If we write $z=a+bi$, then the real part of $z$ is $a$ and the real part of $iz$ is $-b$. The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$, and the red dots represent those ... |
2011-II-9 | 2,011 | 9 | Let $x_1$ , $x_2$ , $\dots$ , $x_6$ be nonnegative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 1$ , and $x_1x_3x_5 + x_2x_4x_6 \ge {\frac{1}{540}}$ . Let $p$ and $q$ be relatively prime positive integers such that $\frac{p}{q}$ is the maximum possible value of $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_6 + x_5x_6x_1 + x_6x_1x_2$ . Find $p + q$ . | 559 | II | [
"Solution 1 Note that neither the constraint nor the expression we need to maximize involves products $x_i x_j$ with $i \\equiv j \\pmod 3$. Factoring out say $x_1$ and $x_4$ we see that the constraint is $x_1(x_3x_5) + x_4(x_2x_6) \\ge {\\frac1{540}}$, while the expression we want to maximize is $x_1(x_2x_3 + x_5x... |
2011-II-10 | 2,011 | 10 | A circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000. | 57 | II | [
"Let $E$ and $F$ be the midpoints of $\\overline{AB}$ and $\\overline{CD}$, respectively, such that $\\overline{BE}$ intersects $\\overline{CF}$. Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$. $B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$. The line through the midpoint of a... |
2011-II-11 | 2,011 | 11 | Let $M_n$ be the $n \times n$ matrix with entries as follows: for $1 \le i \le n$ , $m_{i,i} = 10$ ; for $1 \le i \le n - 1$ , $m_{i+1,i} = m_{i,i+1} = 3$ ; all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n$ . Then $\sum_{n=1}^{\infty} \frac{1}{8D_n+1}$ can be represented as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ . Note: The determinant of the $1 \times 1$ matrix $[a]$ is $a$ , and the determinant of the $2 \times 2$ matrix $\left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] = ad - bc$ ; for $n \ge 2$ , the determinant of an $n \times n$ matrix with first row or first column $a_1$ $a_2$ $a_3$ $\dots$ $a_n$ is equal to $a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n$ , where $C_i$ is the determinant of the $(n - 1) \times (n - 1)$ matrix formed by eliminating the row and column containing $a_i$ . | 73 | II | [
"\\[D_{1}=\\begin{vmatrix} 10 \\end{vmatrix} = 10, \\quad D_{2}=\\begin{vmatrix} 10 & 3 \\\\ 3 & 10 \\\\ \\end{vmatrix} =(10)(10) - (3)(3) = 91, \\quad D_{3}=\\begin{vmatrix} 10 & 3 & 0 \\\\ 3 & 10 & 3 \\\\ 0 & 3 & 10 \\\\ \\end{vmatrix}.\\] Using the expansionary/recursive definition of determinants (also stated i... |
2011-II-12 | 2,011 | 12 | Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . | 97 | II | [
"Use complementary probability and Principle of Inclusion-Exclusion. If we consider the delegates from each country to be indistinguishable and number the chairs, we have \\[\\frac{9!}{(3!)^3} = \\frac{9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4}{6\\cdot6} = 6\\cdot8\\cdot7\\cdot5 = 30\\cdot56\\] total ways to seat the ca... |
2011-II-13 | 2,011 | 13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ , respectively. Given that $AB = 12$ and $\angle O_1PO_2 = 120 ^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ . | 96 | II | [
"Denote the midpoint of $\\overline{DC}$ be $E$ and the midpoint of $\\overline{AB}$ be $F$. Because they are the circumcenters, both Os lie on the perpendicular bisectors of $AB$ and $CD$ and these bisectors go through $E$ and $F$. It is given that $\\angle O_{1}PO_{2}=120^{\\circ}$. Because $O_{1}P$ and $O_{1}B$ ... |
2011-II-14 | 2,011 | 14 | There are $N$ permutations $(a_1, a_2, \dots, a_{30})$ of $1, 2, \dots, 30$ such that for $m \in \{2,3,5\}$ , $m$ divides $a_{n+m} - a_n$ for all integers $n$ with $1 \le n < n+m \le 30$ . Find the remainder when $N$ is divided by 1000. | 440 | II | [
"Solution 1 Be wary of \"position\" versus \"number\" in the solution! Each POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$. The $n$th position is defined by this ordered triple where $i$ is $n \\mod 2$, $j$ is $n \\mod 3$, and $k$ is $n \\mod 5$. There are 2 choices for... |
2011-II-15 | 2,011 | 15 | Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\left\lfloor\sqrt{P(x)}\right\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ , $b$ , $c$ , $d$ , and $e$ are positive integers. Find $a + b + c + d + e$ . | 850 | II | [
"Table of values of $P(x)$: \\begin{align*} P(5) &= 1 \\\\ P(6) &= 9 \\\\ P(7) &= 19 \\\\ P(8) &= 31 \\\\ P(9) &= 45 \\\\ P(10) &= 61 \\\\ P(11) &= 79 \\\\ P(12) &= 99 \\\\ P(13) &= 121 \\\\ P(14) &= 145 \\\\ P(15) &= 171 \\\\ \\end{align*} In order for $\\lfloor \\sqrt{P(x)} \\rfloor = \\sqrt{P(\\lfloor x \\rfloor... |
2012-I-1 | 2,012 | 1 | Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$ . | 40 | I | [
"A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$, there are two possible values for $a$ and $c$, since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$, and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$. There are thus $2 ... |
2012-I-2 | 2,012 | 2 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle term of the original sequence. | 195 | I | [
"Solution 1 If the sum of the original sequence is $\\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \\sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \\rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the te... |
2012-I-3 | 2,012 | 3 | Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person. | 216 | I | [
"Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\\text{BBBCCCFFF}$ respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by $9$ to account for the $9$ different ways in which the pers... |
2012-I-4 | 2,012 | 4 | Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$ . | 279 | I | [
"When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\\frac{1}{6}$ hours per mile, Butch takes $\\frac{1}{4}$ hours per mile, and Sundance takes $\\frac{2}{5... |
2012-I-5 | 2,012 | 5 | Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained. | 330 | I | [
"When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instan... |
2012-I-6 | 2,012 | 6 | The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$ | 71 | I | [
"Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \\neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$nd root of unity, and thus, by De Moivre's theorem, the imaginary ... |
2012-I-9 | 2,012 | 9 | Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | 49 | I | [
"Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that \\[2\\log_{x}(2y) = 2\\log_{2x}(4z) = \\log_{2x^4}(8yz) = 2.\\] Then \\begin{align*} ... |
2012-I-10 | 2,012 | 10 | Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$ . Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$ , where $x$ is in $\mathcal{S}$ . In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$ . | 170 | I | [
"It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16 \\not\\equiv s-16 \\pmod{5},$ one term must... |
2012-I-11 | 2,012 | 11 | A frog begins at $P_0 = (0,0)$ and makes a sequence of jumps according to the following rule: from $P_n = (x_n, y_n),$ the frog jumps to $P_{n+1},$ which may be any of the points $(x_n + 7, y_n + 2),$ $(x_n + 2, y_n + 7),$ $(x_n - 5, y_n - 10),$ or $(x_n - 10, y_n - 5).$ There are $M$ points $(x, y)$ with $|x| + |y| \le 100$ that can be reached by a sequence of such jumps. Find the remainder when $M$ is divided by $1000.$ | 373 | I | [
"First of all, it is easy to see by induction that for any $P(x,y)$ in the frog's jump sequence, $x+y$ will be a multiple of $3$ and $x-y$ will be a multiple of $5.$ The base case $(x,y) = (0,0)$ obviously satisfies the constraints and if $x+y = 3n$ and $x-y = 5m,$ any of the four transformations will sustain this ... |
2012-I-12 | 2,012 | 12 | Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$ | 18 | I | [
"We have $\\angle BCE = \\angle ECD = \\angle DCA = \\tfrac 13 \\cdot 90^\\circ = 30^\\circ$. Drop the altitude from $D$ to $CB$ and call the foot $F$. [asy]import cse5;size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,E0,F; C=origin; B=(10,0); A=(0,5); E0=extension(C,dir(30),A,B); D=extension(C,dir(6... |
2012-I-13 | 2,012 | 13 | Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ | 41 | I | [
"Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^\\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$ [asy]import cse5; size(200); defaultpen(linewidth(0.4)+fontsize(8)); pai... |
2012-I-14 | 2,012 | 14 | Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$ | 375 | I | [
"By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$. Therefore, $\\frac{(a+b+c)}{3}=0$. Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$. Without the loss of generalit... |
2012-II-1 | 2,012 | 1 | Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$ . | 34 | II | [
"",
"Solving for $m$ gives us $m = \\frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \\equiv 503 \\mod 5 \\longrightarrow n \\equiv 1 \\mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\\frac{... |
2012-II-2 | 2,012 | 2 | Two geometric sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ have the same common ratio, with $a_1 = 27$ , $b_1=99$ , and $a_{15}=b_{11}$ . Find $a_9$ . | 363 | II | [
"Call the common ratio $r.$ Now since the $n$th term of a geometric sequence with first term $x$ and common ratio $y$ is $xy^{n-1},$ we see that $a_1 \\cdot r^{14} = b_1 \\cdot r^{10} \\implies r^4 = \\frac{99}{27} = \\frac{11}{3}.$ But $a_9$ equals $a_1 \\cdot r^8 = a_1 \\cdot (r^4)^2=27\\cdot {\\left(\\frac{11}{3... |
2012-II-3 | 2,012 | 3 | At a certain university, the division of mathematical sciences consists of the departments of mathematics, statistics, and computer science. There are two male and two female professors in each department. A committee of six professors is to contain three men and three women and must also contain two professors from each of the three departments. Find the number of possible committees that can be formed subject to these requirements. | 88 | II | [
"There are two cases: Case 1: One man and one woman is chosen from each department. Case 2: Two men are chosen from one department, two women are chosen from another department, and one man and one woman are chosen from the third department. For the first case, in each department there are ${{2}\\choose{1}} \\times... |
2012-II-4 | 2,012 | 4 | Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$ , where $p$ , $q$ , and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$ . | 61 | II | [
"[asy]draw((1.2,0)--(0,0)--(0,1.4)--(6,1.4)--(6,0)--(1.2,0)--(6,1.4)); label(\"$D$\", (1.2,0),dir(-90)); dot((6,1.4)); dot((1.2,0)); label(\"$a$\", (0.6,0),dir(-90)); label(\"$b$\", (3.6,0),dir(-90)); label(\"$c$\", (6,0.7),dir(0));[/asy] Let $a,b,c$ be the labeled lengths as shown in the diagram. Also, assume WLOG... |
2012-II-6 | 2,012 | 6 | Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$ . Find $c+d$ . | 125 | II | [
"Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have $|z^3| * |z^2 - (1+2i)|$ $=|z|^3 * |z^2 - (1+2i)|$ $=125|z^2 - (1+2i)|$ Thus we only need to maximize the value of $|z^2 - (1+2i)|$. To maximize this value, we must have that $z^2$ is in the ... |
2012-II-7 | 2,012 | 7 | Let $S$ be the increasing sequence of positive integers whose binary representation has exactly $8$ ones. Let $N$ be the 1000th number in $S$ . Find the remainder when $N$ is divided by $1000$ . | 32 | II | [
"Okay, an exercise in counting (lots of binomials to calculate!). In base 2, the first number is $11111111$, which is the only way to choose 8 1's out of 8 spaces, or $\\binom{8}{8}$. What about 9 spaces? Well, all told, there are $\\binom{9}{8}=9$, which includes the first 1. Similarly, for 10 spaces, there are $\... |
2012-II-8 | 2,012 | 8 | The complex numbers $z$ and $w$ satisfy the system \[z + \frac{20i}w = 5+i\] \[w+\frac{12i}z = -4+10i\] Find the smallest possible value of $\vert zw\vert^2$ . | 40 | II | [
"Multiplying the two equations together gives us \\[zw + 32i - \\frac{240}{zw} = -30 + 46i\\] and multiplying by $zw$ then gives us a quadratic in $zw$: \\[(zw)^2 + (30-14i)zw - 240 =0.\\] Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \\pm \\sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i ... |
2012-II-9 | 2,012 | 9 | Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . | 107 | II | [
"Examine the first term in the expression we want to evaluate, $\\frac{\\sin 2x}{\\sin 2y}$, separately from the second term, $\\frac{\\cos 2x}{\\cos 2y}$. The First Term Using the identity $\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, we have: $\\frac{2\\sin x \\cos x}{2\\sin y \\cos y} = \\frac{\\sin x \\cos x}{\... |
2012-II-10 | 2,012 | 10 | Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$ . Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ . | 496 | II | [
"Solution 1 We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational. Let $x = a + \\frac{b}{c}$ where $a,b,c$ are nonnegative integers and $0 \\le b < c$ (essentially, $x$ is a mixed number). Then, \\[n ... |
2012-II-11 | 2,012 | 11 | Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 8 | II | [
"After evaluating the first few values of $f_k (x)$, we obtain $f_4(x) = f_1(x) = \\frac{2}{3} - \\frac{3}{3x+1} = \\frac{6x-7}{9x+3}$. Since $1001 \\equiv 2 \\mod 3$, $f_{1001}(x) = f_2(x) = \\frac{3x+7}{6-9x}$. We set this equal to $x-3$, i.e. $\\frac{3x+7}{6-9x} = x-3 \\Rightarrow x = \\frac{5}{3}$. The answer i... |
2012-II-12 | 2,012 | 12 | For a positive integer $p$ , define the positive integer $n$ to be $p$ -safe if $n$ differs in absolute value by more than $2$ from all multiples of $p$ . For example, the set of $10$ -safe numbers is $\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}$ . Find the number of positive integers less than or equal to $10,000$ which are simultaneously $7$ -safe, $11$ -safe, and $13$ -safe. | 958 | II | [
"We see that a number $n$ is $p$-safe if and only if the residue of $n \\mod p$ is greater than $2$ and less than $p-2$; thus, there are $p-5$ residues $\\mod p$ that a $p$-safe number can have. Therefore, a number $n$ satisfying the conditions of the problem can have $2$ different residues $\\mod 7$, $6$ different... |
2012-II-13 | 2,012 | 13 | Equilateral $\triangle ABC$ has side length $\sqrt{111}$ . There are four distinct triangles $AD_1E_1$ , $AD_1E_2$ , $AD_2E_3$ , and $AD_2E_4$ , each congruent to $\triangle ABC$ ,
with $BD_1 = BD_2 = \sqrt{11}$ . Find $\sum_{k=1}^4(CE_k)^2$ . | 677 | II | [
"Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\\sqrt{111}$ from point $A$ and $\\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\\sqrt{111}$ and a circle centered at $B$ with radius $\\sqrt{11}$ intersect. Let $D_1$ b... |
2012-II-14 | 2,012 | 14 | In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$ . | 16 | II | [
"Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, ... |
2012-II-15 | 2,012 | 15 | Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$ , $BC=7$ , and $AC=3$ . The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $\overline{DE}$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 919 | II | [
"Use the angle bisector theorem to find $CD=\\tfrac{21}{8}$, $BD=\\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\\angle CAD = \\tfrac{\\pi} {3}$, hence $\\angle BAD = \\tfrac{\\pi}{3}$ as well, an... |
2013-I-1 | 2,013 | 1 | The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling? | 150 | I | [
"Let $r$ represent the rate Tom swims in miles per minute. Then we have $\\frac{1/2}{r} + \\frac{8}{5r} + \\frac{30}{10r} = 255$ Solving for $r$, we find $r = 1/50$, so the time Tom spends biking is $\\frac{30}{(10)(1/50)} = 150 minutes."
] |
2013-I-3 | 2,013 | 3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | 18 | I | [
"It's important to note that $\\dfrac{AE}{EB} + \\dfrac{EB}{AE}$ is equivalent to $\\dfrac{AE^2 + EB^2}{(AE)(EB)}$ We define $a$ as the length of the side of larger inner square, which is also $EB$, $b$ as the length of the side of the smaller inner square which is also $AE$, and $s$ as the side length of $ABCD$. S... |
2013-I-4 | 2,013 | 4 | In the array of $13$ squares shown below, $8$ squares are colored red, and the remaining $5$ squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated $90^{\circ}$ around the central square is $\frac{1}{n}$ , where $n$ is a positive integer. Find $n$ . [asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); draw((2,0)--(2,2)--(3,2)--(3,0)--(3,1)--(2,1)--(4,1)--(4,0)--(2,0)); draw((1,2)--(1,4)--(0,4)--(0,2)--(0,3)--(1,3)--(-1,3)--(-1,2)--(1,2)); draw((-1,1)--(-3,1)--(-3,0)--(-1,0)--(-2,0)--(-2,1)--(-2,-1)--(-1,-1)--(-1,1)); draw((0,-1)--(0,-3)--(1,-3)--(1,-1)--(1,-2)--(0,-2)--(2,-2)--(2,-1)--(0,-1)); size(100);[/asy] | 429 | I | [
"When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there e... |
2013-I-5 | 2,013 | 5 | The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$ , where $a$ , $b$ , and $c$ are positive integers. Find $a+b+c$ . | 98 | I | [
"We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$. Therefore, we have that $9x^3 = (x+1)^3$, so it follows that $x\\sqrt[3]{9} = x+1$. Solving for $x$ yields $\\frac{1}{\\sqrt[3]{9}-1} = \\frac{\\sqrt[3]{81}+\\sqrt[3]{9}+1}{8}$, so the answer is $98.",
"Let $r$ be the real root ... |
2013-I-6 | 2,013 | 6 | Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . | 47 | I | [
"The total ways the textbooks can be arranged in the 3 boxes is $12\\textbf{C}3\\cdot 9\\textbf{C}4$, which is equivalent to $\\frac{12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7\\cdot 6}{144}=12\\cdot11\\cdot10\\cdot7\\cdot3$. If all of the math textbooks are put into the box that can hold $3$ textbooks, there are ... |
2013-I-7 | 2,013 | 7 | A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ . | 41 | I | [
"Let the height of the box be $x$. After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{\\left(\\frac{x}{2}\\right)^2 + 64}$, and $\\sqrt{\\left(\\frac{x}{2}\\right)^2 + 36}$. Since the area of the triangle is $30$, the altitude of the triangle from the ... |
2013-I-8 | 2,013 | 8 | The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by $1000$ . | 371 | I | [
"We know that the domain of $\\text{arcsin}$ is $[-1, 1]$, so $-1 \\le \\log_m nx \\le 1$. Now we can apply the definition of logarithms: \\[m^{-1} = \\frac1m \\le nx \\le m\\] \\[\\implies \\frac{1}{mn} \\le x \\le \\frac{m}{n}\\] Since the domain of $f(x)$ has length $\\frac{1}{2013}$, we have that \\[\\frac{m}{n... |
2013-I-9 | 2,013 | 9 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ , $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$ . [asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy] | 113 | I | [
"Let $M$ and $N$ be the points on $\\overline{AB}$ and $\\overline{AC}$, respectively, where the paper is folded. Let $D$ be the point on $\\overline{BC}$ where the folded $A$ touches it. [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP(\"B\", (0,0), dir(200... |
2013-I-10 | 2,013 | 10 | There are nonzero integers $a$ , $b$ , $r$ , and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$ . For each possible combination of $a$ and $b$ , let ${p}_{a,b}$ be the sum of the zeros of $P(x)$ . Find the sum of the ${p}_{a,b}$ 's for all possible combinations of $a$ and $b$ . | 80 | I | [
"Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have $(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$ Applying difference of squares, and regrouping, we have $(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$ So matchin... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.