problem_id stringlengths 14 18 | problem stringlengths 119 2.73k | answer stringlengths 1 3 | solution stringlengths 243 20.5k |
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problem_2002-I-5 | Let $A_1, A_2, A_3, \ldots, A_{12}$ be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set $\{A_1,A_2,A_3,\ldots,A_{12}\}$ ? Let's think step by step and output the final answer within \boxed{}. | 183 | Solution 1:
There are 66 ways of picking two vertices. Note with any two vertices one can draw three squares (<i>two</i> with the vertices forming a side, another with the vertices forming the diagonal). So so far we have $66(3)=198$ squares, but we have overcounted since some squares have their other two vertices in t... |
problem_2001-I-10 | Let $S$ be the set of points whose coordinates $x,$ $y,$ and $z$ are integers that satisfy $0\le x\le2,$ $0\le y\le3,$ and $0\le z\le4.$ Two distinct points are randomly chosen from $S.$ The probability that the midpoint of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime ... | 200 | Solution 1:
The distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore,
* For $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities.
* For $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$,... |
problem_2002-I-15 | Polyhedron $ABCDEFG$ has six faces. Face $ABCD$ is a square with $AB = 12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF = AG = 8,$ and $GF = 6;$ and face $CDE$ has $CE = DE = 14.$ The other three faces are $ADEG, BCEF,$ and $EFG.$ The distance from $E$ to face $ABCD$ is 12. Given ... | 163 | Solution 1:
[asy] size(200); import three; import graph; defaultpen(linewidth(0.7)+fontsize(8)); currentprojection=orthographic(-30,50,40); triple A=(-6,-6,0), B = (-6,6,0), C = (6,6,0), D = (6,-6,0), E = (2,0,12), H=(-6+2*sqrt(19),0,12), H1=(-6-2*sqrt(19),0,12), F, G, E1 = (6,0,12); F = 1/2*H+1/2*B; G = 1/2*H+1/2*A; d... |
problem_2012-I-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. Let's think step by step and output ... | 330 | Solution 1:
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... |
problem_2003-II-14 | Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct e... | 51 | Solution 1:
The y-coordinate of $F$ must be $4$. All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting $F = (f,4)$, and knowing that $\angle FAB = 120^\circ$, we can use rewrite $F$ using complex numbers:
$f + 4 i = (b + 2 i)\left(e^{i(2 \pi / 3)}\right) = (b + 2 i)\l... |
problem_2018-I-1 | Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$ . Let's think step by step ... | 600 | Solution 1:
Let the linear factors be $(x+c)(x+d)$.
Then, $a=c+d$ and $b=cd$.
We know that $1\le a\le 100$ and $b\ge 0$, so $c$ and $d$ both have to be non-negative
However, $a$ cannot be $0$, so at least one of $c$ and $d$ must be greater than $0$, ie positive.
Also, $a$ cannot be greater than $100$, so $c+d$ must be ... |
problem_2010-I-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 $A$ , $B$ suc... | 243 | Solution 1:
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 $A$, an... |
problem_1996-13 | In triangle $ABC$ , $AB=\sqrt{30}$ , $AC=\sqrt{6}$ , and $BC=\sqrt{15}$ . There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ , and $\angle ADB$ is a right angle. The ratio \[\dfrac{\text{Area}(\triangle ADB)}{\text{Area}(\triangle ABC)}\] can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$... | 65 | Solution 1:
[asy] pointpen = black; pathpen = black + linewidth(0.7); pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E); D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B); D(MP("E",E)); MP("\sqrt{30}",(A+B)/2,NW); MP("\sqrt{6}",(A+C)/2,SE); MP("\frac{\sqrt{15}}2",(E+C)/2); D(righta... |
problem_2007-II-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 $PQ.$ Let's think step by step and out... | 259 | 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 congruent triang... |
problem_1996-15 | In parallelogram $ABCD,$ let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA,$ and angle $ACB$ is $r$ times as large as angle $AOB$ . Find the greatest integer that does not exceed $1000r$ . Let's think step by step and output the f... | 777 | Solution 1:
[asy]size(180); pathpen = black+linewidth(0.7); pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP("A",A,N)--MP("B",B)--MP("C",C)--MP("D",D,N)--cycle); D(B--D); D(A--C); D(MP("O",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5));D(anglemark(B,A,C,4.5));D(an... |
problem_2002-II-9 | Let $\mathcal{S}$ be the set $\lbrace1,2,3,\ldots,10\rbrace$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$ . (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$ . Let's think step by step and output the final answ... | 501 | Solution 1:
Let the two disjoint subsets be $A$ and $B$, and let $C = \mathcal{S}-(A+B)$. For each $i \in \mathcal{S}$, either $i \in A$, $i \in B$, or $i \in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$.
However, there are $2^{10}$ ways to organize the elements of $\m... |
problem_2002-I-14 | A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer. Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is ... | 30 | Solution 1:
Let the sum of the integers in $\mathcal{S}$ be $N$, and let the size of $|\mathcal{S}|$ be $n+1$. After any element $x$ is removed, we are given that $n|N-x$, so $x\equiv N\pmod{n}$. Since $1\in\mathcal{S}$, $N\equiv1\pmod{n}$, and all elements are congruent to 1 mod $n$. Since they are positive integers, ... |
problem_1993-14 | A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter ... | 448 | Solution 1:
Put the rectangle on the coordinate plane so its vertices are at $(\pm4,\pm3)$, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$.
Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrant... |
problem_2004-II-9 | A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms ... | 973 | Solution 1:
Let $x = a_2$; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5$ $= (2x-1)^2,\ a_6$ $= (2x-1)(3x-2)$, and in general, $a_{2n} = f(n-1)f(n)$, $a_{2n+1} = f(n)^2$, where $f(n) = nx - (n-1)$.<sup><span id="ref_1">[1]</span></sup>
From \[a_9 + a_{10} = f(4)^2 + f(4)f(5) = ... |
problem_2021-II-5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ . Let's think step by step and outpu... | 736 | Solution 1:
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$rd side is between $6$ and $14$, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less... |
problem_2014-II-8 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle $D$ is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in th... | 254 | Solution 1:
[asy] import graph; size(7.99cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079; draw(circle((7.780000000000009,-1.320000000000002), 2.0... |
problem_2012-I-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 rout... | 279 | Solution 1:
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{... |
problem_2009-II-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 determ... | 96 | Solution 1:
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 tria... |
problem_2019-I-5 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ , $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\tfrac{1}{3}$ , independently of its previous moves. Th... | 252 | Solution 1:
One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\] for $x,y \geq 1,$ and the base cases are
$P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero.
... |
problem_2014-II-11 | In $\triangle RED$ , $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$ . $RD=1$ . Let $M$ be the midpoint of segment $\overline{RD}$ . Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$ . Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$ . Then $... | 56 | Solution 1:
Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use ... |
problem_2020-I-9 | Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are ... | 77 | Solution 1:
[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); ... |
problem_2021-I-7 | Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] Let's think step by step and output the final answer within \boxed{}. | 63 | Solution 1:
It is trivial that the maximum value of $\sin \theta$ is $1$, is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$.
This implies that $\sin(mx) = \sin(nx) = 1$, and that $mx = \frac{\pi}{2}+2a\pi$ and $nx = \frac{\pi}{2}+2b\pi$, for integers $a, b$.
Taking their ratio, we have \[\frac{mx}{nx} ... |
problem_2006-I-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. Let's think step by step and output... | 899 | Solution 1:
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 $2^{n-1}-2^{n-2} = 2^{n-2}$ ... |
problem_1997-6 | Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle. What is the largest value of $n$ for which $A_1$ , $A_n$ , and $B$ are consecutive vertices of a regular polygon? Let's think step by step and output the final answer within \boxed{}. | 42 | Solution 1:
Let the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\angle A_2A_1A_n = \frac{(n-2)180}{n}$, $\angle A_nA_1B = \frac{(m-2)180}{m}$, and $\angle A_2A_1B = 60^{\circ}$. Since those three angles add up to $360^{\circ}$,
\begin{eqnarray*} \frac{(n-2)180}{... |
problem_1994-14 | A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle... | 71 | Solution 1:
At each point of reflection, we pretend instead that the light continues to travel straight.
[asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta *... |
problem_2019-I-10 | For distinct complex numbers $z_1,z_2,\dots,z_{673}$ , the polynomial \[(x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3\] can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017}+g(x)$ , where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016$ . The value of \[\left| \sum_{1 \le j <k \le 673} z_jz_k... | 352 | Solution 1:
In order to begin this problem, we must first understand what it is asking for. The notation
\[\left| \sum_{1 \le j <k \le 673} z_jz_k \right|\]
simply asks for the absolute value of the sum of the product of the distinct unique roots of the polynomial taken two at a time or
\[(z_1z_2+z_1z_3+ \dots + z_1z_... |
problem_2018-I-7 | A right hexagonal prism has height $2$ . The bases are regular hexagons with side length $1$ . Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). Let's think step by step and output the final answer within \boxed{}. | 52 | Solution 1:
We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.
Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \sqrt{3}$ with 6 cases or $\sqrt{3}, \sqrt{3}, \sqrt{3}$ with 2 ca... |
problem_2022-II-12 | Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$ Let's think step by step and output the final answer within \boxed{}. | 23 | Solution 1:
Denote $P = \left( x , y \right)$.
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$.
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equ... |
problem_2011-II-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 $a + b$ . Let's think ste... | 96 | Solution 1:
This takes a slightly different route than Solution 1.
Solution 1 proves that $\angle{DPB}=120^{\circ}$ and that $\overline{BP} = \overline{DP}$.
Construct diagonal $\overline{BD}$ and using the two statements above it quickly becomes clear that $\angle{BDP} = \angle{DBP} = 30^{\circ}$ by isosceles triangle... |
problem_2004-II-6 | Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | 408 | Solution 1:
Denote the number of bananas the first monkey took from the pile as $b_1$, the second $b_2$, and the third $b_3$; the total is $b_1 + b_2 + b_3$. Thus, the first monkey got $\frac{3}{4}b_1 + \frac{3}{8}b_2 + \frac{11}{24}b_3$, the second monkey got $\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3$, and t... |
problem_2020-II-7 | Two congruent right circular cones each with base radius $3$ and height $8$ have axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$ , wher... | 298 | Solution 1:
[asy] unitsize(0.6cm); // Coordinates pair A = (0,0); pair B = (6,0); pair C = (0,6); // Calculate point C pair D = (3,8); pair E = (8,3); pair F = (144/73,384/73); // Draw triangles (cones) draw(A--B--D--cycle); draw(A--C--E--cycle); draw(incircle(A,E,F)); pair EE = foot(C, A, B); real radius = arclength(C... |
problem_2012-II-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 $A... | 919 | Solution 1:
Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$, hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $... |
problem_2005-I-15 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ Let's think step by step and output the final answer within \boxed{}. | 38 | Solution 1:
[asy] size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10); pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir); D(MP("A",A,s)--MP("B",B,s)--MP... |
problem_2019-II-12 | For $n\ge1$ call a finite sequence $(a_1,a_2,\ldots,a_n)$ of positive integers progressive if $a_i<a_{i+1}$ and $a_i$ divides $a_{i+1}$ for $1\le i\le n-1$ . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360.$ Let's think step by step and output the final answer wi... | 47 | Solution 1:
If the first term is $x$, then dividing through by $x$, we see that we can find the number of progressive sequences whose sum is $\frac{360}{x} - 1$, and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the a... |
problem_2004-II-8 | How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers? Let's think step by step and output the final answer within \boxed{}. | 54 | Solution 1:
The prime factorization of 2004 is $2^2\cdot 3\cdot 167$. Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$.
We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization. Fo... |
problem_1995-3 | Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n... | 67 | Solution 1:
It takes an even number of steps for the object to reach $(2,2)$, so the number of steps the object may have taken is either $4$ or $6$.
If the object took $4$ steps, then it must have gone two steps <tt>N</tt> and two steps <tt>E</tt>, in some permutation. There are $\frac{4!}{2!2!} = 6$ ways for these fou... |
problem_1985-11 | An ellipse has foci at $(9, 20)$ and $(49, 55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis? Let's think step by step and output the final answer within \boxed{}. | 85 | Solution 1:
An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant. Let $F_1 = (9, 20)$, $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$-axis. Then $X$ must be the point on the axis such that the sum $F_1X + ... |
problem_2001-II-8 | A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1 - |x - 2|$ for $1\leq x \leq 3$ . Find the smallest $x$ for which $f(x) = f(2001)$ . Let's think step by step and output the final answer within \boxed{}. | 429 | Solution 1:
Iterating the condition $f(3x) = 3f(x)$, we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$. We know the definition of $f(x)$ from $1 \le x \le 3$, so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$. Inde... |
problem_2015-I-14 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with ... | 483 | Solution 1:
Let $n\ge 2$ and define $a(n) = \left\lfloor \sqrt n \right\rfloor$. For $2\le n \le 1000$, we have $1\le a(n)\le 31$.
For $a^2 \le x < (a+1)^2$ we have $y=ax$. Thus $A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n$ (say), and $\Delta_n$ is an integer if $a$ is even; otherwise $\Delta_n$ is an integer plus $\tfrac 1... |
problem_2017-II-10 | Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$ . Point $M$ is the midpoint of $\overline{AD}$ , point $N$ is the trisection point of $\overline{AB}$ closer to $A$ , and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$ . Point $P$ lies on the quadrilateral $BCON$ , and $\overline{BP}$ bisects ... | 546 | Solution 1:
[asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,blue); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,W); label("$N$... |
problem_2007-II-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 $f(5).$ Let's think step by step and output the final answer within \boxed{}. | 676 | Solution 1:
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)... |
problem_1998-12 | Let $ABC$ be equilateral , and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $... | 83 | Solution 1:
WLOG, let $\Delta ABC$ have side length $2.$ Then, $DE = EF = FD = 1.$ We also notice that $\angle CEP = \angle DEF = 60^{\circ},$ meaning $\angle CEF = \angle CEP + \angle DEF = 120^{\circ}.$
Let $EP = x.$ Since $FQ = x$ by congruent triangles $\Delta EPC$ and $\Delta FQA,$ $EQ = EF - FQ = 1-x.$ We can now... |
problem_2004-I-4 | A square has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k.$ Find $100k.$ Let's think step by step and output the final answer... | 86 | Solution 1:
Without loss of generality, let $(0,0)$, $(2,0)$, $(0,2)$, and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$. Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$. Because the segment has l... |
problem_1988-15 | In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delive... | 704 | Solution 1:
Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$, or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given tha... |
problem_2001-I-15 | The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where 8 and 1 are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relati... | 85 | Solution 1:
Choose one face of the octahedron randomly and label it with $1$. There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.
Clearly, the ... |
problem_2008-I-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 seg... | 432 | 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(rightan... |
problem_2002-II-14 | The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ . Let's think step ... | 98 | Solution 1:
Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have
\[\frac{19}{AM} = \frac{152-2... |
problem_2016-II-8 | Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,$ and $61$ . Let's think step by step and output the final answer within \boxed{}. | 728 | Solution 1:
Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$. Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$, and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ... |
problem_2006-I-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... | 65 | Solution 1:
The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts thro... |
problem_2013-II-6 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. Let's think step by step and output the final answer within \boxed{}. | 282 | Solution 1:
Let us first observe the difference between $x^2$ and $(x+1)^2$, for any arbitrary $x\ge 0$. $(x+1)^2-x^2=2x+1$. So that means for every $x\ge 0$, the difference between that square and the next square have a difference of $2x+1$. Now, we need to find an $x$ such that $2x+1\ge 1000$. Solving gives $x\ge \fr... |
problem_2005-II-13 | Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$ Let's think step by step and output the final answer within \boxed{}. | 418 | Solution 1:
We define $Q(x)=P(x)-x+7$, noting that it has roots at $17$ and $24$. Hence $P(x)-x+7=A(x-17)(x-24)$. In particular, this means that
$P(x)-x-3=A(x-17)(x-24)-10$. Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$, where $A$, $(x-17)$, and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ be... |
problem_2021-I-4 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. Let's think step by step and output the final answer within \boxed{}. | 331 | Solution 1:
Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the fi... |
problem_1986-10 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ , $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ , $(bca)$ , $(bac)$ , $(cab)$ , and $(cba)$ , to add these five numbers, and... | 358 | Solution 1:
Let $m$ be the number $100a+10b+c$. Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$. But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$.
Recall that $a, b, c$ refer to the digits the three digit numbe... |
problem_2022-II-8 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$ , $\left\lfloor\frac n5\right\rfloor$ , and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real... | 80 | Solution 1:
We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$, $5$, and/or $6$ which are also multiples of $4$, $5$, and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$, $5$, and $6$ is $60$. We can therefore simplify the problem by finding all such numbers d... |
problem_1987-6 | Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm. AIME 1987 Problem 6.png Let's think step by step and output the final answe... | 193 | Solution 1:
Since $XY = WZ$, $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$. This number is also equal to one quarter the area of the entire rectan... |
problem_2014-I-11 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a po... | 391 | Solution 1:
Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$. Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$, $\langle 1, 1 \rangle$, $\langle -1, -1 \rangle$, $\langle -1, 1 \rangle$ respectively. Moreover, the transformation... |
problem_2015-I-6 | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$ . [asy] pair A,B,C,D,E... | 58 | Solution 1:
Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore... |
problem_2012-I-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 numb... | 170 | Solution 1:
Notice that is $16^2=256$, $1016^2$ ends in $256$. In general, if $x^2$ ends in $256$, $(x+1000)^2=x^2+2000x+1000000$ ends in 256 because $1000|2000x$ and $1000|2000000$. It is clear that we want all numbers whose squares end in $256$ that are less than $1000$.
Firstly, we know the number has to end in a $4... |
problem_1994-15 | Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppo... | 597 | Solution 1:
Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$, and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$. According to the problem statement, the ci... |
problem_2004-II-14 | Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression. For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$ ? Let's... | 108 | Solution 1:
Suppose we require $a$ $7$s, $b$ $77$s, and $c$ $777$s to sum up to $7000$ ($a,b,c \ge 0$). Then $7a + 77b + 777c = 7000$, or dividing by $7$, $a + 11b + 111c = 1000$. Then the question is asking for the number of values of $n = a + 2b + 3c$.
Manipulating our equation, we have $a + 2b + 3c = n = 1000 - 9(b ... |
problem_2004-II-15 | A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the... | 593 | Solution 1:
Number the squares $0, 1, 2, 3, ... 2^{k} - 1$. In this case $k = 10$, but we will consider more generally to find an inductive solution. Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$.
Now, consider the strip of length $1024$. The problem asks f... |
problem_2018-II-8 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ , $(x + 2, y)$ , $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ . Let's think step by step ... | 556 | Solution 1:
We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$. This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$, recording down the number of ways to get to each point recursively.
... |
problem_2017-II-3 | A triangle has vertices $A(0,0)$ , $B(12,0)$ , and $C(8,10)$ . The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . Let's think step by ste... | 409 | Solution 1:
[asy] pair A,B,C,D,X,Z,P; A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4); fill(B--X--P--Z--cycle,lightgray); draw(A--B--C--cycle); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); draw(X--P,dashed); draw(Z--P,dashed); dot(X); label("$X$",X,NE); dot(Z); label("$Z$",Z... |
problem_2023-II-7 | Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. Let's think step by step and output the final answe... | 928 | Solution 1:
Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start wi... |
problem_2001-I-13 | In a certain circle, the chord of a $d$ -degree arc is 22 centimeters long, and the chord of a $2d$ -degree arc is 20 centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers. Find $... | 174 | Solution 1:
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$-degree arcs and one chord of one $3d$-degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$-degree arcs. Let $AB$, $AC$, and $BD$ be the chords of the $d$-degree ar... |
problem_2023-I-11 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ Let's think step by step and output the final answer within \boxed{}. | 235 | Solution 1:
Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the cond... |
problem_2023-I-12 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\a... | 75 | Solution 1:
By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles)$\text{*}$. The law of cosines can be used to compute $DE=42$, $EF=35$, and $FD=13$. Toss the points on the coordinate plane; let $B=(-7, 0)$, $D=(0, 0)$, and $C=(48, 0)$, where we will find $\tan^{2}\left(\measuredangle CDP\right)$ w... |
problem_2021-II-9 | Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$ . Let's think step by step and output the final answer within \boxed{}. | 295 | Solution 1:
This solution refers to the <b>Remarks</b> section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\]
We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives
\begin{align*} \gcd\left(2^m+1,2^n-1\right)\... |
problem_2020-II-6 | Define a sequence recursively by $t_1 = 20$ , $t_2 = 21$ , and \[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\] for all $n \ge 3$ . Then $t_{2020}$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . Let's think step by step and output the final answer within \boxed{}. | 626 | Solution 1:
Let $t_n=\frac{s_n}{5}$. Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$. By substitution, we find $s_3 = \frac{53}{50}$, $s_4=\frac{103}{105\cdot50}$, $s_5=\frac{101}{105}$, $s_6=100$, and $s_7=105$. So $s_n$ has a period of $5$. Thus $s_{2020}=s_5=\frac{101}{105}$. So, $\fr... |
problem_2012-I-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| \l... | 373 | Solution 1:
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 sustai... |
problem_2000-I-13 | In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of... | 731 | Solution 1:
Let the intersection of the highways be at the origin $O$, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\fr... |
problem_2008-I-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 $\triangle GEM$ . Let's think step by step and output the final answer within \boxed{}. | 25 | Solution 1:
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)... |
problem_2002-II-15 | Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$ , and the product of the radii is $68$ . The x-axis and the line $y = mx$ , where $m > 0$ , are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$ , where $a$ , $b$ , and $c$ are positiv... | 282 | Solution 1:
Let the smaller angle between the $x$-axis and the line $y=mx$ be $\theta$. Note that the centers of the two circles lie on the angle bisector of the angle between the $x$-axis and the line $y=mx$. Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$. Let $\... |
problem_2004-I-7 | Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$ Let's think step by step and output the final answer within \boxed{}. | 588 | Solution 1:
Let our polynomial be $P(x)$.
It is clear that the coefficient of $x$ in $P(x)$ is $-1 + 2 - 3 + \ldots + 14 - 15 = -8$, so $P(x) = 1 -8x + Cx^2 + Q(x)$, where $Q(x)$ is some polynomial divisible by $x^3$.
Then $P(-x) = 1 + 8x + Cx^2 + Q(-x)$ and so $P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)$, where $R(x)$ ... |
problem_2004-I-11 | A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum-shaped solid $F,$ in such a way that t... | 512 | Solution 1:
Our original solid has volume equal to $V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi$ and has surface area $A = \pi r^2 + \pi r \ell$, where $\ell$ is the slant height of the cone. Using the Pythagorean Theorem, we get $\ell = 5$ and $A = 24\pi$.
Let $x$ denote the radius of the small cone. Let $... |
problem_2018-II-15 | Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ , $f(6) = 12$ , and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$ . Let's think step by step and output the final answer within \boxed{}. | 185 | Solution 1:
First, suppose $x = y + 1$. Then, the inequality becomes $1 \leq |f(y + 1) - f(y)| \leq 3$. In other words, the (positive) difference between consecutive function values is $1$, $2$, or $3$. Let $d_k := f(k) - f(k - 1)$. Note that
$f(x) - f(y) = \sum_{k=y+1}^{x} d_k.$
Thus, $\sum_{k=1}^{6} d_k = f(6) - f(0)... |
problem_2001-II-15 | Let $EFGH$ , $EFDC$ , and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$ , and let $A$ be the eighth vertex of the cube. Let $I$ , $J$ , and $K$ , be the points on $\overline{EF}$ , $\overline{EH}$ , and $\overline{EC}$ , respectively, so that $EI = EJ = EK = 2$ . A solid $S$ is obtained by drillin... | 417 | Solution 1:
[asy] import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)... |
problem_2016-I-7 | For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left(\frac{\sqrt{|a+b|}}{ab+100}\right)i.\] Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number. Let's think step by step and output the final answer within \boxed{}. | 103 | Solution 1:
We consider two cases:
<b>Case 1:</b> $ab \ge -2016$.
In this case, if
\[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}\]
then $ab \ne -100$ and $|a + b| = 0 = a + b$. Thus $ab = -a^2$ so $a^2 < 2016$. Thus $a = -44,-43, ... |
problem_2009-I-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 a... | 150 | Solution 1:
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 BO_1I_B... |
problem_2014-II-7 | Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which $\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.$ Let's think step by step and output the final answer within \boxed{}. | 21 | Solution 1:
First, let's split it into two cases to get rid of the absolute value sign
$\left |\sum_{k=1}^n\log_{10}f(k)\right|=1 \iff \sum_{k=1}^n\log_{10}f(k)=1,-1$
Now we simplify using product-sum logarithmic identites:
$\log_{10}{f(1)}+\log_{10}{f(2)}+...+\log_{10}{f(n)}=\log_{10}{f(1)\cdot f(2) \cdot ... \cdot f(... |
problem_2019-II-7 | Triangle $ABC$ has side lengths $AB=120,BC=220$ , and $AC=180$ . Lines $\ell_A,\ell_B$ , and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$ , and $\overline{AB}$ , respectively, such that the intersections of $\ell_A,\ell_B$ , and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,4... | 715 | Solution 1:
Let the points of intersection of $\ell_A, \ell_B,\ell_C$ with $\triangle ABC$ divide the sides into consecutive segments $BD,DE,EC,CF,FG,GA,AH,HI,IB$. Furthermore, let the desired triangle be $\triangle XYZ$, with $X$ closest to side $BC$, $Y$ closest to side $AC$, and $Z$ closest to side $AB$. Hence, the ... |
problem_2019-II-4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . Let's think step by step and output the final answer within \boxed{}. | 187 | Solution 1:
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are... |
problem_2014-II-9 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. Let's think step by step and output the final answer within \boxed{}. | 581 | Solution 1:
We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have... |
problem_2019-II-9 | Call a positive integer $n$ $k$ - pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of the positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ . Let's think step by step and output the final answer within \boxed{... | 472 | Solution 1:
Every 20-pretty integer can be written in form $n = 2^a 5^b k$, where $a \ge 2$, $b \ge 1$, $\gcd(k,10) = 1$, and $d(n) = 20$, where $d(n)$ is the number of divisors of $n$. Thus, we have $20 = (a+1)(b+1)d(k)$, using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of ... |
problem_2017-I-6 | A circle circumscribes an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the larg... | 48 | Solution 1:
The probability that the chord doesn't intersect the triangle is $\frac{11}{25}$. The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\frac{2x}{360}=\f... |
problem_1995-14 | In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form $m\pi-... | 378 | Solution 1:
Let the center of the circle be $O$, and the two chords be $\overline{AB}, \overline{CD}$ and intersecting at $E$, such that $AE = CE < BE = DE$. Let $F$ be the midpoint of $\overline{AB}$. Then $\overline{OF} \perp \overline{AB}$.
[asy] size(200); pathpen = black + linewidth(0.7); pen d = dashed+linewidth... |
problem_1992-14 | In triangle $ABC^{}_{}$ , $A'$ , $B'$ , and $C'$ are on the sides $BC$ , $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ , $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'... | 94 | Solution 1:
Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, \[\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.\] Therefore, we have
\[\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}\] \[\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}\] \[\frac{CO}{OC'}=\frac{K_A+K_... |
problem_2007-I-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 $p$ . Let's think step by step and output the final answer within \boxed{}. | 80 | 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 intersection $(-1... |
problem_2019-I-9 | Let $\tau(n)$ denote the number of positive integer divisors of $n$ . Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$ . Let's think step by step and output the final answer within \boxed{}. | 540 | Solution 1:
In order to obtain a sum of $7$, we must have:
* either a number with $5$ divisors (<i>a fourth power of a prime</i>) and a number with $2$ divisors (<i>a prime</i>), or
* a number with $4$ divisors (<i>a semiprime or a cube of a prime</i>) and a number with $3$ divisors (<i>a square of a prime</i>). (No in... |
problem_2022-I-3 | In isosceles trapezoid $ABCD,$ parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650,$ respectively, and $AD=BC=333.$ The angle bisectors of $\angle A$ and $\angle D$ meet at $P,$ and the angle bisectors of $\angle B$ and $\angle C$ meet at $Q.$ Find $PQ.$ Let's think step by step and output th... | 242 | Solution 1:
We have the following diagram:
[asy] size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = i... |
problem_2018-I-14 | Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that ... | 351 | Solution 1:
This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line $E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E$. We can count the number of left/r... |
problem_2022-I-12 | For any finite set $X,$ let $|X|$ denote the number of elements in $X.$ Define \[S_n = \sum |A \cap B|,\] where the sum is taken over all ordered pairs $(A,B)$ such that $A$ and $B$ are subsets of $\{1,2,3,\ldots,n\}$ with $|A|=|B|.$ For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A,B) \in ... | 245 | Solution 1:
Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$. The problem states that for $n=2, S_n$ is $4$. Let's try it out for $n=3$.
Let's perform casework on the number of elements in $A, B$.
$\textbf{Case 1:} |A| = |B| = 1$
In this case, the only possible equ... |
problem_2017-II-14 | A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ , $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points. Let's think step by step and output the final answer within \boxed{}. | 168 | Solution 1:
$Case \textrm{ } 1:$ The lines are not parallel to the faces
A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.
We look at the one from $(1,1,1)$ to $(... |
problem_2002-II-8 | Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ . (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$ .) Let's think step by step and output the final answer within \boxed{}. | 49 | Solution 1:
Note that if $\frac{2002}n - \frac{2002}{n+1}\leq 1$, then either $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor$,
or $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1$. Either way, we won't skip any natural numbers.
The greatest $n$ suc... |
problem_2009-II-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 $m \cdot n$ . Let's think step by step and output the final answer within \boxed{}. | 125 | Solution 1:
We have $\log m - \log k = \log \left( \frac mk \right)$, hence we can rewrite the inequality as follows:
\[- \log n < \log \left( \frac mk \right) < \log n\]
We can now get rid of the logarithms, obtaining:
\[\frac 1n < \frac mk < n\]
And this can be rewritten in terms of $k$ as
\[\frac mn < k < mn\]
From ... |
problem_1998-6 | Let $ABCD$ be a parallelogram . Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$ Let's think step by step and output the final answer within \boxed{}. | 308 | Solution 1:
We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$. We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$. Equating the two results gives $\frac{112}{RC} = \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$ |
problem_2020-I-11 | For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$ Let's think step by step and output the final answer within \boxed{}. | 510 | Solution 1:
There can be two different cases for this problem, either $f(2)=f(4)$ or not. If it is, note that by Vieta's formulas $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. ... |
problem_1983-15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | 175 | Solution 1:
[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label("$O$",O,SW); pair M = (4,0);dot(M);label("$M$",M,SE); pair N = (4,2);dot(N);label("$N$",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label("$B$",B,NE); pair C = (4,-3);dot(C);label("$C$",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);l... |
problem_2006-II-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. Let's think step by step and output the final answer within \boxed{}. | 834 | 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}) - \left(\sum^{28}_... |
problem_2007-I-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 sequen... | 169 | Solution 1:
Another way would be to use a table representing the number of ways to reach a certain number
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\ \hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\ \end{tabular}$
How we... |
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