## Problem 1 Zou and Chou are practicing their 100-meter sprints by running 6 races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac{2}{3}$ if they won the previous race but only $\frac{1}{3}$ if they lost the previous race. The probability that Zou will win exactly 5 of the 6 races is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Problem 2 In the diagram below, $A B C D$ is a rectangle with side lengths $A B=3$ and $B C=11$, and $A E C F$ is a rectangle with side lengths $A F=7$ and $F C=9$, as shown. The area of the shaded region common to the interiors of both rectangles is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ### Solution Let $G=AD\cap FC$ - $\triangle AGF\sim\triangle CGD$ by [[AA similarity]] - $\angle AGF\cong\angle CGD$ - by [[vertical angles]] - $\angle AFG\cong\angle CDG$ - since $AECF$ and $ABCD$ are [[rectangle | rectangles]], and right angles are congruent The [[scale factor]] is found from the given $AF$ and $FC$, and applied to $FD$ and $DG$ $\begin{aligned}\frac{AF}{CD}=\frac{FG}{DG}=\frac{GA}{GC}\\ \frac{7}{3}=\frac{FG}{11-x}=\frac{x}{GC}\end{aligned} $ ![[aime2021_p2-sol.svg]] ## Problem 3 Find the number of positive integers less than 1000 that can be expressed as the difference of two integral powers of 2 ### Solution $0<2^n-2^m<1000;m,n\in\mathbb{N}\begin{cases}0\leq m< n\leq9\\n=10, 5\leq m \leq9\end{cases}$ $\frac{9(10)}{2}+5=\boxed{50}$ ## Problem 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. ### Solution Group based on how many are in the smallest stack. (1,2,63),(1,3,62),...,(1,32,33) - 31 ways (2,3,61),(2,4,60),...,(2,31,33) - 29 ways (3,4,59),(3,4,58),...,(3,31,32) - 28 ways ... (20,21,25),(20,22,24) - 2 ways (21,22,23) - 1 way $31+29+28+26+\ldots+4+2+1$ $(30+1)+(30-1)+(27+1)+(27-1)+\ldots+(3+1)+(3-1)+1$ $2(30+27+\ldots+3)+1=6\left(\sum_{1\leq n\leq10}n\right)+1$ $6\cdot\frac{10\cdot11}{2}+1=\boxed{331}$ ## Problem 5 Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. ### Solution $(a_0,a_1,a_2)=(a_1-d, a_1,a_1+d)$ Sum of squares of three terms $(a_1-d)^2+(a_1)^2+(a_1+d)^2 $ $3(a_1)^2+2d^2$ Middle term times square of common difference $a_1\cdot d^2$ Setting them equal $\begin{aligned} 3(a_1)^2+2d^2&=a_1\cdot d^2\\ 3(a_1)^2-d^2(a_1)+2d^2&=0 \end{aligned}$ Then using quadratic formula to find $a_1$ in terms of $d$ $a_1=\frac{d^2\pm\sqrt{d^4+24d^2}}{6}=\frac{d^2\pm d\sqrt{d^2+24}}{6}$ Since $a_1\in\mathbb{N}$ the discriminant must be a perfect square. Since $d\in\mathbb{N}$ as well, $d^2$ must be a perfect square. ## Problem 6 Segments $\overline{A B}, \overline{A C}$, and $\overline{A D}$ are edges of a cube and $\overline{A G}$ is a diagonal through the center of the cube. Point $P$ satisfies $P B=60 \sqrt{10}, P C=60 \sqrt{5}, P D=120 \sqrt{2}$, and $P G=36 \sqrt{7}$. What is $P A$ ? ### Solution Each of these given lengths are ones that are square roots of sums of squares of lengths parallel to edges. Keep in mind that $PA$ can also be written this way Let $A(0,0,0),$ $G(s,s,s),$ and $P(p_x,p_y,p_z)$ >(when squaring the original lengths, it is wise to keep the common factor 12 separate to make the calculations simpler) $\begin{array}{lcccccccc} PB^2=12^2\cdot250&=&(s-{p_x})^2&+&{p_y}^2&+&{p_z}^2\\ PC^2=12^2\cdot125&=&{p_x}&+&(s-{p_y})^2&+&{p_z}^2\\ PD^2=12^2\cdot200&=&{p_x}&+&{p_y}^2&+&(s-{p_z})^2\\ PG^2=12^2\cdot63&=&(s-{p_x})^2&+&(s-{p_y})^2&+&(s-{p_z})^2\\ PA^2&=&{p_x}^2&+&{p_y}^2&+&{p_z}^2\\ \end{array}$ From here, we see $PA^2=\frac12(PB^2+PC^2+PD^2-PG^2)$, or numerically: $PA^2=\frac12\cdot12^2\cdot(250+125+200-63)=12^2\cdot256$ Taking the square root is a little easier now. $PA=12\cdot16=\boxed{192}$ ## Problem 7 Find the number of pairs $(m, n)$ of positive integers with $1 \leq m<n \leq 30$ such that there exists a real number $x$ satisfying $ \sin (m x)+\sin (n x)=2 $ ### Solution For this equation to be true, both $\sin$ components will need to be $1$ at the same time. This forces both $mx$ and $nx$ to reach the top of the unit circle simultaneously at any point $mx=2k_m\pi+\frac\pi2$ $nx=2k_n\pi+\frac\pi2$ If $m=1$ then $n\equiv1\pmod4$ will generate a solution at $x=\frac\pi2$ For all other $m$, we can see $n\equiv m\pmod {4m}$ will generate $x=\frac{\pi}{2m}$ $(1,5),(1,9),(1,13),(1,17),(1,21),(1,25),(1,29)$ $(2,10),(2,18),(2,26)$ $(3,15),(3,27)$ $(4,20)$ For a total of 13 pairs that generate solutions ## Problem 8 ### Solution ## Problem 9 ### Solution ## Problem 10 ### Solution ## Problem 11 ### Solution ## Problem 12 ### Solution ## Problem 13 ### Solution ## Problem 14 ### Solution ## Problem 15 ### Solution ## Problem 16 ### Solution ## Problem 17 ## Problem 18 ## Problem 19 ## Problem 20 ## Problem 21 ## Problem 22 ## Problem 23 ## Problem 24 ## Problem 25 ## Problem 26 ## Problem 27 ## Problem 28 ## Problem 29 ## Problem 30