2021 JMC 10 Problems/Problem 10

Problem

Let $ABCD$ be a square with sides of length $4.$ Point $X$ is on side ${CD}$ and point $Y$ is on side ${BC}$ such that $AX = 5$ and angle ${AYX}$ is right. What is $AY \cdot XY?$

$\textbf{(A) } 10 \qquad\textbf{(B) } 9\sqrt{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 6\sqrt{5} \qquad\textbf{(E) } 7\sqrt{5}$

Solution

By Pythagoras, $DX= \sqrt{5^2 - 4^2} = 3 \implies CX = CD - DX = 4-3=1.$ Furthermore, $\angle{BAY} = \angle{CYD},$ implying that $\triangle{YBA} \sim \triangle{YXC}.$ Let $YB = x.$ Because of the similarity, $\tfrac{1}{4-x} = \tfrac{x}{4} \implies x = 2.$ So $YC = 4-x = 2,$ and by Pythagoras, $DY = \sqrt{5},$ and $AY = 2\sqrt{5}.$ The answer is $2\sqrt{5} \cdot \sqrt{5} = 10.$

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2021 JMC 10 Problems/Problem 10

Problem

Solution