# 2017 AMC 10A Problems/Problem 21

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A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\tfrac{x}{y}$?

$\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}$

## Solution

Analyze the first right triangle.

$[asy] pair A,B,C; pair D, e, F; A = (0,0); B = (4,0); C = (0,3); D = (0, 12/7); e = (12/7 , 12/7); F = (12/7, 0); draw(A--B--C--cycle); draw(D--e--F); label("x", D/2, W); label("A", A, SW); label("B", B, SE); label("C", C, N); label("D", D, W); label("E", e, NE); label("F", F, S); [/asy]$

Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$. This can be written as $\frac{4-x}{x}=\frac{4}{3}$. Solving, $x = \frac{12}{7}$.

Now we analyze the second triangle.

$[asy] pair A,B,C; pair q, R, S, T; A = (0,0); B = (4,0); C = (0,3); q = (1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label("y", (q+R)/2, NW); label("A'", A, SW); label("B'", B, SE); label("C'", C, N); label("Q", (q-(0,0.3))); label("R", R, NE); label("S", S, NE); label("T", T, W); [/asy]$

Similary, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, so $RB' = \frac{4}{3}y$, and $C'S = \frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5$. Solving for $y$, we get $y = \frac{60}{37}$. Thus, $\frac{x}{y} = \frac{37}{35}$.