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Difference between revisions of "2017 AMC 12A Problems/Problem 19"

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==Problem==
  
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A square with side length <math>x</math> is inscribed in a right triangle with sides of length <math>3</math>, <math>4</math>, and <math>5</math> so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length <math>y</math> is inscribed in another right triangle with sides of length <math>3</math>, <math>4</math>, and <math>5</math> so that one side of the square lies on the hypotenuse of the triangle. What is <math>\frac{x}{y}</math>?
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<math> \textbf{(A)}\ \frac{12}{13}
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\qquad \textbf{(B)}\ \frac{35}{37}
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\qquad\textbf{(C)}\ 1
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\qquad\textbf{(D)}\ \frac{37}{35}
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\qquad\textbf{(E)}\ \frac{13}{12}</math>

Revision as of 17:12, 8 February 2017

Problem

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 $\frac{x}{y}$?

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

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