Question

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 $\dfrac{x}{y}?$

$\dfrac{12}{13}$

$\dfrac{35}{37}$

$1$

$\dfrac{37}{35}$

$\dfrac{13}{12}$

Solution:

We can see that $\triangle ABC$ and $\triangle FBE$ are similar (angle-angle). This gives us $\dfrac{BF}{FE} = \dfrac{AB}{AC}$ $\dfrac{4 - x}{x} = \dfrac{4}{3}.$

Cross-multiplying yields $12 - 3x = 4x$ $x = \dfrac{12}{7}.$

Here, we have that $\triangle ABC,$ $\triangle RBQ,$ and $\triangle STC$ are similar (angle-angle).

This means that $RB = \dfrac{4}{3}y$ and $CS = \dfrac{3}{4}y.$ This gives us the equation $\dfrac{4}{3}y + \dfrac{3}{4}y + y = 5$ $\dfrac{37}{12}y = 5.$

Finally, we get that $y = \dfrac{60}{37}.$ The desired ratio is $\dfrac{\frac{12}{7}}{\frac{60}{37}} = \dfrac{37}{35}.$

Thus, D is the correct answer.

2017 AMC 10A Problem 21

Solution:

Problem 21 in Other Years