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:

For the first square, the two smaller triangles it cuts off are similar to the whole triangle, giving $\dfrac{x}{3-x}=\dfrac{4-x}{x},$ so $x=\dfrac{12}{7}.$ (Equivalently, a square in the right angle has side $\dfrac{3\cdot4}{3+4}.$ )

For the second square, take the hypotenuse of length $5$ as base; the altitude to it is $h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}.$ A square with a side on a base $b$ and height $h$ has side $\dfrac{bh}{b+h},$ so $y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}.$

Therefore $\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}.$

Thus, the correct answer is D.

2017 AMC 12A Problem 19

Solution:

Problem 19 in Other Years