2017 AMC 10A Problem 21

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}

Use similar triangles in the first square placement

In the second placement, express the two leftover side pieces in terms of

y

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.

Problem 21 in other years