2017 AMC 12A Problem 19

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}

A square with its corner at the right angle has side

x=\dfrac{3\cdot4}{3+4}

A square with a side on a base of length

b

and height

h

to that base has side

\dfrac{bh}{b+h};

use the hypotenuse as the base

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.

Problem 19 in other years