2007 AMC 10B Problem 21

Problem 21 of 25HarderGeometry

Right

\triangle ABC

has

AB=3,

BC=4,

and

AC=5.

Square

XYZW

is inscribed in

\triangle ABC

with

X

and

Y

on

\overline{AC},

W

on

\overline{AB},

and

Z

on

\overline{BC}.

What is the side length of the square?

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Solution

Let

s

be the side of the square and

h

the altitude from

B

to

AC.

Then

h=\dfrac{AB\cdot BC}{AC}=\dfrac{3\cdot 4}{5}=\dfrac{12}{5}.

The small triangle above the square is similar to

\triangle ABC

with the square’s top side as its base, giving

\dfrac{h-s}{h}=\dfrac{s}{AC},

so

s=\dfrac{AC\cdot h}{AC+h}.

Substituting,

s=\dfrac{5\cdot\frac{12}{5}}{5+\frac{12}{5}}=\dfrac{12}{\frac{37}{5}}=\dfrac{60}{37}.

Thus, the correct answer is B.

More practice

Concepts: similarity · right triangle · altitude

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.