# 2007 AMC 10B Problems/Problem 21

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## Problem

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? $\textbf{(A) } \frac{3}{2} \qquad\textbf{(B) } \frac{60}{37} \qquad\textbf{(C) } \frac{12}{7} \qquad\textbf{(D) } \frac{23}{13} \qquad\textbf{(E)} 2$

## Solution 1

There are many similar triangles in the diagram, but we will only use $\triangle WBZ \sim \triangle ABC.$ If $h$ is the altitude from $B$ to $AC$ and $s$ is the sidelength of the square, then $h-s$ is the altitude from $B$ to $WZ.$ By similar triangles, \begin{align*} \frac{h-s}{s}&=\frac{h}{5}\\ 5(h-s)&=hs\\ 5h-5s&=hs\\ 5h&=s(h+5)\\ s&=\frac{5h}{h+5} \end{align*}

Find the length of the altitude of $\triangle ABC.$ Since it is a right triangle, the area of $\triangle ABC$ is $\frac{1}{2}(3)(4) = 6.$

The area can also be expressed as $\frac{1}{2}(5)(h),$ so $\frac{5}{2}h=6 \longrightarrow h=2.4.$

Substitute back into $s.$ $$s=\frac{5h}{h+5} = \frac{12}{7.4} = \boxed{\mathrm{(B) \ } \frac{60}{37}}$$

## Solution 2

Let $l$ be the side length of the inscribed square. Note that $\triangle ZYC \sim \triangle WBZ \sim \triangle ABC$.

Then we can setup the following ratios: $$\frac{CZ}{l} = \frac{5}{3} \rightarrow CZ = \frac{5}{3}l$$ $$\frac{ZB}{l} = \frac{4}{5} \rightarrow ZB = \frac{4}{5}l$$

But then $\frac{5}{3}l+\frac{4}{5}l = CZ+ZB = CB = 4 \longrightarrow \frac{37}{15}l=4 \longrightarrow l = \frac{60}{37} \Longrightarrow \boxed{\mathrm{(B)}\frac{60}{37}}$

## Video Solution by OmegaLearn

~ pi_is_3.14

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