2007 AMC 10B Problem 21

The square sits on the hypotenuse $AC$, with its top side $WZ$ parallel to $AC$ and its top corners on the legs. Let the side length be $s$.

Draw the altitude from $B$ to $AC$. Its length is
$$h = \frac{AB \cdot BC}{AC} = \frac{3 \cdot 4}{5} = \frac{12}{5}.$$

Because $WZ \parallel AC$, triangle $BWZ$ is similar to triangle $BAC$. The altitude of $BWZ$ from $B$ is the part of $h$ above the square, namely $h - s$, and its base is $WZ = s$. Corresponding bases and altitudes are in the same ratio:
$$\frac{s}{5} = \frac{h - s}{h}.$$
Cross-multiplying, $sh = 5h - 5s$, so $s(h + 5) = 5h$ and
$$s = \frac{5h}{h + 5} = \frac{5 \cdot \frac{12}{5}}{\frac{12}{5} + 5} = \frac{12}{\frac{37}{5}} = \frac{60}{37}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{60}{37}}$.

A segment parallel to a side of a triangle creates a similar triangle, and the cleanest comparison is base against altitude. For a square on a side of length $b$ with opposite altitude $h$, the formula $s = \frac{bh}{b + h}$ always results; here $b = 5$, $h = \frac{12}{5}$. Compute the altitude to the hypotenuse first whenever a figure is built on it.

Work along the hypotenuse. Triangle $AXW$ is right-angled at $X$ with $\tan A = \frac{BC}{AB} = \frac43$, so $AX = \frac{s}{\tan A} = \frac{3s}{4}$. Likewise $\tan C = \frac34$ gives $CY = \frac{4s}{3}$. Then
$$AC = AX + XY + YC = \frac{3s}{4} + s + \frac{4s}{3} = \frac{37s}{12} = 5,$$
so $s = \frac{60}{37}$.

• Solving the more familiar configuration where the square's corner sits at the right angle, which gives 12/7.
• Using a leg ($3$ or $4$) as the base of the similar triangle instead of the hypotenuse $5$ with its altitude $\frac{12}{5}$.