Question

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?

$\dfrac{3}{2}$

$\dfrac{60}{37}$

$\dfrac{12}{7}$

$\dfrac{23}{13}$

$2$

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

Thus, the correct answer is B.

2007 AMC 10B Problem 21

Solution:

Problem 21 in Other Years