2007 AMC 10B Problem 21

Here the square sits on the long side $AC$ (length $5$), with its top two corners $W$ and $Z$ touching the two short sides. Let the side of the square be $s$.

1. Height of the big triangle over $AC$. The area of the right triangle is $\frac12 \times 3 \times 4 = 6$. Using $AC = 5$ as the base instead, $\frac12 \times 5 \times h = 6$, so
$$h = \frac{12}{5}.$$
2. The small triangle on top. The top side $WZ$ of the square is parallel to $AC$. So triangle $BWZ$ has the same shape as triangle $BAC$, just smaller.
3. Compare base and height. Triangle $BWZ$ has base $WZ = s$. Its height is the part of $h$ above the square, $h - s$. In same-shaped triangles, base divided by height is the same:
$$\frac{s}{h - s} = \frac{5}{h}.$$
4. Put in $h = \frac{12}{5}$ and solve. Cross-multiply: $s \times h = 5(h - s)$, so
$$\frac{12}{5}s = 12 - 5s.$$
Multiply by $5$: $12s = 60 - 25s$, so $37s = 60$ and
$$s = \frac{60}{37}.$$
5. Check the size: $\frac{60}{37} \approx 1.62$, a bit smaller than $\frac{12}{7} \approx 1.71$ from the corner square, which makes sense.

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

A line parallel to one side of a triangle cuts off a smaller triangle of the same shape, and the cleanest comparison is base against height. For a square standing on a side of length $b$ whose opposite height is $h$, this always gives $s = \frac{bh}{b + h}$; here $\frac{5 \times 12/5}{5 + 12/5} = \frac{12}{37/5} = \frac{60}{37}$. When a figure is built on the long side of a right triangle, find the height to that side first.

Walk along the long side $AC$. It is split into three pieces: $AX$, the square's bottom $XY = s$, and $YC$.

1. Triangle $AXW$ has a right angle at $X$ and the same shape as the big triangle (they share angle $A$). In the big triangle, the side across from $A$ is $4$ and the side next to $A$ is $3$. So $WX : AX = 4 : 3$, which means $AX = \frac{3}{4}s$.
2. In the same way, triangle $CYZ$ shares angle $C$, where the ratio is $3 : 4$, so $CY = \frac{4}{3}s$.
3. Add up the pieces:
$$\frac34 s + s + \frac43 s = \frac{9 + 12 + 16}{12}s = \frac{37}{12}s = 5,$$
so $s = \frac{60}{37}$.

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