2024 AMC 10B Problem 11

Let $ZA = t$, so $AY = 4 - t$. Triangle $WXM$ has legs $WX = 4$ and $XM$; triangle $WAZ$ has legs $WZ = 8$ and $ZA = t$. Equal areas give $\frac12 \cdot 4 \cdot XM = \frac12 \cdot 8 \cdot t$, so $XM = 2t$ and $MY = 8 - 2t$.

At $M$ the angles $\angle XMW$ and $\angle YMA$ add to $90^\circ$ (they flank the right angle $\angle WMA$ on a straight line). In right triangle $WXM$, $\angle XWM$ is also complementary to $\angle XMW$, so $\angle XWM = \angle YMA$ and the right triangles $WXM$ and $MYA$ are similar, with $WX \leftrightarrow MY$ and $XM \leftrightarrow YA$:
$$\frac{WX}{XM} = \frac{MY}{YA} \quad\Longrightarrow\quad \frac{4}{2t} = \frac{8 - 2t}{4 - t} = 2.$$
So $2t = 2$ and $t = 1$. Then $ZA = 1$, $XM = 2$, $MY = 6$, $AY = 3$.

Subtract the three corner triangles from the rectangle:
$$[WMA] = 32 - \tfrac12(4)(2) - \tfrac12(8)(1) - \tfrac12(6)(3) = 32 - 4 - 4 - 9 = 15.$$

The answer is $\boxed{\textbf{(C)}\ 15}$.

A right angle whose vertex sits on a side of a rectangle always produces a pair of similar right triangles in the two adjacent corners; that is the standard "folded right angle" configuration. The equal-area condition supplies one relation between the unknown segments, the similarity supplies the other, and the target triangle is easiest to measure by subtraction rather than directly.

With $XM = 2t$ and $ZA = t$, apply the Pythagorean theorem to the right triangle $WMA$: $WM^2 + MA^2 = WA^2$ becomes $(4t^2 + 16) + \big((8-2t)^2 + (4-t)^2\big) = t^2 + 64$, which simplifies to $t^2 - 5t + 4 = 0$. The root $t = 4$ would put $A$ at $Y$ and collapse the figure, so $t = 1$, and $[WMA] = \frac12 \sqrt{20} \cdot \sqrt{45} = 15$.

• Solving the similarity proportion but then reporting the area of a corner triangle (like MYA = 9) instead of the rectangle minus all three corners.
• Setting $XM = ZA$ from "equal areas" without accounting for the different legs $4$ and $8$; the correct relation is $XM = 2\,ZA$.