2024 AMC 10B Problem 11

In the figure, $WX = 4$ is the left side, $XY = 8$ is the bottom, $M$ is on the bottom and $A$ is on the right side $YZ$. Triangle $WMA$ is hard to measure directly, but the rectangle minus three corner triangles is easy. So the plan is to find the lengths $XM$ and $ZA$.

1. Use the equal areas. Call $ZA = t$. Triangle $WAZ$ is a right triangle with legs $WZ = 8$ and $ZA = t$, so its area is $\frac12 \cdot 8 \cdot t = 4t$. Triangle $WXM$ has legs $WX = 4$ and $XM$, so its area is $\frac12 \cdot 4 \cdot XM = 2 \cdot XM$. Setting them equal gives $XM = 2t$.
2. Write the other pieces. Then $MY = 8 - 2t$ and $AY = 4 - t$. Notice that $MY = 2(4 - t)$, so $MY$ is exactly twice $AY$.
3. Find two triangles with the same shape. Along the bottom line at $M$, the three angles $\angle XMW$, $\angle WMA = 90^\circ$ and $\angle AMY$ add to $180^\circ$. So $\angle XMW + \angle AMY = 90^\circ$. In triangle $WXM$ the two sharp angles also add to $90^\circ$, so $\angle XWM + \angle XMW = 90^\circ$. Comparing, $\angle XWM = \angle AMY$.
4. So right triangles $WXM$ and $MYA$ have the same angles, which means one is a scaled copy of the other. In each, compare the side next to the matching angle with the side across from it:
$$\frac{WX}{XM} = \frac{MY}{YA}.$$
5. The right side is $2$ by step 2. So $\frac{4}{2t} = 2$, which gives $t = 1$.
6. Now all the lengths are known: $ZA = 1$, $AY = 3$, $XM = 2$, $MY = 6$.
7. Subtract the corners. The rectangle has area $4 \cdot 8 = 32$. The three corner triangles have areas
- $WXM$: $\frac12 \cdot 4 \cdot 2 = 4$
- $WZA$: $\frac12 \cdot 8 \cdot 1 = 4$
- $MYA$: $\frac12 \cdot 6 \cdot 3 = 9$

So the area of $WMA$ is $32 - 4 - 4 - 9 = 15$.

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

When a right angle has its corner on one side of a rectangle, the two triangles beside it are always the same shape. That gives one equation, and the equal-area condition gives another, which is enough to find the unknown lengths. The tilted triangle in the middle is then easiest to find by subtracting the corners from the rectangle.

Use the Pythagorean theorem on the right triangle $WMA$ instead of the matching shapes. With $XM = 2t$ and $ZA = t$:

• $WM^2 = 4^2 + (2t)^2 = 16 + 4t^2$
• $MA^2 = (8 - 2t)^2 + (4 - t)^2 = 80 - 40t + 5t^2$
• $WA^2 = 8^2 + t^2 = 64 + t^2$

Since $WM^2 + MA^2 = WA^2$, we get $96 - 40t + 9t^2 = 64 + t^2$, or $8t^2 - 40t + 32 = 0$, which is $t^2 - 5t + 4 = 0$. So $t = 1$ or $t = 4$. But $t = 4$ would put $A$ at $Y$ and flatten the picture, so $t = 1$ and the answer is again $15$.

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 $\boxed{\textbf{(C)}\ 15}$.