2024 AMC 12B Problem 7

Set $X = (0,0),$ $W = (0,4),$ $Y = (8,0),$ $Z = (8,4),$ with $M = (m, 0)$ on $\overline{XY}$ and $A = (8, a)$ on $\overline{YZ}.$ Since $\angle WMA = 90^\circ,$ $\overrightarrow{MW} \cdot \overrightarrow{MA}$ $= (-m)(8-m)$ $+ 4a = 0,$ so $4a = m(8-m).$ The areas give $[\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m$ and $[\triangle WAZ]$ $= \tfrac12 \cdot 8 \cdot (4 - a)$ $= 4(4 - a).$ Setting these equal yields $m = 8 - 2a.$ Substituting $a = \tfrac{8-m}{2}$ into $4a = m(8-m)$ gives $2(8-m) = m(8-m),$ so $m = 2$ and $a = 3.$ The other algebraic root, $m=8,$ gives $M=A=Y$ and no defined angle $\angle WMA,$ so it is invalid. Then with $W = (0,4),$ $M = (2,0),$ $A = (8,3),$ $$\begin{aligned} [\triangle WMA] &= \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| \\ &= \tfrac12 (30) = 15. \end{aligned}$$ Thus, the correct answer is C.