2022 AIME II Problem 11

Reflect $B$ over the bisector from $A$ and $C$ over the bisector from $D:$ both reflections land on $\overline{AD}$

Since $MB = MC,$ the midpoint $M$ is equidistant from the two reflections, which lie on $\overline{AD}$ at distances $2$ from $A$ and $3$ from $D$

Place $A = (0, 0)$ and $D = (7, 0)$ with $B, C$ above the axis, and let $M$ be the midpoint of $\overline{BC}.$ Reflecting $B$ over the bisector line $AM$ carries ray $AB$ to ray $AD,$ so $B$ maps to $B' = (2, 0),$ and reflecting $C$ over the bisector $DM$ gives $C' = (4, 0).$ Since $M$ lies on both mirror lines, $MB' = MB = MC = MC',$ so $M$ is equidistant from $B'$ and $C'$ and hence $M = (3, h)$ for some $h \gt 0.$

Write $\angle DAB = 2\alpha$ and $\angle ADC = 2\delta,$ so $\tan\alpha = \frac{h}{3}$ and $\tan\delta = \frac{h}{4}.$ Then $B = (2\cos 2\alpha,\, 2\sin 2\alpha)$ and $C = (7 - 3\cos 2\delta,\, 3\sin 2\delta),$ and the midpoint condition on the $x$-coordinates reads $2\cos 2\alpha - 3\cos 2\delta = -1.$ Substituting $\cos 2\alpha = \frac{9 - h^2}{9 + h^2}$ and $\cos 2\delta = \frac{16 - h^2}{16 + h^2}$ and clearing denominators gives $2h^4 = 10h^2,$ so $h^2 = 5.$ (The $y$-coordinate condition is then satisfied automatically: $2\sin 2\alpha$ ${}+ 3\sin 2\delta$ $= \frac{6\sqrt{5}}{7} + \frac{8\sqrt{5}}{7}$ $= 2h.$)

Now $\cos 2\alpha = \frac{2}{7},$ $\sin 2\alpha = \frac{3\sqrt{5}}{7},$ $\cos 2\delta = \frac{11}{21},$ $\sin 2\delta = \frac{8\sqrt{5}}{21},$ so $B = \left(\frac{4}{7}, \frac{6\sqrt{5}}{7}\right)$ and $C = \left(\frac{38}{7}, \frac{8\sqrt{5}}{7}\right).$ The shoelace formula on $A, B, C, D$ gives area $6\sqrt{5},$ whose square is $180.$