2021 AMC 12B Problem 15

The right interior point is distance $2$ from $A,$ $B,$ and $C,$ so these three points lie on a circle of radius $2.$ Chords $AB$ and $BC$ also have length $2,$ so each subtends $60^\circ$ at the center. Thus $AC=2\sqrt3.$ Similarly, the other half is its mirror image. Put $C=(-1,0),$ $D=(1,0),$ and $A=(0,\sqrt{11});$ then $AC=AD=2\sqrt3.$ The other intersection of the radius-$2$ circles centered at $A$ and $C$ is $$B=\left(-\tfrac12-\tfrac{\sqrt{11}}{2\sqrt3}, \tfrac{\sqrt{11}}2+\tfrac{1}{2\sqrt3}\right),$$ and $E$ is its reflection across the $y$-axis. Applying the shoelace formula to $A,B,C,D,E$ gives $[ABCDE]=\sqrt{11}+2\sqrt3,$ which equals $\sqrt{11}+\sqrt{12}.$ So $m+n=11+12=23.$ Thus, the correct answer is D.