2022 AMC 12B Problem 25

Center the square at the origin with vertices $\left(\pm\tfrac12, \pm\tfrac12\right).$ Each hexagon shares one edge with the square and extends across to the opposite side; the hexagon on the bottom edge, for instance, has its far (top) edge from $\left(-\tfrac12, \sqrt3 - \tfrac12\right)$ to $\left(\tfrac12, \sqrt3 - \tfrac12\right).$ The outer boundary is a $12$-gon with flat edges at distance $\sqrt3 - \tfrac12$ from the center, convex vertices such as $\left(\sqrt3 - \tfrac12, \tfrac12\right),$ and four reflex notches where adjacent hexagons’ slanted edges meet, at $\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)$ and its symmetric images. Applying the shoelace formula to these $12$ vertices gives area $16\sqrt3 - 23,$ so $m = 16,$ $n = 3,$ $p = -23,$ and $m + n + p = -4.$ Thus, the correct answer is B.