2024 AMC 10A Problem 14

The equilateral triangle has side $16\sqrt3.$ Put $\ell$ on the $x$-axis with base vertex $V = (16\sqrt3, 0)$; the slanted side lies on $\sqrt3\,x + y = 48.$ The circle sits on $\ell,$ has radius $12,$ and touches that side externally, so its center is $O = (20\sqrt3, 12).$ Let $T = (20\sqrt3, 0)$ be its tangency point on $\ell,$ and let $P$ be the tangency point on the slanted side. The two tangent lengths from $V$ satisfy $VT = VP = 4\sqrt3,$ so kite $VTOP$ has area $4\sqrt3 \cdot 12 = 48\sqrt3.$ The angle at $V$ is $120^\circ,$ so the removed sector has angle $60^\circ$ and area $\tfrac16 \pi (12)^2 = 24\pi.$ The region has area $48\sqrt3 - 24\pi,$ giving $a + b + c = 48 + 3 + 24 = 75.$ Therefore, the answer is D.