Question

One side of an equilateral triangle of height $24$ lies on line $\ell.$ A circle of radius $12$ is tangent to $\ell$ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line $\ell$ can be written as $a\sqrt{b} - c\pi,$ where $a,$ $b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is $a + b + c?$

$72$

$73$

$74$

$75$

$76$

Solution:

The equilateral triangle has side $16\sqrt3.$ Put $\ell$ on the $x$ -axis with base vertex $V = (16\sqrt3, 0);$ the slanted side then lies on $\sqrt3\,x + y = 48.$ The circle sits on $\ell$ (center height $12$ ) and touches that side from outside, so its center is $(20\sqrt3, 12)$ and it meets $\ell$ at $T = (20\sqrt3, 0).$ Our region is bounded by the two tangent segments out of $V$ (one along $\ell,$ one along the triangle's side) and the near arc. The tangent length is $VT = 4\sqrt3,$ so the kite $V$ - $T$ - $\text{center}$ - $P$ has area $4\sqrt3 \cdot 12 = 48\sqrt3.$ The angle at $V$ is $120^\circ,$ so the removed sector is $60^\circ,$ with area $\tfrac16 \pi (12)^2 = 24\pi.$ The region is $48\sqrt3 - 24\pi,$ giving $a + b + c = 48 + 3 + 24 = 75.$ Therefore, the answer is D.

2024 AMC 10A Problem 14

Solution:

Problem 14 in Other Years