2024 AMC 10A Problem 14

Let $B$ be the vertex of the triangle on $\ell$ nearest the circle, let the circle (center $O$) touch $\ell$ at $T$ and the triangle's side at $S$. The region we want is bounded by segment $BT$, arc $TS$, and segment $SB$.

Outside the triangle, the side through $B$ makes an angle of $180^\circ - 60^\circ = 120^\circ$ with $\ell$. The circle is inscribed in this angle, so $O$ lies on its bisector and $\angle OBT = 60^\circ$. In right triangle $OTB$ with $OT = 12$,
$$BT = \frac{12}{\tan 60^\circ} = 4\sqrt3, \qquad BS = BT = 4\sqrt3.$$
(The triangle's side is $24/\tfrac{\sqrt3}{2} = 16\sqrt3 > 4\sqrt3$, so $S$ really lies on the side.)

Draw $OB$, $OT$, $OS$. The kite $BTOS$ consists of two right triangles with legs $4\sqrt3$ and $12$, so its area is $2\cdot\frac12\cdot 4\sqrt3\cdot 12 = 48\sqrt3$. Its angle at $O$ is $360^\circ - 90^\circ - 90^\circ - 120^\circ = 60^\circ$, so the part of the disk inside the kite is a $60^\circ$ sector of area $\frac{60}{360}\pi\cdot 12^2 = 24\pi$.

The region is the kite minus the sector: $48\sqrt3 - 24\pi$. Hence $a + b + c = 48 + 3 + 24 = 75$.

The answer is $\boxed{\textbf{(D)}\ 75}$.

A circle tangent to two lines is inscribed in the angle they form, so its center sits on the bisector and the two tangent lengths from the vertex are equal. The curved region "between the circle and the corner" is always a kite (vertex, two tangent points, center) minus a sector whose angle is $180^\circ$ minus the corner angle. Here the corner is the exterior $120^\circ$, which is the step to get right.

• Using the triangle's 60° interior angle (center angle 120°, sector 48π) instead of the 120° angle between the side and ℓ outside the triangle.
• Computing $BT$ as $12\tan 60^\circ = 12\sqrt3$ by mixing up which angle is at $B$.