2024 AMC 10A Problem 14

Draw it first. The triangle sits on line $\ell$. The circle sits on $\ell$ too, just to the side of the triangle, leaning against one of its slanted sides. The region we want is the small curved corner squeezed between the triangle, the circle and the line.

1. Name the points. Let $B$ be the bottom corner of the triangle nearest the circle. Let $O$ be the circle's center, $T$ the point where the circle touches $\ell$, and $S$ the point where it touches the triangle's side. The region is bounded by $BT$ (along the line), the arc from $T$ to $S$, and $SB$ (along the triangle's side).
2. Find the corner angle. Inside the triangle the angle at $B$ is $60^\circ$. The circle is outside the triangle, so it sits in the angle on the other side, between the triangle's side and the line: $180^\circ - 60^\circ = 120^\circ$.
3. Use the center line. A circle touching both sides of a corner has its center on the line that cuts the corner exactly in half. So $\angle OBT = 120^\circ \div 2 = 60^\circ$.
4. Find $BT$. Triangle $OTB$ has a right angle at $T$ (a radius meets a tangent line at $90^\circ$), a $60^\circ$ angle at $B$, and so a $30^\circ$ angle at $O$. In a $30$-$60$-$90$ triangle, the side across from $60^\circ$ is $\sqrt3$ times the side across from $30^\circ$. Here $OT = 12$ is across from $60^\circ$, so
$$BT = \frac{12}{\sqrt3} = 4\sqrt3.$$
The two tangent lengths from $B$ are equal, so $BS = 4\sqrt3$ as well. (The triangle's side is $16\sqrt3$ long, so $S$ really is on it.)
5. Area of the kite $BTOS$. It is made of two copies of triangle $OTB$, each with legs $4\sqrt3$ and $12$:
$$2 \cdot \tfrac12 \cdot 4\sqrt3 \cdot 12 = 48\sqrt3.$$
6. Area of the circle slice inside the kite. The kite's angles at $T$ and $S$ are $90^\circ$ each and at $B$ is $120^\circ$, so the angle at $O$ is $360^\circ - 90^\circ - 90^\circ - 120^\circ = 60^\circ$. That is $\frac{60}{360} = \frac16$ of the circle:
$$\tfrac16 \cdot \pi \cdot 12^2 = 24\pi.$$
7. The region is the kite with the circle slice removed: $48\sqrt3 - 24\pi$. So $a = 48$, $b = 3$, $c = 24$, and $a + b + c = 75$.

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

When a circle is tucked into a corner, draw the center, the two touching points and the corner point. They make a kite, and the curved gap in the corner is always that kite minus a slice of the circle. The slice's angle is $180^\circ$ minus the corner angle. The step to get right here is which corner: the circle is outside the triangle, so the corner is $120^\circ$, not $60^\circ$.

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 $\boxed{\textbf{(D)}\ 75}$.