2017 AMC 10A Problem 22

The answer is a fraction, so the size does not matter. Let the circle have radius $1$ and center $O$.

1. Right angles at $B$ and $C$. A radius drawn to a point where a line touches a circle meets that line at a right angle. So $\angle ABO = 90^\circ$ and $\angle ACO = 90^\circ$.
2. The angle at the center. The four angles of shape $ABOC$ add to $360^\circ$. The angle at $A$ is $60^\circ$ (the triangle is equilateral), so
$$\angle BOC = 360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ.$$
3. Where the center sits. $\angle ABC$ is only $60^\circ$ but $\angle ABO$ is $90^\circ$, so $O$ lies past side $BC$, on the opposite side from $A$. The extra angle is $\angle OBC = 90^\circ - 60^\circ = 30^\circ$, and likewise $\angle OCB = 30^\circ$.
4. Length of $BC$. Triangle $OBC$ has $OB = OC = 1$ and a $120^\circ$ angle at $O$. Draw the line from $O$ straight down to the middle $M$ of $BC$. This splits it into two $30$-$60$-$90$ triangles with long side $1$. So $OM = \tfrac12$ and $BM = \tfrac{\sqrt3}{2}$, which gives $BC = \sqrt3$.
5. Area of triangle $ABC$. An equilateral triangle with side $s$ has area $\tfrac{\sqrt3}{4}s^2$:
$$\tfrac{\sqrt3}{4} \cdot 3 = \tfrac{3\sqrt3}{4}.$$
6. Part of the triangle inside the circle. The circle touches $AB$ and $AC$ without crossing them, and $O$ is past $BC$. So the only part of the circle inside the triangle is the slice between chord $BC$ and the arc on $A$'s side. That slice is the $120^\circ$ pie wedge $OBC$ minus triangle $OBC$:
- Wedge: $\tfrac{120}{360} \cdot \pi \cdot 1^2 = \tfrac{\pi}{3}$.
- Triangle $OBC$: $\tfrac12 \cdot BC \cdot OM = \tfrac12 \cdot \sqrt3 \cdot \tfrac12 = \tfrac{\sqrt3}{4}$.
- Slice: $\tfrac{\pi}{3} - \tfrac{\sqrt3}{4}$.
7. Part outside the circle is the triangle minus the slice:
$$\tfrac{3\sqrt3}{4} - \left(\tfrac{\pi}{3} - \tfrac{\sqrt3}{4}\right) = \sqrt3 - \tfrac{\pi}{3}.$$
8. As a fraction of the triangle:
$$\frac{\sqrt3 - \frac{\pi}{3}}{\frac{3\sqrt3}{4}} = \frac{4\sqrt3}{3\sqrt3} - \frac{4\pi}{9\sqrt3} = \frac43 - \frac{4\pi}{9\sqrt3}.$$
Multiply the last fraction by $\tfrac{\sqrt3}{\sqrt3}$: $\frac{4\pi}{9\sqrt3} = \frac{4\sqrt3\pi}{27}$. So the fraction is $\frac43 - \frac{4\sqrt3\pi}{27}$.

The answer is $\boxed{\textbf{(E)}\ \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}}$.

When a line touches a circle, the radius to that point makes a right angle with the line. Those right angles fix the angle at the center, and from it every length follows. The area between a chord and its arc is always "pie wedge minus triangle," a piece worth recognizing whenever a circle and a straight edge overlap.

Estimate to check. The answer $\sqrt3 - \frac{\pi}{3} \approx 1.732 - 1.047 = 0.685$ is the outside area, and the triangle's area is $\frac{3\sqrt3}{4} \approx 1.299$. The fraction is about $0.685 \div 1.299 \approx 0.53$. Choice (E) is $1.333 - 0.806 \approx 0.53$, which matches; choice (A) is about $0.47$, the fraction inside instead.