2017 AMC 10A Problems/Problem 22

Problem

Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$

Solution

Let the radius of the circle be r, and let its center be O. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle O, then $\angle OBA = \angle OCA = 90^{\circ}$ , so $\angle BOC = 120^{\circ}$ . Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to r, then (pick your favorite method) $\overline{BC} = r\sqrt{3}$ . The area of the equilateral triangle is $\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4$ , and the area of the sector we are subtracting from it is $\frac 13 \pi r^2 - \frac 12 r \cdot r \cdot \frac{\sqrt{3}}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$ . The area outside of the triangle is $\frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3$ . Therefore, the answer is $\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}$

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2017 AMC 10A Problems/Problem 22

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