2017 AMC 10A Problem 22

Let the radius of the circle be $r.$ To find the area of the triangle outside of the circle, we can find the area of the triangle inside the circle and subtract it. We get that $\angle BOC = 120^{\circ}$ since $\angle ABO$ and $\angle ACO$ are right angles. This means that the area of sector $OBC$ is $$\dfrac{120}{360} \cdot \pi r^2 = \dfrac{\pi r^2}{3}.$$ Now, we need to find the area of $\triangle BOC.$ Using the formula for the area of a triangle with sine, we get the area to be $$\dfrac{1}{2} \sin (120^{\circ}) \cdot r^2 = \dfrac{r^2\sqrt{3}}{4}.$$ Then the area of the triangle inside the circle is $$\dfrac{\pi r^2}{3} - \dfrac{r^2\sqrt{3}}{4} = \dfrac{r^2(4\pi - 3\sqrt{3})}{12}.$$ The area of $\triangle ABC$ is $$\dfrac{(r\sqrt{3})^2\sqrt{3}}{4} = \dfrac{3r^2\sqrt{3}}{4}.$$ The fraction of the triangle lying inside the circle is $$\dfrac{r^2(4\pi-3\sqrt3)}{12}\cdot \dfrac{4}{3r^2\sqrt3},$$ which simplifies to $$\dfrac{4\pi-3\sqrt3}{9\sqrt3}.$$ Hence the desired fraction is $$\begin{aligned} &1-\dfrac{4\pi-3\sqrt3}{9\sqrt3}\\ &=1-\dfrac{4\pi\sqrt3}{27}+\dfrac13\\ &=\dfrac43-\dfrac{4\pi\sqrt3}{27}. \end{aligned}$$ Thus, E is the correct answer.