2017 AMC 10A Problem 22

Question

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?

\dfrac{4\sqrt{3}\pi}{27}-\dfrac{1}{3}

\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{8}

\dfrac{1}{2}

\sqrt{3}-\dfrac{2\sqrt{3}\pi}{9}

\dfrac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}

Let the circle radius be

r

Compare the circular segment inside the triangle with the full area of

\triangle ABC

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.

Problem 22 in other years