Circle and Equilateral Triangle | AMC 10A, 2017| Problem No 22

Pre College Mathematics

AMC-10A, 2017 Problem-22

$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

Given that ABC is a equilateral triangle whose \(AB\) & \(AC\) are the tangents of the circle whose centre is \(O\). We have to find out the fraction of the area of $\triangle A B C$ lies outside the circle

we have to find out thr ratio of the areas of Blue colour : Red colour area. Therefore we have to findout the area of the circle and Triangle ABC.

Later we have to find out red area and subtract from the Triangle ABC.

Now can you finish the problem?

Let the radius of the circle be $r$, and let its center be $O$. since $\overline{A B}$ and $\overline{A C}$ are tangent to circle $O$, then $\angle O B A=\angle O C A=90^{\circ}$, so $\angle B O C=120^{\circ} .$ Therefore, since $\overline{O B}$ and $\overline{O C}$ are equal to $r$, then $\overline{B C}=r \sqrt{3}$. The area of the equilateral triangle is $\frac{(r \sqrt{3})^{2} \sqrt{3}}{4}=\frac{3 r^{2} \sqrt{3}}{4},$ and the area of the sector we are subtracting from it is $\frac{1}{3} \pi r^{2}-\frac{1}{2} r \cdot r \cdot \frac{\sqrt{3}}{2}=\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4} .$

Now Can you finish the Problem?

Therefore the area outside the circle is $\frac{3 r^{2} \sqrt{3}}{4}-\left(\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4}\right)=r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}$

Therefore the Required fraction is $\frac{r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}}{\frac{3 r^{2} \sqrt{3}}{4}}$