2003 AMC 10A Problem 17

The perimeter uses the side of the triangle and the area uses the radius of the circle, so first write the side in terms of the radius.

1. Let $r$ be the radius. Put the circle's center $O$ in the middle of the triangle, and let $A$ and $B$ be two corners of the triangle. Then $OA = OB = r$.
2. The three corners split the full turn at $O$ into three equal parts, so the angle $AOB$ is $360^\circ \div 3 = 120^\circ$.
3. Drop a line from $O$ to the midpoint $M$ of side $AB$. It cuts the $120^\circ$ angle in half, so angle $AOM = 60^\circ$, and it meets $AB$ at a right angle. Triangle $AOM$ is a $30$-$60$-$90$ triangle.
4. In a $30$-$60$-$90$ triangle the sides are in the ratio $1 : \sqrt{3} : 2$. The longest side, $OA = r$, plays the role of $2$. The side $AM$ is opposite the $60^\circ$ angle, so it plays the role of $\sqrt{3}$:
$$AM = \frac{\sqrt{3}}{2} r.$$
5. The whole side is twice that: $AB = \sqrt{3}\, r$.
6. The perimeter is $3 \times \sqrt{3}\, r = 3\sqrt{3}\, r$. The area of the circle is $\pi r^2$. They are equal:
$$3\sqrt{3}\, r = \pi r^2.$$
7. The radius is not zero, so divide both sides by $r$:
$$3\sqrt{3} = \pi r, \qquad r = \frac{3\sqrt{3}}{\pi}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{3\sqrt{3}}{\pi}}$.

The fact "side $= r\sqrt{3}$" for an equilateral triangle inside a circle comes from cutting the $120^\circ$ angle at the center into two $30$-$60$-$90$ triangles. It is worth remembering. Once everything is written with the same letter $r$, the condition becomes a one-step equation.

Look at the shape of the equation before computing. The perimeter is some number times $r$, and the area is $\pi$ times $r^2$. Setting (number)$\times r = \pi r^2$ always gives $r =$ (number)$\,/\,\pi$. So the answer must be a number divided by $\pi$: that leaves (A), (B) and (D). The number is $3$ times the side-to-radius ratio, which involves $\sqrt{3}$ for an equilateral triangle, pointing to (B).

Let the circle have radius $r$ and the triangle have side $s$. Draw segments from the center $O$ to two adjacent vertices and to the midpoint $M$ of the side between them. The central angle over a side is $120^\circ$, so $\angle MOA = 60^\circ$ and triangle $OMA$ is a $30$-$60$-$90$ triangle with hypotenuse $r$. Hence $AM = r \sin 60^\circ = \tfrac{\sqrt{3}}{2} r$ and $s = 2 \cdot AM = r\sqrt{3}$.

Setting perimeter equal to circle area:
$$3 r\sqrt{3} = \pi r^2 \quad\Longrightarrow\quad 3\sqrt{3} = \pi r \quad\Longrightarrow\quad r = \frac{3\sqrt{3}}{\pi}.$$
(Dividing by $r$ is fine since $r > 0$.)

The answer is $\boxed{\textbf{(B)}\ \frac{3\sqrt{3}}{\pi}}$.