2003 AMC 10A Problem 17

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

The relation $s = r\sqrt{3}$ between an equilateral triangle and its circumradius (equivalently $r = s/\sqrt{3}$, and inradius $s/(2\sqrt{3})$) is worth memorizing; it comes from the $120^\circ$ central angle split into two $30$-$60$-$90$ halves. Once the perimeter is expressed in $r$, the condition is a one-variable equation.

Sanity check with units: perimeter is linear in $r$ and area is quadratic, so the equation must be $c \cdot r = \pi r^2$ for a constant $c$, and $r = c/\pi$. Only choices (A), (B) and (D) have this form, and $c = 3\sqrt{3}$ selects (B).

• Using a wrong side-to-circumradius relation (such as s = 2r or s = r/sqrt(3)) from a misdrawn 30-60-90 triangle.
• Equating the perimeter to the circle's circumference $2\pi r$ instead of its area.