The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Express your answer in terms of pi and in simplest radical form.
Let the triangle have vertices $A$, $B$, and $C$, let $O$ be the center of the circle, and let $D$ be the midpoint of $
\overline{BC}$. Triangle $COD$ is a $30 - 60 - 90$ degree triangle. If $r$ is the radius of the circle, then the sides of $\triangle COD$ are $r$, $\frac{r}{2}$, and $\frac{r\sqrt{3}}{2}$. The perimeter of $\triangle ABC$ is $6\displaystyle\left(\frac{r \sqrt{3}}{2}\displaystyle\right)=3r\sqrt{3}$, and the area of the circle is $\pi r^{2}$. Thus $3r\sqrt{3} = \pi r^{2}$, and $r =
\boxed{\frac{3\sqrt{3}}{\pi}}$.

[asy]
pair A,B,C,D,O;
O=(0,0);
A=(0,1);
B=(0.87,-0.5);
C=(-0.87,-0.5);
D=(0,-0.5);
draw(Circle(O,1),linewidth(0.7));
draw(C--O--D--cycle,linewidth(0.7));
draw(A--B--C--cycle,linewidth(0.7));
label("$\frac{r}{2}$",(0,-0.25),E);
label("$\frac{r \sqrt{3}}{2}$",(-0.43,-0.5),S);
label("$r$",(-0.43,-0.25),NW);
label("$O$",O,N);
label("$A$",A,N);
label("$B$",B,SE);
label("$C$",C,SW);
label("$D$",D,S);
[/asy]
\frac{3\sqrt{3}}{\pi}
Level 5
Geometry
Problem: The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Express your answer in terms of pi and in simplest radical form.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the radius of the circumscribed circle of an equilateral triangle, we start by letting the side length of the triangle be denoted as s. The perimeter of the triangle is then 3s. 

Next, we calculate the area of the circumscribed circle. The radius R of the circumscribed circle for an equilateral triangle is given by the formula R = s / (sqrt(3)). The area A of the circle can be expressed as A = πR^2. Substituting the expression for R, we have:

A = π(s / (sqrt(3)))^2 = π(s^2 / 3).

According to the problem, the perimeter of the triangle equals the area of the circumscribed circle. Therefore, we set the perimeter equal to the area:

3s = π(s^2 / 3).

To eliminate the fraction, we multiply both sides by 3:

9s = πs^2.

Next, we can rearrange this equation to isolate s:

πs^2 - 9s = 0.

Factoring out s gives us:

s(πs - 9) = 0.

This yields two solutions: s = 0 or πs - 9 = 0. Since s cannot be zero in this context, we solve for s:

πs = 9,
s = 9/π.

Now, we substitute s back into the formula for the radius R:

R = s / (sqrt(3)) = (9/π) / (sqrt(3)) = 9 / (πsqrt(3)).

To express the radius in terms of pi and in simplest radical form, we rationalize the denominator:

R = 9 / (πsqrt(3)) * (sqrt(3)/sqrt(3)) = 9sqrt(3) / (3π) = 3sqrt(3) / π.

Thus, the radius of the circumscribed circle is expressed as 3sqrt(3) / π.<|END_OF_LATENT|> The final answer is $\frac{3\sqrt{3}}{\pi}$.