$20 \cdot 65$ If $p$ is the perimeter of an equilateral triangle, then the area of its circumscribed circle is
(A) $\frac{\pi p^{2}}{3}$.
(B) $\frac{\pi p^{2}}{9}$.
(C) $\frac{\pi p^{2}}{27}$.
(D) $\frac{\pi p^{2}}{81}$.
(E) $\frac{\sqrt{3} \pi p^{2}}{27}$.
(10th American High School Mathematics Examination, 1959)
[Solution] Let the side length of the equilateral triangle be $a$, the height be $h$, and the radius of the circumscribed circle be $R$. Then, by the given, we have $a=\frac{p}{3}, \quad h=\frac{\sqrt{3}}{2} a=\frac{\sqrt{3}}{6}$.

Also, $h=\frac{3}{2} R$, so $R=\frac{\sqrt{3}}{9} p$.
Therefore, $S_{\odot}=\pi R^{2}=\frac{\pi p^{2}}{27}$.
Hence, the answer is $(C)$.