1959 AHSME Problems/Problem 21

Problem 21

If $p$ is the perimeter of an equilateral $\triangle$ inscribed in a circle, the area of the circle is: $\textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27}$

Solution

A side length of the triangle is $\frac{p}3$ . An altitude of the triangle, by 30-60-90 triangles, is $\frac{p\sqrt{3}}{6}$ . Because all classical triangle centers coincide on an equilateral triangle, by centroid properties a circumradius is $\frac{p\sqrt{3}}{9}$ . Finally, the area of the circumcircle is $\frac{\pi p^2}{27}\rightarrow\boxed{\textbf{C}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1959 AHSME Problems/Problem 21

Problem 21

Solution

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