1959 AMC 12 Problem 21

Attempt Problem 21 of the 1959 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1959 AMC 12 solutions, or check the answer key.

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21.

p

is the perimeter of an equilateral triangle inscribed in a circle, the area of the circle is:

\dfrac{\pi p^2}{3}

\dfrac{\pi p^2}{9}

\dfrac{\pi p^2}{27}

\dfrac{\pi p^2}{81}

\dfrac{\pi p^2\sqrt3}{27}

The triangle side length is

\frac{p}{3}

An equilateral triangle with side

s

has circumradius

\frac{s}{\sqrt3}

The side length is

\frac{p}{3},

so the circumradius is

R=\frac{\frac{p}{3}}{\sqrt3}=\frac{p}{3\sqrt3}.

Therefore the circle’s area is

\pi R^2=\frac{\pi p^2}{27}.

Thus, the correct answer is C.

Problem 21 in other years