1952 AHSME Problems/Problem 27

Problem

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:

$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$

Solution

If the radius of the circle is $r$ , then the perimeter of the first triangle is $3\left(\frac{2r}{\sqrt3}\right)=2r\sqrt3$ , and the perimeter of the second is $3r\sqrt3$ . So the ratio is $\boxed{\frac23{\textbf{ (E)}}}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1952 AHSME Problems/Problem 27

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