1967 AHSME Problems/Problem 3

Problem

The side of an equilateral triangle is $s$ . A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:

$\text{(A)}\ \frac{s^2}{24}\qquad\text{(B)}\ \frac{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3}$

Solution

The radius of our circle is $\frac{\sqrt{3}}{6}s$ . This means the square has side length $\frac{\sqrt{6}}{6}s$ which has area of $\frac{s^2}{6}=\fbox{B}$ .

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1967 AHSME Problems/Problem 3

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