1967 AMC 12 Problem 3

Question

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:

\dfrac{s^2}{24}

\dfrac{s^2}{6}

\dfrac{s^2\sqrt2}{6}

\dfrac{s^2\sqrt3}{6}

\dfrac{s^2}{3}

The inradius of an equilateral triangle of side

s

is

\frac{s\sqrt3}{6}

The square’s diagonal is the circle’s diameter

The inradius is

r=\frac{s\sqrt3}{6}.

The inscribed square has diagonal

2r,

so its area is half the square of that diagonal:

\begin{aligned} \frac{(2r)^2}{2} &=2r^2\\ &=2\left(\frac{s\sqrt3}{6}\right)^2\\ &=\frac{s^2}{6}. \end{aligned}

Therefore, the correct answer is B.

Problem 3 in other years