1967 AMC 12 Problem 20

Question

A circle is inscribed in a square of side

m,

then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If

S_n

is the sum of the areas of the first

n

circles so inscribed, then, as

n

grows beyond all bounds,

S_n

approaches:

\dfrac{\pi m^2}{2}

\dfrac{3\pi m^2}{8}

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

\dfrac{\pi m^2}{4}

\dfrac{\pi m^2}{8}

The first circle has area

\frac{\pi m^2}{4}

Each successive circle has half the preceding circle’s area

The first circle has area

\pi(\frac{m}{2})^2=\frac{\pi m^2}{4}.

Each inscribed square reduces the next circle’s squared radius, and hence its area, by a factor of

\frac{1}{2}.

Thus the limiting sum is

\begin{aligned} &\frac{\pi m^2}{4} \left(1+\frac12+\frac14+\cdots\right)\\ &\qquad=\frac{\pi m^2}{2}. \end{aligned}

Therefore, the correct answer is A.

Problem 20 in other years