1972 AMC 12 Problem 23

Question

The radius of the smallest circle containing the symmetric figure composed of

3

unit squares shown is:

\sqrt2

\sqrt{1.25}

1.25

\dfrac{5\sqrt{17}}{16}

None of these

By symmetry, place the circle’s center on the vertical axis of the figure

Equate its distances to a lower outer corner and an upper outer corner

Put the midpoint of the bottom edge at

O=(0,0).

A lower outer corner is

A=(1,0),

and an upper outer corner is

B=(\frac{1}{2},2).

By symmetry the center of the smallest enclosing circle is

P=(0,k).

At the optimum both types of outer corner lie on the circle, so

1+k^2=\left(\frac12\right)^2+(2-k)^2.

This gives

k=\frac{13}{16}.

Hence

\begin{aligned} r^2&=1+\left(\frac{13}{16}\right)^2 =\frac{425}{256},\\ r&=\frac{5\sqrt{17}}{16}. \end{aligned}

Therefore, the correct answer is D.

Problem 23 in other years