3500.31 – Circle Around Three Squares

Consult the figure below.

We'll design a circle passing through $A$, $B$, $C$, and $D$. Its center $O$ will be on the line $EF$ bisecting the symmetry line of the figure. Its radius will be $x$.

Note that $x$ is the hypotenuse of two right triangles, OEB and OFD. Let OF = $t$. Then $OE = 1 + (1 - t) = 2 - t$. Now $x^2 = t^2 + 1^2$ and $x^2 = (2 – t)^2 + (1/2)^2$

Here we go:

$$\begin{aligned}t^2 + 1 &= 4 - 4t + t^2 + \dfrac{1}{4} \\4t &= 3 + \dfrac{1}{4} \\t &= \dfrac{13}{16}\end{aligned}$$

So, in $\triangle OFD$,

$$\begin{aligned}x^2 &= \left(\dfrac{13}{16}\right)^2 + 1^2 \\[0.5em]&= \dfrac{169}{256} + \dfrac{256}{256} \\[0.5em]&= \dfrac{425}{256} \\[0.5em]&= \dfrac{25\cdot17}{256}\\[0.5em]x &= \dfrac{5\sqrt 17}{16}\end{aligned}$$

We see that $x$ is approximately 1.29, not so very different from answer choice (c), but (d) is exact. Were you to draw the figure with sides of one inch and make a careful measurement you'd get just about an inch and a quarter for the radius.