Problem
As shown in the figure, two equal circles touch each other on a straight line, and a square can be placed between them. If the diameter of the equal circle is $10 \ inches$, what is the length of one side of the square?
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Solution
Let the diameter of the equal circle be $d$, and the length of one side of the square be $x$. Considering the right triangle as shown in the figure,
$$AB^2+BO^2=OA^2,$$
$$(\frac{d}{2}-\frac{x}{2})^2+(\frac{d}{2}-x)^2=(\frac{d}{2})^2,$$
$$(d-x)^2+(d-2x)^2=d^2,$$
$$5x^2-6dx+d^2=0,$$
$$(5x-d)(x-d)=0,$$
$$∴ \ x=\frac{d}{5}, \ d.$$
Since obviously $x<d$,
$$x=\frac{d}{5}.$$
Then, when $d=10$,
$$x=\frac{10}{5}=2.$$
$$(Answer) \quad 2 \ inches.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.39; pp.354-355.