Problem
As shown in the figure, there are three equal circles inside the outer circle. If the diameter of the outer circle is $10 \ inches$, what is the diameter of the equal circle?
$$ $$
$$ $$
$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
$$ $$
Solution
Draw a right triangle as shown.
If the diameter of the outer circle is $10$ and the diameter of the equal circle is $d$, then the length $l$ of the hypotenuse of this right triangle is
$$l=\frac{10-d}{2}. \qquad [1]$$
Since this is a right triangle of $30°-60°-90°$ type, the ratio of the base to the hypotenuse is $\sqrt{3}∶2$,
$$\sqrt{3}∶2=\frac{d}{2}∶l,$$
$$∴ \ l=\frac{d}{\sqrt{3}}. \qquad [2]$$
From $[1]$ and $[2]$,
$$\frac{10-d}{2}=\frac{d}{\sqrt{3}},$$
$$∴ \ d=\frac{10 \sqrt{3}}{2+\sqrt{3}}=20 \sqrt{3}-30≒4.6410… \qquad [3]$$
$$(Answer) \quad approximately \ 4.641 \ inches.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.40; pp.348-349.