Problem
As shown in the figure, draw two diagonal lines inside an equilateral triangle and insert two equal circles. If the length of one side of the equilateral triangle is 10 inches, find the length of the diameter of the circle.
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Solution
If the length of the diameter of the circle to be sought is $d$, then
$$△OAH=\frac{1}{2}×\frac{d}{2}×\frac{\sqrt{3} d}{2}=\frac{\sqrt{3} d^2}{8}.$$
$$△OCH (=△OCI)=\frac{1}{2}×\frac{d}{2}×(10-\frac{\sqrt{3} d}{2})=\frac{d}{8} (20-\sqrt{3} d).$$
$$△O’ CJ (=△O’ CF)=\frac{1}{2}×\frac{d}{2}×\frac{10}{2}=\frac{5d}{4}.$$
$$△ACF=\frac{1}{2}×5×5 \sqrt{3}=\frac{25 \sqrt{3}}{2}.$$
Then,
$$△ACF=△OAH+△OCH+△OCI-△OGI+△O’ CJ+△O’ GJ+△O’ CF,$$
$$△ACF=△OAH+△OCH+△OCI+△O’ CJ+△O’ CF,$$
$(∵ \ △OGI≡△O’ GJ)$
$$\frac{25 \sqrt{3}}{2}=\frac{\sqrt{3} d^2}{8}+\frac{d}{8} (20-\sqrt{3} d)+\frac{d}{8} (20-\sqrt{3} d)+\frac{5d}{4}+\frac{5d}{4},$$
$$\sqrt{3} d^2-60d+100 \sqrt{3}=0,$$
$$∴ \ d=\frac{30±\sqrt{900-300}}{\sqrt{3}}=10(\sqrt{3}±\sqrt{2}).$$
Then, since $d<10$,
$$d=10(\sqrt{3}-\sqrt{2})≒3.17837245195782.$$
$$(Answer) \quad approximately \ 3.178 \ inches.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.39; pp.355-356.