Problem
As shown in the figure, in an outer circle with a diameter of $1 \ m$, there is a square, $4$ equal equilateral triangles, and $4$ equal circles. Then, find the diameter of the equal circle.
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Solution
As shown in the figure above, the red line represents the diameter of the outer circle. When a side of the square is $a$,
$$\frac{\sqrt{3}}{2} a+a+\frac{\sqrt{3}}{2} a=1,$$
$$(\sqrt{3}+1)a=1,$$
$$∴ \ a=\frac{(\sqrt{3}-1)}{2}. \qquad [1]$$
Here, as shown in the figure below, the angle made by the straight line passing through the centers of the square and the equicircle, and the perpendicular line drawn from the center of the equicircle to a side of the equilateral triangle is $15°$.
Therefore,
$$cos15°=\frac{d⁄2}{x}=\frac{d}{2x},$$
$$∴ \ x=\frac{d}{2 cos15°}.$$
The value of $cos15°$ is given by the double angle formula:
$$cos2θ=2 cos^2 θ-1.$$
Then,
$$cos30°=2 cos^2 15°-1=\frac{\sqrt{3}}{2},$$
$$cos^215°=\frac{\sqrt{3}+2}{4}=\frac{\sqrt{3}+1}{8},$$
$$∴ \ cos15°=\frac{\sqrt{3}+1}{2 \sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4},$$
$$∴ \ x=\frac{d}{(\sqrt{6}+\sqrt{2})/2}=\frac{2d}{\sqrt{6}+\sqrt{2}}.$$
Therefore, from the blue line segment, which is another expression of the diameter of the outer circle,
$$\frac{d}{2}+\frac{2d}{\sqrt{6}+\sqrt{2}}+\sqrt{2} a+\frac{2d}{\sqrt{6}+\sqrt{2}}+\frac{d}{2}=1,$$
$$d+\frac{4d}{\sqrt{6}+\sqrt{2}}+\sqrt{2} a=1,$$
$$(1+\frac{4}{\sqrt{6}+\sqrt{2}}) d+\sqrt{2} a=1. \qquad [2]$$
Substituting $[1]$ into $[2]$ and summarizing, we have
$$(1+\frac{4}{\sqrt{6}+\sqrt{2}}) d+\frac{\sqrt{6}-\sqrt{2}}{2}=1,$$
$$∴ \ d=\frac{2-\sqrt{6}+\sqrt{2}}{2}×\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}+4}=\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3}+2 \sqrt{2}+1}≒0.237000714909251.$$
$$(Answer) \quad approximately \ 0.237 \ m.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.39; pp.352-354.