Problem
As shown in the figure, there is an equilateral triangle, one large circle, two medium circles, and one small circle within a square.
If the diameter of the large circle is $9.5$ inches, find the diameter of the medium circle.
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Solution
Let $a$ be the side of an equilateral triangle, $b$ be the diameter of the large circle, and $x$ be the diameter of the middle circle.
For $△BFO_2$,
$$BO_2=\sqrt{\left( \dfrac{x}{2} \right)^2+\left( \dfrac{a-x}{2} \right)^2}.$$
Also, since $∠O_2 BF=15°$,
$$cos∠O_2 BF=cos15°=\frac{BF}{BO_2}=\frac{\cfrac{a-x}{2}}{\sqrt{\left( \dfrac{x}{2} \right)^2+\left( \dfrac{a-x}{2} \right)^2}}. \qquad [1]$$
The value of $cos15°$ is calculated from the double-angle formula: $cos2θ=2 cos^2θ-1$,
$$cos30°=2 cos^2 15°-1=\frac{\sqrt{3}}{2},$$
$$cos^2 15°=\frac{\sqrt{3}+2}{4}=\frac{(\sqrt{3}+1)^2}{8},$$
$$∴ \ cos15°=\frac{\sqrt{3}+1}{2 \sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}. \qquad [2]$$
Therefore, from $[1]$ and $[2]$,
$$\frac{\cfrac{a-x}{2}}{\sqrt{\left( \dfrac{x}{2} \right)^2+\left( \dfrac{a-x}{2} \right)^2}}=\frac{\sqrt{6}+\sqrt{2}}{4},$$
$$\frac{\left( \dfrac{a-x}{2} \right)^2}{\left( \dfrac{x}{2} \right)^2+\left( \dfrac{a-x}{2} \right)^2}=\frac{2+\sqrt{3}}{4},$$
$$(2+\sqrt{3}) \left( \dfrac{x}{2} \right)^2=(2-\sqrt{3}) \left( \dfrac{a-x}{2} \right)^2,$$
$$\left( \dfrac{a-x}{2} \right)^2=\frac{(2+\sqrt{3})}{(2-\sqrt{3})} \left( \dfrac{x}{2} \right)^2,$$
$$4a^2-4x-(6+4 \sqrt{3}) x^2=0,$$
$$\begin{eqnarray} ∴ \ a=\frac{2x±\sqrt{4x^2+4 (6+4 \sqrt{3}) x^2}}{4} \\ = \frac{x±\sqrt{(7+4 \sqrt{3}) x^2}}{2} \\ = \frac{x±\sqrt{(2+\sqrt{3})^2 x^2}}{2} \\ = \frac{x±(2+\sqrt{3})x}{2}. \end{eqnarray}$$
Since $a>0$,
$$a=\frac{x+(2+\sqrt{3})x}{2}=\frac{(3+\sqrt{3})x}{2}. \qquad [3]$$
For the equilateral triangle $EBC$, the following relationship holds according to Heron’s formula:
$$\frac{1}{2}×a×\frac{\sqrt{3}}{2} a=\frac{1}{2}×3a×\frac{b}{2},$$
$$∴ \ a=b \sqrt{3}. \qquad [4]$$
If you connect the right-hand sides of $[3]$ and $[4]$ with an equal sign,
$$\frac{(3+\sqrt{3})x}{2}=b \sqrt{3},$$
$$(3+\sqrt{3})x=2b \sqrt{3},$$
$$∴ \ x=\frac{2b \sqrt{3}}{3+\sqrt{3}}=\frac{2b \sqrt{3} (3-\sqrt{3})}{6}=\frac{2b(\sqrt{3}-1)}{2}. \qquad [5]$$
If you substitute $b=9.5$ into $[5]$,
$$x=\frac{19(\sqrt{3}-1)}{2}≒6.95448267190433…$$
$$(Answer) \qquad approximately \ 6.95 \ inches.$$
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Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.41; pp.340-342.