Problem

As shown in the figure, there is a line segment that comes out from one vertex of the square and reaches the opposing side, and two large circles and one small circle are inscribed in the square.

However, the large and small circles are assumed to be tangent to each other on the line segment mentioned above.

When you know the length of one side of the square, find the diameters of the large and small circles.

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Solution

Let the length of one side of the square be $a$, and the diameters of the large and small circles be $D$ and $d$, respectively.
The Pythagorean theorem holds between the sides of the right triangle $∆OIO’$.

Thus, from the diagram,
$$OI^2+IO’^2=O’ O^2,$$
$$\left( \dfrac{D}{2} \right)^2+(a-D)^2=D^2,$$
$$D^2-2×4aD+4a^2=0,$$
$$∴ \ D=(4±2 \sqrt{3})a.$$
Here, $(4+2 \sqrt{3})a$ is larger than one side of the square and is inappropriate.

Therefore,

$$D=(4-2 \sqrt{3})a.$$
From the diagram,

Since $OI=OH, \quad IO’=HB \quad and \quad ∠OIO’=∠OHB=90°$,

$$∆OIO’≡∆OHB.$$
Thus,

$$OB=D.$$

Since $∆OHB$ is a right triangle of the $2∶\sqrt{3}∶1$ type,
$$∠OBH \ (=∠OBK) \ = 30°.$$

Therefore, since $∠EBA=30°$, $∆EAB$ is also a right triangle of the $2∶\sqrt{3}∶1$ type.

Since $AB∶BE=\sqrt{3}∶2$,
$$a∶BE=\sqrt{3}∶2,$$
$$∴ \ BE=\frac{2 \sqrt{3}}{3} a.$$

$HB=a-D=(2 \sqrt{3}-3)a \qquad and \qquad HB=KB=JB$.
Since $∠O^” BJ=15°$,
$$\frac{d}{2}=JB×tan15°,$$
$$d=(4 \sqrt{3}-6)a×tan15°.$$
Here, from the half-angle formula,
$$tan^2⁡15°=\frac{1-cos30°}{1+cos30°}=\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}=7-4 \sqrt{3}=(2-\sqrt{3})^2,$$
$$∴ \ tan15°=2-\sqrt{3}.$$

Therefore,
$$d=(4 \sqrt{3}-6)×a×(2- \sqrt{3})=(14 \sqrt{3}-24)a.$$

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Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.44; pp.329-330.