Tanokuma Hachiman Shrine is located 4.2km southeast of Takano Station on the JR Inbi Line and 4.2km north of Nishi Katsumata Station on the JR Kishin Line.
Problem
There are three circles in a right triangle $ABC$ as shown in the figure.
When the product of the square of the side $CA$ and the square of the diameter of the medium circle is 40.96, and the difference between the diameter of the large circle and the length of $CD$ is 1.2 inches, what is the diameter of the large circle?
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Solution
Let $m$ be the diameter of the medium circle, and $x$ be the diameter of the large circle. And also let:
$CA^2×m^2=p —[1],$ and $CD-x=k —[2]$.
From [1] and [2], it follows that
$CA=\frac{\sqrt{p}}{m} —[1]’$, and $x=CD-k —[2]’.$
From $∆ABC∼∆DAC$, we have
$$\frac{x}{CA}=\frac{m}{CD},$$
$$\frac{CD-k}{\frac{\sqrt{p}}{m}}=\frac{m}{CD},$$
$$CD×(CD-k)=\sqrt{p},$$
$$CD^2-k×CD-\sqrt{p}=0,$$
$$CD=\frac{k±\sqrt{k^2+4\sqrt{p}}}{2},$$
$$\therefore CD=\frac{k+\sqrt{k^2+4\sqrt{p}}}{2} —[3]$$
$(\because \frac{k-\sqrt{k^2+4\sqrt{p}}}{2}<0).$
Substituting [3] into [2]’, we have
$$x=\frac{k+\sqrt{k^2+4\sqrt{p}}}{2}-k=\frac{\sqrt{k^2+4\sqrt{p}} -k}{2}.$$
Since $p = 40.96, k = 1.2$ from the problem statement, we see that
$$x=\frac{\sqrt{1.2^2+4\sqrt{40.96}} -1.2}{2}=2.$$
(Answer) 2 inches.
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.55; pp.292-294.