Problem

There are three circles in a right triangle $ABC$ as shown in the figure.

When the product of the diameter of the middle circle and the length of $BD$ is 2.88, and the sum of the length of $AD$ and the diameter of the small circle is 3.6 inches, what is the diameter of the small circle?

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Solution

Let $m$ be the diameter of the medium circle and $z$ be the diameter of the small circle.

And also let:

$BD×m=p —[1],$ and $AD+z=k —[2].$

From [1] and [2], it follows that

$BD=\frac{p}{m} —[1]’, $ and $AD=k-z —[2]’.$

From $∆ADC∼∆BDA$, we have

$$\frac{m}{AD}=\frac{z}{BD},$$

$$\frac{m}{k-z}=\frac{z}{\frac{p}{m}},$$

$$p=z(k-z),$$

$$z^2-kz+p=0,$$

$$z=\frac{k±\sqrt{k^2-4p}}{2},$$

$$\therefore z=\frac{k-\sqrt{k^2-4p}}{2}$$

$(\because \frac{k+\sqrt{k^2-4p}}{2}>\frac{k}{2}>z).$

Since $p=2.88, k=3.6$ from the problem statement, we see that

$$z=\frac{3.6-\sqrt{3.6^2-4×2.88}}{2}=1.2.$$

(Answer) 1.2 inches.

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Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.55; pp.292-294.