Problem
When two equal circles are placed inside a right triangle with a length of $3 \ ins.$ and a width of $4 \ ins.$ as shown in the figure, find the diameter of the equal circles.
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Solution
First, let $AB=a$ and $BC=b$, and let $d$ be the diameter of the equal circles. From the Pythagorean theorem, we know that
$$CA=\sqrt{a^2+b^2}.$$
If we set $AG (=AL)$ and $CJ (=CK)$ as $x$ and $y$, we can see the following from the diagram:
$$DE=a-\frac{d}{2}-x,\qquad EF=b-\frac{d}{2}-y,\qquad FD=d$$
From $∆ABC〜∆DEF$, we know that
$$DE∶EF=a∶b,$$
$$a-\frac{d}{2}-x∶b-\frac{d}{2}-y=a∶b,$$
$$a×(b-\frac{d}{2}-y)=b×(a-\frac{d}{2}-x),$$
$$ab-\frac{ad}{2}-ay=ab-\frac{bd}{2}-bx,$$
$$\frac{ad}{2}+ay=\frac{bd}{2}+bx,$$
$$ad+2ay=bd+2bx,$$
$$∴ \quad (b-a)d+2bx-2ay=0. \qquad [1]$$
$$DE∶FD=a∶\sqrt{a^2+b^2 },$$
$$a-\frac{d}{2}-x∶d=a∶\sqrt{a^2+b^2 },$$
$$ad=\sqrt{a^2+b^2}×(a-\frac{d}{2}-x),$$
$$∴ \quad (2a+\sqrt{a^2+b^2})d+2x \sqrt{a^2+b^2}-2a \sqrt{a^2+b^2}. \qquad [2]$$
Also, from the figure,
$$d+x+y=\sqrt{a^2+b^2 } \ (=CA),$$
$$∴ \quad d+x+y-\sqrt{a^2+b^2 }=0. \qquad [3]$$
From $[1]+2a×[3]$,
$$(b-a)d+2bx-2ay+2a×(d+x+y-\sqrt{a^2+b^2})=0$$
$$ ∴ \quad (a+b)d+2(a+b)x-2a \sqrt{a^2+b^2}=0. \qquad [4]$$
From $(a+b)×[2]-\sqrt{a^2+b^2}×[4],$
$$\begin{eqnarray}& &(a+b){(2a+\sqrt{a^2+b^2})d+2x \sqrt{a^2+b^2}-2a \sqrt{a^2+b^2}}\\[5pt]& &-\sqrt{a^2+b^2}{(a+b)d+2(a+b)x-2a \sqrt{a^2+b^2}}=0,\end{eqnarray}$$
$$2a(a+b)d-2a(a+b) \sqrt{a^2+b^2}+2a(a^2+b^2 )=0,$$
$$(a+b)d=(a+b) \sqrt{a^2+b^2}-(a^2+b^2),$$
$$∴ \quad d=\sqrt{a^2+b^2}-\frac{a^2+b^2}{a+b}. \qquad [5]$$
Since a=3 and b=4 from the problem statement, by substituting these into [5], we get the following:
$$d=\sqrt{3^2+4^2}-\frac{3^2+4^2}{3+4}=5-\frac{25}{7}=\frac{35-25}{7}=\frac{10}{7},$$
$$∴ \quad d=\frac{10}{7}.$$
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$$Ans. \frac{10}{7} \ ins.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.52; p.300.