Problem
As shown in the figure, when a large circle with a diameter of $450 \ cm$ and a small circle with a diameter of $288 \ cm$ are inscribed within a right triangle, find the lengths of the three sides of the right triangle.
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Solution
Let the diameter of the large circle be $d_1$ and the diameter of the small circle be $d_2$.
First, find the length of $DE \ (=D’O’)$ as follows:
$$OD’^2+D’O’^2=O’O^2,$$
$$(\frac{d_1}{2}-\frac{d_2}{2})^2+D’O’^2=(\frac{d_1}{2}+\frac{d_2}{2})^2,$$
$$D’O’^2=d_1d_2,$$
$$∴ \ D’O’=\sqrt{d_1d_2}.$$
Here, if we set $EC \ (=CG)=a$,from $∆ODC∼∆O’EC$, we can see that
$$\frac{DC}{OD}=\frac{EC}{O’E},$$
$$\frac{\sqrt{d_1d_2}+a}{d_1/2}=\frac{a}{d_2/2},$$
$$d_2(\sqrt{d_1d_2}+a)=d_1 a,$$
$$∴ \ a=\frac{d_2 \sqrt{d_1d_2}}{d_1-d_2}. \qquad [1]$$
If we put $AH \ (=FA)=b$, we have
$$AB^2+BC^2=CA^2,$$
$$(b+\frac{d_1}{2})^2+(\frac{d_1}{2}+\sqrt{d_1d_2}+\frac{d_2 \sqrt{d_1d_2}}{d_1-d_2})^2=(\frac{d_2 \sqrt{d_1d_2}}{d_1-d_2}+\sqrt{d_1d_2}+b)^2,$$
$$b(4 \sqrt{d_1d_2}+\frac{4 d_2\sqrt{d_1d_2}}{d_1-d_2}-2d_1)=d_1(2 \sqrt{d_1d_2}+\frac{2d_2 \sqrt{d_1d_2}}{d_1-d_2}+d_1),$$
$$∴ \ b=\frac{d_1(2 \sqrt{d_1d_2}+\frac{2d_2 \sqrt{d_1d_2}}{d_1-d_2}+d_1)}{4 \sqrt{d_1d_2}+\frac{4d_2\sqrt{d_1d_2}}{d_1-d_2}-2d_1}. \qquad [2]$$
From the diagram, it can be seen that
$$AB=AH+HB=b+\frac{d_1}{2}. \qquad [3]$$
$$BC=BD+DE+EC=\frac{d_1}{2}+\sqrt{d_1d_2}+a. \qquad [4]$$
$$CA=CG+GF+FA=a+\sqrt{d_1d_2}+b. \qquad [5]$$
From the problem statement, $d_1=450$ and $d_2=288$, so by substituting these into $[1]$ and $[2]$, we get the following:
$$a=\frac{288× \sqrt{450×288}}{450-288}=\frac{103680}{162}=640. \qquad [6]$$
$$b=\frac{450×(2 \sqrt{450×288}+\frac{2×288 \sqrt{450×288}}{450-288}+450)}{4 \sqrt{450×288}+\frac{4×288\sqrt{450×288}}{450-288}-2×450}=\frac{11025}{31}. \qquad [7]$$
Thus, by substituting $d_1=450, \ d_2=288$ and $[6]$ and $[7]$ into [3], [4] and [5], we have
$$AB=\frac{11025}{31}+\frac{450}{2}≒580.645,$$
$$BC=\frac{450}{2}+\sqrt{450×288}+640=1225,$$
$$CA=640+\sqrt{450×288}+\frac{11025}{31}≒1355.645.$$
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$$Ans. \quad AB≒580.645 \ cm, \ BC=1225 \ cm, \ CA≒1355.645 \ cm.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.52; pp.301-302.