Problem
As shown in the figure, parts of the long side and the hypotenuse of a right triangle are used as the two sides of a rhombus, and after cutting out the rhombus, each of the two remaining right triangles contains a circle inscribed on its three sides. Assume that the diameter of the upper circle is $d_2=10 \ cm$, and the diameter of the lower circle is $d_3=8 \ cm$. Although it is not shown in the figure, let $d_1$ be the diameter of the circle inscribed on the three sides of the original right triangle. In this case, find the length $l$ of one side of the rhombus.
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Solution
Let the diameters of circles $O_2$ and $O_3$ be $d_2=x$ and $d_3=y$ respectively. Let the diameter of $O_1$ be $d_1$. Then we get
$$d_2=AE+ED-DA=AE+l-(CA-l)=x,$$
$$∴ \ AE-CA+2l=x. \quad [1]$$
$$d_3=EB+BF-FE=EB+(BC-l)-l=y,$$
$$∴ \ EB+BC-2l=y. \quad [2]$$
From $[1]+[2]$, we know that
$$(AE-CA+2l)+(EB+BC-2l)=x+y,$$
$$AE+EB+BC-CA=x+y,$$
$$∴ \ AB+BC-CA=x+y \ (=d_1 ). \quad [3]$$
($∵ \ AE+EB=AB$)
From the similarity of the three right triangles, we get the following:
$$\frac{BC}{d_1}=\frac{ED}{d_2}=\frac{BF}{d_3},$$
$$∴ \ \frac{BC}{x+y}=\frac{l}{x}=\frac{BC-l}{y}. \quad [4]$$
From the left and middle terms of $[4]$, we can say the following:
$$BC×x=(x+y)×l,$$
$$∴ \ BC=\frac{x+y}{x}l. \quad [5]$$
Also, from the similarity of right triangles, the following can be said:
$$\frac{CA}{d_1}=\frac{DA}{d_2}=\frac{FE}{d_3},$$
$$\frac{CA}{x+y}=\frac{CA-l}{x}=\frac{l}{y}. \quad [6]$$
From the left and right terms of $[6]$, we have
$$y×CA=(x+y)×l,$$
$$∴ \ CA=\frac{x+y}{y}l. \quad [7]$$
From the Pythagorean theorem, we have the following:
$$AB^2+BC^2=CA^2,$$
$$AB^2=CA^2-BC^2=(\frac{x+y}{y})^2l^2-(\frac{x+y}{x})^2l^2=\frac{(x ^2-y^2)(x+y)^2}{x^2y^2} l^2,$$
$$∴ \ AB=\frac{\sqrt{x^2-y^2}(x+y)}{xy} l. \quad [8]$$
Therefore, from $[3],[5],[7],[8]$, we see that
$$\begin{eqnarray} & & AB+BC-CA=\frac{\sqrt{x^2-y^2}(x+y)}{xy} l+\frac{x+y}{x} l-\frac{x+y}{y} l \\[5pt] &=& \frac{(\sqrt{x^2- y^2}+y-x)(x+y)}{xy} l \\&=& x+y,\end{eqnarray}$$
$$(\sqrt{x^2-y^2}+y-x)l=xy,$$
$$∴ \ l=\frac{xy}{\sqrt{x^2-y^2}+y-x}. \quad [9]$$
According to the problem text, $x=10$ and $y=8$, so substituting these into $[9]$ yields the following:
$$l=\frac{10×8}{\sqrt{10^2-8^2}+8-10}=\frac{80}{\sqrt{36}+8-10}=\frac{80}{4}=20.$$
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$$Ans.\quad 20 \ cm.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.51; pp.304-306.