Sangaku - Find diameter of congruent circles in a $9$-$12$-$15$ right triangle

Solution:

From the figure below, $\triangle AJI \cong \triangle KJC$. So $9+4z = 12-3z$, and $z = \frac 37$.

So one of the circles is the inscribed circle of the triangle which have side lengths $\frac{75}{7}, \frac{60}{7}$ and $\frac{45}{7}$, so the radius is

$$ \frac 12 \left(\frac{45}{7} + \frac{60}{7} - \frac{75}{7}\right) = \frac{15}{7}. $$

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Let $M$, $N$ be the centers of circles respectively closer to $A$, $C$. Let parallel to $BC$ trough $N$ and parallel to $AB$ trough $M$ intersect at $L$. Let circles touch $AC$ at $P$ and $Q$ ($P$ closer to $A$). Then $MNQP$ is rectangle so $PQ = 2r$ and let $AP=x$ and $CQ=y$. Let circles touch $BC$ at $R$ and $AB$ at $S$.

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Then $CR = y$ and $AS=x$. Since $AC= 15$ we have $$x+y+2r =15$$ where $r$ is radius of circles. Since $\triangle ABC\sim \triangle MLN$ we have $${2r\over 15} = {9-x-r\over 9} = {12-y-r\over 12}$$ we get $$ x= {15-11r\over 5}\;\;\;\;{\rm and}\;\;\;\;y={60-13r\over 5}$$ Pluging this into $x+y+2r=15$ we get $$r={15\over 7}$$