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My attention was brought to a sangaku problem in this book by Ubukata Tou. It shows this figure: enter image description here

The question asks us to find the diameter of the circles (both circles are congruent) in a right triangle ($∠ABC = 90$), where $AB = 9$ and $BC = 12$. It also says that the diameter of the two circles is $30/7$. How would you solve this problem. In the book, it also states that this was a problem from the early Edo period suggesting that trigonometry may not have been around in Japan then. It would be very interesting to see a solution without the use of trigonometry then.

Jean-Claude Arbaut
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Eames
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  • An extra challenge: prove that the largest rectangle formed under the two circles has area $432/49$! – TheSimpliFire May 30 '18 at 19:31
  • Then ChristianF's solution is quicker for your extra challenge. Though the much harder part is to formally prove that the largest rectangle is actually the one which have sides parallel to the right triangle. – Hw Chu May 30 '18 at 19:58

3 Answers3

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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

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

enter image description here

Hw Chu
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  • Nice solution +1 – nonuser May 30 '18 at 19:36
  • +1. It's not difficult to show (even for arbitrary right triangles) that $\overline{BJ}$ is the bisector of $\angle B$, so your identification of the circles as incircles of adjacent right triangles effectively provides convenient a way to construct the configuration. – Blue May 31 '18 at 06:55
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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 $S$.

enter image description here

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 $$r={15\over 7}$$

nonuser
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Using Aqua's proof, the general solution for any right triangle is:

$$ x = \frac{S_3}{2} \cdot \frac{S_1 - S_2 + S_3}{S_1 + S_2}\\ y = \frac{S_3}{2} \cdot \frac{-S_1 + S_2 + S_3}{S_1 + S_2}\\ r = \frac{S_3}{2} \cdot \frac{S_1 + S_2 - S_3}{S_1 + S_2} $$ for adjacent sides $S_1$ and $S_2$, and hypotenuse $S_3$.

JAJ
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