A Restored Sangaku Problem
The sangaku discussed below was originally hung [Fukagawa and Rothman, pp. 194195] in 1806 in the Atsuta shrine by Ehara Masanori, a disciple of Kusaka Makoto (17641833). The tablet, which contained only this problem and no solution, was subsequently lost. However, at an unknown date the mathematician Kitagawa Moko (17631839) visited the shrine and recorded the sangaku in his note "Kyuka Sankei," or "Nine Flowers Mathematics," along with his solution. More recently the shrine constructed a replica from Kitagawa's manuscript. Fukagawa and Rothman reproduce Kitagawa's solution with a warning: This may be the most involved exploitation of the Pythagorean theorem you have ever seen.
The problem is this: ΔABC is isosceles. Two lines BD and CH divide it into three smaller triangles, each of which circumscribes a circle of the same radius r. Find r in terms of CH.
Following [Fukagawa and Rothman, pp. 212216], I shall give Kitagawa's solution and then, on a separate page, a solution by J. M. Unger.
First, notice from the figure above that triangle CDB is isosceles. Therefore let
T and T'' are equidistant from D and therefore
(1)  s = (CD''  CD) / 2 = (a  b) / 2. 
Further, by examining the tangents in the figure above, we see that
(2)  b = (k  r) + (a/2  r) = a/2 + k  2r. 
Hence, from equation (1)
(3)  s = a / 4  (k  2r) / 2. 
Also, with the Pythagorean theorem, from either ΔBCH or ΔCDH,
(4)  a = 4r(k  r) / (k  2r). 
Next, as shown below, inscribe a circle of radius R in triangle BCD.
With the help of the figure it is easy to show that
(5)  CQ = √(k  R)²  R² = √k²  2Rk = b  a/2 = k  2r, 
where the last equality follows from equation (2). Squaring equation (5) gives
(6)  R = 2r (k  r) / k. 
The strategy is now to eliminate R in favor of r and k. Applying the Pythagorean theorem directly to the figure below gives
(R + r)² + (a/2  s)² = (R  r)² + (s + a/2)²,
or, after simplifying,
(7)  2Rr = as. 
The problem is almost solved. We now replace R on the left of equation (7) by equation (6) and as on the right by the expressions in equations (3) and (4) to get
4r²(k  r) / k = 4r(k  r)/(k  2r) × [(r(k  r)/(k  2r)  (k  2r)/2],
which gives the cubic
k³  4rk²  2kr² + 8r³ = (k²  2r²)(k  4r) = 0.
From any of the figures, the relevant root is
r = CH / 4.
References
 H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008
Sangaku

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