A Restored Sangaku Problem
The sangaku discussed below was originally hung [Fukagawa and Rothman, pp. 194-195] in 1806 in the Atsuta shrine by Ehara Masanori, a disciple of Kusaka Makoto (1764-1833). The tablet, which contained only this problem and no solution, was subsequently lost. However, at an unknown date the mathematician Kitagawa Moko (1763-1839) 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.
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.
- H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008
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Copyright © 1996-2018 Alexander Bogomolny