Out of Pentagon Sangaku
Many sangaku problems include circle and ellipses, but quite a few do not. An elegant one [Fakagawa & Pedoe, p. 49] with a rather computational solution is presented below.
6 congruent right triangles fan out along the sides of a regular pentagon of side a. Find the length of the hypotenuse t of these triangles in terms of a.
This sangaku appears on a extant tablet in the Miyagai prefecture. Unless there is a misprint, [Fakagawa & Pedoe, p. 134] dates the tablet from 1912.
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989
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Copyright © 19962018 Alexander Bogomolny
6 congruent right triangles fan out along the sides of a regular pentagon of side a. Find the length of the hypotenuse t of these triangles in terms of a.
Solution
We can imagine a couple of right triangles with sides computable from the Pythagorean theorem that combine into the altitude of one of the congruent triangles:
Given that the internal angle of a regular pentagon is 108°, one of the triangles has angles
The altitude in question equals
h = a·(sin(36°) + sin(72°)).
The fan forming triangles are also of the
t = h·(tan(36°) + tan(54°)).
Combining the two we get
t = a·(tan(36°) + tan(54°))·(sin(36°) + sin(72°)).
The solution t = a·(1 + √5) given by [Fakagawa & Pedoe] tells us that the expression for t, if correct, is amenable to a simplification effort.
Let's denote c = cos(36°) and s = sin(36°). By the Pythagorean theorem
t  = a·(tan(36°) + tan(54°))·(sin(36°) + sin(72°)) 
= a·(s/c + c/s)·(s + 2sc)  
= a·s/sc·(s^{2} + c^{2})·(1 + 2c)  
= a·(1 + 2c)/c 
This must be transformed further taking into account that
t  = a·(1 + 2c)/c 
= a·(1 + (1 + √5)/2) / (1 + √5)·4  
= a·(3 + √5)) / (1 + √5)·2  
= a·(3 + √5))·(1 + √5) / 2  
= a·(3 + 5  √5 + 3√5) / 2  
= a·(2 + 2√5) / 2  
= a·(1 + √5). 
Sangaku

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Copyright © 19962018 Alexander Bogomolny