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.
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 36°- 54°- 90°, the other 18°- 72°- 90°.
The altitude in question equals
h = a·(sin(36°) + sin(72°)).
The fan forming triangles are also of the 36°- 54°- 90° variety since two of their smallest angles supplement 108°. In such a triangle the hypotenuse t is expressable in terms of the altitude:
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 c2 + s2 = 1. In addition, tan(36°) = s/c, and tan(54°) = cot(36°) = c/s, since 36° and 54° are supplementary angles. Also, sin(72°) = 2sc. We see that t can be written as