Maximal Properties of the Pythagorean Relation
A 1893 sangaku from the Fukusima prefecture [Fukagawa, Pedoe, problem 2.3.6] admits a relatively straightforward computational solution, but also an elegant solution that practically avoids any kinds of computations.
Rectangles cut off from a right triangle leave three triangles with inradii r_{1}, r_{2}, r_{3}, in the increasing order. Show that when the area of the rectangle is a maximum possible then
(r_{1})² + (r_{2})² = (r_{3})².
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989
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Copyright © 19962018 Alexander Bogomolny
Rectangles cut off from a right triangle leave three triangles with inradii r_{1}, r_{2}, r_{3}, in the increasing order. Show that when the area of the rectangle is a maximum possible then
(r_{1})² + (r_{2})² = (r_{3})².
Solution
The four triangles ABC, EDC, AEF, DBG are obviously similar. Their inradii relate exactly like any of the corresponding sides. For the definiteness sake, let's focus on the hypotenuse.
If the sides of ΔABC are a = BC, b = AC, c = AB, then, for some
ABC  a  b  c  
EDC  ka  kb  kc  
AEF  ...  ...  (1k)b  
DBG  ...  ...  (1k)a 
We want to show that, for k, for which the area of the rectangle DEFG is a maximum,
[(1  k)a]² + [(1  k)b]² = (kc)²,
or
(1)  (1  k)²·a² + (1  k)²·b² = k²·c². 
However, we also have
a² + b² = c².
Multiplying (2) by k² and subtracting from (1) we get
(1  2k)·(a² + b²) = 0
which is only possible when k = 1/2. This is then bound to be the optimal configuration implied by the sangaku. To proceed, we need to fill some of the empty cells in the table above:
DG / AC = BD / AB,
so that DG = (1  k)ab/c. Since DE = kc, the area S of the rectangle DEFG is given by
S = DG × DE = (1  k)k·ab
which attains its maximum along with
Sangaku

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Contact Front page Contents Geometry Up
Copyright © 19962018 Alexander Bogomolny