A Hard but Important Sangaku
ABC is a given triangle, and circle O(R) passes through A and B, C lying within the circle. The circle O'(x) lies within O(R), touches AC and BC and also touches
Show that
x = r + 2d(s  a)(s  b) / cs, 
where s is the semiperimeter of ΔABC and r is the radius of the incircle of ΔABC.
The problem is listed as 2.2.8 by Fukagawa and Pedoe and said to be hard but important because of its relevance to a variety of other sangaku. It comes from a third volume of a 1781 collection Seiyo Sanpo by Teisi Fujita (17341807).
The solution below is by a New York physicist who posted it under the pseudonym yetti at the MathLinks Forums.
An alternative solution was found by James M. Unger, Professor of East Asian Languages & Literatures, The Ohio State University, Columbus, Ohio.
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, pp. 101102
Write to:
Charles Babbage Research Center
P.O. Box 272, St. Norbert Postal Station
Winnipeg, MB
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Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny
First extend, say, BC to cut O(R) in P:
This brings the problem into the framework of one of V. Thébault's theorems and suggests a change of notations.
Further (re)define a = BC,
s_{b} + 
The incenter of ΔABC will be denoted I, the incircle (I). For ΔABD, the incenter is I_{b}, the incircle (I_{b}) and the inradius r_{b}. We also consider the curvilinear triangles ABD and ACD with the incenters J_{b} and J_{c} and incircles (J_{b}) and (J_{c}). Note that the unknown x is the radius of (J_{b}).
In these notations, we want to show that
(1)  x/r_{b} = 1 + 2·MN·(s_{b}  u)(s_{b}  d) / cr_{b}s_{b}. 
Using Heron's formula for ΔABD, viz.,
Area( ΔABD)² = s_{b}(s_{b}  u)(s_{b}  d)(s_{b}  c) 
we transform (1) into the equivalent
(2) 

Let now T, T_{b}, S_{b} denote the points of tangency of (I), (I_{b}) and (J_{b}) with BC and Q_{b} that of (J_{b}) and AD. By Sawayama's Lemma, I, S_{b} and Q_{b} are collinear. By definition,
If
TS_{b} = IT· tan(φ) = r · tan(φ). 
With this in mind,
DT_{b} = s_{b}  c, 
whereas,

Using x = DS_{b}tan(φ) and r_{b} = DT_{b}tan(φ),

Using

we see that
(3)  x / r_{b}  1 = [r · tan(φ)  (s_{c}  b)] / (s_{b}  c). 
We have now to show that the right hand side in (3) coincides with the right hand side in (2):
[r · tan(φ)  (s_{c}  b)] / (s_{b}  c) = 2·MN· r_{b} / c(s_{b}  c) 
which is the same as showing that
(4)  r · tan(φ) = 2·MN· r_{b} / c + (s_{c}  b). 
Observe that
∠MAN = ∠BAN = ∠BCN = ∠ACB / 2 
implying

from ΔICT. Also, from ΔABD,
tan(φ) = r_{b} / (s_{b}  c), 
the combination of which reduces (4) to the equivalent
(5)  rr_{b} / (s_{b}  c) = rr_{b} / (s  c) + s_{c}  b. 
Using (s  c)  (s_{b}  c) = s  s_{b} = s_{c}  d (5) becomes
(6)  rr_{b}(s_{c}  d) / (s_{b}  c)(s  c) = s_{c}  b. 
But now again r / (s  c) = tan(∠ACB / 2) = r_{c} / (s_{c}  d) which further upgrades (6) to
r_{b}r_{c} / (s_{c}  b)(s_{b}  c) = 1 
which is the same as
(7)  r_{b} / (s_{b}  c) × r_{c} / (s_{c}  b) = 1. 
However, r_{c} / (s_{c}  b) = tan(ψ) where ψ = ∠ADC/2. So that (7) merely asserts that
Since (7) holds, the equivalent (1) is also true.
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Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny
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