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 O(R) internally. M is the midpoint of AB and N the midpoint of the arc AB, and MN = d.

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 (1734-1807).

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

1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, pp. 101-102

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   Charles Babbage Research Center
   P.O. Box 272, St. Norbert Postal Station
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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, b = CA, c = AB, d = AD, u = BD, v = DC, where u + v = a. s = (a + b + c)/2 is semiperimeter of ΔABC; s_b = (u + c + d)/2 and s_c = (v + b + d)/2 the semiperimeters of ΔABD and ΔACD,

sb + sc = s + d.

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/rb = 1 + 2·MN·(sb - u)(sb - d) / crbsb.

Using Heron's formula for ΔABD, viz.,

Area( ΔABD)² = sb(sb - u)(sb - d)(sb - c)

we transform (1) into the equivalent

(2)	x/rb - 1 = 2·MN·(sb - u)(sb - d) / crbsb   = 2·MN·(sb - u)(sb - d) / c·Area( ΔABD)   = 2·MN·Area( ΔABD) / csb(sb - c)   = 2·MN· rb / c(sb - c).

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, DQ_b = DS_b, so that ΔDQ_bS_b is isosceles.

If φ = ∠ADB / 2, then ∠IS_bD = ∠IS_bT = π/2 - φ. Thus in right ΔITS_b,

TSb = IT· tan(φ) = r · tan(φ).

With this in mind,

DTb = sb - c,

whereas,

DSb = DT + TSb   = DB - TB + TSb   = u - (s - b) + r · tan(φ).

Using x = DS_btan(φ) and r_b = DT_btan(φ),

x / rb = DSb / DTb   = [u - (s - b) + r · tan(φ)] / (sb - c).

Using

u - (s - b) = u - (sb + sc - d - b)   = (sb - c) - (sc - b)

we see that

(3)	x / rb - 1 = [r · tan(φ) - (sc - b)] / (sb - c).

We have now to show that the right hand side in (3) coincides with the right hand side in (2):

[r · tan(φ) - (sc - b)] / (sb - c) = 2·MN· rb / c(sb - c)

which is the same as showing that

(4)	r · tan(φ) = 2·MN· rb / c + (sc - b).

Observe that

∠MAN = ∠BAN = ∠BCN = ∠ACB / 2

implying

2·MN / c = MN / (c/2)   = tan(∠ACB / 2)   = r / (s - c)

from ΔICT. Also, from ΔABD,

tan(φ) = rb / (sb - c),

the combination of which reduces (4) to the equivalent

(5)	rrb / (sb - c) = rrb / (s - c) + sc - b.

Using (s - c) - (s_b - c) = s - s_b = s_c - d (5) becomes

(6)	rrb(sc - d) / (sb - c)(s - c) = sc - b.

But now again r / (s - c) = tan(∠ACB / 2) = r_c / (s_c - d) which further upgrades (6) to

rbrc / (sc - b)(sb - c) = 1

which is the same as

(7)	rb / (sb - c) × rc / (sc - b) = 1.

However, r_c / (s_c - b) = tan(ψ) where ψ = ∠ADC/2. So that (7) merely asserts that tan(φ)·tan(ψ) = 1 which is obvious in view of φ + ψ = 90°.

Since (7) holds, the equivalent (1) is also true.

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6. A 49^th Degree Challenge
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35. Japanese Art and Mathematics
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37. Maximal Properties of the Pythagorean Relation
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39. Out of Pentagon Sangaku
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52. Sangaku with Quadratic Optimization
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68. Another Sangaku in Square
69. Sangaku via Peru
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Copyright © 1996-2018 Alexander Bogomolny