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

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.

References

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

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Copyright © 1996-2008 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, 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,

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

Using Heron's formula for ΔABD, viz.,

we transform (1) into the equivalent

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,

With this in mind,

whereas,

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

Using

we see that

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

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

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

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

which is the same as

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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Copyright © 1996-2008 Alexander Bogomolny