Sangaku with Versines
The following Sangaku has been written in 1825 in the Tokyo prefecture but has since disappeared. It relates the rarely used nowadays versine quantities and the distance from a vertex to the inclircle.
Assuming that MP and NQ are the perpendicular bisectors of AB and AC, respectively, we have
(1) | 4·MP·NQ = AI2. |
It sheds some light on the appearance of the inradius in a different formula from a related problem.

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Assuming that MP and NQ are the perpendicular bisectors of AB and AC, respectively, we have
(1) | 4·MP·NQ = AI2. |
Solution
Let A', B', C' be the points of tangency of the incircle with BC, AC, and AB, respectively. Introduce
a' = AB' = AC' = p - a,
b' = BA' = BC' = p - b,
c' = CA' = CB' = p - c,
where p is the semiperimeter p = (a + b + c)/2. Note that
Lemma 1
Proof
By Heron's formula, area S can be obtained from
S2 = pa'b'c',
but also S = rp, which gives (2).
Lemma 2
Proof
We use a known relation between the angles in a triangle
(4) | cot(A/2) + cot(B/2) + cot(C/2) = cot(A/2)·cot(B/2)·cot(C/2). |
Observe that AQN = (π - B)/2, so that
a'/r + (a' + c')/(2d1) + c'/r = (a'/r)((a' + c')/(2d1))(c'/r),
which leads to
1/r + 1/(2d1) = a'c'/(2d1r2).
So that we get a'c' = r2 + 2rd1, the first equation in (3). The second one is obtained similarly.
Proof of (1)
Add the equations in (3):
a'(b' + c') = 2r2 + 2r(d1 + d2),
and multiplying the equations in (3) we get
a'b'c' = 2r2(r + 2d1)(r + 2d2) / a'.
Lemma 1 in the form r(a' + b' + c') = a'b'c' gives
r2(a' + (2r2 + 2r(2d1 + 2d2) / a')) = r2(r + 2d1)(r + 2d2)/ a',
which leads to (a')2 + r2 = 4d1d2, but
Remark
If we use k, l, m (as in a related problem) for MP, NQ and the altitude of the circular segment cut off by BC, then the product of the three identities implied by (1) shows that
klm = 8·AI·BI·CI.

References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, pp. 101-102
Write to:
Charles Babbage Research Center
P.O. Box 272, St. Norbert Postal Station
Winnipeg, MB
Canada R3V 1L6

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- Sangakus with a Mixtilinear Circle
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- Two Sangaku with Equal Incircles
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- FJG Capitan's Sangaku

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Copyright © 1996-2018 Alexander Bogomolny
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