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

It sheds some light on the appearance of the inradius in a different formula from a related problem.

Solution

Copyright © 1996-2009 Alexander Bogomolny

(1)	4·MP·NQ = AI2.

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 p = a' + b' + c'. The proof is based on two lemmas.

Lemma 1

By Heron's formula, area S can be obtained from

but also S = rp, which gives (2).

Lemma 2

If NQ = d₁ and MP = d₂ then

We use a known relation between the angles in a triangle

Observe that AQN = (π - B)/2, so that cot(B/2) = AN/NQ. Thus substituting into (4),

which leads to

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):

and multiplying the equations in (3) we get

Lemma 1 in the form r(a' + b' + c') = a'b'c' gives

which leads to (a')² + r² = 4d₁d₂, but (a')² + r² = AI².

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

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 © 1996-2009 Alexander Bogomolny