A Relation in Triangle
A triangle cuts off from the circumcircle three circular segments. Greg Markowsky has discovered a relation linking the altitudes of the segments with the in- and circumradii of the triangle. With a reference to the following diagram
the relationship reads
where R is the circumradius, r the inradius, and k, l, and m are the three segment altitudes.
Solution

Copyright © 1996-2009 Alexander Bogomolny
Solution
Let O be the circumcenter of ΔABC, K the midpoint of BC. Then in right triangle OCK, COK = BAC, which I'll also denote A. Thus
The altitude of the corresponding circular segment is
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k = R - OK = R(1 - cosA) = 2R·sin2(A/2).
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Similarly, in obvious notations,
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l = 2R·sin2(B/2) and m = 2R·sin2(C/2).
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Therefore,
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klm = 8R3·sin2(A/2)·sin2(B/2)·sin2(C/2).
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Now taking into account that
easily yields the desired result.
Remark
If we denote α = COK = A, and assume the circumradius of ΔABC equals 1, then CK = sin(α) and OK = cos(α). If S is the midpoint of arc OK below K, then KS also is a named quantity, viz., versine (versed sine) of α:
As we saw,
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| vers(α) | = 1 - cos(α) |
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(Curiously, there is a sangaku problem that relates the altitudes of the circumsegments as above to the distance between a vertex and the incenter.)

Copyright © 1996-2009 Alexander Bogomolny
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