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

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Copyright © 1996-2012 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

OK = R·cos(A).

The altitude of the corresponding circular segment is

k = R - OK = R(1 - cosA) = 2R·sin2(A/2).

Similarly, in obvious notations,

l = 2R·sin2(B/2) and m = 2R·sin2(C/2).

Therefore,

klm = 8R3·sin2(A/2)·sin2(B/2)·sin2(C/2).

Now taking into account that

r = 4Rsin(A/2)·sin(B/2)·sin(C/2)

easily yields the desired result.

(Curiously, there is a sangaku problem that relates the altitudes of the circumsegments as above to the distance between a vertex and the incenter.)

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Copyright © 1996-2012 Alexander Bogomolny