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
2klm = Rr^{2}, 
where R is the circumradius, r the inradius, and k, l, and m are the three segment altitudes.
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Copyright © 19962018 Alexander Bogomolny
Solution
Let O be the circumcenter of ΔABC, K the midpoint of BC. Then in right triangle OCK,
OK = R·cos(A). 
The altitude of the corresponding circular segment is
k = R  OK = R(1  cosA) = 2R·sin^{2}(A/2). 
Similarly, in obvious notations,
l = 2R·sin^{2}(B/2) and m = 2R·sin^{2}(C/2). 
Therefore,
klm = 8R^{3}·sin^{2}(A/2)·sin^{2}(B/2)·sin^{2}(C/2). 
Now taking into account that
r = 4Rsin(A/2)·sin(B/2)·sin(C/2) 
easily yields the desired result.
Remark
If we denote α = COK = A, and assume the circumradius of ΔABC equals 1, then
KS = vers(α). 
As we saw,

(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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