## 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 2klm = Rr2,

where R is the circumradius, r the inradius, and k, l, and m are the three segment altitudes.

Solution ### 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.

### 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:

 KS = vers(α).

As we saw,

 vers(α) = 1 - cos(α) = 2·sin2(α/2).

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