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A problem with incircles and circumcircles: What is it about?
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Copyright © 1996-2008 Alexander Bogomolny
Explanation
Let I be the incenter of 
ABC. Consider 3 triangles: IBC, ICA, and IAB. An interesting fact is that the circumcenters of these triangles lie on the circumcircle of ABC.
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Indeed, let D be the circumcenter of IBC. D lies at the intersection of the perpendicular bisectors MaD, MbD, and McD of sides BC, IC, and IB. We want to show that D lies on the circumcircle of ABC.
Angles MaDMc and IBC have pairwise perpendicular sides. Therefore they are equal. And similarly for angles MaDMb and ICB:
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 MaDMc = IBC = B/2MaDMb = ICB = C/2
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From (1) we obtain
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Comparing this to (1) gives CDB = C + B. In other words, CDB + A = 180o, which exactly means that D lies on the circumcircle of ABC.
It is also clear that the circumcenter D of ΔBIC lies on bisector AI of BAC.
Remark
A weaker variant of the problem has been offered at the 1988 USA Olympiad where it was required to prove that the circumcircles of ΔABC and ΔO1O2O3 are concentric.
In [Johnson, p. 185-185, 292°] the problem appears thus: Let D be the intersection of the A-bisector of ΔABC with the circumcircle. Then the circle centered at D with the radius equal to DB (= DC) passes through the incenter I.
References
- R. Honsberger, From Erdös To Kiev, MAA, 1996, pp. 56-57.
- R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960
Copyright © 1996-2008 Alexander Bogomolny
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