A problem with incircles and circumcircles
What is it about?
A Mathematical Droodle

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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

Explanation

Let I be the incenter of ΔABC. Consider 3 triangles: IBC, ICA, and IAB. An interesting fact is that the circumcenters of these triangles lie on the circumcircle of ΔABC.

Indeed, let D be the circumcenter of ΔIBC. D lies at the intersection of the perpendicular bisectors M_aD, M_bD, and M_cD of sides BC, IC, and IB. We want to show that D lies on the circumcircle of ΔABC.

Angles M_aDM_c and IBC have pairwise perpendicular sides. Therefore they are equal. And similarly for angles M_aDM_b and ICB:

(1)	∠MaDMc = ∠IBC = ∠B/2 ∠MaDMb = ∠ICB = ∠C/2

From (1) we obtain

∠MaDMc + ∠MaDMb	= ∠MbDMc
	= ∠IDMb + ∠IDMc
	= ∠IDC/2 + ∠IDB/2
	= ∠CDB/2

Comparing this to (1) gives ∠CDB = ∠C + ∠B. In other words, ∠CDB + ∠A = 180°, which exactly means that D lies on the circumcircle of ΔABC.

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 ΔO_AO_BO_C 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

1. R. Honsberger, From Erdös To Kiev, MAA, 1996, pp. 56-57.
2. R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960

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Copyright © 1996-2018 Alexander Bogomolny