A problem with incircles and circumcircles
What is it about?
A Mathematical Droodle
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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_{a}D, M_{b}D, and M_{c}D of sides BC, IC, and IB. We want to show that D lies on the circumcircle of ΔABC.
Angles M_{a}DM_{c} and IBC have pairwise perpendicular sides. Therefore they are equal. And similarly for angles M_{a}DM_{b} and ICB:
(1) |
∠M_{a}DM_{c} = ∠IBC = ∠B/2 ∠M_{a}DM_{b} = ∠ICB = ∠C/2 |
From (1) we obtain
∠M_{a}DM_{c} + ∠M_{a}DM_{b} | = ∠M_{b}DM_{c} |
= ∠IDM_{b} + ∠IDM_{c} | |
= ∠IDC/2 + ∠IDB/2 | |
= ∠CDB/2 |
Comparing this to (1) gives ∠CDB = ∠C + ∠B. In other words,
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_{A}O_{B}O_{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
- R. Honsberger, From Erdös To Kiev, MAA, 1996, pp. 56-57.
- R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960
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