Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Austria

Let ABCABC be a triangle. Let PP be the point on the extension of BCBC beyond BB such that BP=BABP = BA. Let QQ be the point on the extension of BCBC beyond CC such that CQ=CACQ = CA. Prove that the circumcenter OO of the triangle APQAPQ lies on the angle bisector of the angle BAC\angle BAC.

Figure 1
Figure 3: Problem 10

Solution

Since ACQACQ is an isosceles triangle, the perpendicular bisector of AQAQ is the angle bisector of QCA\angle QCA. But the perpendicular bisector of AQAQ also passes through the circumcenter OO of the triangle APQAPQ.

Therefore, OO lies on the angle bisector of QCA\angle QCA which is the exterior angle bisector of ACB\angle ACB by definition of QQ.

Analogously, the point OO lies also on the exterior angle bisector of CBA\angle CBA. Therefore, the point OO is the intersection of the two exterior angle bisectors which makes it the excenter of the excircle of ABCABC tangent to BCBC. This excenter lies on the angle bisector of BAC\angle BAC as desired.

(Theresia Eisenkölbl) \square

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