Geometry · Isosceles triangle · Inscribed angle · Circumcircle · Centroid · Right triangle circumcentre · Point reflection

Problem 2, 2015

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NationalProof

A B C P Q R
The data of the problem. The point \(Q\) need not lie inside the triangle.

Let \(ABC\) be an isosceles triangle with \(AB = AC\). A point \(P\) is taken inside the triangle so that

\[ \angle BPC = 90^\circ + \tfrac{1}{2}\angle BAC , \]

and a point \(Q\) is taken so that \(\angle BPQ = \angle PQA = 90^\circ\). Let \(R\) be the point of the segment \(QB\) for which \(BR = 2RQ\). Prove that the points \(R\), \(P\) and \(C\) are collinear.

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Serbian National Competition (Drzavno takmicenje) 2015, high school grade I, category A, problem 2. Organized by the Mathematical Society of Serbia (DMS). Source

National problem · Geometry · Lemma