Proving two line is Parallel from the circumcircle and circumcentre (Solution Problem 3 Shortlist IMO 2014)

Problem 3:

Let \Omega and O be the circumcircle and the circumcentre of an acute-angled triangle ABC with AB a BC. The angle bisector of \angle ABC intersects \Omega at M\neq B. Let \Gamma be the circle with diameter BM. The angle bisectors of \angle AOB and \angle BOC intersect \Gamma at points P and Q, respectively. The point R is chosen on the line PQ so that BR = MR. Prove that BR || AC. (Here we always assume that an angle bisector is a ray).


Also Read:

Let K be the midpoint of BM, i.e., the centre of G. Notice that AB\neq BC implies K\neq O. Clearly, the lines OM and OK are the perpendicular bisectors of AC and BM, respectively. Therefore, R is the intersection point of PQ and OK.

Let N be the second point of intersection of \Gamma with the line OM. Since BM is a diameter of \Gamma, the lines BN and AC are both perpendicular to OM. Hence BN || AC, and it suffices to prove that BN passes through R. Our plan for doing this is to interpret the lines BN, OK, and PQ as the radical axes of three appropriate circles.

Let \omega be the circle with diameter BO. Since \angle BNO=\angle BKO=90^0, the points N and K lie on \omega.

Next we show that the points O, K, P, and Q are concyclic. To this end, let D and E be the midpoints of BC and AB, respectively. Clearly, D and E lie on the rays OQ and OP, respectively. By our assumptions about the triangle ABC, the points B, E, O, K, and D lie in this order on \omega. It follows that \angle EOR=\angle EBK=\angle KBD=\angle KOD, so the line KO externally bisects the angle POQ. Since the point K is the centre of \Gamma, it also lies on the perpendicular bisector of PQ. So K coincides with the midpoint of the arc POQ of the circumcircle γ of triangle POQ.

Thus the lines OK, BN, and PQ are pairwise radical axes of the circles \omega ,\gamma ,\text{ and }\Gamma. Hence they are concurrent at R, as required.

Also Read : Geometry shortlist problem in IMO 2014

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