Proving that two line intersect on the circumcircle of the triangle (Solution Shortlist Geometry problem in IMO 2014)

Problem 1

The points P and Q are chosen on the side BC of an acute-angled triangle ABC so that \angle PAB=\angle ACB and \angle QAC=\angle CBA. The points M and N are taken on the rays AP and AQ, respectively, so that AP = PM and AQ = QN. Prove that the lines BM and CN intersect on the circumcircle of the triangle ABC.

Solution 1.

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Denote by S the intersection point of the lines BM and CN. Let moreover \beta =\angle QAC =\angle CBA and \gamma =\angle PAB =\angle ACB. From these equalities it follows that the triangles ABP and CAQ are similar (see Figure 1). Therefore we obtain

\frac{BP}{PM}=\frac{BP}{PA}=\frac{AQ}{QC}=\frac{NQ}{QC}

Moreover,

\angle BPM=\beta +\gamma =\angle CQN.

Hence the triangles BPM and NQC are similar. This gives \angle BMP=\angle NCQ, so the triangles BPM and BSC are also similar. Thus we get

\angle CSB=\angle BPM=\beta +\gamma =180^0-\angle BAC,

which completes the solution.

Solution 2.

As in the previous solution, denote by S the intersection point of the lines BM and NC. Let moreover the circumcircle of the triangle ABC intersect the lines AP and AQ again at K and L, respectively (see Figure 2).

Note that \angle LBC=\angle LAC=\angle CBA and similarly \angle KCB=\angle KAB=\angle BCA. It implies that the lines BL and CK meet at a point X, being symmetric to the point A with respect to the line BC. Since AP = PM and AQ = QN, it follows that X lies on the line MN.

Therefore, using Pascal’s theorem for the hexagon ALBSCK, we infer that S lies on the circumcircle of the triangle ABC, which finishes the proof.

Comment.

Both solutions can be modified to obtain a more general result, with the equalities

AP = PM and AQ = QN

replaced by

\frac{AP}{PM}=\frac{QN}{AQ}.

Also Read : Geometry shortlist problem in IMO 2014

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