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Problem 2453

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it Team selection test for 47. IMO · Bulgaria

Let $D$ and $E$ be points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $DE \parallel BC$ . The circumcircle $k$ of $\triangle ADE$ meets the segments $BE$ and $CD$ at points $M$ and $N$ . The lines $AM$ and $AN$ meet $BC$ at points $P$ and $Q$ such that $BC = 2PQ$ and $P$ lies between $B$ and $Q$ . Prove that the circle $k$ , the line $BC$ and the bisector of $\angle BAC$ are concurrent.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:
Since $\angle PBM = \angle MED = \angle BAP$ we have $PB^{2} = PM \cdot PA$ . Analogously $QC^{2} = QN \cdot QA$ . Since $BC = 2PQ$ and $P$ lies between $B$ and $Q$ , there exists a point $L$ on $PQ$ such that $PB = PL$ and $QC = QL$ . Thus, $PL^{2} = PM \cdot PA$ , i.e. $M$ lies on the circle $k'$ through $A$ , tangent to $BC$ at $L$ . Analogously $N \in k'$ , and therefore $k' = k$ . Finally, we obtain that

$\frac{BL^{2}}{CL^{2}} = \frac{BD \cdot BA}{CE \cdot CA} = \frac{BA^{2}}{CA^{2}}, \text{ i.e. } \angle BAL = \angle CAL$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.