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Team Selection Test for IMO

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Turkey

Let $ABC$ be a triangle with $m(\hat{B}) > m(\hat{C})$ . Interior and exterior angle bisectors at vertex $A$ intersect $BC$ at $D$ and $E$ respectively. A variable point $P$ lies on the ray $[EA$ such that $A$ is closer to $E$ than $P$ . Lines $DP$ and $AC$ intersect at point $M$ and lines $ME$ and $AD$ intersect at point $Q$ . Prove that all lines $PQ$ when a variable point $P$ changes intersect at a unique point. (Ali Adali).

Solution

Let $PQ \cap BC = N$ . Menelaus' theorem applied to $\triangle (AEC)$ and collinear points $P, M, D$ yields:
$\frac{|PA|}{|PE|} \cdot \frac{|ED|}{|DC|} \cdot \frac{|CM|}{|MA|} = 1.$
Menelaus' theorem applied to $\triangle (ADC)$ and collinear points $E, Q, M$ yields:
$\frac{|ED|}{|EC|} \cdot \frac{|CM|}{|MA|} \cdot \frac{|AQ|}{|QD|} = 1.$
Side by side division gives:
$\frac{|PA|}{|PE|} \cdot \frac{|EC|}{|DC|} \cdot \frac{|QD|}{|AQ|} = 1.$
Finally, Menelaus' theorem applied to $\triangle (AED)$ and collinear points $P, Q, N$ yields:
$\frac{|PA|}{|PE|} \cdot \frac{|EN|}{|ND|} \cdot \frac{|QD|}{|AQ|} = 1.$
Therefore,
$\frac{|EN|}{|ND|} = \frac{|EC|}{|DC|} = \frac{|EB|}{|BD|}$
(the last equality follows from $\frac{|EB|}{|EC|} = \frac{|AB|}{|AC|} = \frac{|BD|}{|DC|}$ ).
Thus, $N = B$ .

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