Track / Stage 6 / 89 of 400 #1089 of 1964

Problem 1089

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Vijetnam · Vietnam · 2007

Let $ABCD$ be a trapezium with the bottom edge $BC$ ( $BC \parallel AC$ and $BC > AD$ ) and inscribed in the circle $(O)$ ( $O$ is the center of $(O)$ ). Let $P$ be a point moving on the line $BC$ outside the segment $BC$ such that $PA$ doesn't touch the circle $(O)$ . The circle with the diameter $PD$ intersects $(O)$ in $E$ ( $E \neq D$ ). Let $M$ be the point of intersection of $BC$ and $DE$ , and $N$ ( $N \neq A$ ) be the second point of intersection of $PA$ and $(O)$ . Prove that the line $MN$ passes through a fixed point.

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

Let $A'$ be the image of the point $A$ by the reflection through the point $O$ . We prove that $N$ , $M$ , $A'$ are collinear and therefore the line $MN$ passes the fixed point $A'$ . First, $DE$ is the radical axis of the circle $(O)$ and the circle ( $\gamma_1$ ) with the diameter $PD$ .

Since $\angle PNA' = 90^\circ$ the line $NA'$ is the radical axis of the circle $(O)$ and the circle ( $\gamma_2$ ) with the diameter $PA'$ .

The line $DA'$ meets the line $BC$ at the point $F$ ; since $\angle PFA' = 90^\circ$ , $\angle ADA' = 90^\circ$ $\Rightarrow \angle PFA' = 90^\circ$ , therefore $BC$ is the axis of the circle ( $\gamma_1$ ) and ( $\gamma_2$ ), thus the radical axis $DE$ , $BC$ and $NA'$ are concurrent at the radical center $M$ , thus the points $M$ , $N$ and $A'$ are collinear.

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