Let $P$ and $Q$ be points on a circle $k$. A chord $AC$ of $k$ passes through the midpoint $M$ of $PQ$. Consider a trapezoid $ABCD$ inscribed in $k$ with $AB \parallel PQ \parallel CD$. Prove that the intersection point $X$ of $AD$ and $BC$ depends only on $k$ and $P,Q.$
Problem
Source: Mediterranean MO 2001
Tags: geometry, trapezoid, symmetry, parallelogram, rectangle, perpendicular bisector, geometry unsolved
oneplusone
07.11.2010 05:42
If this problem is true, then by symmetry, $X$ must lie on the perpendicular bisector of $PQ$, which is not always true.
dgreenb801
28.11.2010 05:54
I think he means $AB \parallel PQ \parallel CD$
oneplusone
28.11.2010 06:02
If $AB\parallel PQ$, then since $\angle CMQ=\angle AMP=\angle BMQ$, we have $BQCP$ is harmonic. Also by symmetry, $X$ must lie on the perpendicular bisector of $PQ$. Thus $XP$ and $XQ$ are tangent to the circle by properties of harmonic quads. Thus $X$ is fixed.
Goutham
28.11.2010 20:36
Lemma: For a point $P$ inside a circle, if distinct chords $AC$ and $BD$ have their midpoint $P$, then $P$ is the centre of the circle. Proof: If so, $ABCD$ is a parallelogram and any cyclic parallelogram is a rectange$\Longrightarrow P$ is the centre of the circle as required. Now, in the problem, let $XP', XQ'$ be the tangents from $X$ to the circle. Since $AC, BD$ intersect at $M$, the polar of $X$ with respect to $k$ passes through $M$ and hence, $P'Q'$ passes through $M$. Moreover, we have $P'M=MQ'$ but it is given that $PM=MQ$ and so, $M$ is the centre of the circle or $PQ$ coincides with $P'Q'$. If $M$ is the centre of the circle, $ABCD$ is a rectange as $AC, BD$ intersect at $M$ and so, $X$ lies at infinity which is fixed. If $PQ$ coincides with $P'Q'$, we have $PMQ$ to be the polar of $X$ with respect to $k$ and since $PQ$ is fixed, it's pole $X$ is fixed as well.