1994 IMO Problems/Problem 2

Let $ABC$ be an isosceles triangle with $AB = AC$ . $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$ . $Q$ is an arbitrary point on $BC$ different from $B$ and $C$ . $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E, Q, F$ are distinct and collinear. Prove that $OQ$ is perpendicular to $EF$ if and only if $QE = QF$ .

Solution

Let $E'$ and $F'$ be on $AB$ and $AC$ respectively such that $E'F'\perp OQ$ . Then, by the first part of the problem, $QE'=QF'$ . Hence, $Q$ is the midpoint of $EF$ and $E'F'$ , which means that $EE'FF'$ is a parallelogram. Unless $E=E'$ and $F=F'$ , this is a contradiction since $EE'$ and $F'F$ meet at $A$ . Therefore, $E=E'$ and $F=F'$ , so $OQ\perp EF$ , as desired.

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1994 IMO Problems/Problem 2

Solution