Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

25. $X$ is a point on side $BC$ of $\triangle ABC$ , and $Y$ is the intersection of $AX$ with the circumcircle of $\triangle ABC$ . Prove or disprove: the length of segment $XY$ is maximized when point $X$ lies between the median from $A$ and the angle bisector of $\angle BAC$ .

Solution

25. Let $A B \leqslant A C$ . If $P$ and $M$ are points on $B C$ such that $A P$ and $A M$ are the angle bisector and median from point $A$ , respectively. Then $A P$ intersects the midpoint $Q$ of the arc $B C$ . If $A M$ intersects the circumcircle at $N$ , and $X$ is between $M$ and $C$ , then
$\begin{aligned} A X \cdot X Y & =B X \cdot X C A M$, so $X Y \parallel A P$, then $X Y < P Q$. If $A X \leqslant A P$, and if $t$ is the tangent line at point $\mathrm{Q}$ of the circle, then $t \parallel B C$, $(.4 X$ and $t$ are parallel). Since $A X \leqslant A P$, we have$
\because Y < X Z \leqslant P Q \text {. }

From this, it follows that if $X Y$ is maximized, $X$ must lie on the segment $P M$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.