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Taiwan IMO Selection Camp

Geometry Difficulty 4.5 AIME Prove it Taiwan

Let $ABC$ be an isosceles triangle with $AB = AC$ , and let $M$ be the midpoint of side $BC$ . Let $P$ be a point in the plane distinct from $A$ , satisfying $PB < PC$ , and such that $PA$ is parallel to $BC$ . Let point $X$ lie on line $PB$ and point $Y$ lie on line $PC$ , such that $B$ lies on segment $PX$ , $C$ lies on segment $PY$ , and $\angle PXM = \angle PYM$ . Prove that the four points $A, P, X, Y$ are concyclic.

Solution

Since $AB = AC$ , we know that $AM$ is the perpendicular bisector of side $BC$ , so
$\angle PAM = \angle AMC = 90^\circ$

Now draw through point $Y$ a line perpendicular to $PC$ , and let it meet line $AM$ at point $Z$ . (Note: point $M$ lies between the two points $A, Z$ .) We can see that
$\angle PAZ = \angle PYZ = 90^\circ.$
Hence the four points $P, A, Y, Z$ are concyclic.

$\angle CMZ = \angle CYZ = 90^\circ,$
so we get that $C, Y, Z, M$ are four concyclic points, hence $\angle CZM = \angle CYM$ .

From the problem's assumption we know $\angle CYM = \angle BXM$ , and since $B, C$ are symmetric with respect to $ZM$ , we have
$\angle CZM = \angle BZM.$

Combining the above, we obtain $\angle BXM = \angle BZM$ , so $B, X, Z, M$ are four concyclic points. Therefore we get
$\angle BXZ = 180^\circ - \angle BMZ = 90^\circ.$

$\angle PXZ = \angle PYZ = \angle PAZ = 90^\circ,$
hence $P, A, X, Y, Z$ are five concyclic points. Of course it follows that $A, P, X, Y$ are four concyclic points, which completes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.