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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.6 AIME Prove it Saudi Arabia

Let $ABC$ be a triangle and let $P$ be a point on $BC$ . Points $M$ and $N$ lie on $AB$ and $AC$ , respectively, such that $MN$ is not parallel to $BC$ and $AMPN$ is a parallelogram. Line $MN$ meets the circumcircle of $ABC$ at $R$ and $S$ . Prove that the circumcircle of triangle $RPS$ is tangent to $BC$ .

Solution

Because $BM$ and $NP$ are parallel, and $AN$ and $MP$ are parallel, we have
$\frac{QP}{QB} = \frac{NP}{MB} = \frac{AM}{MB} = \frac{CP}{PB},$
and
$\frac{QC}{QP} = \frac{CN}{PM} = \frac{CN}{NA} = \frac{CP}{PB}.$

We deduce that
$QP^2 = QB \cdot QC.$
But from the power of point $Q$ with respect to the circumcircle of $ABC$ , we have $QB \cdot QC = QR \cdot QS$ . We deduce that
$QP^2 = QR \cdot QS,$
that is, line $BC$ is tangent to the circumcircle of triangle $PRS$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.