IMO 1996 SL G1

Origin: GBR | Category: Geometry

Problem

Let $\triangle ABC$ have orthocenter $H$, and let $P$ be a point on its circumcircle, distinct from $A$, $B$, $C$. Let $E$ be the foot of the altitude from $B$, let $PAQB$ and $PARC$ be parallelograms, and let $AQ$ meet $HR$ in $X$. Prove that $EX\parallel AP$.

Solution

We first show that $H$ is the common orthocenter of the triangles $\triangle ABC$ and $\triangle AQR$.

Let $G$, $G'$, $H'$ be respectively the centroid of $\triangle ABC$, the centroid of $\triangle PBC$, and the orthocenter of $\triangle PBC$. Since the triangles $\triangle ABC$ and $\triangle PBC$ have a common circumcenter, from the properties of the Euler line we get

$$\overrightarrow{HH'}=3\overrightarrow{GG'}=\overrightarrow{AP}.$$

But $\triangle AQR$ is exactly the image of $\triangle PBC$ under translation by $\overrightarrow{AP}$; hence the orthocenter of $\triangle AQR$ coincides with $H$.

(Figure: $A$, $B$, $C$, $P$, $E$, $H$, $Q$, $R$, $X$)

Remark: This can also be shown by noting that $AHBQ$ is cyclic.

Now we have that $RH\perp AQ$; hence

$$\angle AXH=90^\circ=\angle AEH.$$

It follows that $AXEH$ is cyclic; hence

$$\angle EXQ = 180^\circ-\angle AHE = 180^\circ-\angle BCA = 180^\circ-\angle BPA = \angle PAQ$$

(as oriented angles). Hence $EX\parallel AP$.