LeetCode Problem 3312

Difficulty: 🔴 Hard
Topics: Array, Hash Table, Math, Binary Search, Combinatorics, Counting, Number Theory, Prefix Sum

Solution

=== 1996-G1 ===

IMO 1996 Shortlist G1

Origin: GBR

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 $BH$, let $PAQB$ and $PARC$ be parallelograms, and let $AQ$ meet $HR$ in $X$. Prove that $EX$ is parallel to $AP$.

Solution

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

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 $ABC$ and $PBC$ 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 $AQR$ coincides with $H$. (Remark: This can be shown by noting that $AHBQ$ is cyclic.)

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

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 $EX \parallel AP$.