Track / Stage 7 / 114 of 300 #1514 of 2000

Problem 1514

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.2 Prove it

Points $P$ and $Q$ are taken sides $AB$ and $AC$ of a triangle $ABC$ respectively such that $\hat{APC}=\hat{AQB}=45^{0}$ . The line through $P$ perpendicular to $AB$ intersects $BQ$ at $S$ , and the line through $Q$ perpendicular to $AC$ intersects $CP$ at $R$ . Let $D$ be the foot of the altitude of triangle $ABC$ from $A$ . Prove that $SR\parallel BC$ and $PS,AD,QR$ are concurrent.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Given Conditions and Setup:
- Points $P$ and $Q$ are on sides $AB$ and $AC$ of triangle $ABC$ respectively.
- $\angle APC = \angle AQB = 45^\circ$ .
- Line through $P$ perpendicular to $AB$ intersects $BQ$ at $S$ .
- Line through $Q$ perpendicular to $AC$ intersects $CP$ at $R$ .
- $D$ is the foot of the altitude from $A$ to $BC$ .

2. **Prove $SR \parallel BC$ :**
- Since $\angle APC = 45^\circ$ and $\angle AQB = 45^\circ$ , we have:
$\angle SPR = 90^\circ + 45^\circ = 135^\circ$
- Similarly, $\angle SQR = 180^\circ - \angle QSR$ because $S, P, R, Q$ are concyclic.
- Therefore, $\angle QSR = 45^\circ$ and $\angle QBC = \angle QPC = 45^\circ$ .
- Hence, $\angle QPR = \angle QSR$ , which implies:
$SR \parallel BC$

3. **Prove $PS, AD, QR$ are concurrent:**
- Let $AD \cap SR = F$ .
- Since $\angle APS = 90^\circ$ , $A, F, P, S$ are concyclic.
- Similarly, $\angle AQR = 90^\circ$ , $A, F, Q, R$ are concyclic.
- By the Radical Center Theorem, the radical axes of the three circles (which are $QR, AF, SP$ ) concur at a single point.

$\blacksquare$

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