Competition problem — Geometry — Maths Olympiad Prep

Library / /26 of 27

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Croatia

Let $\triangle ABC$ be an acute triangle with $|AC| > |AB|$ . Let $N$ be the foot of the altitude from point $A$ to side $\overline{BC}$ . Let $P$ be a point on the extension of $\overline{AB}$ over point $B$ , and let $Q$ be a point on the extension of $\overline{AC}$ over point $C$ such that $BPQC$ is a cyclic quadrilateral.
If $|NP| = |NQ|$ , show that $N$ is the centre of the circle circumscribed to triangle $APQ$ . (Japan)

Solution

The sum of opposite angles in a cyclic quadrilateral is $180^\circ$ , therefore $\angle CQP = 180^\circ - \angle PBC$ and $\angle QPB = 180^\circ - \angle BCQ$ , from which it follows that $\angle AQP = \angle CBA$ and $\angle QPA = \angle ACB$ .
Let $O$ be the centre of the circle circumscribed to $APQ$ . Then $\angle AOP = 2\angle AQP$ . Since the triangle $AOP$ is isosceles, we have that $\angle PAO = 90^\circ - \angle AQP = 90^\circ - \angle ABC$ .

Let $p$ be the line which closes an angle of $90^\circ - \angle ABC$ with line $AP$ , and which intersects the segment $\overline{PQ}$ . Since $\angle ANB = 90^\circ$ , it follows that $\angle ANB = 90^\circ - \angle ABC$ , which implies that $A, N$ and $O$ all lie on $p$ .
Since $|AB| \ne |AC|$ , it follows that $\angle AQP = \angle ABC \ne \angle ACB = \angle APQ$ , i.e. $A$ does not lie on the bisector of segment $\overline{PQ}$ . Therefore, the intersection of line $p$ and bisector of $\overline{PQ}$ is unique and it follows that $N = O$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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