THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

If $AB=AC$, the statement is obvious. In the following, we assume $AB<AC$, the other case being similar.

Triangles $PAM$ and $PBM$ are congruent (SSS), hence $\angle PMA\equiv \angle PMB$. Similarly, $\angle QMA\equiv \angle QMC$, and this leads rapidly to $\angle PMQ=90^{\circ }$.

As $\tan B=\frac{AH}{BH}=\frac{AC}{AB}=\frac{AQ}{BP}$ and $\angle HAQ=\angle HBP=60^{\circ }+\angle B$, triangles $HAQ$ and $HBP$ are similar (SAS).

It follows that $\angle QHA=\angle PHB$, hence $\angle QHP=\angle QHA+\angle AHP=\angle PHB+\angle AHP=\angle AHB=90^{\circ }$.

In conclusion, the points $P,Q,M$ and $H$ are co-cyclic, hence $\angle HPQ\equiv \angle CMQ\equiv \angle AMQ$.

Let $\{D\}=HQ\cap MP$. The angle $\angle MNQ$ is exterior to the triangle $PMN$, therefore $\angle MNQ=\angle MPN+\angle NMP=\angle MHQ+\angle BMP=\angle MDQ$. (1)

It follows that the quadrilateral $NDMQ$ is cyclic, hence $\angle DNQ=90^{\circ }$. This means that $NDHP$ is also cyclic, therefore $\angle PNH\equiv \angle PDH$. (2)

From (1) and (2) it follows that $\angle PNH\equiv \angle PDH\equiv \angle MDQ\equiv \angle MNQ$, i.e. ($ND$ is the bisector of angle $\angle HNM$). For the triangle $HMN$ ray ($NP$ is the external bisector, while ($MP$ is an internal bisector, which means that $P$ is the excenter opposite to the vertex $M$). Then ($HP$ is the internal bisector of angle $\angle NHB$.

Finally, $\angle NHP\equiv \angle PHB\equiv \angle AHQ$.

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Alternative solution.

(given in the contest by Paul Bécsi*)

Let $\{S\}=AH\cap PQ$. An easy computation shows that $\angle SAQ=\angle NAP=120^{\circ }-\angle B$, which shows that the rays ($AN$ and ($AS$ are isogonal in the angle $\angle PAQ$). From Steiner's Theorem it follows that $\frac{PN}{NQ}\cdot \frac{PS}{SQ}={\left(\frac{PA}{AQ}\right)}^2$.

The conclusion means the rays ($HN$ and ($HS$ are isogonal in the angle $\angle PHQ$, which, according to Steiner's Theorem, is equivalent to $\frac{PN}{NQ}\cdot \frac{PS}{SQ}={\left(\frac{PH}{HQ}\right)}^2$.

In conclusion, we need to prove that $\frac{PA}{AQ}=\frac{PH}{HQ}$.



But $\Delta BHA\sim \Delta BAC$ leads to $\frac{BH}{HA}=\frac{BA}{CA}=\frac{BP}{CQ}$. As $\angle PBH=60^{\circ }+\angle B=\angle QAH$, it follows that $\Delta PBH\sim \Delta QAH$, i.e., $\frac{PH}{QH}=\frac{PB}{AQ}=\frac{PA}{AQ}$ and the conclusion.