XXVII Olimpiada Matemática Rioplatense

Let $S$ be the intersection of the lines $QR$ and $BC$.

First, note that the homothety with center $Q$ which transforms $\omega$ into $\Gamma$ maps $P$ to a point in $\Gamma$ whose tangent is parallel to $BC$, namely, $M$. Then, $Q$, $P$, $M$ are collinear. Then, the equality $BQM=MQC$ implies that $QP$ is the internal bisector of $BQC$ and, since $PQR=90^{\circ }$, it follows that $QS$ is the external bisector of $BQC$.

Setting $BC=a$, $CA=b$, $AB=c$, and $2p=a+b+c$, we have that $BP=P^{\prime }C=p-b$, $PC=p-c$ and $P{P}^{\prime }=b-c$ (in case $b>c$; the other case is similar), and, by the angle bisector theorem, we have: $$\frac{BS}{CS}=\frac{BQ}{CQ}=\frac{BP}{CP}=\frac{p-b}{p-c}$$ Then, $$\frac{BS}{BC}=\frac{BS}{CS-BS}=\frac{p-b}{b-c}$$ or, equivalently, $BS=\frac{a(p-b)}{b-c}$.

Let $S$ be the area, $r$ the inradius, and $r_A$ the $A$-exradius of the triangle $ABC$. We have that $RPS=PQS=P^{\prime }J=90^{\circ }$; hence, $PQS=RPS=P^{\prime }JP$. Therefore, $PSR\sim P^{\prime }JP$, which implies that: $$\frac{SP}{RP}=\frac{P^{\prime }J}{P{P}^{\prime }}\text{or, equivalently,}\frac{\frac{a(p-b)}{b-c}+(p-b)}{2r}=\frac{P^{\prime }J}{b-c}$$ and, as a consequence, $$P^{\prime }J=\frac{(a+b-c)(p-b)}{2r}=\frac{(p-c)(p-b)}{S/p}=\frac{p(p-b)(p-c)}{S}=\frac{S}{p-a}=r_A$$ (here, we use that $S=\sqrt{p(p-a)(p-b)(p-c)}$).

Therefore, $J$ is the center of the excircle of $ABC$ relative to the vertex $A$.

Similarly, it can be proved that $P^{\prime }{Q}^{\prime }$ passes through the incenter $I$ of $ABC$. (To do this, it suffices to show that $Q^{\prime }{P}^{\prime }$ is the internal bisector of $B{Q}^{\prime }C$ and that, if $R^{\prime }$ is the point diametrically opposite to $P^{\prime }$ in ${\omega }^{\prime }$, then $R^{\prime }{Q}^{\prime }$ is the external bisector of $B{Q}^{\prime }C$. Then, if $S^{\prime }$ is the intersection point of $R^{\prime }{Q}^{\prime }$ and $BC$ we can show as before that $P^{\prime }IP\sim P^{\prime }{S}^{\prime }{R}^{\prime }$).

Now we will prove that both lines $JP$ and $I{P}^{\prime }$ pass through the midpoint $T$ of the altitude $AH$ of the triangle $ABC$. This finishes the problem, since $T=N$ would be the intersection of the lines $JP=PQ$ and $I{P}^{\prime }=P^{\prime }{Q}^{\prime }$, and $AT$ is perpendicular to $BC$. The homothety with center $A$ that transforms $\omega$ into ${\omega }^{\prime }$, transforms the diameter $RP$ into the diameter $P^{\prime }{R}^{\prime }$. Since $AHP\sim R^{\prime }{P}^{\prime }P$ (since $AH\parallel P^{\prime }{R}^{\prime }$), and $PT$, $PJ$ are medians relative to the corresponding parallel sides $AH$ and $P^{\prime }{R}^{\prime }$, then $T$, $P$ and $J$ are collinear. In addition, since $AH{P}^{\prime }\sim RP{P}^{\prime }$ (because $AH\parallel RP$) and $P^{\prime }I$, $P^{\prime }T$ are medians relative to the corresponding parallel sides $AH$ and $RP$, then $T$, $P^{\prime }$, $I$ are also collinear. Therefore, $JP$ and $I{P}^{\prime }$ both pass through $T$, as we wanted to prove.