41st Balkan Mathematical Olympiad

Consider $i$ = the inversion of pole $A$ and $k=\frac{AB\cdot AC}{2}$ followed by the reflection with respect to angle bisector of $\angle BAC$. Denote $X^{\prime }=i(X)$ for any $X$ in the plane. Notice that $B^{\prime }=N$, $C^{\prime }=M$, the midpoints of $AC$ and $AB$ respectively and that Euler's circle of the triangle $ABC$ is $(DMN)$, so it's 'inverse' is the circle $(D^{\prime }{M}^{\prime }{N}^{\prime })$. But now, $M^{\prime }=C$, $N^{\prime }=B$ and $D^{\prime }$ is the circumcenter of $ABC$, denoted by $O$: Indeed, the line $B-D-C$ is sent to the circle $(AN{D}^{\prime }C)$, which is the circle with diameter $AO$. And since $AO$ and $AD$ are isogonals, it follows that $D^{\prime }=O$. Hence the circle $(DMN)$ is sent to $(OBC)$. At the same time circle $ABC$ is sent to the line $MN$. As $PQ\parallel BC$, it follows that arcs $PB$ and $QC$ are equal, so $AP$ and $AQ$ are isogonals, $P^{\prime }=AQ\cap MN$, $Q^{\prime }=AP\cap MN$ and the circle $PDQ$ is sent to $P^{\prime }O{Q}^{\prime }$.

$P_1^{\prime }=Q_1$ and $Q_1^{\prime }=Q_1$ and then the circles ($O{P}_1{Q}_1$) and ($OBC$) are tangent as they are isosceles with $OB=OC$ and $OP=OQ$. Hence ($P_{1}D{Q}_1$) and ($MDN$) are tangent at $D$ and $M\in P_1{Q}_1$. Now ($DPQ$) and ($DMN$) are tangent if and only if ($O{P}^{\prime }{Q}^{\prime }$) and ($OBC$) are tangent, so if and only if the tangent at $O$ to ($OBC$) is the tangent at $O$ to ($O{P}^{\prime }{Q}^{\prime }$) which happens if and only if the triangle $O{P}^{\prime }{Q}^{\prime }$ is isosceles with base $P^{\prime }{Q}^{\prime }$ which is equivalent to $O{P}^{\prime }=O{Q}^{\prime }$. So we have $O{P}^{\prime }=O{Q}^{\prime }$ and also $O{P}_1=O{Q}_1$. It follows that $$\begin{array}{rl}{Q}^{\prime }{P}_1=Q_1{P}^{\prime }& \iff Q^{\prime }{Q}_1^{\prime }=P^{\prime }{P}_1^{\prime }\iff Q{Q}_1\cdot \frac{k}{AQ\cdot A{Q}_1}=P{P}_1\cdot \frac{k}{AP\cdot A{P}_1}\\ & \iff AQ\cdot A{Q}_1=AP\cdot A{P}_1\iff S_{Q{Q}_1}=S_{AP{P}_1}\iff \mathrm{dist}(A,Q{Q}_1)=\mathrm{dist}(A,P{P}_1).\end{array}$$ But this means that $A$ lies on the segment bisector of $P_1{Q}_1$ and $PQ$ respectively. So the minor arcs $AB$ and $AC$ are equal and the triangle is isosceles, contradiction. The conclusion follows now. $\square$