49th Mathematical Olympiad in Ukraine

Let us draw the line $l\parallel BC$ through the point $A$ and denote $W=BP\cap l$. Then

$\angle BPC=180^{\circ }-(\angle PBC+\angle PCB)=$ $180^{\circ }-(\angle PCA+\angle PCB)=180^{\circ }-\angle BCA\Rightarrow$ $\angle CPW=\angle CAW$. And so points $A,P,C,W$ are cyclic. Thus $\angle AWC=180^{\circ }-\angle APC$, $\angle APC=180^{\circ }-(\angle PAC+\angle PCA)=$ $180^{\circ }-(\angle PAC+\angle PAB)=180^{\circ }-\angle BAC$, and it follows that (fig.17) $\angle AWC=$ $180^{\circ }-\angle APC=\angle BAC$. Now $AW\parallel CB$ implies that $\angle WAC=\angle BCA$. And so $△ABC\sim △ACW$. Since $M,N$ are the midpoints of the corresponding sides of similar triangles we have that $\angle WNA=\angle AMC\Rightarrow$ $\angle WNA+\angle ANP=\angle AMC+\angle AMB=180^{\circ }$. And so $B,P,N,W$ lie on the same line. Therefore $\angle BNA=\angle BMA$, which implies that $A,B,M,N$ are cyclic. Since $MN\parallel AB$ as the centerline, $ABMN$ is an isosceles trapezoid, whence $AN=BM\Rightarrow AC=BC$, what was to be proved.

Fig.17

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Alternative solution.

Let $Q$ be a point of the median $AM$ such that $\angle QCB=\angle PCA$. Since $\angle QMB=\angle PNA$ we have that $\angle QMC=\angle PNC$ and so $△PNC\sim △QMC$. From this similarity $\frac{PC}{CQ}=\frac{CN}{CM}=\frac{BC}{AC}$ and thus $△QCB\sim △PCA$. Then we can obtain that (fig.18) $\angle BQC=\angle APC=$ Fig.18

$\angle PAC-\angle PCA=180^{\circ }-\angle PAC+\angle PAB-\angle PCA=180^{\circ }-\angle BAC$. Now let $Q^{\prime }$ be symmetric to $Q$ with respect to $M$. Then $BQC{Q}^{\prime }$ is a parallelogram and $\angle B{Q}^{\prime }C=\angle BQC=180^{\circ }-\angle BAC$. From this it also follows that the quadrilateral $AB{Q}^{\prime }C$ is cyclic and so $\angle MAC=\angle Q^{\prime }AC=\angle Q^{\prime }BC=\angle OBC$. Denote $T=NP\cap BC$. Then the triangles $NTC$ and $MAC$ are similar and thus $\angle PTC=\angle NTC=\angle MAC=\angle PBC$. Points $T$ and $B$ lie on the circumcircle of the triangle $PBC$, and also on the line $BC$. Therefore $T$ coincides with one of the points $B$ or $C$. It is evident that $T$ cannot coincide with $C$, so $T=B$, which means that the points $N,P$ and $B$ lie on a line. Since $\angle ANB=\angle ANP=\angle AMB$, points $A,N,M$ and $B$ are cyclic, and it follows that $CN\cdot NA=CM\cdot CB$, which implies that $CA=CB$, i.e. that $ABC$ is an isosceles triangle.