SILK ROAD MATHEMATICAL COMPETITION

We will make several uses of the fact that the lines connecting a vertex of a triangle with the circumcentre and the orthocentre are symmetrical with respect to the angle bisector at this vertex. Let $AC>BC$, $A{A}_1$ and $B{B}_1$ be the altitudes of the triangle $ABC$, $O$ its circumcentre. The points $A_1$, $B_1$ and $R$ lie on the circle with diameter $CH$. The triangles $A_1{B}_{1}C$ and $ABC$ are similar, and their similarity ratio is $\frac{A_{1}C}{AC}=\cos C$. Then the ratio of the circumradii of these triangles is also $\cos C$, that is, $\frac{CN}{AO}=\cos C$, hence $CN=AO\cos C=OM$. Since $OM\parallel CN$, $COMN$ is a parallelogram. Let $\angle ACO=\angle BCH=x$, $\angle OCM=\angle NMC=y$, $\angle RCT=\angle RNT=z$. Then, since $NC=NR$, we have $z+\angle TCN=\angle RCN=\angle CRN=y+z$, whence, $\angle TCN=y$ and the lines $CT$ and $CM$ are symmetrical with respect to the bisector of $ACB$. Since $\angle AH{B}_1=\angle ACB$ and $\angle AHP=RH{B}_1=\angle B_{1}CR=\angle BCT$, we have $\angle RCT=\angle RHP$, that is, $C$, $R$, $Q$ and $H$ belong to the same circle with diameter $CH$ (here $Q=CT\cap PH$). The desired perpendicularity follows immediately.

Note. We have proved in this solution that $CT$ lies on the symmedian of the triangle $ABC$.



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Alternative solution.

Let $A{A}_1$, $B{B}_1$, and $C{C}_1$ be the altitudes of the triangle $ABC$, and $T_1$ the midpoint of $A_1{B}_1$. Then $M{A}_1=M{B}_1=AB/2$, $N{A}_1=N{B}_1=CH/2$, therefore $MN$ is the perpendicular bisector of $A_1{B}_1$. We will prove that $T_1=T$. Since $CM$ and $C{T}_1$ are respective medians in similar triangles $ABC$ and $A_1{B}_{1}C$, while $AB$ and $A_1{B}_1$ are anti-parallel, $C{T}_1$ lies on the symmedian $CS$ of the triangle $ABC$. Let $\angle NRC=\angle RCN=\delta$. Then $90^{\circ }-\delta =\angle CMB=\angle C{T}_1{B}_1=\angle S{T}_1{A}_1$, hence $\delta =\angle M{T}_{1}S=\angle C{T}_{1}N$, that is, $\angle NRC=\angle C{T}_{1}N$, and it follows that $T_1$ belongs to the circumcircle of the triangle $CNR$. Next, as in the first solution, we can prove $HP\perp CT$. The proof is indeed complete.