Note that <NPQ=<NBQ=60 so BPQN is cyclic. In similar way CMPQ is cyclic. So 180-<PQM=180-<PCM=180-<ABP=<PQN so <PQM+<PQN=180 so P,Q and R are collinear.

Just use the generalited Simson's line and it's done.

What is the generalized Simson line?

Dear Mathlinkers, when the perpendiculars are remplaced by isoclines i.e. lines making the same angle wrt to the side of the triangle... Sincerely Jean-Louis

Little bit about generalized Simson's line here: http://www.math.uoc.gr/~pamfilos/eGallery/problems/SimsonGeneral.html

It suffices to prove that $\measuredangle PQN = \measuredangle PQM$. Note that aside from $\triangle ABC$, quadrilaterals $QBNP$, $PNMA$, and $PMCQ$ are also cyclic. These are very easy to prove by angle-chasing and with the help of pairs of parallel lines. Now, observe that $\measuredangle PQM = \measuredangle PCM = \measuredangle PBA = \measuredangle PQN$ and we're done.

Attachments:

Let $S=PN\cap BC$ and $T=PM\cap AB$. Note that by construction, $PMCS$ is a parallelogram so $\angle MPS=\angle C=60^{\circ}.$ Similarly, $PTBQ$ is a parallelogram so $\angle PQB=\angle ABC=60^{\circ}.$ Note that $QPMC$ is a trapezoid because $PM\parallel QC$ so $\angle QPM=120^{\circ}.$ Hence, $\angle QPS=60^{\circ}.$ Because $\angle MPN=60^{\circ}=\angle MAN$, $MAPN$ is cyclic. Similarly, $\angle NPQ=60^{\circ}=\angle NBC$ so $NPQB$ is cyclic. Hence, $$\angle MNQ=\angle MNP+\angle PNQ=180^{\circ}-\angle PAM+\angle PBQ=180^{\circ}-\angle PAM+\angle PAM=180^{\circ},$$as desired.$\blacksquare$ Note that in the last equality, we used the fact that $PACB$ is cyclic and $\angle QBA$ is an exterior angle of $\angle B$ by construction.

прикол в том что самая 1 решениялучше всех