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Similar with Simson's line !

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#1 h Feb 4, 2015, 2:40 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an equilateral triangle , and $P$ be a point on the circumcircle of the triangle but distinct from $A$ , $B$ and $C$ . The lines through $P$ and parallel to $BC$ , $CA$ , $AB$ intersect the lines $CA$ , $AB$ , $BC$ at $M$ , $N$ and $Q$ respectively .Prove that $M$ , $N$ and $Q$ are collinear .

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Omez

#2 h Feb 4, 2015, 3:19 PM • 2 Y

Y by Adventure10, Mango247

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.

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Radar

#3 h Feb 10, 2015, 1:06 PM • 2 Y

Y by Adventure10, Mango247

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

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DrMath

#4 h Feb 10, 2015, 1:13 PM • 2 Y

Y by Adventure10, Mango247

What is the generalized Simson line?

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9777 posts

jayme

#5 h Feb 10, 2015, 1:58 PM • 3 Y

Y by DrMath, Adventure10, Mango247

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

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Radar

#6 h Feb 11, 2015, 6:28 AM • 1 Y

Y by Adventure10

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

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#11 h Apr 16, 2016, 7:26 AM • 2 Y

Y by Adventure10, Mango247

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.

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#12 h Sep 27, 2018, 1:42 AM • 1 Y

Y by Adventure10

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.

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#13 h Oct 2, 2020, 3:37 AM

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прикол в том что самая 1 решениялучше всех :first:

:first:

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