HYACINTHOS 25606

From "Romantics of Geometry" #574

[Tran Quang Hung‎]

Hexagon flower, I inspire this problem from a problem of Mr Kostas Vittas on AoPS quite long time ago.
Let ABC be a triangle. Construct out side ABC four regular hexagons as in the figure. QM cuts RN at P. Prove that A,P,X are collinear.

[APH]:

Similarly there are Y, Z corresponding to sides CA, AB, resp.
I am wondering if AX, BY, CZ are concurrent.

[Tran Quang Hung‎]

Yes you are right, dear Mr Antreas :) beautiful flower with a triangle and six hexagons. I think this is a consequence of Kiepert points.
I conjecture this problem with 2n-regular polygon but I can't try all the cases.

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Which is the point of concurrence? (see figure)

APH

 

[César Lozada]:

 

Z = 1/(S+3*sqrt(3)*SA) : :  (Barycentrics)

= 7*X(17)-4*X(5238) = X(17)+8*X(5350) = X(5238)+14*X(5350)

= on Kiepert hyperbola and these lines: {2,10646}, {3,10188}, {5,10187}, {13,3830}, {14,3845}, {16,5066}, {17,30}, {18,381}, {62,3839}, {98,5470}, {383,7608}, {395,3860}, {531,11122}, {532,5487}, {542,11602}, {671,6778}, {1080,7607}, {2043,10195}, {2044,10194}, {3412,3627}, {5071,5237}, {5485,5863}, {8781,9116}, {10159,11303}

= [ -1.860007742164223, -2.02472583763082, 5.900862865881513 ]

If hexagons are built inward ABC, then:

 

Z’ = 1/(-S+3*sqrt(3)*SA) : :  (Barycentrics)

7*X(18)-4*X(5237) = X(18)+8*X(5349) = X(5237)+14*X(5349)

= on Kiepert hyperbola and these lines: {2,10645}, {3,10187}, {5,10188}, {13,3845}, {14,3830}, {15,5066}, {17,381}, {18,30}, {61,3839}, {98,5469}, {383,7607}, {396,3860}, {530,11121}, {533,5488}, {542,11603}, {671,6777}, {1080,7608}, {2043,10194}, {2044,10195}, {3411,3627}, {5071,5238}, {5485,5862}, {8781,9114}, {10159,11304}

= [ 126.149931257987800, 155.65192057981470, -162.341402653958600 ]

 

César Lozada