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25607Re: "Hexagon flower'

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  • Antreas Hatzipolakis
    Mar 10, 2017

      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.

      ****************

      Which is the point of concurrence? (see figure)

      I do not know why the attached figure do not appear in the archive.

      Here is it:

      Let ABC be a triangle.

      Construct a regular hexagon outwardly BC. Construct another regular hexagon outwardly the side of the first hexagon which is parallel to BC. Let X be its center. Similarly Y, Z the centers corresponding to CA, AB.

      AX, BY, CZ are concurrent.

      Which is the point of concurrence?

      APH



      [César Lozada]:



      >> I conjecture this problem with 2n-regular polygon but I can't try all the cases.

       

      Suppose you build m adjacent n-regular polygons, where n is even and n>=4, on each side of a triangle ABC (outwardly ABC. In the sketch of the  hexagon flower m=2 and n=6). ABC and the triangle formed by the centers of the last polygons are perspective.      

       

      The perspector Zo(m,n) is:

      Zo(n,m) = 1/xo : 1/yo : 1/zo  (barycentrics)

       

      where

       

      xo = SA*SW*cos(θ)^2*(2*m-1)^2+S*(3* SA+SW)*(2*m-1)*sin(θ)*cos(θ)+ 3*sin(θ)^2*S^2)

       

      and θ = π/n.

       

      If the polygons are built inwardly ABC, perspector Z(m,n) is:

      Zi(n,m) = 1/xi : 1/yi : 1/zi  (barycentrics)

       

      where

       

      xi = SA*SW*cos(θ)^2*(2*m-1)^2-S*(3* SA+SW)*(2*m-1)*sin(θ)*cos(θ)+ 3*sin(θ)^2*S^2)

       

      and θ = π/n.


      Zi(n,m) = (S->-S in Zo(n,m) )

       

      Both perspectors lie on the Kiepert hyperbola.

       

      Examples (all barycentrics):

       

      One-square:  Zo(4,1) = X(485) and Zi(4,1) = X(486)

      Two-squares:  Zo(4,2) = X(1327) and Zi(4,2) = X(1328)

      Three-squares:

      Zo(2,3) = 1/(S+5*SA) : :

      = on Kiepert hyperbola and lines: {5,6434}, {372,3591}, {382,485}, {486,546}, {550,10195}, {1131,6561}, {1132,6436}, {1152,11737}, {1327,6470}, {1328,3070}, …

      = [ -1.786974027736880, -1.94145489754157, 5.809505885314797 ]

       

      Zi(2,3) = 1/(-S+5*SA) : :

      = on Kiepert hyperbola and lines: {5,6433}, {371,3590}, {382,486}, {485,546}, {550,10194}, {1131,6435}, {1132,6560}, {1151,11737}, {1327,3071}, {1328,6471},…

      = [ 74.077586938299460, 91.68749232910599, -94.024947255919540 ]

       

      ---------------

       

      One-hexagon:  Zo(6,1) = X(13) and Zi(6,1) = X(14)

      Two-hexagons: Zo(6,2) = 1/(S+3*sqrt(3)*SA) and Zi(6,2) = 1/(-S+3*sqrt(3)*SA) (See https://groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/25606)

       

      Three-hexagons:

      Zo(6,3) = 1/(S+5*sqrt(3)*SA) : : 

      = on Kiepert hyperbola and lines: {17,382}, {18,546}, {383,11669}, {550,10188},…

      = [ -2.807407424964899, -3.11998814576167, 7.096382778957028 ]

       

      Zi(6,3) = 1/(-S+5*sqrt(3)*SA) : : 

      = on Kiepert hyperbola and lines: {17,546}, {18,382}, {550,10187}, {1080,11669},…

      = [ -18.304176959068590, -21.91580668343714, 27.261227705395600 ]

       

      --------------

       

      One-octagono:   Zo(8,1) = X(3387) and Zi(8,1) = X(3374)

       

      Two-octagonos:

      Zo(8,2) = 1/((3*(1+sqrt(2)))*SA+S) : :

      = on Kiepert hyperbola and lines: {30,3373}, {381,3388},…

      = [ -2.483600128990801, -2.74289044676768, 6.685865619973145 ]

       

      Zi(8,2) = 1/((3*(1+sqrt(2)))*SA-S) : :

      = on Kiepert hyperbola and lines: {30,3388}, {381,3373},…

      = [ -28.515834338462180, -34.43416988067517, 40.640859478588200 ]

       

      Regards,

      César Lozada

       



       




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