- 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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