X(370)

Created at: Sun, Nov 3, 2024 at 12:26 PM
From: Antreas Hatzipolakis
To: euclid@groups.io, Chris van Tienhoven
Subject: Re: [euclid] Homothetic to Morley

Dear Chris
1. X(5390)
The point X(5390) was listed "coordinates-less"

X(5390) = EULER-MORLEY-ZHAO POINT
Barycentrics (unknown)
Let DEF be the classical Morley triangle. The Euler lines of the three triangles AEF, BFD, CDE appear to concur in a point for which barycentric coordinates remain to be discovered.
Construction by Zhao Yong of Anhui, China, October 2, 2012.

Then you, with your fruitful "fields method", managed to find the trilinears of the point with trigonometric expressions (Hyacinthos 21902)

2. X(370)
Jiang Huanxin proposed in the American Mathematical Monthly the following problem
In triangle ABC find all points P such that the cevian triangle of P is equilateral (my wording)
The problem was solved analytically by David Goering.
(I have scanned the solution and can be found in my blog here CEVIAN TRIANGLES ) Jean-Pierre Ehrmann computed the barycentrics as unique solutions of a system of equations.

I am wondering if the trlinears of the point can be computed with trigonometric expressions, as in the point X(5390) The same for the center of the equilateral triangle in question

Greetings from sunny Athens
APH

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Created at: Sun, Nov 3, 2024 at 11:48 PM
From: Chris van Tienhoven
To: Antreas Hatzipolakis
Subject: RE: Homothetic to Morley

Dear Antreas,

I am not so sure if X(370) is suitable to tackle with Perspective Fields.
In 2010 I corresponded with Francisco about X(370) and in 2012 with Peter Moses.
I had already calculated the coordinates of X(370) and I found there are 3 solutions for the specifications of X(370), two of which can be imaginary. Peter told me there even may be 6 solutions.
He wrote in 2012 to me:
There are 2 sets of 3 solutions, depending on the external Fermat (giving X(370) and a maximum of 2 imaginaries) or internal Fermat (giving 3 reals) construction.
X(370) pertains to the equilateral cevian point that is inside the triangle. Each set of solutions comes from intersecting 3 conics.
As to it being a center .. I think it probably is, but the test is to see if, when symmetrically written, the coordinates remain unchanged under a bicentric exchange. It doesn't necessarily mean that a point is not a center if the coordinates are not symmetric. It may well be possible they can be made so.
See attachment for the one real solution I found in 2012. It is pretty long.
The expression is checked with figures to be correct.
Best regards,
Chris

Chris-solution-X370-in 2012

Mail Antreas P. Hatzipolakis