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

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

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

CEVIAN TRIANGLES

Jiang Huanxin and David Goering, EquilateralCevian Triangles
The American Mathematical Monthly
Vol. 104, No. 6 (Jun. - Jul., 1997), pp. 567-570
David Goering
X(370) = EQUILATERAL CEVIAN TRIANGLE POINT

Mowaffaq Hajja, The Arbitrariness of the Cevian Triangle
The American Mathematical Monthly
Vol. 113, No. 5 (May, 2006), pp. 443-447
Mowaffaq Hajja

Mail Antreas P. Hatzipolakis

FEUERBACH POINT PROBLEM

Lemma 1

Let ABC be a triangle. The Oi line intersects AB at C'.The circle (Oa) with diameter AI intersects the circumcircle again at A". The line C'A" intersects (Oa) again at Ca.Ca is the orthogonal projection of A on IC.

Lemma 2

Let ABC be a triangle and Ba, Ca the orthogonal projeections of A on IB, IC, resp. The line BaCa intersects AC, AB at B'a, C'a, resp. B'a,C'a are the midpoints of AC, AB, resp.

Lemma 3

Let ABC be a triangle. Denote C' = the midpoint of AB, Ma, Mb = the midpoints of AI, BI, resp. and Ba = the orthogonal projection of A on BI. The points Ma, C', Mb, Ba are concyclic. The circle is the NPC of AIB.

Problem
By David Nguyen

Rewritten with cyclic notation
Let ABC be a triangle. The OI line intersects BC, CA, AB at A', B', C', resp.  
Denote
(Oa) = the circle with diameter AI
A" = the 2nd intersection of the circumcircle and (Oa)
Ba = the 2nd intersection of B'A" and (Oa)
Ca = the 2nd intersection of C'A" and (Oa)
B'a = BaCa ∩ AC  , C'a = BaCa ∩ AB

(Oab) = the circumcircle of OaB'aCa
(Oac)= the circumcircle of OaBaC'a
The 2nd intersection of (Oab), (Oac) is the Feuerbach point

Proof
By the lemmata, the problem is equivalent to the following problem
Let ABC be a triangle and A'B'C' the medial triangle
Denote
A", B", C" = the midpoints of IA, IB, IC, resp.
The circumcircles of A'B"C", B'C"A", C'A"B" concur at the Feuerbach point
True, since the circumcircles of A'B"C", B'C"A", C'A"B" are the NPCs of IBC ICA, IAB resp. and they concur at the Poncelet point of ABCI = the Feuerbach point.

Mail Antreas P. Hatzipolakis

PANAKIS' PSEUDOISOSCELES TRIANGLE

Let ABC be a trangle and D1, D2, D3 the feet of the internal angle bisectors [D1D2D3 = the cevian triangle of the incenter I]
Prove that the triangle D1D2D3 can be isosceles without ABC being isosceles.
Ioannis Panakis, Plane Trigonometry, vol. B, Athens (1973), p. 110 [in Greek]

I call this triangle ABC as Panakis pseudoisosceles triangle

Properties of ABC (in the same book pp. 109-111)
1. The A, D1, D2, D3 are concyclic [the cevian circle of I passes through A]
2. (a + b + c)*(-a^2 + b^2 + c^2) + abc = 0
3. a / (b + c) = b / (c + a) + c / (a + b)
4. (r1 - r) / (r1 + r) = ((r2 - r) / (r2 + r)) + ((r3 - r) / (r3 + r))
where r = the inradius and r1, r2, r3 the exradii.

PDF of the pages of the book Panakis

Mail Antreas P. Hatzipolakis

THEBAULT'S PSEUDOISOSCELES TRIANGLE

Victor Thebault published the following theorem as an exercise:

Si le cercle qui passe par les pieds des bissectrices intérieures d'un triangle est tangent à l'un des côtés, le triangle est isocèle, et réciproquement. 
(If the circle passing through the feet of the interior bisectors of a triangle is tangent to one of the sides, the triangle is isosceles, and vice versa.)

Solution by (A.M.)  [false]
Journal de mathématiques élémentaires.  
75e Annee - No 1 -  1er Octobre 1950, p. 3, #14250

Joseph Andersonn proved that the triangle is not necessarily isosceles.
CERCLE PASSANT PAR LES PIEDS DES BISSECTRICES INTÉRIEURES D'UN TRIANGLE ET TANGENT À L'UN DES CÔTÉS
par A. Monjallon. 
Journal de mathématiques élémentaires.  
75e Annee - No 20, 15 Juillet 1951, pp. 153 - 4 - 

PDF File Victor Thebault

Francisco Javier García Capitán's Solution (in Spanish)
wrong-thebault

Mail Antreas P. Hatzipolakis

ETC

X(66089) = PERSPECTOR OF THESE TRIANGLES: 1st MORLEY AND HATZIPOLAKIS-VAN TIENHOVEN EQUILATERAL TRIANGLE

Trilinears    Cos[A] + (Cos[A/3] + 2 Cos[B/3] Cos[C/3]) (1 + 2 Sin[(2 A)/3 - Pi/6] + 2 Sin[(2 B)/3 - Pi/6] + 2 Sin[(2 C)/3 - Pi/6])

See Antreas Hatzipolakis and Chris van Tienhoven, euclid 7157.

X(66089) lies on this line: {3, 356}

---

FJGC A confluence of concepts

Francisco Javier García Capitán,A confluence of concepts
FJGC