HYACINTHOS 26283

[APH]:

 

Dear Mr rodinos (*), I have two new problems (and synthetic proofs):

Probelm 1. Let Ia, Ib, Ic  be A, B, C-excenters of a triangle ABC, resp.
Let Ka, Kb, Kc  be the incenters of triangles IaBC, IbCA, IcAB, resp.
Then the Euler lines of the triangles IaKbKc, IbKcKa, IcKaKb  are concurrent. 

Which is the intersection point ?

Problem 2. Let Ia, Ib, Ic  be A, B, C-excenters of a triangle ABC, resp.

Let Jab, Jac, Jbc, Jba, Jca, Jcb  be the incenters of the triangles ABIb, ACIc, BCIa, BAIa, CAIa, CBIb, resp.
Let Ka = JbcJba /\ JcbJca, Kb = JcaJcb /\ JacJab, Kc = JabJac /\ JbaJbc.
Then, the Euler lines of the triangles IaKbKc, IbKcKa, IcKaKb are concurrent. 

Which is the intersection point ?

Thank you very much.
Sincerely
Le Viet An

 

(*) rodinos is my nickname in AoPS

[Peter Moses]:

 

Hi Antreas,

 

1). X(10023).

 

2). on {40,164}.

 

I do have some barys, but they do seem rather treacherous and simplifications, if any are to be hard, would probably take an age!

 

Best regards,

Peter Moses.

 

[Seiichi Kirikami]:

Dear Antreas, dear Peter,

 

As for X(10023)(Problem 1), I feel very awkward because it has no name and no coordinates in ETC. I tried to compute them several times, but I met “no memory available” messages or the computation simply did not finish. I suppose that Peter has the coordinates of X(10023), but it is too long for ETC.

 

Best regards, Seiichi Kirikami.

 

 

[Peter Moses]:

Hi Antreas and Seiichi,

 

It so happens that X(10023) manages to simplify quite considerably to reveal barys of:

2 a^2 (-a^2+b^2+c^2)+(-a+b+c) ((a+b-c) Sqrt[a b (-a+b+c) (a-b+c)]+(a-b+c) Sqrt[a c (-a+b+c) (a+b-c)])::

 

Actually, I found this initially by recasting the problem in terms of the orthic triangle (now, ABC is the excentral to the orthic). We end up with this center:

 

P = a (a^5 b-a^4 b^2-2 a^3 b^3+2 a^2 b^4+a b^5-b^6+a^5 c-2 a^2 b^3 c-a b^4 c+2 b^5 c-a^4 c^2-4 a^2 b^2 c^2+b^4 c^2-2 a^3 c^3-2 a^2 b c^3-4 b^3 c^3+2 a^2 c^4-a b c^4+b^2 c^4+a c^5+2 b c^5-c^6):: 

on lines {{1,3},{5,1858},{7,10629},{10, 343},{12,912},{72,498},{90, 6913},{226,5884},{377,5086},{ 381,1898},{405,920},{407,1785} ,{442,1737},{499,5439},{518, 10039},{758,13411},{960,7483}, {971,3585},{974,2779},{1006, 7098},{1046,3074},{1064,1393}, {1071,1478},{1210,5883},{1254, 4303},{1479,12711},{1770,9943} ,{1776,6920},{1781,2182},{ 1788,6889},{1837,6917},{1864, 10826},{1905,4185},{2252,2294} ,{2771,8068},{3085,3868},{ 3485,6833},{3486,6934},{3487, 10321},{3555,12647},{3583, 5806},{3753,5794},{3827,5135}, {3869,6910},{4295,6836},{4299, 10167},{4333,5918},{5219,5693} ,{5530,10974},{5691,12671},{ 5728,5880},{5777,7951},{5887, 6862},{6001,6831},{6738,10122} ,{7354,13369},{7686,10391},{ 9612,12664},{10043,11045},{ 10106,12005},{10107,10609},{ 10320,11374},{10590,12528},{ 10948,11019},{11013,11032},{ 11015,11020},{11570,13407}}.

Midpoint of X(65) and X(2646).

X[35] + 3 X[5902], 3 X[354] - X[11011].

{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1,46,11507),(1,57,8071),(1, 942,5570),(46,3612,165),(46, 5902,65),(942,9957,6583),( 7686,10391,10572),(10122, 12736,6738).

Now, X(10023) = P of the excentral triangle.

 

Best regards,

Peter Moses.