Πέμπτη 24 Οκτωβρίου 2019

HYACINTHOS 26194

[Antreas P. Hatzipolakis]
 
Let ABC be a triangle and A'B'C' the pedal triangle of I.

Denote:

A", B", C" = the reflections of A, B, C in I, resp.

Ab, Ac = the orthogonal projections of A" on BI, CI, resp.

Na, N1 = the NPC centers of A"AbAc, IAbAc, resp.
Similarly Nb, N2 and Nc, N3.

The perpendicular bisectors of NaN1, NbN2, NcN3 bound a triangle A*B*C*.
 
1. NaNbNc, N1N2N3 are perspective. Perspector (on the OI line) ?
 
2. ABC, A*B*C* are orthologic.
The orthologic center (ABC, A*B*C*) is the I
Which point is the other orthologic center (A*B*C*, ABC)  (on the OI line)? It is the circumcenter of A*B*C*
 
3. A'B'C', A*B*C* are homothetic.
Homothetic center (on the OI line) ?

4. The I of ABC lies on the Euler line of A*B*C* = Euler line of A'B'C' .
Which point is it wrt triangle A*B*C*?
 
 
[Peter Moses]:
 


Hi Antreas,
 
1). X(9957).
 
2). >Which point is the other orthologic center (A*B*C*, ABC)  (on the OI line)? 
 
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-8 a^4 b c+12 a^3 b^2 c+6 a^2 b^3 c-13 a b^4 c+2 b^5 c-a^4 c^2+12 a^3 b c^2-24 a^2 b^2 c^2+12 a b^3 c^2+b^4 c^2-2 a^3 c^3+6 a^2 b c^3+12 a b^2 c^3-4 b^3 c^3+2 a^2 c^4-13 a b c^4+b^2 c^4+a c^5+2 b c^5-c^6):: 
on lines {{1,3},{5,6736},{8,6939} ,{145,12528},{388,12700},{392, 12245},{519,5777},{946,3880},{ 952,9856},{1071,3241},...}.
midpoint of X(i) and X(j) for these {i,j}: {{145, 12672}, {3057, 7982}, {10284, 11278}}.
reflection of X(i) in X(j) for these {i,j}: {{942,10222},{5836,13464},{ 12645,9947},{12675,3635}}.
{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1,10310,1385).
 
>It is the circumcenter of A*B*C*
Yes.
 
3).
a (a+b-c) (a-b+c) (a^3 b-a^2 b^2-a b^3+b^4+a^3 c-4 a^2 b c+7 a b^2 c-a^2 c^2+7 a b c^2-2 b^2 c^2-a c^3+c^4):: 
on lines {{1,3},{8,12709},{144, 7672},{226,5836},{388,10914},{ 392,1788},{960,4848},{971, 10950},...}.
{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1,13370,1319),(65,2099,942),( 65,3057,57),(65,3340,5173),( 1466,2099,1).
 
 
4). X(235).

Best regards,
Peter Moses.
 

 

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