Πέμπτη 31 Οκτωβρίου 2019

ADGEOM 3874 * ADGEOM 3878

#3874
 
Dear geometers,
 

Let ABC be a triangle with three Feuerbach points Fa,Fb,Fc.

 
H is orthocenter.
 
Fha is H-Feuebrach point of triangle HBC.
 
Fhb is H-Feuebrach point of triangle HCA.
 
Fhc is H-Feuebrach point of triangle HAB.
 
FaFha meets BC at A'. Similarly, we have B',C'.
 
Then the lines AA',BB',CC' are concurrent. Which is this point ?
 
Best regards,
Tran Quang Hung.
 
------------------------------------------------------

#3878
 
Dear Tran Quang Hung

The lines AA',BB',CC' are concurrent at

W =  ( 2 a^6-a^5 (b+c)-a^4 (3 b^2+2 b c+3 c^2)+2 a^3 (b^3+b^2 c+b c^2+c^3)-a (b-c)^2 (b+c)^3+(b-c)^2 (b+c)^4+4 (-a^4+a^3 (b+c)-a^2 (b+c)^2-a (b-c)^2 (b+c)+2 (b^2-c^2)^2) r s :  ... : ...).

W is the midpoint of X(176) and X(10405).
 
 W lies on lines X(i)X(j) for these {i, j} : {1,1336}, {2,175}, {4,9}, {176,10405}, {189,13389}, {219,1378},{226,13459}, {278,3535}, {388,6204, {497,7347}, {637,1944}, {1123,1785}, {1146,3070}, .....
 
   (6 - 9 - 13) - search numbers  of W : (7.51986707668436, 5.97337161677889, -3.96545451894763).
  
   Best regards,
   Angel Montedeoca
 
 

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου