HYACINTHOS 29196

[Antreas P. Hatzipolakis]:


Let ABC be a triangle and A'B'C' the antipedal traungle of O (tangential triangle)

Let Oa be the center of the circle passing through B', C' and tangent to circumcircle.

Similarly Ob, Oc

The centroid of OaObOc lies on the Euler line of ABC.

Point?


[Angel Montesdeoca]:


***   The centroid of OaObOc is

W = X(2)X(3) ∩ X(524)X(10282)  

 = -10 a^10+21 a^8 (b^2+c^2)-2 a^6 (b^4+4 b^2 c^2+c^4)-4 a^4 (5 b^6-b^4 c^2-b^2 c^4+5 c^6)+4 a^2 (b^2-c^2)^2 (3 b^4+b^2 c^2+3 c^4)-(b^2-c^2)^4 (b^2+c^2) : :

= lies on these lines: {2,3}, {524,10282}, {5907,32267}, {6146,32225}, {10182,29181}.

= midpoint of X(i) and X(j), for these {i, j}:  {549,9909}, {10154,14070}

= reflection of X(5066) in X(10201)

 (6 - 9 - 13) - search numbers  of W: (-0.188644917013375, -1.05766718985670, 4.45996249812211).
 
 Angel Montesdeoca