HYACINTHOS 28942

 [Vu Thanh Tung]:

Dear all, 

 

Let's first define a point on the Euler line.

Definition: Given a triangle XYZ with circumcenter O and orthocenter H.

The α-Euler point of ABC is the point X in the Euler line such that: OX = α OH   (signed distance)

 

 

Then I have a conjecture:

Problem: Let DEF be a pedal triangle of a point P wrt a triangle ABC, α be a real number.

A1 , B1 , C1 are the α-Euler point of triangles AEF , BFD, CDE respectively.

Then 

    (1)    the triangle A1B1C1 are orthologic to both triangles ABC and DEF.

    (2)    when P fixed and α varies the orthologic centers move in  fixed lines .

 

Best regards,

 

Vu Thanh Tung

 

[Angel Montesdeoica]:

 

Dear Vu Thanh Tung,

A particular case:

 Let DEF be the pedal triangle of a point P wrt a triangle ABC.
Oa and Ha are the circumcenter and the orthocenter  of AEF, resp.
Let  α be a real number and  A1 be the point such that: OaA1 = α OaHa. Define B1 and C1 cyclically.

    (1)    The triangle A1B1C1 are orthologic to both triangles ABC and DEF, with orthologic centers V and V', resp.
    (2)    When P fixed and  α varies the orthologic centers move in  fixed lines, L and L'.

   

If P=(u:v:w), then the baricentric equation of L is:

 

Ciclic sum[(b^2 c^2 (b^2-c^2)^3 u (v+w)+a^8 (c^2 v (u+2 v+w)-b^2 w (u+v+2 w))+a^4 (3 c^6 v (u+2 v+w)-3 b^6 w (u+v+2 w)+b^4 c^2 (2 u v+2 v^2-3 u w-3 v w-4 w^2)+b^2 c^4 (3 u v+4 v^2-2 u w+3 v w-2 w^2))+a^6 (-3 c^4 v (u+2 v+w)+3 b^4 w (u+v+2 w)-2 b^2 c^2 (v-w) (u+2 (v+w)))+a^2 (2 b^2 c^6 (u-v) w+2 b^6 c^2 v (-u+w)-c^8 v (u+2 v+w)+b^8 w (u+v+2 w)+b^4 c^4 (v-w) (3 u+2 (v+w))))x]=0.

 

The equation of the L' is:

 

(c^2 u v^2-b^2 u w^2) x+(-c^2 u^2 v+a^2 v w^2) y+(b^2 u^2 w-a^2 v^2 w) z=0.

 

***    When the point P traverses the Euler line of ABC the line L envelope the parabola tangent to Euler line at X(5), tangent to line X(4)X(51) at X(185), and tangent to line X(389)X(546) at X(10095).

Barycentric equation of this parabola:

Cyclic sum [(b-c)^2 (b+c)^2 (a^16+b^12 c^4-4 b^10 c^6+6 b^8 c^8-4 b^6 c^10+b^4 c^12+a^14 (-6 b^2-6 c^2)+a^12 (15 b^4+48 b^2 c^2+15 c^4)+a^10 (-20 b^6-130 b^4 c^2-130 b^2 c^4-20 c^6)+a^8 (15 b^8+160 b^6 c^2+299 b^4 c^4+160 b^2 c^6+15 c^8)+a^6 (-6 b^10-90 b^8 c^2-264 b^6 c^4-264 b^4 c^6-90 b^2 c^8-6 c^10)+a^4 (b^12+16 b^10 c^2+77 b^8 c^4+132 b^6 c^6+77 b^4 c^8+16 b^2 c^10+c^12)+a^2 (2 b^12 c^2+2 b^10 c^4-4 b^8 c^6-4 b^6 c^8+2 b^4 c^10+2 b^2 c^12)) x^2-2 (a-b) (a+b) (a-c) (a+c) (-b^14 c^2+6 b^12 c^4-15 b^10 c^6+20 b^8 c^8-15 b^6 c^10+6 b^4 c^12-b^2 c^14+a^12 (b^4+2 b^2 c^2+c^4)+a^10 (-5 b^6+9 b^4 c^2+9 b^2 c^4-5 c^6)+a^8 (10 b^8-31 b^6 c^2+46 b^4 c^4-31 b^2 c^6+10 c^8)+a^6 (-10 b^10+20 b^8 c^2-10 b^6 c^4-10 b^4 c^6+20 b^2 c^8-10 c^10)+a^4 (5 b^12+6 b^10 c^2-73 b^8 c^4+124 b^6 c^6-73 b^4 c^8+6 b^2 c^10+5 c^12)+a^2 (-b^14-5 b^12 c^2+21 b^10 c^4-15 b^8 c^6-15 b^6 c^8+21 b^4 c^10-5 b^2 c^12-c^14)) y z ] = 0.

The focus is X(125)X(546)∩X(133)X(1594)

= 2 a^14 (b^2+c^2) +a^12 (-6 b^4+8 b^2 c^2-6 c^4) +a^10 (b^6-3 b^4 c^2-3 b^2 c^4+c^6) +a^8 (15 b^8-44 b^6 c^2+62 b^4 c^4-44 b^2 c^6+15 c^8) -a^6 (b^2-c^2)^2 (20 b^6+b^4 c^2+b^2 c^4+20 c^6) +a^4 (b^2-c^2)^2 (8 b^8+31 b^6 c^2-42 b^4 c^4+31 b^2 c^6+8 c^8) +a^2 (b^2-c^2)^4 (b^6-14 b^4 c^2-14 b^2 c^4+c^6) -(b^2-c^2)^6 (b^4+5 b^2 c^2+c^4) : : 

 

= lies on these lines:  {125, 546}, {133, 1594}  

*** If P=X(5) then X(11801) is  the intersection of  L and L'.

 

Best regards,

Angel Montesdeoca