HYACINTHOS 28760

[Tran Quang Hung]:

 

Dear Mr Antreas,

 

{Qa,Qb,Qc} is Perspective to the 4th Euler triangle [Hyacinthos 28747] can be generalized as follows:

 

Let ABC be a triangle and P a point. 

 

IaIbIc is the excentral triangle.

 

AaAbAc is the pedal triangle of Ia.

 

The reflection of the line IaP in the sides of the triangle AaAbAc bound a triangle which has circumcenter Qa..

 

Define similarly the points Qb and Qc.

 

QaQbQc is an Euler triangle.

 

Then the triangle QaQbQc and the 4th Euler triangle are perspective. 

 

Which is the perspector in terms of P?

 

[Peter Moses]:

 

[...]

------------------------------------------------------------------------

P = X(1740) -> 

=  EULER LINE INTERCEPT OF X(12)X(3743)

a^5*b^2 - a^3*b^4 + a^2*b^5 - b^7 + 2*a^5*b*c - a^3*b^3*c - a*b^5*c + a^5*c^2 + 2*a^3*b^2*c^2 + a^2*b^3*c^2 + 2*b^5*c^2 - a^3*b*c^3 + a^2*b^2*c^3 + 2*a*b^3*c^3 - b^4*c^3 - a^3*c^4 - b^3*c^4 + a^2*c^5 - a*b*c^5 + 2*b^2*c^5 - c^7  :: 

 

= lies on these lines: {2, 3}, {12, 3743}, {80, 1834}, {119, 137}, {1089, 3704}, {1211, 3878}, {1213, 16548}, {3454, 11813}, {5443, 17056}, {20625, 25640}

= {X(5),X(1904)}-harmonic conjugate of X(11113)
= X(26711)-Ceva conjugate of X(523)
------------------------------------------------------------------------

 

CORRECTION:

= EULER LINE INTERCEPT OF X(12)X(986)

Barycentrics    a^5*b^2 - a^3*b^4 + a^2*b^5 - b^7 + 2*a^5*b*c - a^3*b^3*c - a*b^5*c + a^5*c^2 + 2*a^3*b^2*c^2 + a^2*b^3*c^2 + 2*b^5*c^2 - a^3*b*c^3 + a^2*b^2*c^3 + 2*a*b^3*c^3 - b^4*c^3 - a^3*c^4 - b^3*c^4 + a^2*c^5 - a*b*c^5 + 2*b^2*c^5 - c^7 : :

= 5 X[1698] - X[21375]

= lies on these lines: {2, 3}, {12, 986}, {119, 5518}, {517, 2887}, {970, 3454}, {1698, 21375}, {3821, 3822}, {4417, 9567}, {5254, 22380}, {6211, 26446}, {7680, 29243}, {7951, 17596}, {13323, 20083}

= {X(6881),X(30444)}-harmonic conjugate of X(5)