HYACINTHOS 29373

[Tran Quang Hung]:

 

Let ABC be a triangle and P any point.

Ha, Hb, Hc are the orthocenters of triangles PBC, PCA, PAB resp.

Then the triangles AHbHc, BHcHa, CHaHb have the same orthocenter.

Is this theorem known before and which is this common orthocenter in terms of P?

 

[APH]:

Which is the locus of P such that the common orthocenter lies on the Euler line ?

 

[Peter Moses]:


Hi Antreas,

>Then the triangles AHbHc, BHcHa, CHaHb have the same orthocenter.

The Kirikami-Euler image of P{p,q,r}.
1/(q r ((a^2-b^2-c^2) p^2+(a^2-b^2+c^2) p q+(a^2+b^2-c^2) p r+2 a^2 q r)) : :

Example P = X(113) -->

KIRIKAMI-EULER IMAGE OF X(113) = 

= (2*a^4 - a^2*b^2 - b^4 - a^2*c^2 + 2*b^2*c^2 - c^4)*(a^4*b^2 - 2*a^2*b^4 + b^6 + a^4*c^2 + 2*a^2*b^2*c^2 - b^4*c^2 - 2*a^2*c^4 - b^2*c^4 + c^6)^2 : :

= lies on these lines: {2,3}, {3258,23097}, {3580,14264}, {15454,16319}

= X(403)-Ceva conjugate of X(113).
= crossdifference of every pair of points on line {647, 15470}.
= barycentric product X(113)*X(3580).
= barycentric quotient X(i)/X(j) for these {i,j}: {113, 2986}, {3003, 10419}

>Which is the locus of P such that the common orthocenter lies on the Euler line ?

K059

 

Best regards,
Peter Moses.