## Τρίτη, 17 Ιανουαρίου 2012

### Point on the Euler Line

Let ABC be a triangle, A1B1C1 the pedal triangle of O = cevian triangle of G, G1G2G3 the circumcevian triangle of G with respect A1B1C1 (ie G1 is the other than A1 intersection of the NPC and AA1 etc), O1O2O3 the circumcevian triangle of O with respect A1A2A3 (ie O1 is the other than A1 intersection of A1O and the NPC etc) and GaGbGc the circumcevian triangle of O with respect G1G2G3 (ie Ga is the other than G1 intersection of OG1 and the NPC etc).

The triangles O1O2O3 and GaGbGc are perspective.

Perspector (on the Euler line of ABC) ?

APH 17 January 2012

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The perspector Q lies on the OH line and satisfies OQ:QH = -2 cosA cosB cosC

The coordinates are:

{(a^2 - b^2 - c^2) (a^4 - b^4 - 2 a^2 b c + 2 b^2 c^2 - c^4) (a^4 -
b^4 + 2 a^2 b c + 2 b^2 c^2 - c^4), -(a^2 - b^2 + c^2) (a^4 -
b^4 - 2 a b^2 c - 2 a^2 c^2 + c^4) (a^4 - b^4 + 2 a b^2 c -
2 a^2 c^2 + c^4), -(a^2 + b^2 - c^2) (a^4 - 2 a^2 b^2 + b^4 -
2 a b c^2 - c^4) (a^4 - 2 a^2 b^2 + b^4 + 2 a b c^2 - c^4)}

This is not in ETC.

Generalization for point P instead of O:

P a point (instead of O)
A1B1C1 = the cevian triangle of G (medial triangle)
P1P2P3 = the circumcevian triangle of P with respect A1B1C1
G1G2G3 = the circumcevian triangle of G with respect A1B1C1
GaGbGc = the circumcevian triangle of P with respect G1G2G3

P1P2P3 and GaGbGc are perspective.
The perspector S is on line GP and satisfies the ratio:

k = GS:SP = (PN^2 - R^2/4) (a^2x+b^2 y+ c^2z)/(x+y+z), where N is the nine point center (and R/2 is the radius of the nine point circle)

Francisco Javier García Capitán
18 January 2012

#### 1 σχόλιο:

1. For a synthetic proof see here:- http://www.artofproblemsolving.com/blog/65848