Σάββατο 25 Μαΐου 2013

EULER LINES

NOTATIONS:

Let La, Lb, Lc be three lines, A,B,C three respective points and 1,2,3 three respective lines.

Denote:

rLa, rLb, rLc = the reflections of La, Lb, Lc in 1,2,3, resp.

pLa, pLb, pLc = the parallels to La, Lb, Lc through A, B, C, resp.

CONCURRENT EULER LINES.

Let ABC be a triangle and A'B'C' the cevian triangle I.

Denote:

Ab,Ac: = the reflections of A' in BB', CC', resp.

Bc,Ba: = the reflections of B' in CC', AA', resp.

Ca,Cb: = the reflections of C' in AA', BB', resp.

1,2,3: = the cevians AA', BB', CC' of I (bisectors)

1. L, La,Lb,Lc: = the Euler lines of ABC, AAbAc, BBcBa, CCaCb, resp.

1.1. La,Lb,Lc concur at the infinity point of L = X30 (ie L, La,Lb,Lc are parallel)

1.2. rLa, rLb, rLc are concurrent.

1.3. prLa, prLb, prLc are concurrent at a point on the circumcircle. The point is the antipode of the Euler line reflection point (ie the point of concurrence of the reflections of the Euler in the sidelines of ABC) = the isogonal conjugate of the infinity point of the Euler line (X30) = X74.

2. Da, Db, Dc: = the Euler lines of the triangles A'AbAc, B'BcBa, C'CaCb, resp.

2.1. Da, Db, Dc are concurrent at I. (I is the common circumcenter of the triangles)

2.2. pDa, pDb, pDc are concurrent.

Antreas P. Hatzipolakis, 24-25 May 2013

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1.2

The lines intersect at:

( a(a^9 - a^8(b+c) - a^7(b-c)^2 + a^6(2b^3-b^2c-b*c^2+2c^3) - a^5(3b^4+b^3c-7b^2c^2+b*c^3+3c^4) + 4a^4b*c(b-c)^2(b+c) + a^3(b^2-c^2)^2(5b^2-4b*c+5c^2) - a^2(b-c)^2(2b^5+5b^4c+b^3c^2+b^2c^3+5b*c^4+2c^5) - a(b^2-c^2)^2(2b^4-3b^3c+5b^2c^2-3b*c^3+2c^4) + (b-c)^4(b+c)^3(b^2+c^2)) : ... : ...),

with (6-9-13)-search number: 5.63864638926896001044233914

Angel Montesdeoca Anopolis, #301

On lines {{1,2779},{21,104},{36,1725},{65,74},{125,860}}

(2 r + R) X[110] - 4 (r + R) X[1385]

2 R X[65] + (2 r + R) X[74]

Peter Moses 27 May 2013

2.2

The lines are concurrent in X(80), reflection of incenter in Feuerbach point.

Angel Montesdeoca Anopolis, #308

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