Let ABC be a triangle and P a point. Let (O1),(O2), (O3) be the reflections of the circumcircle (O) in BC, CA, AB, resp.
Let (O11), (O22), (O33) be the reflections of (O1), (O2), (O3) in AP, BP, CP, resp., L1,L2,L3 the radical axes of [(O22),(O33)], [(O33),(O11)], [(O11), (O22)], resp. and M1,M2,M3 the parallels to L1,L2,L3 through A,B,C resp.
Which is the locus of P such that M1,M2,M3 are concurrent?
APH, 5 December 2011
***********************************
Locus:
(-a - b + c) (a - b + c) (-a + b + c) (a + b + c) (x + y +
z) (-a^4 c^4 x^3 y^2 + a^2 b^2 c^4 x^3 y^2 + 2 a^2 c^6 x^3 y^2 +
b^2 c^6 x^3 y^2 - c^8 x^3 y^2 - a^2 b^2 c^4 x^2 y^3 +
b^4 c^4 x^2 y^3 - a^2 c^6 x^2 y^3 - 2 b^2 c^6 x^2 y^3 +
c^8 x^2 y^3 - a^6 b^2 x^3 y z + 3 a^4 b^4 x^3 y z -
3 a^2 b^6 x^3 y z + b^8 x^3 y z + a^6 c^2 x^3 y z -
b^6 c^2 x^3 y z - 3 a^4 c^4 x^3 y z + 3 a^2 c^6 x^3 y z +
b^2 c^6 x^3 y z - c^8 x^3 y z - a^8 x^2 y^2 z +
2 a^6 b^2 x^2 y^2 z - 2 a^2 b^6 x^2 y^2 z + b^8 x^2 y^2 z +
2 a^6 c^2 x^2 y^2 z - 2 b^6 c^2 x^2 y^2 z - a^4 c^4 x^2 y^2 z +
b^4 c^4 x^2 y^2 z - a^8 x y^3 z + 3 a^6 b^2 x y^3 z -
3 a^4 b^4 x y^3 z + a^2 b^6 x y^3 z + a^6 c^2 x y^3 z -
b^6 c^2 x y^3 z + 3 b^4 c^4 x y^3 z - a^2 c^6 x y^3 z -
3 b^2 c^6 x y^3 z + c^8 x y^3 z + a^4 b^4 x^3 z^2 -
2 a^2 b^6 x^3 z^2 + b^8 x^3 z^2 - a^2 b^4 c^2 x^3 z^2 -
b^6 c^2 x^3 z^2 + a^8 x^2 y z^2 - 2 a^6 b^2 x^2 y z^2 +
a^4 b^4 x^2 y z^2 - 2 a^6 c^2 x^2 y z^2 - b^4 c^4 x^2 y z^2 +
2 a^2 c^6 x^2 y z^2 + 2 b^2 c^6 x^2 y z^2 - c^8 x^2 y z^2 -
a^4 b^4 x y^2 z^2 + 2 a^2 b^6 x y^2 z^2 - b^8 x y^2 z^2 +
2 b^6 c^2 x y^2 z^2 + a^4 c^4 x y^2 z^2 - 2 a^2 c^6 x y^2 z^2 -
2 b^2 c^6 x y^2 z^2 + c^8 x y^2 z^2 - a^8 y^3 z^2 +
2 a^6 b^2 y^3 z^2 - a^4 b^4 y^3 z^2 + a^6 c^2 y^3 z^2 +
a^4 b^2 c^2 y^3 z^2 + a^2 b^6 x^2 z^3 - b^8 x^2 z^3 +
a^2 b^4 c^2 x^2 z^3 + 2 b^6 c^2 x^2 z^3 - b^4 c^4 x^2 z^3 +
a^8 x y z^3 - a^6 b^2 x y z^3 + a^2 b^6 x y z^3 - b^8 x y z^3 -
3 a^6 c^2 x y z^3 + 3 b^6 c^2 x y z^3 + 3 a^4 c^4 x y z^3 -
3 b^4 c^4 x y z^3 - a^2 c^6 x y z^3 + b^2 c^6 x y z^3 +
a^8 y^2 z^3 - a^6 b^2 y^2 z^3 - 2 a^6 c^2 y^2 z^3 -
a^4 b^2 c^2 y^2 z^3 + a^4 c^4 y^2 z^3) (-a^2 c^4 x^4 y^2 +
b^2 c^4 x^4 y^2 + c^6 x^4 y^2 - c^6 x^3 y^3 + a^2 c^4 x^2 y^4 -
b^2 c^4 x^2 y^4 + c^6 x^2 y^4 + a^6 x^4 y z - 2 a^4 b^2 x^4 y z +
a^2 b^4 x^4 y z - 2 a^4 c^2 x^4 y z + 2 b^4 c^2 x^4 y z +
a^2 c^4 x^4 y z + 2 b^2 c^4 x^4 y z + 2 a^6 x^3 y^2 z -
3 a^4 b^2 x^3 y^2 z + b^6 x^3 y^2 z - 4 a^4 c^2 x^3 y^2 z +
4 a^2 b^2 c^2 x^3 y^2 z - 6 b^2 c^4 x^3 y^2 z + 2 c^6 x^3 y^2 z +
a^6 x^2 y^3 z - 3 a^2 b^4 x^2 y^3 z + 2 b^6 x^2 y^3 z +
4 a^2 b^2 c^2 x^2 y^3 z - 4 b^4 c^2 x^2 y^3 z -
6 a^2 c^4 x^2 y^3 z + 2 c^6 x^2 y^3 z + a^4 b^2 x y^4 z -
2 a^2 b^4 x y^4 z + b^6 x y^4 z + 2 a^4 c^2 x y^4 z -
2 b^4 c^2 x y^4 z + 2 a^2 c^4 x y^4 z + b^2 c^4 x y^4 z -
a^2 b^4 x^4 z^2 + b^6 x^4 z^2 + b^4 c^2 x^4 z^2 +
2 a^6 x^3 y z^2 - 4 a^4 b^2 x^3 y z^2 + 2 b^6 x^3 y z^2 -
3 a^4 c^2 x^3 y z^2 + 4 a^2 b^2 c^2 x^3 y z^2 -
6 b^4 c^2 x^3 y z^2 + c^6 x^3 y z^2 + 3 a^6 x^2 y^2 z^2 -
3 a^4 b^2 x^2 y^2 z^2 - 3 a^2 b^4 x^2 y^2 z^2 +
3 b^6 x^2 y^2 z^2 - 3 a^4 c^2 x^2 y^2 z^2 -
6 a^2 b^2 c^2 x^2 y^2 z^2 - 3 b^4 c^2 x^2 y^2 z^2 -
3 a^2 c^4 x^2 y^2 z^2 - 3 b^2 c^4 x^2 y^2 z^2 +
3 c^6 x^2 y^2 z^2 + 2 a^6 x y^3 z^2 - 4 a^2 b^4 x y^3 z^2 +
2 b^6 x y^3 z^2 - 6 a^4 c^2 x y^3 z^2 + 4 a^2 b^2 c^2 x y^3 z^2 -
3 b^4 c^2 x y^3 z^2 + c^6 x y^3 z^2 + a^6 y^4 z^2 -
a^4 b^2 y^4 z^2 + a^4 c^2 y^4 z^2 - b^6 x^3 z^3 + a^6 x^2 y z^3 -
6 a^2 b^4 x^2 y z^3 + 2 b^6 x^2 y z^3 + 4 a^2 b^2 c^2 x^2 y z^3 -
3 a^2 c^4 x^2 y z^3 - 4 b^2 c^4 x^2 y z^3 + 2 c^6 x^2 y z^3 +
2 a^6 x y^2 z^3 - 6 a^4 b^2 x y^2 z^3 + b^6 x y^2 z^3 +
4 a^2 b^2 c^2 x y^2 z^3 - 4 a^2 c^4 x y^2 z^3 -
3 b^2 c^4 x y^2 z^3 + 2 c^6 x y^2 z^3 - a^6 y^3 z^3 +
a^2 b^4 x^2 z^4 + b^6 x^2 z^4 - b^4 c^2 x^2 z^4 +
2 a^4 b^2 x y z^4 + 2 a^2 b^4 x y z^4 + a^4 c^2 x y z^4 +
b^4 c^2 x y z^4 - 2 a^2 c^4 x y z^4 - 2 b^2 c^4 x y z^4 +
c^6 x y z^4 + a^6 y^2 z^4 + a^4 b^2 y^2 z^4 - a^4 c^2 y^2 z^4)
Francisco Javier García Capitán
5 December 2011
Εγγραφή σε:
Σχόλια ανάρτησης (Atom)
REGULAR POLYGONS AND EULER LINES
Let A1A2A3 be an equilateral triangle and Pa point. Denote: 1, 2, 3 = the Euler lines of PA1A2,PA2A3, PA3A1, resp. 1,2,3 are concurrent. ...
-
Let ABC be a trangle and D1, D2, D3 the feet of the internal angle bisectors [D1D2D3 = the cevian triangle of the incenter I] Prove that th...
-
Thanasis Gakopoulos - Debabrata Nag, Morley Theorem ̶ PLAGIOGONAL Approach of Proof Abstract: In this work, an attempt has been made b...
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου