## Τετάρτη, 29 Μαΐου 2013

### CONCURRENT CIRCLES

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

Denote:

L11 = the perpendicular line to AA' at A'

L22 = the perpendicular line to BB' at B'

L33 = the perpendicular line to CC' at C'

(ie the lines L11,L22,L33 bound the antipedal triangle of I wrt A'B'C')

L12 = the reflection of L11 in BB'

L13 = the reflection of L11 in CC'

M12 = the parallel to L12 through B'

M13 = the parallel to L13 through C'

A" = line M12 /\ line M13

Similarly B" and C".

O1 = the circumcenter of the triangle A"B'C'

O2 = the circumcenter of the triangle B"C'A'

O3 = the circumcenter of the triangle C"A'B'

Conjecture 1:

The points I, O1, O2, O3 are concyclic. Center of the circle?

Conjecture 2: The circumcircles (O1), (O2), (O3) are concurrent. Point?

Antreas P. Hatzipolakis, 29 May 2013

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The points I, O1, O2, O3 are concyclic. Center of the circle (barycentrics)

( a(b+c)(a^5-2a^3(b^2+c^2) + a(b^4-b^2c^2+c^4) - a^2b c(b+c)+ b(b-c)^2c(b+c) ) : ... : ... ).

The circumcircles (O1), (O2), (O3) are concurrent.

Point of concurrence:

( a(a^6-a^5(b+c) - a^4(b+c)^2 + (b^2-c^2)^2(b^2+c^2) + a^3(2b^3+b^2c+b c^2+2c^3) - a^2(b^4-b^3c-3b^2c^2-b c^3+c^4) - a(b^5+b^4c+b c^4+c^5)) : ... : ... )

Angel Montedeoca, Anopolis #327

Points X(5496), X(5497) in ETC