## Τρίτη, 28 Μαΐου 2013

### CONCYCLIC CIRCUMCENTERS

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'

L21 = the reflection of L22 in AA'

L23 = the reflection of L22 in CC'

L31 = the reflection of L33 in AA'

L32 = the reflection of L33 in BB'

O1 = the circumcenter of the triangle bounded by the lines (L11,L12,L13)

O2 = the circumcenter of the triangle bounded by the lines (L21,L22,L23)

O3 = the circumcenter of the triangle bounded by the lines (L31,L32,L33)

Conjecture 1:

O1, O2, O3 and O [circumcenter of ABC] are concyclic.

Denote:

Oa = the circumcenter of the triangle bounded by the lines (L11,L21,L31)

Ob = the circumcenter of the triangle bounded by the lines (L12,L22,L32)

Oc = the circumcenter of the triangle bounded by the lines (L13,L23,L33)

Conjecture 2:

The circumcenter of the triangle OaObOc is the I.

Antreas P. Hatzipolakis, 28 May 2013

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Conjecture 1:

Yes, O1, O2, O3 and O are concyclic. The center of the circle is:

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

with (6-9-13)-search number: 0.025111873257385778374122883

Angel Montesdeoca, Anopolis #321

Point X(5495) in ETC