ANOPOLIS 321

[Antreas P. Hatzipolakis]

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

Denote:

1 = the perpendicular line to AA' at A'
2 = the perpendicular line to BB' at B'
3 = the perpendicular line to CC' at C'
(ie the lines 1,2,3 bound the antipedal triangle of I wrt A'B'C')

Now let's reflect 1,2,3 in AA', BB', CC':

11 = the reflection of 1 in AA' (it is identical to 1)
12 = the reflection of 1 in BB'
13 = the reflection of 1 in CC'

21 = the reflection of 2 in AA'
22 = the reflection of 2 in BB' (it is identical to 2)
23 = the reflection of 2 in CC'

31 = the reflection of 3 in AA'
32 = the reflection of 3 in BB'
33 = the reflection of 3 in CC' (it is identical to 3)

O1 = the circumcenter of the triangle bounded by
the lines (11,12,13)
O2 = the circumcenter of the triangle bounded by
the lines (21,22,23)
O3 = the circumcenter of the triangle bounded by
the lines (31,32,33)

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

 

APH

 

Angel Montesdeoca]:

**** 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