Antreas P. Hatzipolakis
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'
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 of concurrence?
Figure:
http://anthrakitis.blogspot.gr/2013/05/concurrent-circles.html
APH
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'
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 of concurrence?
Figure:
http://anthrakitis.blogspot.gr/2013/05/concurrent-circles.html
APH
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου