Sorry, an error occurred while loading the content.

## Concurrent circles

Expand Messages
• Let ABC be a triangle and A B C the cevian triangle of I. Denote: B a, C a = the reflections of B ,C in AA , resp. A a = (perpendicular to BB from B a) /
Message 1 of 42 , Feb 26, 2013
Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

B'a, C'a = the reflections of B',C' in AA', resp.

A'a = (perpendicular to BB' from B'a) /\ (perpendicular to CC' from
C'a).

C'b, A'b = the reflections of C',A' in BB', resp.

B'b = (perpendicular to CC' from C'b) /\ (perpendicular to AA' from
A'b).

A'c, B'c = the reflections of A',B' in CC', resp.

C'c = (perpendicular to AA' from A'c) /\ (perpendicular to BB' from
B'c).

Are the circumcircles of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c
concurrent?.

Let Ia,Ib,Ic be the incenters of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c,
resp.

Are the triangles ABC, IaIbIc perspective?

APH
• Let ABC be a triangle, P a point and A B C the pedal triangle of P. Denote: L = the Euler line of A B C A , B , C = orthogonal projections of A, B, C, on
Message 42 of 42 , Jan 30

Let ABC be a triangle, P a point andÂ  A'B'C' the pedal triangle of P.

Denote:

L = the Euler line of A'B'C'

A", B", C" = orthogonal projections of A, B, C, on L, resp.

AaAbAc, BaBbBc, CaCbCc = the pedal triangles of A", B", C", wrt triangle A'B'C' resp.

The circumcircles of AaAbAc, BaBbBc, CaCbCc are concurrent on the NPC circle of A'B'C'.
The point of concurrence is the reflection point of L wrt the medial triangle of A'B'C'.

Problem:

Let ABC be a triangle and L a line.

For which points P's the L is the Euler line of the pedal triangle of P ?

APH
Your message has been successfully submitted and would be delivered to recipients shortly.