Let ABC be a triangle.
Denote: A1, B1, C1 = The NPC centers of NBC, NCA, NAB, resp.
A2, B2 ,C2 = The NPC centers of A1BC, B1CA, C1AB, resp.
A3, B3, C3 = The NPC centers of A2BC, B2CA, C2AB
An, Bn, Cn = The NPC centers of A_n-1BC, B_n-1CA, C_n-1AB.
On = the circumcenter of the triangle AnBnCn.
The points O1, O2, O3, .... On,.... lie on the Euler line (see HERE)
Denote:
O11 = the reflection of O1 in BC
O111 = the reflection of O11 in AN
O112 = the reflection of O111 in BC
Similarly:
O12 = the reflection of O1 in CA
O121 = the reflection of O12 in BN
O122 = the reflection of O121 in CA
and
O13 = the reflection of O1 in AB
O131 = the reflection of O13 in CN
O132 = the reflection of O131 in AB
The circumcenter of O112O122O132 coincides with the cirumcenter of ABC
The same if we take O2:
O21 = the reflection of O2 in BC
O211 = the reflection of O21 in AN
O212 = the reflection of O211 in BC
Similarly:
O22 = the reflection of O2 in CA
O221 = the reflection of O22 in BN
O222 = the reflection of O221 in CA
and
O23 = the reflection of O2 in AB
O231 = the reflection of O23 in CN
O232 = the reflection of O231 in AB
The circumcenter of O212O222O232 coincides with the cirumcenter of ABC
and so on......
The circumcenter of On12On22On32 coincides with the cirumcenter of ABC
So we have a sequence of concentric circles.
Antreas P. Hatzipolakis, 22 June 2016