## Τετάρτη, 22 Ιουνίου 2016

### CONCENTRIC CIRCLES

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