HYACINTHOS 25724

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

A1B1C1 = the pedal triangle of I

A2,B2,C2 = the reflections of I in BC, CA, AB, resp.

S1 = the point of concurrence of AA1, BB1, CC1 (Gergonne point)

S2 = the point of concurrence of AA2, BB2, CC2.

(N11), (N21), (N31) = the NPCs of AIS1, BIS1, CIS1, resp.

(N12), (N22), (N32) = the NPCs of AIS2, BIS2, CIS2, resp.

(N13), (N23), (N33) = the NPCs of AS1S2, BS1S2, CS1S2

1. (N11), (N21), (N31) are coaxial
2. (N12), (N22), (N32) are coaxial
3. (N12), (N22), (N32) are coaxial.

The second point of intersections (other than the midpoints of IS1, IS2, S1S2) is the Feuerbach point Fe

The lines N11N21N31, N12N22N32, N12N22N32 are concurrent.

Equivalently:

Let M1, M2, M3 be the midpoints of IS1, IS2, S1S2, resp.

The perpendicular bisectors of FeM1, FeM2, FeM3  are concurrent.

Or:
M1, M2, M3, Fe are concyclic.

Center of the circle?

 

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Simpler:

The NPC of IS1S2 passes through the Feuerbach point.

Which point is its center?

 

[Peter Moses]:

Hi Antreas,

 

>The NPC of IS1S2 passes through the Feuerbach point.

and X(3649),X(5542).

X(3649) is the midpoint of X(1) and X(79)

X(5542) is the midpoint of X(1) and X(7)

 

center:

(b-c) (-a^8+5 a^6 b^2-2 a^5 b^3-7 a^4 b^4+4 a^3 b^5+3 a^2 b^6-2 a b^7+4 a^6 b c-4 a^5 b^2 c-a^4 b^3 c-a^3 b^4 c+2 a^2 b^5 c-a b^6 c+b^7 c+5 a^6 c^2-4 a^5 b c^2+3 a^4 b^2 c^2+3 a^3 b^3 c^2-2 a^2 b^4 c^2-3 a b^5 c^2-2 b^6 c^2-2 a^5 c^3-a^4 b c^3+3 a^3 b^2 c^3-6 a^2 b^3 c^3+6 a b^4 c^3-b^5 c^3-7 a^4 c^4-a^3 b c^4-2 a^2 b^2 c^4+6 a b^3 c^4+4 b^4 c^4+4 a^3 c^5+2 a^2 b c^5-3 a b^2 c^5-b^3 c^5+3 a^2 c^6-a b c^6-2 b^2 c^6-2 a c^7+b c^7)::

 

Another point on the circle is
2 a^6-a^5 b-4 a^4 b^2+4 a^2 b^4+a b^5-2 b^6-a^5 c-6 a^4 b c-5 a^3 b^2 c+8 a b^4 c+4 b^5 c-4 a^4 c^2-5 a^3 b c^2-8 a^2 b^2 c^2-9 a b^3 c^2+2 b^4 c^2-9 a b^2 c^3-8 b^3 c^3+4 a^2 c^4+8 a b c^4+2 b^2 c^4+a c^5+4 b c^5-2 c^6::
the midpoint of X(7) and X(79)
on lines {{7,79},{9,6701},{11,553},{30, 5542},...}.

 

So, as you mention, the circle is the NP circle of X(1,7,79).

 

Best regards,

Peter Moses.