Κυριακή, 19 Μαΐου 2013

ANOPOLIS CIRCLE

Let ABC be a triangle.

Denote:

(N1),(N2), (N3) = the NPCs of IBC, ICA, IAB, resp.

(12), (13) = the reflections of (N1) in BI, CI, resp.

(23), (21) = the reflections of (N2) in CI, AI, resp.

(31), (32) = the reflections of (N3) in AI, B1, resp.

The six centers 12,13,23,21,31,32 are concyclic.

The circle has diameter IO.

Perspectivity:

The circles (I), (21), (31) concur at a point A'.

The circles (I), (32), (12) concur at a point B'.

The circles (I), (13), (23) concur at a point C'.

The triangles ABC, A'B'C' are perspective (??)

Antreas P. Hatzipolakis, 19 May 2013

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Midpoint of OI = X(1385)

12 = {a (a^2-b^2+a c-2 b c-c^2),-a^3+a b^2+a^2 c+2 a b c-2 b^2 c+a c^2-3 b c^2-c^3,-c^2 (b+c)} 13 = {a (a^2+a b-b^2-2 b c-c^2),-b^2 (b+c),-a^3+a^2 b+a b^2-b^3+2 a b c-3 b^2 c+a c^2-2 b c^2}

ETC points on the Anopolis circle {1,3,1083,3109,3110}

The orthogonal projection of I on a line through O is also on the circle .. hence X(3109) & X(3110). So too X(1083) as it is the orthogonal projection of I on line X(3) X(667)

Also the orthogonal projection of O on a line through I is also on the circle

A non ETC point on the Anopolis circle

a^2 (a^4 b^2-a^2 b^4-2 a^4 b c+a b^4 c+a^4 c^2+a^2 b^2 c^2-b^3 c^3-a^2 c^4+a b c^4) ::

on lines {{1,667},{3,238},{35,1083},{41,813}}

A’ = reflection of midpoint of AI in 21 31 = {(a-b-c) (b-c)^2 (a+b-c) (a-b+c),-(a-b)^2 b^2 (a+b-c),-(a-c)^2 c^2 (a-b+c)}

A’B’C’ is perspective to ABC at X(59)

Also to the tangential triangle at:

a^2 (a^9-3 a^8 b+8 a^6 b^3-6 a^5 b^4-6 a^4 b^5+8 a^3 b^6-3 a b^8+b^9-3 a^8 c+12 a^7 b c-12 a^6 b^2 c-12 a^5 b^3 c+30 a^4 b^4 c-12 a^3 b^5 c-12 a^2 b^6 c+12 a b^7 c-3 b^8 c-12 a^6 b c^2+33 a^5 b^2 c^2-21 a^4 b^3 c^2-21 a^3 b^4 c^2+33 a^2 b^5 c^2-14 a b^6 c^2+2 b^7 c^2+8 a^6 c^3-12 a^5 b c^3-21 a^4 b^2 c^3+48 a^3 b^3 c^3-21 a^2 b^4 c^3-4 a b^5 c^3+2 b^6 c^3-6 a^5 c^4+30 a^4 b c^4-21 a^3 b^2 c^4-21 a^2 b^3 c^4+18 a b^4 c^4-2 b^5 c^4-6 a^4 c^5-12 a^3 b c^5+33 a^2 b^2 c^5-4 a b^3 c^5-2 b^4 c^5+8 a^3 c^6-12 a^2 b c^6-14 a b^2 c^6+2 b^3 c^6+12 a b c^7+2 b^2 c^7-3 a c^8-3 b c^8+c^9)::

= (2r - R) (r^2 + 6 r R + 8 R^2 - s^2) X[1486] – 4 r (r^2 + 5 r R + 4 R^2 - s^2) X[1618]

On line {1486, 1618}

Search = 0.89807482985690351940

Peter J. C. Moses, 19 May 2013

3 σχόλια:

  1. The circles (I), (21), (31) concur at a point A':
    May you specify what are circles (21) and (31)? I suppose you give their centers, but I can't see the radii. Thank you.

    ΑπάντησηΔιαγραφή
  2. The notation is (X) = a circle with center X.
    So (I) = the circle with center the Incenter I (=incircle).
    The circles (21), (31) are defined in the starting.
    (21) is the reflection of the circle (N2) in CI.
    [N2 is the Nine point Circle (NPC) of ICA] etc.

    ΑπάντησηΔιαγραφή

 

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