ANOPOLIS 115

Let ABC be a triangle and A'B'C' its orthic triangle.

Denote:

Ab := The Orthogonal Projection of A' on AC

aB := The Orthogonal Projection of A' on CC'

Ac := The Orthogonal Projection of A' on AB

aC := The Orthogonal Projection of A' on BB'.

[The 4 points are collinear]

1. DON QUIXOTE POINT:

La := The Euler Line of A'AbAc

Ma := The Euler Line of A'aBaC

Similarly Lb, Lc and Mb, Mc.

We know that La,Lb,Lc concur at X(973) and Ma,Mb,Mc
at the Feuerbach point of A'B'C'

Now, denote:

A1 := La /\ Ma, B1 := Lb /\ Mb, C1 := Lc /\ Mc

The triangles ABC, A1B1C1 are homothetic (perspective).

Antreas P. Hatzipolakis, Hyacinthos Message #17901

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The perspector is a point S with coordinates:

{-a^2 (a^6 b^2 - 3 a^4 b^4 + 3 a^2 b^6 - b^8 + a^6 c^2 -
a^4 b^2 c^2 - 3 a^2 b^4 c^2 + 3 b^6 c^2 - 3 a^4 c^4 -
3 a^2 b^2 c^4 - 4 b^4 c^4 + 3 a^2 c^6 + 3 b^2 c^6 -
c^8),
-b^2 (-a^8 + 3 a^6 b^2 - 3 a^4 b^4 + a^2 b^6 + 3 a^6 c^2 -
3 a^4 b^2 c^2 - a^2 b^4 c^2 + b^6 c^2 - 4 a^4 c^4 -
3 a^2 b^2 c^4 - 3 b^4 c^4 + 3 a^2 c^6 + 3 b^2 c^6 -
c^8),
-c^2 (-a^8 + 3 a^6 b^2 - 4 a^4 b^4 + 3 a^2 b^6 - b^8 +
3 a^6 c^2 - 3 a^4 b^2 c^2 - 3 a^2 b^4 c^2 + 3 b^6 c^2 -
3 a^4 c^4 - a^2 b^2 c^4 - 3 b^4 c^4 + a^2 c^6 + b^2 c^6)}

This point satisfies AS:SA* = BS:SB* = CS:SC* = 4 and I'S:SG'=-6,
where I'(=H), G' are the incenter and the centroid of the orthic
triangle A'B'C'.

Francisco Javier Garcia Capitan, Hyacinthos Message #17902

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2. SANCHO PANZA POINT

Pa := The Euler Line of A'AbaB

Qa := The Euler Line of A'AcaC

Similarly Pb, Qb and Pc, Qc.

A2 := Pa /\ Qa, B2 := Pb /\ Qb, C2 := Pc /\ Qc.

The Triangles ABC, A2B2B2 are Homothetic (perspective).

Antreas P. Hatzipolakis, 27 June 2009

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It is a point S with coordinates:

{-a^2 (a^2 b^2 - b^4 + a^2 c^2 + b^2 c^2 - c^4),
-b^2 (-a^4 + a^2 b^2 + a^2 c^2 + b^2 c^2 - c^4),
-c^2 (-a^4 + a^2 b^2 - b^4 + a^2 c^2 + b^2 c^2)

It satisfies A2S:SA = B2S:SB = C2S:SC = -1/4 (ABC and A2B2C2)
are homotetic. And, S is also the point X(1699) of the orthic
triangle.


Francisco Javier Garcia Capitan, 28 June 2009

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3. MIGUEL DE CERVANTES POINT

The Triangles A1B1C1, A2B2C2 are homothetic (perspective)

Antreas P. Hatzipolakis, 27 June 2009

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It is point a point Z with coordinates:

{-a^2 (a^2 b^2 - b^4 + a^2 c^2 + 2 b^2 c^2 - c^4) (a^4 - 2 a^2 b^2 + b^4 - 2 a^2 c^2 - b^2 c^2 + c^4),
-b^2 (-a^4 + a^2 b^2 + 2 a^2 c^2 + b^2 c^2 - c^4) (a^4 - 2 a^2 b^2 + b^4 - a^2 c^2 - 2 b^2 c^2 + c^4),
-c^2 (-a^4 + 2 a^2 b^2 - b^4 + a^2 c^2 + b^2 c^2) (a^4 - a^2 b^2 + b^4 - 2 a^2 c^2 - 2 b^2 c^2 + c^4)}

Z is in fact the point X(143) of ABC, the Nine Point Circle
center of the orthic triangle.

In fact we have also that A2B2C2 is the reflection of A1B1C1
on the point Z.


Francisco Javier Garcia Capitan, 28 June 2009

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