Σάββατο, 14 Ιανουαρίου 2012

PICASSO POINT [X(4550]

Let ABC be a triangle, A1B1C1 the pedal triangle of H, A2B2C2 the antipodal triangle of A1B1C1 with respect the circumcircle of A1B1C1 (= NPC of ABC), A'B'C' the antipodal triangle of ABC (= circumcevian triangle of O).

Denote:

A" = The other than A' intersection of A'A2 and the circumcircle of ABC

B" = The other than B' intersection of B'B2 and the circumcircle of ABC

C" = The other than C' intersection of C'C2 and the circumcircle of ABC

N1 = The NPC Center of A"BC

N2 = The NPC center of B"CA

N3 = The NPC center of C"AB

The four NPC centers N, N1,N2,N3 are concyclic (??).

Center of the circle ?

APH
14 January 2012

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Yes. The center is the point:

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

The radius of the circle has a long expression:

-(1/(4 (p^2 - r^2 - 4 r R)^2))(3 p^6 - 5 p^4 r^2 + 5 p^2 r^4 -
3 r^6 - 36 p^4 r R + 56 p^2 r^3 R - 36 r^5 R - 46 p^4 R^2 +
224 p^2 r^2 R^2 - 190 r^4 R^2 + 368 p^2 r R^3 - 560 r^3 R^3 +
216 p^2 R^4 - 952 r^2 R^4 - 864 r R^5 - 324 R^6)

where p= semiperimeter.

Francisco Javier García Capitán
15 January 2012

Now in ETC X(4550)
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I name this new point as PICASSO Point in honor of great Pablo Picasso, whose a painting was recently stolen from the Greek National Gallery. See HERE and also my article in my blog Madara HERE

Antreas

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This point P lies on line G-X74 and satisfies the ratio

GP:PX74= (2 p^2 - 2 r^2 - 8 r R - 9 R^2)/(9 R^2).

Francisco, Hyacinthos 20683

Like Segovia point, also lies in line H-X1209, moreover Picasso point is the midpoint of H and Segovia point.

Francisco, Hyacinthos 20684

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The four nine-point centers are concyclic on a circle with center

(a^2(S^2+3S_{AA})(S^2+3S_{BC}) : ... : ...)

in barycentric coordinates.

This point lies on the line joining the centroid to X(74), the fourth intersection of the circumcircle with the Jerabek hyperbola.

Paul Yiu

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Denote:

Na = The NPC Center of AB"C"

Nb = The NPC center of BC"A"

Nc = The NPC center of CA"B"

The radical axes of the NPCs (N1) and (Na) - (N2) and (Nb) - (N3) and (Nc) are concurrent [See NPCs and Radical Axes], or equivalently the triangles N1N2N3, NaNbNc are perspective.

Antreas