- A simple property of X(5020):

Let ABC be an triangle, DEF is the medial triangle. Let (Ca) be a circle through E and F that is tangent to the circumcircle at a point A' different of A. Defined similarly the circles (Cb) and (Cc), and the points B' and C'.

The triangles ABC and A'B'C' are perspective and the perspector is X(25)

The orthic triangle and A'B'C' are perspective and the perspector is X(2)

The radical center of the circles (Ca), (Cb) and (Cc) is X(5020)= a^2(a^4-b^4-c^4+6b^2c^2): ... : ...

A' = (a^4-(b^2-c^2)^2: -2b^2(a^2+b^2-c^2): -2c^2(a^2-b^2+c^2)).

(Ca) = c^2x*y + b^2x*z + a^2y*z -

(x+y+z)/(2(a^2+(b-c)^2)(a^2+(b+c)^2))

(4a^2b^2c^2x + c^2(a^2-b^2+c^2)^2y + b^2(a^2+b^2-c^2)^2z).

Angel M. - Dear Angel,

Here are a couple more properties of X(5020):

X(5020) = homothetic center of medial triangle and 3rd antipedal triangle of X(3)

X(5020) = trilinear pole of polar of X(3527) wrt 2nd Lemoine circle

Best regards,

Randy

>________________________________

[Non-text portions of this message have been removed]

> From: Angel <amontes1949@...>

>To: Hyacinthos@yahoogroups.com

>Sent: Wednesday, January 30, 2013 6:29 AM

>Subject: [EMHL] The triangle center X(5020) as radical center

>

>

>

>A simple property of X(5020):

>

>Let ABC be an triangle, DEF is the medial triangle. Let (Ca) be a circle through E and F that is tangent to the circumcircle at a point A' different of A. Defined similarly the circles (Cb) and (Cc), and the points B' and C'.

>

>The triangles ABC and A'B'C' are perspective and the perspector is X(25)

>The orthic triangle and A'B'C' are perspective and the perspector is X(2)

>

>The radical center of the circles (Ca), (Cb) and (Cc) is X(5020)= a^2(a^4-b^4-c^4+6b^2c^2): ... : ...

>

>A' = (a^4-(b^2-c^2)^2: -2b^2(a^2+b^2-c^2): -2c^2(a^2-b^2+c^2)).

>

>(Ca) = c^2x*y + b^2x*z + a^2y*z -

>(x+y+z)/(2(a^2+(b-c)^2)(a^2+(b+c)^2))

>(4a^2b^2c^2x + c^2(a^2-b^2+c^2)^2y + b^2(a^2+b^2-c^2)^2z).

>

>Angel M.

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>

>

>

>