[size=85]We construct triangle [math]ABC[/math] and then K001 cubic. After that we take point [math]M[/math] which can be any point from the cubic then we cojugate that point isogonal and we get point [math]N[/math] which is also on the same cubic. And by using GeoGebra we conclude that K001 is isogonal transformed of itself.[/size]

[math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]x\left(c^2y^2-b^2y^2\right)=0[/math]

[size=85] Let the barycentric coordinates of a point are [math]x[/math], [math]y[/math] and [math]z[/math]. The barycentric coordinates of its isogonal conjugate are [math]\frac{a^2}{x}[/math], [math]\frac{b^2}{y}[/math], [math]\frac{c^2}{z}[/math] or [math]a^2zy[/math], [math]b^2xz[/math] and [math]z^2xy[/math]. And now we substitude [math]x[/math], [math]y[/math] and [math]z[/math] in the equation with [math]a^2zy[/math], [math]b^2xz[/math] and [math]z^2xy[/math].[br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]x\left(c^2y^2-b^2y^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]a^2zy\left(c^2{(c^2xz)}^2-b^2(z^2xy)^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]a^2zy\left(c^2b^4x^2z^2-b^2z^4x^2y^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]a^2b^2c^2xyzx\left(c^2y^2-b^2z^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]a^2b^2c^2xyzx\left[-(c^2y^2-b^2z^2\right)]=0[/math][br][math]-a^2b^2c^2xyz\sum_{cyclic}\left[a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)-2a^4\right]x\left(c^2y^2-b^2z^2\right)=0[/math][br]In this we get the same equqtion as that in the beginig and this means that Neuberg cubic is isogonal transform of itself.[/size][br][br][br][br][br]