Cubics in the Triangle Plane
By definition, triangle cubic is a curve that can be expressed in trilinear coordinates such that the highest degree term is of order three. The first main part of the project will be to use dynamic mathematics software such as GeoGebra in order to investigate properties of these cuves and look for connections amog them. The second more difficult part of the project will be to supportall findings from part one with proofs - either already existing ones, or work on own author's proofs.
Table of Contents
- Definitions
- Isogonal transform
- Barycentric coordinate system
- Trilinear coordinates
- Cubics isogonal transform of themselves
- Neuberg cubic (K001)
- Thomson cubic (K002)
- McCay cubic(K003)
- Darboux cubic (K004)
- Napoleon cubic (K005)
- Orthocubic (K006)
- Isogonal transform cubics
- Lucas cubic (K007)
- Droussent cubic K(008)
- Lemoine cubic (K009)
- Simson cubic K(010)
- Future development
- Future development
- Bibliography
- References
- Acknowledgements
- Acknowledgements
Isogonal transform
The object obtained by collecively taking the conjugates of all its points.
Neuberg cubic (K001)
[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]
Barycentric equation
[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]
Proof
[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]
Lucas cubic (K007)
[size=85]Here we construct a triangle [math]ABC[/math]. After that we construct Lucas cubic for this triangle and we take point [math]M[/math] on it then we conjugate [math]M[/math] isogonal and we get point [math]N[/math]. By using GeoGebra we get another cubic, the blue one one the figure. Lucas cubic is isogonal transfom of K172.[/size]
Barycentric equation
[math]\sum_{cyclic}\left(b^2+c^2-a^2\right)x\left(y^2-z^2\right)=0[/math]
[size=85] Here we again sustitude [math]x[/math], [math]y[/math] and [math]z[/math] with [math]a^2zy[/math], [math]b^2xz[/math] and [math]z^2xy[/math].[br][math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2yz\left(\left(b^2xz\right)^2-\left(c^2xy\right)^2\right)=0[/math][br][br][math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2yz\left(b^4x^2z^2-c^4x^2y^2\right)=0[/math][br][br][math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2x^2yz\left[-\left(b^4z^2-c^4y^2\right)\right]=0[/math][br][br][math]xyz\sum_{cyclic}a^2\left(b^2+c^2-a^2\right)x\left(b^4z^2-c^4y^2\right)=0[/math][br][br]But the barycentric equation of the isogonal transform cubic is [math]\sum_{cyclic}a^2S_ax\left(b^4z^2-c^4y^2\right)=0[/math] where [math]S_A=\frac{b^2+c^2-a^2}{2}[/math] and in order to get the same equation we have to multiply [math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2x^2yz\left[-\left(b^4z^2-c^4y^2\right)\right]=0[/math] by [math]\frac{2}{2}[/math] and the result is-[br][br][math]2xyz\sum_{cyclic}a^2\frac{\left(b^2+c^2-a^2\right)}{2}x\left(b^4z^2-c^4y^2\right)=0[/math] and this is the same equation as that of the isogonal transform cubic which means that Lucas cubic is isogonal transform of K172.[/size][br][br][br]
Future development
[size=85]In future plans we can try to make the same constructions but for other cojugations as isotomic and cyclocevian conjugation. For isotomic cojugation we change the barycentric coordinates [math]x[/math], [math]y[/math], [math]z[/math] with [math]\frac{1}{x}[/math], [math]\frac{1}{y}[/math], [math]\frac{1}{z}[/math] or [math]yz, xz, xy[/math] but for cylocevian conjugation I have not found what is the change of the barycentric coordinates.[/size]
References
http://bernard.gibert.pagesperso-orange.fr/ctc.html[br]
Trilinear coordinates
Barycentric coordinate system
Geometry2013Chapter12
Acknowledgements
SRS18[br]HSSIMI[br]Nadezhda Aplakova[br]Valcho Milchev[br]Milko Bagdasarov