BarycentricsProjectively

Barycentric Coordinates as a projective mapping

Here I continue some thoughts initiated in BarycentricCoordinates.html . Given the triangle of reference ABC, the corresponding Barycentric coordinates represent a mapping between two copies of the projectified euclidean plane. Denote by P² and Q² the two projective planes and by F: P² --> Q² the corresponding map. F is completely determined by the conditions:
1) There is a distinguished equilateral triangle A*B*C* with centroid M*.
2) F maps the vertices of the equilateral to corresponding vertices of the triangle (A*->A, B*->B, C*->C) and the centroid M* to the corresponding centroid M of ABC.

This map could be geometrically realized through the projectification of an affine transformation of two planes in the euclidean three space. The first plane, representing the affine part of P², would be the affine plane given by the equation x+y+z=1 . A*, B*, C* would be the endpoints of the three standard unit coordinate vectors e₁, e₂, e₃. The other affine plane, representing the affine part of Q², would be defined by z= 1 and A, B, C would be defined by the endpoints of three linearly independent vectors emanating from the origin. F would then be defined by the linear application represented by the matrix:

Here (p_x,p_y,p_z) denote the coordinates of P* and (P₁, P₂, P₃) denote the homogeneous coordinates of P.
[1] The mapping sends the line at infinity of the first plane, defined by p_x+p_y+p_z=0, to the line at infinity of the second plane defined by P₃=0.
[2] Denote by F' the affine restriction of F to the corresponding affine planes. F' being affine, preserves the parallelity of lines, the middles of segments and, more generally, the ratios of three points on a line. Thus, in the above picture: P*A*/P*D* = PA/PD and D*B*/D*C* = DB/DC.
[3] Besides, F preserves the nature of conics, mapping ellipses to ellipses, hyperbolas to hyperbolas and parabolas to parabolas correspondingly.
[4] From the linearity of F' follows also the formula for the area of a triangle DEF, as explained in BarycentricsFormulas.html .
[5] Barycentric coordinates use the standard projective coordinates of point P* as parameters for P. Thus, lines are represented by linear equations like k*p_x+m*p_y+n*p_z = 0, and conics with quadratic equations:

[6] In particular, conics passing through the vertices of triangle ABC have their equations satisfied by the coordinate vectors (1,0,0), (0,1,0) and (0,0,1), implying a₁₁=a₂₂=a₃₃=0. They are images under F of conics circumscribing A*B*C*.
[7] In particular, the circumcircle of A*B*C* is mapped by F onto the outer Steiner ellipse of triangle ABC. To see this, notice first that the circumconics of ABC are completely determined through their tangents at the three vertices (see PascalOnTriangles.html ). The Steiner ellipse is the one for which the tangents to the vertices are parallel to the corresponding opposite sides.

Barycentric Coordinates as a projective mapping

See Also