## Generation of the Artzt parabola

Given a triangle ABC, the Artzt parabola relative to side BC (see Artzt.html ) is the parabola tangent to sides AB, AC at points B and C respectively. Here it is shown that this parabola is the envelope of the diagonals B'C' of paralellograms AB'PC' for a point P gliding on BC.

The proof given here is modeled as an exercise in the use of trilinear coordinates (see Trilinears.html ). The (generalized) trilinears used here are defined through the projective base {A,B,C,D}, D being the middle of the median AM of the triangle.
In these system these four basic points have coordinates {A=(1,0,0), B=(0,1,0), C=(0,0,1), D=(1,1,1)}. Side-lines of the triangles correspond to equations BC(x=0), CA(y=0), AB(z=0). A point on AB has coordinates (0,s,t). The line at infinity corresponds to equation 2x+y+z=0 and its intersection with line ax+by+cz=0, is the point at infinity of the later and has coordinates (b-c, 2c-a, a-2b). These remarks allow the determination of the two lines from P parallel to the side-lines of the triangle.
Lines PB', PC' assumed in the form ax+by+cz=0 must be satisfied by (0,s,t), thus giving bs+ct=0, a solution of later being (b=-t, c=s).
PB' is parallel to AC(y=0), hence its point at infinity is (1,0,-2). Since this satisfies also the line equation, we get for it the coefficients (a=2s, b=-t, c=s).
Analogously PC' is parallel to AB(z=0), and has point at infinity (-1,2,0), consequently (a=-2t, b=-t, c=s). Thus, the two lines are PB'(2sx-ty+sz=0) and PC'(-2tx-ty+sz=0). Their intersection points B' with z=0, and C' with y=0 satisfy correspondingly 2sx-ty=0, satisfied by (x=t,y=2s) and -2tx+sz=0, satisfied by (x=s,z=2t). Thus B'(t,2s,0) and C'(s,0,2t) and their line B'C' is given by (2st)x-(t2)y-(s2)z=0.
Calculating the envelope of these lines we come to equation x2-yz=0, which represents a conic tangent to AB, AC at points B, C correspondingly and passing through D.
To see that this is a parabola, find its points at infinity i.e. solve the system {x2=yz, 2x+y+z=0}, which setting x=-(y+z)/2 in the first gives (y-z)2=0 and finally the unique point at infinity (-1,1,1). This shows already that the conic is a parabola. Next paragraph could be omitted but serves as test for the calculations.
The tangents to a point of the curve have coefficients proportional to (2x,-z,-y), hence for its point at infinity (-1,1,1) the tangent has coefficients (-2,-1,-1) = -(2,1,1), hence coincides with the line at infinity. This completes the proof that the curve is a parabola and identifies it with the Artzt parabola as claimed.

For an alternative synthetic proof of the same fact see the file Artzt_Generation2.html .