[alogo] 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.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

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 .

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