ArtztIsosceles

1. Artzt parabola in isosceles triangle

Consider an isosceles triangle ABC and its circumcircle O(r). Draw line a' parallel to its basis a to the same side with point A and at distance r from a. Take point P varying on the basis a and let P' be the intersection point with a' of the line orthogonal to a and passing through P.
[1] Draw OP' and the chord QQ' through P and parallel to OP'. Q' is a fixed point on the circle.
[2] OAP'P and OP'PQ' are parallelograms. Quadrangle OPQP' is an isosceles trapezium.
[3] The intersection point W of the diagonals of the trapezium describes a parabola c.
[4] This parabola is the Artzt parabola passing through {B,C} and tangent there to the sides of the triangle.

The proofs are easy.
[1] follows from the fact that drawing OQ' orthogonal to a, the resulting OP'PQ' is a parallelogram.
[2] follows from [1] and the equality OQ = OA = P'P = r.
[3] follows from the fact that the medial line t_W of OP' passes through W, which coincides with the intersection point of this line t_W and the vertical line P'W to a' at P'. This is a standard definition of the parabola. W describes a parabola with focus at O and directrix the line a'.
[4] follows from the fact that t_W obtains the positions of {AB,AC} when OP' is respectively orthogonal to these two lines (see Artzt.html for the definition of these parabolas).

2. Homography mapping circle to parabola

As is well known (see Projectivity.html ) a projectivity is defined by prescribing the image points {Y_i, i=1,..4} of four points {X_i, i=1,...,4} in general position. Thus there is a unique projectivity F with the properties:
(i) F fixes points {B,C, O} and
(ii) F maps A to W₀.
- It follows easily (exercise) that F fixes every point of line BC and also leaves invariant every line passing through O (maps such a line into itself fixing O and its intersection point with BC).
- Using this and the invariance of line AO one sees easily that every line parallel to BC maps to a line also parallel to BC. In particular the parallel to BC through A maps to the parallel to BC through W₀.
- Using cross ratios allong line AO one sees easily that the tangent line t of the circle at Q' maps via F to the line at infinity.
- From the previous result follows also immediately that the tangents {t_B, t_C} to the circle, map via F to sides {BA, CA} respectively.
- From these facts follows, that points {A,B,C} and the tangents to the circle at these points map to {W₀,B,C} and the tangents to the parabola at these points correspondingly. Thus, also the circle maps via F to the Artzt parabola c.

3. Artzt parabola and circumcircle

The previous properties indicate a relation between the Artzt parabola and the circumcircle of the triangle. This relation does not generalize immediately to arbitrary triangles, their circumcircles and their Artzt parabolas. There is though a generalization involving the Steiner outer ellipse. The discussion related to these matters continues to ArtztSteiner.html .

1. Artzt parabola in isosceles triangle

2. Homography mapping circle to parabola

3. Artzt parabola and circumcircle

See Also