Constructions by Ruler Alone, plus a Parabola

1. Introduction. Constructions by compass alone, or by straight edge alone plus a given circle and its center, are excellent subjects to stimulate high school and college students into creativity, because first, they are fun, and second, many known algorithms can be improved. For example, given 2 points A,B on the plane, what is the least number of arcs to construct by compass alone a point C on the invisible line AB, so that AC is say 17 times AB? The record to this problem has been broken again and again by my students. Now this subject of "construction by ruler alone plus a parabola" (henceforth referred to as ruler+parabola constructions), assumes that your ruler is a infinitely long unmarked straight edge. The parabola and its focus and its axis of symmetry must be all given. We can develop computer tools to help students do these constructions, so that intersection points between lines or between a line and the parabola can be very accurately computed and displayed.

People like myself are not interested to know the theoretical proof that ruler+parabola constructions are as powerful as compass+ruler constructions. We are interested to see how each problem can be solved, and how to do it with minimum no. of lines. This paper is only a first cut at the subject, so I welcome readers' future improvements on all the algorithms.

Notations: We write x^2 for "x squared", "//" for "parallel to", "//gm" for "parallelogram", "pt" for "point".

2. Fundamental Theorems needed.
These theorems can be easily taught without introducing concepts of projectivities, homogeneous coordinates or the real projective plane. Let us assume a given parabola K: y^2 = 4 p x where p is a constant. Then x-axis is its axis of symmetry, its focus F is (p,0), its directrix is x = - p. The point O(0,0) is called its vertex. The set {(x,y) : y^2 < 4 p x } is called the interior of K. {(x,y) : y^2 > 4 p x } is its exterior. The positive x-axis is called the eastern direction.

Thm. 1. (Pole and polar). Given a point A not on K. For each line L through A, if it cuts K at 2 points B,C, then the 4-th harmonic point A' shall be called the harmonic image of A on L. Then for all lines L through A which also cut K at 2 points, the set of all harmonic images of A will form a subset of a straight line L' called the polar of A. A is called the pole of L'.
A nice proof: (I am happy to found this proof without too much algebra.) In 3-D space, let the parabola K be drawn only on the plane P1: z = 2 p. Then one can prove that the cone with the origin as vertex, going through the parabola is a circular cone, and a circle K' is cut on the cone by the plane P2: x + z = 2 p. Then the perspectivity f radiated from V(0,0,0) (like a light-bulb) maps points of P1 to P2 (almost a bijection) and lines to lines (except a line on each plane). Using Menelaus and Cevas' theorems we can prove harmonic quadruples are preserved under the projection. In the plane P2, the harmonic images of f(A) on all lines that cut the circle K' can be easily proved to lie on a straight line, called the polar of f(A) relative to K'. Now the inverse mapping f^-1 will take the polar in P2 back to a line in P1. QED

Thm. 2. (Tangents and the polar). If A is in the exterior of K, then the polar of A will cut K at 2 points B,C, and AB, AC are tangents to K.

Thm. 3. (I scratch your back, you scratch mine). Every line in the plane, if not // x-axis, has a unique pole, and every point A in the plane has a unique polar. If a point B lies on the polar of A, then A lies on the polar of B.

Thm. 4. (Parallel chords). Any family X of parallel chords have their mid pts on a line L // x-axis. The polar of any point P on any line L parallel to the x-axis, west of K, will be one of a family of parallel chords each bisected by L.
Proof: Just pull the point A in Thm. 1 along one of the lines in X to infinity. All the lines through A will converge to members of X, and then all the harmonic images will become midpoints of the chords. These images lie on a line L which will become parallel to the x-axis, and meanwhile the polar of any point on L will pass through A. QED

Thm. 5. (Parabolic mirror). At any point A on K, the tangent line bisects the angle between the line thru A // x-axis and the line FA.

Thm. 6. (Tangents of focal chords). Let AFB be a focal chord. Then tangents at A, B cross each other at right angles on the directrix. A nice proof: The right angle follows from Thm. 5. That they intersect on the directrix follows Thm. 3 and the fact that the directrix is the polar of the focus. QED

3. Basic Constructions without the parabola.

Constr. 1. (4th harmonic point). Given A,B,C on a line L1, and B is not mid pt of AC. Construct the 4th harmonic point D. Soln. Pick any point E not on L1. Join EA, EB, EC. Pick any point G on EB. Draw line AG to cross EC at P. Draw line CG to cross EA at Q. Draw line PQ to cross L1 at D. Proof. Exercise using Menelaus and Cevas' theorems.

Constr. 2. (Parallel lines). Given 4 points A,B,C,D with B being the mid pt of AC. Draw a line through D parallel to ABC. Soln. Pick any point E on line AD. Join EB, EC. Draw CD cutting EB at G. Draw line AG cutting EC at H. Then line DH will // line ABC. Proof. Suppose contrary, then DF will cut ABC at some point P forming a harmonic quadruple with A,B,C. Then AB = BC is impossible. QED

Hit here for Section 4 ff.