Pedal Polygons: What Is This About?
A Mathematical Droodle

Explanation

Copyright © 1996-2008 Alexander Bogomolny

The applet suggests the following theorem [Kodokostas]:

Assume points Ai, i = 1, 2, ..., N (N > 1) lie on a circle C with center O and radius R. Let P be another point on the circle. Through point P, draw lines equally inclined to each of the lines CAi, i = 1, 2, ..., N. (The angles are also assumed to be similarly oriented.) Denote the feet of the lines so drawn Mi, i = 1, 2, ..., N. Then the polygon M1M2...MN does not depend on the position of P on the circle.

The proof follows from extending the simplest case - that of N = 2, which in turn, exploits the same features of the construction as the theorems of Wallace and Carnot.

If N = 2, then by construction, angles PM₁O and PM₂O add up to 180^o. The quadrilateral PM₁OM₂ is, therefore, cyclic. When the common angle of inclination is 90^o, OP is a diameter of its circumscribed circle, the radius of this circle is R/2 and does not depend on the position of P. M₁M₂ is a chord subtending the same central angle M₁OM₂ regardless of the position of P. We conclude that M₁M₂ preserves its length for all P on the given circle.

For other angles of inclination, we still have

Angles M₁PM₂ and M₁OM₂, too, add up to 180^o, and, since the latter is fixed, so is the former. To show that the length of the segment M₁OM₂ that subtends a fixed angle M₁PM₂, does not depend on P, suffice it to demonstrate that the radius of the circle that circumscribes PM₁OM₂ does not depend on P. But this is so because the segment OP if fixed length R always subtends the same angle, say, OM₁P. (The situation is reminiscent of the combination of two lighthouses with rotating beams.)

If N = 3, each side of the triangle M₁M₂M₃ maintains its length as P ranges over the circle. By SSS, the triangle is, therefore, the same for all positions of P. (The shape of triangles M₁M₂M₃ is uniquely determined by the points _i, i = 1, 2, ..., N, whereas their size is a function the angle of inclination.)

This is a consequence of the case N = 2, that for N > 3, the side lengths of the polygon M₁M₂...M_N do not change as P moves over the circle. And this is a consequence of the case N = 3 that, for N > 3, the angles of M₁M₂...M_N do not change either. Indeed, each of those angles is an algebraic sum of the angles M_i1M_i2M_i3 for various combinations of indices I₁, I₂, and I₃.

References

1. D. Kodokostas, Projected Rotating Polygons, Math Magazine, vol 77, No 5, Dec. 2004

Copyright © 1996-2008 Alexander Bogomolny