Pedal Polygons: What Is This About?
A Mathematical Droodle


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Pedal Polygons

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.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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°. The quadrilateral PM1OM2 is, therefore, cyclic. When the common angle of inclination is 90°, 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

  PM1O + PM2O = 180°.

Angles M1PM2 and M1OM2, too, add up to 180°, 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...MN 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...MN do not change either. Indeed, each of those angles is an algebraic sum of the angles Mi1Mi2Mi3 for various combinations of indices I1, I2, and I3.

References

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

 63437967

Search by google: