Pedal Polygons: What Is This About?
A Mathematical Droodle
What if applet does not run? |
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Copyright © 1996-2018 Alexander Bogomolny
Pedal Polygons
The applet suggests the following theorem [Kodokostas]:
Assume points Ai, |
What if applet does not run? |
The proof follows from extending the simplest case - that of
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,
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
References
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Copyright © 1996-2018 Alexander Bogomolny
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