Pedal Polygons: What Is This About?
A Mathematical Droodle
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Copyright © 1996-2018 Alexander Bogomolny
Pedal Polygons
The applet suggests the following theorem [Kodokostas]:
Assume points A_{i}, |
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The proof follows from extending the simplest case - that of
If N = 2, then by construction, angles PM_{1}O and PM_{2}O add up to 180°. The quadrilateral PM_{1}OM_{2} 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. M_{1}M_{2} is a chord subtending the same central angle M_{1}OM_{2} regardless of the position of P. We conclude that M_{1}M_{2} preserves its length for all P on the given circle.
For other angles of inclination, we still have
PM_{1}O + PM_{2}O = 180°. |
Angles M_{1}PM_{2} and M_{1}OM_{2}, too, add up to 180°, and, since the latter is fixed, so is the former. To show that the length of the segment M_{1}OM_{2} that subtends a fixed angle M_{1}PM_{2}, does not depend on P, suffice it to demonstrate that the radius of the circle that circumscribes PM_{1}OM_{2} does not depend on P. But this is so because the segment OP if fixed length R always subtends the same angle, say, OM_{1}P. (The situation is reminiscent of the combination of two lighthouses with rotating beams.)
If N = 3, each side of the triangle M_{1}M_{2}M_{3} 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 M_{1}M_{2}M_{3} 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 M_{1}M_{2}...M_{N} 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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