Pedal polygons result by projecting a point P on the sides of a polygon. Steiner proved a basic theorem for these polygons, namely that their area is a function of P whose level-lines are circles: Theorem (Steiner Werke I, p. 15)
Given a polygon c, the geometric locus of points P such that their pedal polygons c' with respect to c have area equal to a constant k is a circle. The centers of these circles, for various values of k are the same.
Discard for the moment the index from C1. The proof amounts to measure the difference:
D = area(ACBP)-2*area(ABP).
area(ACBP) = area(ACB)+area(ABP) = (1/2)*(AC*CB*sin(w)+PA*PB*sin(w)).
2*area(ABP) = PA*PB*sin(w). Hence
D = (1/2)*sin(w)*(AC*CB-PA*PB). Using angle a, this implies:
D = (1/2)*sin(w)*PC2*(cos(a)*cos(C-a)-sin(a)*sin(C-a)) = (1/2)*sin(w)*PC2*cos(C) =>
D = (1/4)*sin(2*C)*PC2.
The sum of areas of all quadrangles like ACBP gives the area of the initial polygon. The sum of the areas of all triangles ABP gives the area of the pedal polygon c'.
Thus summing over all such differences we get for the polygon c, and the pedal c' :
area(c) - 2*area(c') = (1/4)*(PC12*sin(2*C1) + ... + PCn2*sin(2*Cn)).
Thus, when area(c') = k (a constant), then the weighted sum on the right of the previous equation is also constant. But then the locus of points P making such weighted sums of squares constant is a circle with the stated property about its center. This is handled in SquaresCombination.html .
See the file Projection_Polygon.html illustrating the special case where c is a regular polygon. Notice that the Simson lines of a triangle represent also a special case of this theorem. A case for which the pedal triangle has zero area.
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Pedal Polygons ![[0_0]](PedalPolygons_files/PedalPolygons_0_0_0.png)
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