Petr–Douglas–Neumann theorem

In this case the vertices of A1 are the free apices of isosceles triangles with apex angles π/2 erected over the sides of the quadrilateral A0.

So there are three possible values for k, namely 1, 2 and 3, and hence three possible apex angles for isosceles triangles: According to the PDN-theorem, A3 is a regular pentagon.

This list can be thought of as a vector in the n-dimensional complex linear space Cn.

Take an n-gon A and let it be represented by the complex vector Let the polygon B be formed by the free vertices of similar triangles built on the sides of A and let it be represented by the complex vector Then we have This yields the following expression to compute the br ' s: In terms of the linear operator S : Cn → Cn that cyclically permutes the coordinates one place, we have This means that the polygon Aj+1 obtained at the jthe step is related to the preceding one Aj by where ω = exp( 2πi/n ) is the nth primitive root of unity and σj is the jth term of a permutation σ of the integer sequence (1,..., n-2).

The last polygon in the sequence, An−2, which we need to show is regular, is thus obtained from A0 by applying the composition of all the following operators: (These factors commute, since they are all polynomials in the same operator S, so the ordering of the product does not depend on the choice of the permutation σ.)

A polygon P = ( p1, p2, ..., pn ) is a regular n-gon if each side of P is obtained from the next by rotating through an angle of 2π/n, that is, if This condition can be formulated in terms of S as follows: Or equivalently as The main result of the Petr–Douglas–Neumann theorem now follows from the following computations.

Taking the scalar product of both sides of the equation with E, and noting that E is invariant under the cyclic permutation operator S, we obtain so all the centroids are equal.

Diagram illustrating the fact that Napoleon's theorem is a special case of Petr–Douglas–Neumann theorem.
Diagram illustrating Petr–Douglas–Neumann theorem as applied to a pentagon. The pentagon A 0 is ABCDE . A 1 ( = FGHIJ ) is constructed with apex angle 72°, A 2 ( = KLMNO ) with apex angle 144° and A 3 ( = PQRST ) with apex angle 216°.