Petr–Douglas–Neumann theorem

The Petr–Douglas–Neumann theorem asserts the following.[3][5]

If isosceles triangles with apex angles 2kπ/n are erected on the sides of an arbitrary n-gon A₀, and if this process is repeated with the n-gon formed by the free apices of the triangles, but with a different value of k, and so on until all values 1 ≤ k ≤ n − 2 have been used (in arbitrary order), then a regular n-gon A_n−2 is formed whose centroid coincides with the centroid of A₀.

Diagram illustrating the fact that Napoleon's theorem is a special case of Petr–Douglas–Neumann theorem.

In the case of triangles, the value of n is 3 and that of n − 2 is 1. Hence there is only one possible value for k, namely 1. The specialisation of the theorem to triangles asserts that the triangle A₁ is a regular 3-gon, that is, an equilateral triangle.

A₁ is formed by the apices of the isosceles triangles with apex angle 2π/3 erected over the sides of the triangle A₀. The vertices of A₁ are the centers of equilateral triangles erected over the sides of triangle A₀. Thus the specialisation of the PDN theorem to a triangle can be formulated as follows:

If equilateral triangles are erected over the sides of any triangle, then the triangle formed by the centers of the three equilateral triangles is equilateral.

The last statement is the assertion of the Napoleon's theorem.

In the case of quadrilaterals, the value of n is 4 and that of n − 2 is 2. There are two possible values for k, namely 1 and 2, and so two possible apex angles, namely:

(2×1×π)/4 = π/2 = 90° ( corresponding to k = 1 )

(2×2×π)/4 = π = 180° ( corresponding to k = 2 ).

According to the PDN-theorem the quadrilateral A₂ is a regular 4-gon, that is, a square. The two-stage process yielding the square A₂ can be carried out in two different ways. (The apex Z of an isosceles triangle with apex angle π erected over a line segment XY is the midpoint of the line segment XY.)

Images illustrating application of the theorem to quadrilaterals

Petr–Douglas–Neumann theorem as applied to a quadrilateral A0 = ABCD. A1 = EFGH is constructed using apex angle π/2 and A2 = PQRS with apex angle π.	Petr–Douglas–Neumann theorem as applied to a quadrilateral A0 = ABCD. A1 = EFGH is constructed using apex angle π and A2 = PQRS with apex angle π/2.

Petr–Douglas–Neumann theorem as applied to a self-intersecting quadrilateral A0 = ABCD. A1 = EFGH is constructed using apex angle π/2 and A2 = PQRS with apex angle π.	Petr–Douglas–Neumann theorem as applied to a self-intersecting quadrilateral A0 = ABCD. A1 = EFGH is constructed using apex angle π and A2 = PQRS with apex angle π/2.

Diagram illustrating the fact that van Aubel's theorem is a special case of Petr–Douglas–Neumann theorem.

Diagram illustrating Petr–Douglas–Neumann theorem as applied to a pentagon. The pentagon A₀ is ABCDE. A₁ ( = FGHIJ ) is constructed with apex angle 72°, A₂ ( = KLMNO ) with apex angle 144° and A₃ ( = PQRST ) with apex angle 216°.

In the case of pentagons, we have n = 5 and n − 2 = 3. So there are three possible values for k, namely 1, 2 and 3, and hence three possible apex angles for isosceles triangles:

(2×1×π)/5 = 2π/5 = 72°

(2×2×π)/5 = 4π/5 = 144°

(2×3×π)/5 = 6π/5 = 216°

According to the PDN-theorem, A₃ is a regular pentagon. The three-stage process leading to the construction of the regular pentagon A₃ can be performed in six different ways depending on the order in which the apex angles are selected for the construction of the isosceles triangles.

The theorem can be proved using some elementary concepts from linear algebra.[2][6]

The proof begins by encoding an n-gon by a list complex numbers representing the vertices of the n-gon. 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

A = ( a₁, a₂, ... , a_n ).

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

B = ( b₁, b₂, ... , b_n ).

Then we have

α( a_r − b_r ) = a_r+1 − b_r, where α = exp( i θ ) for some θ (here i is the square root of −1).

This yields the following expression to compute the b_r ' s:

b_r = (1−α)⁻¹ ( a_r+1 − αa_r ).

In terms of the linear operator S : Cⁿ → Cⁿ that cyclically permutes the coordinates one place, we have

B = (1−α)⁻¹( S − αI )A, where I is the identity matrix.

This means that the polygon A_n−2 that we need to show is regular is obtained from A₀ by applying the composition of the following operators:

( 1 − ω^k )⁻¹( S − ω^k I ) for k = 1, 2, ... , n − 2, where ω = exp( 2πi/n ). (These commute because they are all polynomials in the same operator S.)

A polygon P = ( p₁, p₂, ..., p_n ) 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

p_{r + 1} − p_r = ω( p_{r + 2} − p_{r + 1} ).

This condition can be formulated in terms of S as follows:

( S − I )( I − ωS ) P = 0.

Or equivalently as

( S − I )( S − ω^{n − 1} I ) P = 0, since ωⁿ = 1.

Petr–Douglas–Neumann theorem now follows from the following computations.

( S − I )( S − ω^{n − 1} I ) A_{n − 2}

= ( S − I )( S − ω^{n − 1} I ) ( 1 − ω )⁻¹ ( S − ω I ) ( 1 − ω² )⁻¹ ( S − ω² I ) ... ( 1 − ω^{n − 2} )⁻¹ ( S − ω^{n − 2} I ) A₀

= ( 1 − ω )⁻¹( 1 − ω² )⁻¹ ... ( 1 − ω^{n − 2} )⁻¹ ( S − I ) ( S − ω I ) ( S − ω² I ) ... ( S − ω^{n − 1} I)A₀

= ( 1 − ω )⁻¹( 1 − ω² )⁻¹ ... ( 1 − ω^{n − 2} )⁻¹ ( Sⁿ − I ) A₀

= 0, since Sⁿ = I.