title: Petr–Douglas–Neumann theorem
text: # Petr–Douglas–Neumann theorem
  
  In geometry, the Petr–Douglas–Neumann theorem (or the PDN-theorem) is a result concerning arbitrary planar polygons. The theorem asserts that a certain procedure when applied to an arbitrary polygon always yields a regular polygon having the same number of sides as the initial polygon. The theorem was first published by Karel Petr (1868–1950) of Prague in 1908. The theorem was independently rediscovered by Jesse Douglas (1897–1965) in 1940 and also by B H Neumann (1909–2002) in 1941. The naming of the theorem as Petr–Douglas–Neumann theorem, or as the PDN-theorem for short, is due to Stephen B Gray. This theorem has also been called Douglas's theorem, the Douglas–Neumann theorem, the Napoleon–Douglas–Neumann theorem and Petr's theorem.
  
  The PDN-theorem is a generalisation of the Napoleon's theorem which is concerned about arbitrary triangles and of the van Aubel's theorem which is related to arbitrary quadrilaterals.
  
  ## Statement of the theorem
  
  The Petr–Douglas–Neumann theorem asserts the following.
  
      If isosceles triangles with apex angles 2kπ/n are erected on the sides of an arbitrary n-gon $A_0,$ 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 $A_0.$
  
  ## Specialisation to triangles
  
  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 $A_1$ is a regular 3-gon, that is, an equilateral triangle.
  
  $A_1$ is formed by the apices of the isosceles triangles with apex angle 2π/3 erected over the sides of the triangle $A_0.$ The vertices of $A_1$ are the centers of equilateral triangles erected over the sides of triangle $A_0.$ 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.
  
  ## Specialisation to quadrilaterals
  
  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 $A_2$ is a regular 4-gon, that is, a square. The two-stage process yielding the square $A_2$ 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.)
  
  ### Construct $A_1$ using apex angle π/2 and then $A_2$ with apex angle π.
  
  In this case the vertices of $A_1$ are the free apices of isosceles triangles with apex angles π/2 erected over the sides of the quadrilateral $A_0.$ The vertices of the quadrilateral $A_2$ are the midpoints of the sides of the quadrilateral $A_1.$ By the PDN theorem, $A_2$ is a square.
  
  The vertices of the quadrilateral $A_1$ are the centers of squares erected over the sides of the quadrilateral $A_0.$ The assertion that quadrilateral $A_2$ is a square is equivalent to the assertion that the diagonals of $A_1$ are equal and perpendicular to each other. The latter assertion is the content of van Aubel's theorem.
  
  Thus van Aubel's theorem is a special case of the PDN-theorem.
  
  ### Construct $A_1$ using apex angle π and then $A_2$ with apex angle π/2.
  
  In this case the vertices of $A_1$ are the midpoints of the sides of the quadrilateral $A_0$ and those of $A_2$ are the apices of the triangles with apex angles π/2 erected over the sides of $A_1.$ The PDN-theorem asserts that $A_2$ is a square in this case also.
  
  ### Images illustrating application of the theorem to quadrilaterals
  
  | Petr–Douglas–Neumann theorem as applied to a quadrilateral $A_0$ = ABCD. $A_1$ = EFGH is constructed using apex angle π/2 and $A_2$ = PQRS with apex angle π. | Petr–Douglas–Neumann theorem as applied to a quadrilateral $A_0$ = ABCD. $A_1$ = EFGH is constructed using apex angle π and $A_2$ = PQRS with apex angle π/2. |
  | - | - |
  | Petr–Douglas–Neumann theorem as applied to a self-intersecting quadrilateral $A_0$ = ABCD. $A_1$ = EFGH is constructed using apex angle π/2 and $A_2$ = PQRS with apex angle π. | Petr–Douglas–Neumann theorem as applied to a self-intersecting quadrilateral $A_0$ = ABCD. $A_1$ = EFGH is constructed using apex angle π and $A_2$ = PQRS with apex angle π/2. |
  
  * Diagram illustrating the fact that van Aubel's theorem is a special case of Petr–Douglas–Neumann theorem.
  
  ## Specialisation to pentagons
  
  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, $A_3$ is a regular pentagon. The three-stage process leading to the construction of the regular pentagon $A_3$ 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.
  
  | Serial number | Apex angle in the construction of $A_1$ | Apex angle in the construction of $A_2$ | Apex angle in the construction of $A_3$ |
  | - | - | - | - |
  | 1 | 72° | 144° | 216° |
  | 2 | 72° | 216° | 144° |
  | 3 | 144° | 72° | 216° |
  | 4 | 144° | 216° | 72° |
  | 5 | 216° | 72° | 144° |
  | 6 | 216° | 144° | 72° |
  
  ## Proof of the theorem
  
  The theorem can be proved using some elementary concepts from linear algebra.
  
  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 $C^n.$ Take an n-gon A and let it be represented by the complex vector
  
      A = ( $a_1,$ $a_2,$ ... , $a_n$ ).
  
  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 = ( $b_1,$ $b_2,$ ... , $b_n$ ).
  
  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 : $C^n$ → $C^n$ 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 $A_0$ 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 = ( $p_1,$ $p_2,$ ..., $p_n$ ) 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 $ω^n$ = 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 − $ω^2$ $)^{−1}$ ( S − $ω^2$ I ) ... ( 1 − $ω^{n−2}$ $)^{−1}$ ( S − $ω^{n−2}$ I ) $A_0$
         = ( 1 − ω $)^{−1}($ 1 − $ω^2$ $)^{−1}$ ... ( 1 − $ω^{n−2}$ $)^{−1}$ ( S − I ) ( S − ω I ) ( S − $ω^2$ I ) ... ( S − $ω^{n−1}$ $I)A_0$
         = ( 1 − ω $)^{−1}($ 1 − $ω^2$ $)^{−1}$ ... ( 1 − $ω^{n−2}$ $)^{−1}$ ( $S^n$ − I ) $A_0$
         = 0, since $S^n$ = I.
  