1000/1000

Hot
Most Recent

Submitted Successfully!

To reward your contribution, here is a gift for you: A free trial for our video production service.

Thank you for your contribution! You can also upload a video entry or images related to this topic.

Do you have a full video?

Are you sure to Delete?

Cite

If you have any further questions, please contact Encyclopedia Editorial Office.

HandWiki. Petr–Douglas–Neumann Theorem. Encyclopedia. Available online: https://encyclopedia.pub/entry/36875 (accessed on 14 June 2024).

HandWiki. Petr–Douglas–Neumann Theorem. Encyclopedia. Available at: https://encyclopedia.pub/entry/36875. Accessed June 14, 2024.

HandWiki. "Petr–Douglas–Neumann Theorem" *Encyclopedia*, https://encyclopedia.pub/entry/36875 (accessed June 14, 2024).

HandWiki. (2022, November 28). Petr–Douglas–Neumann Theorem. In *Encyclopedia*. https://encyclopedia.pub/entry/36875

HandWiki. "Petr–Douglas–Neumann Theorem." *Encyclopedia*. Web. 28 November, 2022.

Copy Citation

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.

quadrilaterals
regular polygon
pdn-theorem

The Petr–Douglas–Neumann theorem asserts the following.^{[1]}^{[2]}

*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}

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.

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*.)

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.

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.

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-intersectingquadrilateral 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-intersectingquadrilateral 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. https://handwiki.org/wiki/index.php?curid=2079397 |

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° |

The theorem can be proved using some elementary concepts from linear algebra.^{[3]}^{[4]}

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}− α*a*_{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*− α*I*)*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( 2π*i*/*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*− ω*S*)*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*.

- = (

- Douglas, Jesse (1946). "On linear polygon transformations". Bulletin of the American Mathematical Society 46 (6): 551–561. doi:10.1090/s0002-9904-1940-07259-3. http://www.ams.org/journals/bull/1940-46-06/S0002-9904-1940-07259-3/S0002-9904-1940-07259-3.pdf. Retrieved 7 May 2012.
- "Petr–Neumann–Douglas Theorem.". From MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/Petr-Neumann-DouglasTheorem.html. Retrieved 8 May 2012.
- Stephen B. Gray (2003). "Generalizing the Petr–Douglas–Neumann Theorem on n-gons". American Mathematical Monthly 110 (3): 210–227. doi:10.2307/3647935. http://www.experimentalmath.info/workshop2004/gray-article.pdf. Retrieved 8 May 2012.
- Omar Antolín Camarena. "The Petr–Neumann–Douglas theorem through linear algebra". http://www.matem.unam.mx/omar/notes/petr.html. Retrieved 10 Jan 2018.

More

Information

Subjects:
Mathematics

Contributor
MDPI registered users' name will be linked to their SciProfiles pages. To register with us, please refer to https://encyclopedia.pub/register
:

View Times:
373

Entry Collection:
HandWiki

Update Date:
28 Nov 2022