Magic Polygons and Degenerated Magic Polygons: Characterization and Properties
Danniel Dias Augusto Josimar da Silva Rocha





ed1]Danniel Dias Augusto
nieldiasguto@gmail.com
ed]Josimar da Silva Rochaau2
jrocha@utfpr.edu.br

[au2]Corresponding author

[ed1]Coordenação da Matemática, Universidade Estadual do Goiás, Unidade Universitária de Formosa, 73807-250, Formosa - GO, Brazil 

[ed]Departamento de Matemática, Universidade Tecnológica Federal do Paraná, 
 Câmpus Cornélio Procópio, 

86300-000, Cornélio Procópio - PR, Brazil





In this work we define Magic Polygons  P(n,k) and Degenerated Magic Polygons D(n,k) and we obtain their main properties, such as the  magic sum and the value  corresponding to the  root vertex. The  existence of magic polygons P(n, k) and degenerated magic polygons D(n, k) are discussed for certain values of  n and k. 




Combinatorics Magic Polygons Degenerated Magic Polygons.
 





§ INTRODUCTION



Magic Squares have been known for a long time in different people and cultures that, sometimes, has attributed mystic meanings <cit.>.  In addiction to being  used for recreational purposes, we can now find applications for magic squares in Physics, in Computer Science, in Image Processing and in Criptography <cit.>, among others.   In this way, have been developed several  methods to construct magic squares that satisfies some  particular properties and some generalization have been created, as we can see in  <cit.>. 

We can see in <cit.> a generalization of the same idea of the representation magic squares of order 3 using vertices, midpoints and the geometric center of a square, where we can find some properties and the condiction of existence of magic polygons and a construction for magic square of order 4, for each  n even. In <cit.> others similar structures was proposed.

In this work we will cover a generalization of the work proposed in  <cit.> and we show that some valid properties of  magic polygons for a given  order sometimes are not valid in general. In addition, we will introduce another class of polygonal structure known as the class of degenerate magic polygons.




§  MAGIC POLYGONS   P(N,K)


Let Ω be a set of  k/2 regular polygons on plane with n sides and corresponding parallel sides and centered in a central point  C. 

A magic polygon  P(n, k) of  n sides and order k+1 is a set of 
  k^2n/2 + 1 points satisfying the following conditions:
  

  (i) Points of magic polygon are labeled by distict values from  1 to  k^2n/2 + 1;  

  (ii) One point of a magic polygon is the central point C;

  (iii) kn/2 points of magic polygon are vertices of the  k/2 regular polygons of  Ω; 

  (iv) The magic polygon has  k-1 intermediate points on each edge of regular polygons in  Ω, which gives a total of  
(k-1)nk/2 intermediate points.


  (v) Segments with  diametrically opposite ends of the larger polygon of  Ω intersecting the central vertex contain k+1 points of the magic polygon;


  (vi) Segments with ends at two adjacent vertices of a polygon of  Ω contains  k+1 points of the magic polygon;

  (vii) The sum of values corresponding  to the k+1 points on each segment defined in  (iv) and  (v) is a fixed value  u, called of  magic sum.  


In Figures <ref> e <ref> we can see examples of Magic Polygons P(4, 4) and P(8,2), respectively.


 




 
  In a magic polygon  P(n,k), we have the following properties:
 
 
  (i) the magic sum is  (k+1)k^2n+4/4;
 
  (ii) the value corresponding to the root vertex is  c = k^2n+4/4;

  (iii) the sum S_j of the values representing to   j-th points partitioning each edge on magic polygon  chosen clockwise is 

    S_j =  kn[k^2n+4]/8







Let  x_(t-1)nk+(i-1)k+j be the value correspondig to the point  P_(t-1)nk+(i-1)k+j,  j-th point of the  i-th edge of the  i-th regular polygon.

Each point  P_(t-1)nk+(i-1)k+j of the magic polygon is labeled for a number 

    x_(t-1)nk+(i-1)k+j

where  t ∈{1, 2, ⋯, k/2}, i ∈{ 1, 2, ⋯, n} and j∈{1, ⋯, k}  and the root vertice is labeled by the number c, then we have 

    k^2n/2 + 1
 
point in the magic polygon.

The sum of values correspondig points on the  i-th edge of  t-th regular polygon is given by 

    x_(t-1)nk+(i-1)k+1 + x_(t-1)nk+(i-1)k+2 + ⋯ + x_(t-1)nk+ik + x_(t-1)nk+ik+1 = u,

where  u is the magic sum.


The sum of values correspondig points on the segments determined by the points  P_j+ik and P_j+ (i-1)k + kn/2 is 


    ∑_t=1^k/2 x_(t-1)nk + (i-1)k+j + ∑_t=1^k/2 x_(t-1)nk+(i-1)k+j+kn/2+ c = u,

 where c is the value assigned to the root vertice and  j + (i-1)k ≤kn/2 - 1.



Let 

    S_j = ∑_i=1^n∑_t=1^k/2 x_(t-1)nk+(i-1)k+j

for  j ∈{1, 2, ⋯, k}. 


By (<ref>) and (<ref>),

    S_j +nc/2 = nu/2,

which implies  that 

    S_j = n(u-c)/2


Adding equations involving points on perimeters of the polygons, we obtain

    2S_1 + S_2 + ⋯ + S_k  = knu/2.


Adding equations involving the root vertex of the magic polygon, we obtain

    S_1 + S_2 + ⋯ + S_k + knc/2 = knu/2

which implies that 


    S_1+S_2 + ⋯ + S_k = kn(u-c)/2


Subtracting  (<ref>) from  (<ref>), we have 


    S_1 = knc/2


By (<ref>) and (<ref>),  

    n(u-c)/2 = knc/2.


Therefore

    u = (k+1)c



As  values corresponding to the points of the magic polygon are distinct values in  {1, 2, ⋯, k^2n/2+1},
it follows that 


    S_1 + S_2 + ⋯ + S_k + c = ∑_i=1^k^2n/2+1 i = (k^2n/2+1)(k^2n/2 + 2)/2


By (<ref>) and (<ref>), it follows that 


    kn(u-c)/2 + c = (k^2n/2+1)(k^2n/2 + 2)/2


By (<ref>) and (<ref>), it follows that 

    (k^2n/2+1)c = (k^2n/2+1)(k^2n/2 + 2)/2.


Therefore

    c = k^2n+4/4



Moreover,

    S_j(<ref>)=n(u-c)/2(<ref>)=nkc/2(<ref>)=nk[k^2n+4]/8





 §.§ Constructing examples for P(n,4)

The following result affords us a construction for  P(n,4), provided that some conditions are satisfied:


Let two regular polygons with n sides of distinct sizes centered on a central point C whose sides are partitioned into 4 segments by points such that 2n segments passing through the center point and cutting the larger polygon intercept the sides of the polygons at these points, satisfying the following conditions:



  (i) 
Each point P_4(t-1)n+4(i-1)+j on partitions is labeled by   
x_4(t-1)n+4(i-1)+j∈{1, 2, ⋯, 8n+1};
  

  (ii) a central point is labeled by   c = 4n+1;


  (iii) x_4(i-1)+j + x_4n + 4(i-1) + j = 2(4n+1);

  (iv) x_4(t-1)n+4(i_1-1)+j_1 + x_4(t-1)n+4(i_2-1)+j_2≠ 2(4n+1), if  (i_1,j_1) ≠ (i_2, j_2); 

  (v) x_4(i-1)+1 + x_4(i-1)+2 + x_4(i-1)+3 + x_4(i-1)+4 + x_4(i-1)+5 = 5(4n+1);

where t ∈{1, 2}, i_1, i_2, i ∈{ 1, 2, ⋯, n} e j_1, j_2, j∈{1, 2, 3, 4}.

In this conditions, we obtain  8n+1 points that define a magic polygon  P(n,4). 


 Let i ∈{1, 2, ⋯, n-1}. 

By  (iii), we have  x_4(i-1)+j + x_4n+4(i-1) + j = 2(4n+1) for j ∈{1, 2, 3, 4}. 

Therefore

    x_4(i-1)+1 + x_4n+4(i-1)+1 + ⋯ + x_4(i-1) + 4 + x_4n + 4(i-1) + 4  + x_4(i-1) + 5 + x_4n + 4(i-1) + 5 = 5· 2 (4n+1)


By   (<ref>) and (v), we get  

    5(4n+1) + x_4n+4(i-1)+1 + x_4n+4(i-1)+2 + x_4n + 4(i-1) + 3 + x_4n+4(i-1) + 4 + x_4n + 4(i-1) + 5 = 10(4n+1)
 

Thus 

    x_4n + 4(i-1)+ 1 + x_4n+4(i-1) + 2 + x_4n + 4(i-1) + 3 + x_4n + 4(i-1) + 4 + x_4n + 4(i-1) + 5 = 5(4n + 1)
 

Consequently, the sum of values corresponding to the points on edges of the polygons is   5(4n + 1). 

Therefore, by (iii) and  (ii), we get  


    x_4(i-1) + j + x_4n + 4(i-1) + j + c + x_4(i-1) + 2n + j + x_4n + 4(i-1) + 2n + j = 2(4n+1) + 4n+1 + 2(4n+1) = 5(4n+1)
 

Consequently, the sum of values corresponding to the points on the segments lying by central point is also 5(4n+1). 


Hence, by definition, we get  P(n, 4).  


In Figure <ref> we have an example of Magic Polygon P(4, 4) constructed by Theorem <ref>. An example of Magic Polygon P (4, 4) that can not be obtained by this construction can be seen in the figure <ref>.








 §.§ The  particular case P(n,2)


Although it is a particular case of the general case seen in the previous section, we will use another reasoning for a demonstration of the properties of the magic polygons P(n,2). 


A magic polygon P(n,2) is formed by 2n+1 points, consisting of the vertices, the midpoints of the edges, and the geometric center of a regular polygon of n sides, which are labeled by numerical values from 1 to 2n+1, so that the sum of the values assigned to any three collinear pointes is constant, called magic sum.



 If n is even, then a magic polygon  P(n,2) has the following properties:
 
 
  (i) the magic sum is  é 3(n+1);
 
  (ii) the value corresponding to the central point is   é n+1;

  (iii) the sum  S_1 of the  values assigned  to  vertices of the magic polygon and the sum  S_2 of the values assigned to  midpoints of  edges of the magic polygon  satisfies S_1 = S_2 = n(n+1).






As a polygon with  n sides has  n vertex and  n midpoints, including a central point, we obtain 2n+1 points labeled using  distinct integers numbers  from  1 to 2n+1.

Let  x_n be the value assign to vertex V_n and  let be   y_n the values assign to the midpoint  M_n whose endpoints are  V_n and V_1.  For each  i ∈{1, ⋯, n-1}, let  x_i be the value assigned to the vertex V_i and let  y_i be the value assigned to midpoint M_i whose endpoints are  V_i e V_i+1. 


For each  i ∈{1, ⋯, n/2}, let x_i^∗ be the value assigned to the point diametrically opposite to vertice  V_i and  y_i^∗ be the value assigned to the point  diametrically opposite to the midpoint  M_i. 

Denoting by  c the value assigned by the  central point  C of the magic polygon and denoting by  u the value of the magic sum, we obtain two sets of equations that define a magic polygon with n sides, for an even  number n: the equations involving points on the same side  and equations involving points on simmetry axes of the magic polygon.


Analyzing equation involving points on the same side of a magic polygon, we obtain
    
    {[ x_1+y_1+x_2=u; x_2+y_2+x_3=u; x_3+y_3+x_4=u;             ⋮; x_n+y_n+x_1=u;               ].

    
Analyzing equations involving points on the same segment in the definition of a magic polygon, we obtain

    {[       x_1+c+x_1^*=u;       y_1+c+y_1^*=u;                   ⋮; x_n/2+c+x_n/2^* = u; y_n/2+c+y_n/2^* = u;                     ].

    
Let   
    S_1=∑_i=1^n x_i    S_2=∑_i=1^n y_i
   
     
Adding the equations of  (<ref>), we get

    2S_1 + S_2 = nu


Adding the equations of (<ref>), we get

 
    S_1+S_2+nc=nu


Subtracting (<ref>) from (<ref>), we get

    S_1=nc


By  (<ref>) and (<ref>), we get

    S_2=nu-2nc


As the values assigned to the   n midpoints of the magic polygon with  n sides are distinct values of the set {1, ⋯, 2n+1} and the  sum of this values is  S_2, then  


    ∑_i=1^n i ≤ S_2 ≤∑_i=n+2^2n+1 i


In addition,

    ∑_i=1^n i = n(n+1)/2

and 

    ∑_i=n+2^2n+1 i = 3n(n+1)/2


Therefore, by (<ref>), (<ref>) and (<ref>), we get

    n(n+1)/2≤ S_2 ≤3n(n+1)/2


By (<ref>) and (<ref>), we obtain


    S_1+S_2+c=nu-c(n-1)


Moreover, 

    S_1+S_2+c = ∑_i=1^2n+1 i = (2n+1)(2n+2)/2 = (2n+1)(n+1).


Therefore, by (<ref>) and (<ref>), we obtain the follow diophantine equation

    nu -c(n-1) = (2n+1)(n+1)

whose general solution is 

    (u,c) = ( (2n+1)(n+1) + (n-1)t, (2n+1)(n+1) + nt), ∀ t ∈ℤ.


By (<ref>) and (<ref>), we obtain 

    [                                               S_2                                        = nu - 2nc;                                                   = n[ (2n+1)(n+1) + (n-1)t] - 2n[(2n+1)(n+1) + nt];                                                                           = -n(2n+1)(n+1) - n(n+1)t;                                                                                   = -(n+1)n(2n+1+t) ]


By (<ref>) and (<ref>), we get 

    n(n+1)/2≤ -(n+1)n(2n+1+t) ≤3n(n+1)/2.


Simplyfing  (<ref>), we obtain


    1/2≤ -(2n+1+t) ≤3/2


It follows from   -(2n+1+t) ∈ℤ and   (<ref>) that  
-(2n+1+t) = 1; hence 

    t = -2(n+1)


By   (<ref>) and (<ref>), 

    [                                               (u,c)          = ((2n+1)(n+1) + (n+1)t, (2n+1)(n+1) + nt);                                                     = ( (2n+1)(n+1) -2(n+1)(n-1), (2n+1)(n+1) -2n(n+1));                                                                                       = (3(n+1), (n+1));                                                     ]


Therefore,  the magic sum is  u = 3(n+1) and the value assigned to the central point is  c = n+1.

Replacing   u e c from    (<ref>) in  (<ref>) and (<ref>),
 we obtain
 
    S_1 = S_2 = n(n+1).

 
 




 
If  n is odd, then there is no magic polygon  P(n,2). 


 Let  n  be odd and suppose that there is magic polygons with  n sides.


If  x_1, ⋯, x_n are integer numbers corresponding  vertices and  y_1, ⋯, y_n, are integer numbers corresponding  midpoints of the magic polygon, where y_i is the midpoint between the vertices  x_i and x_i+1, we have 


    x_i + y_i + x_i+1 = u

and

    x_i + y_m+i-1 + c = u

where m = n+1/2, c is the value assigned to the root vertex of the magic polygon and  x_a = x_b or  y_a = y_b  if  a ≡ b  n. 

By  (<ref>), we obtain 

    x_i = u - c - y_m+i-1


Substituting  (<ref>) into (<ref>), we obtain


    y_i = 2c - u + y_m+i-1 + y_m+i

and

    y_i+1 = 2c - u + y_m+i + y_m+i+1


By  (<ref>) and (<ref>), we obtain


    [                                                 y_i+1 - y_i                                         = y_m+i+1 - y_m+i-1;                                                                                     = y_m+i+1 - y_m+i + y_m+i - y_m+i-1;                                                             = y_m+(m+i)+1 - y_m+(m+i)-1 + y_m+(m+i-1)+1 - y_m+(m+i-1)-1;                                                                                          = y_i+2 - y_i + y_i+1 - y_i-1,;                                                             ]


Taking    n=3 and  i = 1  in the first equality of  (<ref>),  we obtain  
y_2 - y_1 = y_1 - y_2, which implies  y_1 = y_2. 

Taking  n ≥ 5 and i=2 in  (<ref>), we get     
y_1 = y_4.

Therefore, we can not have a magic polygons with n sides, for n odd.


 

  If n is an even number greater than or equal to 4, then there is a magic polygon P(n,2).


 Figure <ref> shows the existence of Magic Polygons P(4, 2).  Let n be even greater than or equal to 6 and consider a regular polygon with n sides whose perimeter is indicated by the sequence of n vertices clockwise V_1, V_2, ⋯, V_n. Thus, if for each j ∈{1, ⋯, n}, x_j is the value assigned to vertex  V_j and  y_j is the value assigned to midpoint of the side whose  ends are the  vertices  V_j and V_j+1 of the magic polygon, then we obtain a magic polygon with  n sides such that, for  i ∈{ 1, 2, ⋯, n/2 - 2}, the values assigned to vertices of the magic polygon satisfy
 
 
    {[                                                                                       x_i            ={[        i + 1,  if  i is odd;    n + i + 2, if  i is even;               ].;                                                                                          ;                                                                                   x_i+n/2 = {[ 2(n+1) - i - 1,     if i is odd;          n - i,   if  i is even;                 ].;                                                                                   x_n/2-1                                                                                   = n + 3;                                                                                     x_n/2                                                                                       = 1;                                                                                           ].

and the values assigned to midpoints of magic polygon satisfy

    y_j = {[     2(n+1) - x_j - x_j+1, if  j ∈{ 1, 2, 3, ⋯, n-1};       2(n+1) - x_1 - x_n,                  if j = n;                           ].


The verification that this construction defines  a magic polygon can be seen in <cit.>. 


In Figure  <ref> we have an example of Magic Polygon  P(4,2) using construction in the proof of the Theorem <ref>.






 
There are no magic polygons whose values assigned to all vertices have odd parity.


If the values assigned to all vertices are odd numbers, then, by definition of Magic Polygon, the values assigned to midpoints are even numbers. This contradicts the fact that the magic sum is an odd number, by Theorems  <ref> and <ref>.



There is no Magic Polygon  P(n,2) whose values assigned to all midpoints are even numbers.


If all values assigned to midpoints of the Magic Polygon are even numbers, then, by definition of Magic Polygon, the values assigned to vertices of the Magic Polygon are odd numbers. This contradicts the  Proposition  <ref>.




§  DEGENERATED MAGIC POLYGONS D(N,K)


Let  Ω be a set of  k regular polygons with distinct sizes and with a common vertex C, called the root vertex.

A degenerated magic polygon  D(n, k) with  n sides and order  k+1 is a set of   k^2(n-2) + k + 1 points that satisfies the following conditions:



  (i)  k^2(n-2) + k + 1 points of degenerated magic polygon are assigned by distinct numbers from  1 to k^2(n-2) + k + 1; 

  (ii) One point is the root vertex C;

  (iii) kn-k+1 points are the vertices of  k regular polygons of Ω; 

  (iv) each of the edges of the regular polygons not adjacent to the root vertices of  Ω have  k-1 intermediate points of the magic polygon beyond the vertices of the polygons, which gives a total of  (n-2)(k-1)k intermediate points;

  (v) Segments with one end in points at the border of the larger polygon of  Ω and other end on root vertex have  k+1 points of degenerated magic polygon;

  (vi) Segments with ends on adjacent vertices of polygons of  Ω that do not contain the root vertex have  k+1 points of degenerated magic polygon;

  (vii) The sum of values assigned to the k+1 points of the degenerated magic polygon in each of the  segments defined in (v) and (vi) is a fixed value   u, called  the  magic sum. 


 Figures  <ref> and <ref> illustrate  the existence of Degenerated Magic Polygons.

 



 



 
 A degenerated magic polygon  D(n,k) has the following properties:
 
 
  (i) the magic sum is  (k+1)k^2(n-2)+k+2/2;
 
  (ii) the value that corresponds to the root vertex is  c = k^2(n-2)+k+2/2;

  (iii) the sum S_j of the values assigns to the  j-th points on the edges in the representation of the degenerated magic polygon satisfies

    S_j = (n-2)kk^2(n-2)+k+2/2






Let P_(t-1)k(n-2)+k(i-1) + j be the   j-th  point of the  i-th edge of the t-th largest polygon that represents the degenerated magic polygon  D(n,k), considering the clockwise direction, let 

    x_(t-1)k(n-2)+k(i-1) + j
 be the value assigned to the point  P_(t-1)k(n-2)+k(i-1) + j,   

    x_tk(n-2) + 1
 
be the value assigned to the point    P_tk(n-2)+1,
where  j, t ∈{ 1, 2, ⋯, k} and i ∈{ 1, 2, ⋯, (n-2)};  the root vertex is labeled by the number   c. So we have  

    k^2(n-2) + k + 1
 
points in the degenerated magic polygon.

The equations involving  segments that don't  contain the root vertex are 

    {[                        x_(t-1)k(n-2)+1 + x_(t-1)k(n-2) + 2 + ⋯ + x_(t-1)k(n-2) + k + 1 = u;              x_(t-1)k(n-2) + k + 1 + x_(t-1)k(n-2) + k + 2 + ⋯ + x_(t-1)k(n-2) + 2k + 1= u;                                                                                          ⋮;     x_(t-1)k(n-2) + (i-1)k+1 + x_(t-1)k(n-2) + (i-1)k + 2 + ⋯ + x_(t-1)k(n-2) + ik + 1 = u;                                                                                          ⋮; x_(t-1)k(n-2)+(n-3)k + 1 + x_(t-1)k(n-2) + (n-3)k + 2 + ⋯ + x_(t-1)k(n-2) + (n-2)k + 1 = u;                                                                                            ].



The equations involving segments containing  the root vertex are 




    ∑_t=1^k (x_(t-1)k(n-2) + (i-1)k + j)  + c = u,

where  
i ∈{ 1, 2, ⋯, n-2} e j ∈{1, 2, ⋯, k} and

    ∑_t=1^k x_tk(n-2) + 1 + c = u


Let 

    S_1 = ∑_i=1^n-1∑_t=1^k x_(t-1)k(n-2)+k(i-1)+1 = (n-1)(u-c)

 and

    S_j = ∑_i=1^n-2∑_t=1^k x_(t-1)k(n-2) + k(i-1)+j = (n-2)(u-c)

for j ∈{2, ⋯, k}. 

By adding equations involving  perimeters in the degenerated magic polygon, we obtain 

    2S_1 + S_2 + ⋯ + S_k + 2c = [k(n-2)+2]u,

or, equivalently,

    2S_1 + S_2 + ⋯ + S_k = [k(n-2)+2]u - 2c


By adding equations involving the root vertex of the degenerated magic polygon, we obtain

    S_1 + S_2 + ⋯ + S_k + (k(n-2)+1)c = (k(n-2)+1)u

or, equivalently,

    S_1 + S_2 + ⋯ + S_k = (k(n-2)+1)(u-c)


By subtracting (<ref>) from  (<ref>), we obtain 


    S_1 = u + (k(n-2)-1)c


By (<ref>) and (<ref>),  

    (n-1)(u-c) = u + (k(n-2)-1)c.


Hence

    u = (k+1)c



As the values assigned to the points of the degenerated magic polygon are distinct values of the set  {1, 2, ⋯, k^2(n-2)+k+1}, we obtain


    S_1 + S_2 + ⋯ + S_k + c = ∑_i=1^k^2(n-2)+k+1 i = (k^2(n-2)+k+1)(k^2(n-2)+k+2)/2


By (<ref>) and (<ref>),

    [k(n-2)+1](u-c) + c = (k^2(n-2)+k+1)(k^2(n-2)+k+2)/2


By (<ref>) and (<ref>), 

    [k^2(n-2)+k+1]c = (k^2(n-2)+k+1)(k^2(n-2)+k+2)/2,

that implies 

    c = k^2(n-2)+k+2/2




 If k and n are positive integer numbers, such that 
k is odd and  n is even, then there is no degenerated magic polygon  D(n,k).


 By Theorem <ref>,  

    c = k^2(n-2) + k + 2/2.

 Therefore,  if  k is odd and  n is even, then c is not a integer number. Hence, there is no degenerated magic polygon of order k+1 with n vertex for k odd and n even.
  


 If n is an integer number greater than or equal to 3,  then there is a degenerated magic polygon D(n,2).


 Let  C be the root vertex of a degenerated magic polygon  D(n, 2),  (C, V_1, ⋯, V_n-1) be the sequence of vertices of the largest polygon and  (C, V_1^∗, ⋯, V_n-1^∗) be the sequence of vertices of smallest polygon, both on clockwise direction.

For each  j ∈{1, 2, ⋯, n-2}, we consider   M_j  the point of degenerated magic polygon between  V_j and  V_j+1 and  M_j^∗  the point of the degenerated magic polygon between  V_j^∗ and V_j+1^∗. 

Thus, if   c is the value assigned to the root vertex C and for each j∈{ 1, 2, ⋯, n-2}, z_j is the value assigned to the vertex  V_j, z_j^∗ is the value assigned to the vertex V_j^∗, m_j is the value assigned to the point M_j and m_j^∗ is the value assigned to the point  M_j^∗,  then, the following conditions are satisfied:


    z_j + m_j + z_j+1 = u
 


    z_j^∗ + m_j^∗ + z_j+1^∗ = u


and

    z_j + z_j^∗ + c = u
 

Setting, for each j ∈{1, 2, ⋯, n-1}, 
 
 
    {[                                                                             z_j  ={[           j ,  if  j is odd;    2n + j -3, if  j is even;               ].;                                                                                ;                                                                           z_j^∗ = {[   4(n-1) - j,   if j is odd;     2n - j-1, if  j is even;               ].;                                                                             m_j                                                                     = 4(n-1)-2j;                                                                           m_j^∗                                                                            = 2j;                                                                                 ].

then, we obtain a degenerated magic polygon P(n,2).

In fact, the condictions on segments are satisfied, because

    z_j+m_j + z_j+1 = {[       j + 2n+j+1-3 + 4(n-1) - 2j = 6(n-1),                                if j is odd; 2n + j - 3 + 4(n-1) - 2j + j + 1 = 6(n-1),                               if j is even;                                            ].

  
 
    z_j^∗ + m_j^∗ + z_j+1^∗ = {[   4(n-1) - j +2j + 2n - j -1 - 1 =  6(n-1),                                 if j is odd; 2n - j - 1 + 2j + 4(n-1) - j - 1 =  6(n-1),                                if j is even;                                             ].
  
  
  
    z_j + z_j^∗ + c = {[           j + (4(n-1) - j) + 2(n-1) = 6(n-1),                                   if j is odd; (2n + j - 3) + (2n - j - 1) + 2(n-1) = 6(n-1)                                  if j is even;                                               ].

  
  
    m_j + m_j^∗ + c = 
      4(n-1) - 2j + 2j + 2(n-1) = 6(n-1)

  
  In addition, all values assigned to the points are different because
  
  
    [                                                                                                                     A_1                                                                              = { z_j|j is odd and  j ∈{1, 2, ⋯, n - 1}};                                                                                                                         = {[ { 1, 3, 5, ⋯, n - 2},           if n is odd; { 1, 3, 5, ⋯, n - 1},          if n is even;                       ].;                                                                                                                         ]

  
    [                                                                                                                                                                            B_1                                                                                                                                    = { z_j|j is even and  j ∈{1, 2, ⋯, n - 1}};                                                                                                                                                                                = {[    { 2n-1, 2n+1, 2n+3, ⋯, 3n-4},                      if n is odd; { 2n-1, 2n+1,  2n+3, ⋯, 3n - 5},                     if n is even;                                  ].;                                                                                                                                                                                ]

  
    [                                                                                                                                                                                 A_2                                                                                                                                        = { z_j^∗|j is odd and  j ∈{1, 2, ⋯, n - 1}};                                                                                                                                                                                     = {[ { 4n-5, 4n - 7, 4n - 9, ⋯, 3n-2},                       if n is odd;  { 4n-5, 4n -7, 4n-9, ⋯, 3n - 3},                      if n is even;                                   ].;                                                                                                                                                                                     ]

  
    [                                                                                                                                                             B_2                                                                                                                   = { z_j^∗|j is even and  j ∈{1, 2, ⋯, n - 1}};                                                                                                                                                                 = {[    { 2n-3, 2n-5, 2n-7, ⋯, n},                   if n is odd; { 2n-3, 2n-5, 2n-7, ⋯,  n+1},                  if n is even;                               ].;                                                                                                                                                                 ]

  
    [                               C_1    = { m_j|  j ∈{1, 2, ⋯, n - 2}};                                   = 
      { 4n-6, 4n-8, 4n-10, ⋯, 2n };                                   ]

    
    [                                 C_2    = { m_j^∗|  j ∈{1, 2, ⋯, n - 2}};                                     = 
      { 2, 4, 6, 8, 10, 12, ⋯, 2n-4};                                     ]

   
    D={ c } = { 2(n-1)}

   satisfy
   
    |A_1| = |A_2| = {[       n-1/2,  if n is odd;         n/2, if n is even;              ].

 
    |B_1| = |B_2| = {[       n-1/2,  if n is odd;       n-2/2, if n is even;              ].

   
    |C_1| = |C_2| = n - 2,

   
    |A_1| + |A_2| + |B_1| + |B_2| + |C_1| + |C_2| + |D| = 4n -5

   
   
    [                                 A_1∪ A_2∪ B_1∪ B_2∪ C_1∪ C_2∪ D;                               =          {j ∈ℤ| 1 ≤ j ≤ 4n - 5};                               =             { 1, 2, ⋯,  4n - 5};                                 ]
   
    and 
   
    |A_1∪ A_2∪ B_1∪ B_2∪ C_1∪ C_2∪ D| = 4n-5
 


In Figure <ref> we have an example of Magic Polygon D(5, 2) constructed in the proof of the Theorem <ref>. An example of Magic Polygon D(5, 2) that can not be obtained by this construction can be seen in the figure <ref>.

 









sssd


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