Lesson Sum of vectors connecting the center of a regular n-sided polygon with its vertices

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Sum of vectors connecting the center of a regular n-sided polygon with its vertices


In this lesson you will learn that for any regular n-sided polygon in a plane the sum of the vectors that connect the center of the polygon with its vertices is equal to zero.

We will prove, one after other, the following statements:

    - The sum of vectors connecting the center of a regular triangle with its vertices is equal to zero.
    - The sum of vectors connecting the center of a square with its vertices is equal to zero.
    - The sum of vectors connecting the center of a regular 2n-sided polygon with its vertices.
    - The sum of vectors connecting the center of a regular n-sided polygon with its vertices.


Problem 1

The sum of the vectors connecting the center of a regular triangle with its vertices is equal to zero.  Prove.

Solution
We have the following geometric facts:

a)  For any triangle the sum of the vectors connecting the triangle's                                    
center of mass with its vertices is equal to zero.
It was proved in the lesson  Sum of vectors connecting
the center of mass of a n-sided polygon with its vertices  in this site.

b)  For any triangle the center of mass  (centroid)  of the triangle
coincides with the intersection point of its medians.
It was proved in the lesson  The Centroid of a triangle
is the Intersection point of its medians  in this site.

c)  For any regular triangle the intersection point of its medians
is the triangle's geometric center.

These facts imply that for any regular triangle the sum of vectors
connecting the center of the triangle with its vertices is equal to zero.

The proof is completed.

Figure 1.  The triangle  PQR,  its geometric
center  C  and the vectors  CP,  CQ and  CR


Problem 2

The sum of the vectors connecting the center of the square with its vertices is equal to zero.  Prove.

Solution
We have the following geometric facts:

a)  For any quadrilateral the sum of the vectors connecting the quadrilateral's                  
center of mass with its vertices is equal to zero.
It was proved in the lesson  Sum of vectors connecting
the center of mass of a n-sided polygon with its vertices  in this site.

b)  For any parallelogram the center of mass  (centroid)  of the parallelogram
coincides with the intersection point of its diagonals.
It was proved in the lesson  The Centroid of a parallelogram
is the Intersection point of its diagonals  in this site.

c)  For any square the intersection point of its diagonals is the square's
geometric center.

These facts imply that for any square the sum of vectors
connecting the center of the square with its vertices is equal to zero.

The proof is completed.

  Figure 2.  The square ABCD,  its center  O
      and the vectors  OA,  OB,  OC  and  OD

Surely,  other proofs are possible,  for example,  by grouping and canceling the opposite vectors  OA  and  OC,  OB  and  OD  (Figure 2),  as it was completed in the lesson Sum of vectors that connect the center of a parallelogram with its vertices.


Problem 3

The sum of the vectors connecting the center of the regular  2n-sided polygon with its vertices is equal to zero.  Prove.

Solution
Let  m = 2n  be our even integer number and  A1A2A3...Am  be a regular  2n-sided          
polygon with the vertices  A1, A2, A3, . . . , Am.
The example of such a polygon is shown in  Figure 3  for  m = 2n = 6.

Let the point  C  be the center of the polygon  A1A2A3...Am.  
Consider the vectors  CA1, CA2, CA3, . . . , CAm  connecting the center
with the vertices  A1, A2, A3, . . . , Am  (Figure 3).

The vector  CA1  has the components  (R, 0),
the vector  CA2  has the components  (R%2Acos%28%282pi%29%2Fm%29,   R%2Asin%28%282pi%29%2Fm%29),
the vector  CA3  has the components  (R%2Acos%28%282%2A2pi%29%2Fm%29, R%2Asin%28%282%2A2pi%29%2Fm%29),
the vector  CA3  has the components  (R%2Acos%28%283%2A2pi%29%2Fm%29, R%2Asin%28%283%2A2pi%29%2Fm%29),  and so on
. . . . . . . . ,

the vector  CAm  has the components  (R%2Acos%28%28m-1%29%2A2pi%2Fm%29%29, R%2Asin%28%28m-1%29%2A2pi%2Fm%29).

where  R  is the radius of the circle circumferenced around the regular polygon.


Figure 3. The polygon A1A2A3...A6,  the center  C
          and the vectors CA1, CA2, CA3, . . . , CA6

The proof is based on the fact that each vector  CAi,  i = 1, 2, 3, . . ., n,  has the opposite vector  CAj  in this set,  where j = n+i-1.

By the way, the fact that the vectors  CAi  (i = 1, 2, 3, . . ., n)  and corresponding opposite vectors    CAj  (j = n+i-1, i = 1, 2, 3, . . ., n)  are different and cover the entire set of vectors  CA1, CA2, CA3, . . . , CAm  without exclusions and repetitions means that  m = 2n  is the even integer number.

Since such opposite vectors cancel each other,  the sum   CA1 + CA2 + CA3 + . . . + CAm   is the zero vector.

This is what has to be proved.


Problem 4

The sum of the vectors connecting the center of the regular  (2n+1)-sided polygon with its vertices is equal to zero.  Prove.

Solution 1
Let  m = 2n+1  be our odd integer number and  A1A2A3...Am  be a regular  2n+1-sided          
polygon with the vertices  A1, A2, A3, . . . , Am.
The example of such a polygon is shown in  Figure 4  for  m = 2n+1 = 7.

Let the point  C  be the center of the polygon  A1A2A3...Am.  
Consider the vectors  CA1, CA2, CA3, . . . , CAm  connecting the center
with the vertices  A1, A2, A3, . . . , Am  (Figure 3).

The vector  CA1  has the components  (R, 0),
the vector  CA2  has the components  (R%2Acos%28%282pi%29%2Fm%29,   R%2Asin%28%282pi%29%2Fm%29),
the vector  CA3  has the components  (R%2Acos%28%282%2A2pi%29%2Fm%29, R%2Asin%28%282%2A2pi%29%2Fm%29),
the vector  CA3  has the components  (R%2Acos%28%283%2A2pi%29%2Fm%29, R%2Asin%28%283%2A2pi%29%2Fm%29),  and so on
. . . . . . . . ,

the vector  CAm  has the components  (R%2Acos%28%28m-1%29%2A2pi%2Fm%29%29, R%2Asin%28%28m-1%29%2A2pi%2Fm%29).

where  R  is the radius of the circle circumferenced around the regular polygon.


Figure 4. The polygon  A1A2A3...A7,  the center  C
          and the vectors CA1, CA2, CA3, . . . , CA7

This time we can not apply the approach used in the solution of the previous problem, because we can not pair the vectors with the opposite ones.
We will proceed by another way.  We will use analogy between vectors and complex numbers.  For this analogy,  see the lessons

    - Complex numbers and arithmetical operations over them
    - Complex plane
    - Addition and subtraction of complex numbers in complex plane
    - Multiplication and division of complex numbers in complex plane
    - Raising a complex number to an integer power
    - How to take a root of a complex number

under the topic  Complex numbers  of the section  Algebra-II  in this site.

Let us write the required sum of the vectors in the component form

CA1 + CA2 + CA3 + . . . + CAm = R * ( (1,0) + (cos%28%282pi%29%2Fm%29, sin%28%282pi%29%2Fm%29) + (cos%28%282%2A2pi%29%2Fm%29, sin%28%282%2A2pi%29%2Fm%29) + (cos%28%283%2A2pi%29%2Fm%29, sin%28%283%2A2pi%29%2Fm%29) + . . . + (cos%28%28m-1%29%2A2pi%2Fm%29%29, sin%28%28m-1%29%2A2pi%2Fm%29)  ).    (1)

According to the analogy between vectors and complex numbers noted above,  the sum in parentheses in the right hand side of the equality  (1)  is the sum of the geometric progression

S  =  1 + epsilon + epsilon%5E2 + epsilon%5E3 + . . . + epsilon%5E%28m-1%29                 (2)

of complex numbers,  where  epsilon  is the complex number  epsilon = cos%28%282%2Api%29%2Fm%29 + i*sin%28%282%2Api%29%2Fm%29,  i  is the  imaginary unit.

This complex number is the complex root of the degree m of unit:  epsilon = root%28m%2C1%29,  i.e.  epsilon%5Em = 1.

Now, apply the formula for summing geometric progression to get the sum  (2).  This formula is

S = %28epsilon%5Em+-1%29%2F%28epsilon-1%29                                                            (3)

(see the lesson  Geometric progressions  under the topic  Sequences of numbers, series and how to sum them  of the section  Algebra-II  in this site).

Since  epsilon%5Em = 1,  the formula  (3)  gives the value of  S  equal to zero.
Hence,  the sum of the vectors in the left hand side of the equality  (1)  is equal to zero.

The proof is completed.


Solution 2
The more geometric solution is presented here.  It does not use complex numbers.

The starting part of the  Solution 2  is the same as in the preceding  Solution 1  till getting the formula  (1).
So,  now we shall again calculate the sum of the vectors

S = CA1 + CA2 + CA3 + . . . + CAm = R * ( (1,0) + (cos%28%282pi%29%2Fm%29, sin%28%282pi%29%2Fm%29) + (cos%28%282%2A2pi%29%2Fm%29, sin%28%282%2A2pi%29%2Fm%29) + (cos%28%283%2A2pi%29%2Fm%29, sin%28%283%2A2pi%29%2Fm%29) + . . . + (cos%28%28m-1%29%2A2pi%2Fm%29%29, sin%28%28m-1%29%2A2pi%2Fm%29)  ).    (1)

Let us apply the anti-clockwise rotation of the plane in the angle of  2%2Api%2Fm  radians.
This means that every vector  u  in the plane rotates in the anti-clockwise direction by the angle of  2%2Api%2Fm  radians.
Particularly,  every vector  (cos%28%282%2Api%2Ak%29%2Fm%29, sin%28%282%2Api%2Ak%29%2Fm%29)  of the family of the vectors  (1),  presented in the  Figure 4,  is transformed by the rotation to its next closest neighboor in this family.

It is clear that the entire finite family of vectors  (1)  remains the same after rotation.

Thus,  from the one side the entire sum of the vectors should rotate in the angle of  2%2Api%2Fm  radians,  because every additive member vector rotates in this angle.
From the other side,  the sum of vectors remains the same after rotation as the entire set of vectors remains the same.
The only vector which remains unchangeable after rotation in the angle  2%2Api%2Fm  radians is the zero vector.
Hence,  the summary vector  S  is the zero vector.

The proof is completed.


Note.
The geometric  Solution 2  uses the geometric transformation rotating the plane in the angle  2%2Api%2Fm,  while the  Solution 1  uses the formula of the sum of the geometric progression  (2).  But the rotation of the complex plane in the angle  2%2Api%2Fm  is nothing else as multiplication by the complex number  epsilon = root%28m%2C+1%29 = cos%28%282%2Api%29%2Fm%29 + i*sin%28%282%2Api%29%2Fm%29.
So,  these two solutions are not too far each from the other.  They simply use different terms,  but express the same idea of symmetry,  each by its own way.


My introductory lessons on vectors in this site are
    - Vectors in a plane
    - Sum of vectors that are coherently oriented sides of a convex closed polygon
    - Sum of vectors that are coherently oriented sides of an unclosed polygon
    - Sum of vectors that connect the center of a parallelogram with its vertices
    - Vectors in a coordinate plane
    - Addition, Subtraction and Multiplication by a number of vectors in a coordinate plane
    - Summing vectors that are coherently oriented sides of a convex closed polygon
    - Summing vectors that are coherently oriented sides of an unclosed polygon
    - The Centroid of a triangle is the Intersection point of its medians
    - The Centroid of a parallelogram is the Intersection point of its diagonals
    - Sum of vectors connecting the center of mass of a triangle with its vertices
    - Sum of vectors connecting the center of mass of a quadrilateral with its vertices
    - Sum of vectors connecting the center of mass of a n-sided polygon with its vertices
    - Sum of vectors connecting the center of a regular n-sided polygon with its vertices                  (this lesson)
    - Solved problems on vectors in a plane
    - Solved problems on vectors in a coordinate plane
    - HOW TO find the length of the vector in a coordinate plane
    - Flying airplane, blowing wind, airspeed, groundspeed etc.

    - OVERVIEW of Introductory lessons on vectors in a plane

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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