The Math Forum

Ask Dr. Math: FAQ

   Regular Polygon Formulas   

Geometric Formulas: Contents || Ask Dr. Math || Dr. Math FAQ || Search Dr. Math

also see Defining Geometric Figures

 Regular Polygon

Number of sides, all equal length a: n
Number of interior angles, all equal measure beta: n
Central angle subtending one side: alpha
Perimeter: P
Area: K
Radius of circumscribed circle: R
Radius of inscribed circle: r
    beta = Pi(n-2)/n radians = 180o(n-2)/n
alpha = 2 Pi/n radians = 360o/n
alpha + beta = Pi radians = 180o
P = na = 2nR sin(alpha/2)
K = na2 cot(alpha/2)/4
    = nR2 sin(alpha)/2
    = nr2 tan(alpha/2)
    = na sqrt(4R2-a2)/4
R = a csc(alpha/2)/2
r = a cot(alpha/2)/2
a = 2r tan(alpha/2) = 2R sin(alpha/2)

To read about regular polygons, visit The Geometry Center.

   Special Cases of the Regular Polygon

n = 3 (equilateral triangle)
n = 4 (square)
n = 5 (regular pentagon)
n = 6 (regular hexagon)
n = 8 (regular octagon)
       The Regular Pentagon    regular pentagon

       (For constructions using straightedge and compass, see examples below by
        Floor van Lamoen and John Conway.)

      Number of sides n = 5
      Internal angles beta = 3/5 radians = 108 degrees
      Central angles alpha = 2/5 radians = 72 degrees

      Perimeter P = 5a = 5R sqrt(10-2 sqrt[5])/2
      Area K = 5a2 sqrt(1+2/sqrt[5])/4
                  = 5R2 sqrt(10+2 sqrt[5])/8
                  = 5r2 sqrt(5-2 sqrt[5])
                  = 5a sqrt(4R2-a2)/4

      Circumradius R = a sqrt(2+2/sqrt[5])/2
      Apothem r = a sqrt(1+2/sqrt[5])/2 = R(1+sqrt[5])/4

      Side a = 2r sqrt(5-2 sqrt[5]) = R sqrt(10-2 sqrt[5])/2

       The Regular Hexagon    regular hexagon

      Number of sides n = 6
      Internal angles beta = 2/3 radians = 120 degrees
      Central angles alpha = /3 radians = 60 degrees

      Perimeter P = 6a = 6R
      Area K = 3a2 sqrt(3)/2
                  = 3R2 sqrt(3)/2
                  = 2r2 sqrt(3)
                  = 3a sqrt(4R2-a2)/2

      Circumradius R = a
      Apothem r = a sqrt(3)/2 = R sqrt(3)/2

      Side a = 2r/sqrt(3) = R

       The Regular Octagon    regular octagon

      Number of sides n = 8
      Internal angles beta = 3/4 radians = 135 degrees
      Central angles alpha = /4 radains = 45 degrees

      Perimeter P = 8a = 8R sqrt(2-sqrt[2])
      Area K = 2a2(sqrt[2]+1)
                  = 2R2 sqrt(2)
                  = 8r2(sqrt[2]-1)
                  = 2a sqrt(4R2-a2)

      Circumradius R = a sqrt(sqrt[2]/2+1)
      Apothem r = a(sqrt[2]+1)/2 = R sqrt(2+sqrt[2])/2

      Side a = 2r(sqrt[2]-1) = R sqrt(2-sqrt[2])

   Constructible Regular Polygons

Here is a table of regular polygons constructible with straightedge and compass whose angles are whole numbers of degrees:








  1.7320508 = sqrt(3)




  1.4142136 = sqrt(2)




  1.1755705 = sqrt(5-sqrt[5])/2




  1.0000000 = 1




  0.7653669 = sqrt(2-sqrt[2])




  0.6180340 = (sqrt[5]-1)/2




  0.5176381 = (sqrt[3]-1)/sqrt(2)




  0.4158234 = (sqrt[3]sqrt[1-sqrt(5)]+sqrt[2]sqrt[5+sqrt(5)])/4




  0.3128689 = (sqrt[2][1+sqrt(5)] - 2 sqrt[5-sqrt(5)])/4




  0.2610524 = sqrt([4-sqrt(2)-sqrt(6)]/2)




  0.2090569 = (sqrt[6]sqrt[5-sqrt(5)]-sqrt[5]-1)/4




  0.1569182 = (sqrt[2]sqrt[2+sqrt(2)][1-sqrt(5)]+
     2 sqrt[2-sqrt(2)]sqrt[5+sqrt(5)])/(4 sqrt[2])




  0.1046719 = ([sqrt(3)+1][sqrt(5)-1]sqrt[2]+




  0.0523539 = ([sqrt(2-sqrt[2])+
     sqrt(6+3 sqrt[2])][1+sqrt(5)-sqrt(30-6 sqrt[5])]+
     [sqrt(2+sqrt[2])-sqrt(6-3 sqrt[2])][sqrt(3)+
     sqrt(15)+sqrt(10-2 sqrt[5])])/16

The following messages were sent by and in answer to Professor John Conway of Princeton University's Mathematics Department to geometry discussion groups (geometry-puzzles, geometry-pre-college, geometry-puzzles) in answer to questions about constructing regular polygons. An alternative simple construction of a regular pentagon has been contributed by Floor van Lamoen ("Dr. Floor"); see also Inscribing a regular pentagon in a circle - and proving it, by Scott E. Brodie.

1. Polygons with compass and straightedge

It's a VERY famous theorem of Gauss that the only regular polygons
with a prime number of sides that can be constructed with straightedge
and compass are those for which the prime is one of the Fermat primes

          3, 5, 17, 257, 65537, ...     

(that is, primes of the form 2^n + 1).  Nobody knows if there are
any Fermat primes larger than 65537.

The only constructible regular polygons with an odd number of
sides are those for which this number is a product of distinct
Fermat primes (so for instance 15 = 3 times 5,  51 = 3 times 17),
and the only ones with an even number of sides are those obtained
by repeatedly doubling these numbers (including 1), thus:-

            (1,2), 4, 8, 16, 32, 64, ...

             3, 6, 12, 24, 48, ...

             5, 10, 20, 40, 80, ...

             15, 30, 60, ...

             17, 34, 68,...

             51, ...


Some people might like the following little observation.
Write out the Pascal triangle modulo 2 :

                     1 1
                    1 0 1
                   1 1 1 1
                  1 0 0 0 1
                 1 1 0 0 1 1
                1 0 1 0 1 0 1
               1 1 1 1 1 1 1 1
              1 0 0 0 0 0 0 0 1

then by reading the first 31 rows as the binary expansions of
numbers, you get   

          1, 3, 5, 15, 17, 51, 85, 255, 257, ...

which give the first few odd-sided constructible polygons (and
very probably all there are).

John Conway

2. Construction of a regular pentagon

Let  N,S,E,W  be the points of a circle with center O in the four
compass directions, M be the midpoint of ON and  MX  (with  X  on  OE) 
the bisector of the angle  OME:

Then the line through  X  perpendicular to  OE  hits the circle
in two points of the regular pentagon that has a vertex at  E.  You
can either get the other two points by stepping around the circle with
a compass set to the edge-length so found for the pentagon, or by
replacing  X  in the above by the point  Y  where the EXTERNAL bisector
of  OME  meets  OW.

How about septagons?

Well, for one thing the proper name is "heptagon", not "septagon."  There
isn't a construction for a regular heptagon using ruler and compass according
to Euclid's rules, but there is a construction using an angle-trisector
which you can find in "The Book of Numbers" that I wrote with Richard Guy.
That book also gives an angle-trisector construction that uses ruler and 
compasses in a manner not sanctioned by Euclid, so you can combine them to
give such a construction for the regular heptagon.  The book also gives
similar constructions for the regular polygons with 13 and 17 sides (for the
regular 11-gon there's a construction using an angle-quinquesector, but it
was too complicated for us to put into the book).

John Conway

3. Construction of a 17-sided regular polygon

The neatest construction I know is due to Richmond - I call it the
"quadruple quadrisection constriction":

   1) quadrisect the perimeter of the circle, by points  N,S,E,W;

   2) quadrisect the radius  ON  by the point A;

   3) quadrisect the angle  OAE  by the line  AB; 

   4) quadrisect the straight angle BAC by the line AD:

               I   |
                   |       J
                C  |

    5) draw the semicircle DFE, cutting  ON  in  F;
    6) draw the semicircle GFH, centred at B;

    7) cut the semicircle WNE by the perpendiculars  GI  and  HJ  to WE.

Then  I  and  J  are points of the regular heptakaidecagon on the
circle ENWS that has one vertex at  E.

    I first saw this in Hardy and Wright's book on The Theory
    of Numbers, which is where I've just checked up on it.  
    H & W confirm my impression of the history.  They say
    that Gauss worked out the general theory in Paragraphs
    335-366 of his Disquisitiones, but that the first explicit
    construction was given by Erchinger, for whom they refer
    to Gauss' Werke, vol II, pp186-187.

  This "Quadruple Quadrisection" construction (my name) is due
  to Richmond, who gave it in the Quarterly Journal of Math, 1893.

  Of the four or five constructions I have seen, it is definitely
  the nicest.  If you intersect the other quadrisectors of that
  straight angle with WE and treat the resulting points similarly,
  you can get more vertices in the same way - but it's easier to
  use your compasses to step around the circle from the ones given,
  for which the constructing points are the most conveniently

John Conway

John, I discovered another nice construction by Henri Lebesgue.
Let me tell the story:

Henri Lebesgue published in 1937 the following paper:

Lebesgue, Henri: Sur une construction du polygone regulier de 17 cotes,
due a Andre-Marie Ampere, d'apres des documents conserves dans les
archives de l'academie des sciences. 
C. R. Acad. Sci., Paris 204(1937) 925-928.
[Republished in: Enseign. Math., II. Ser. 3(1957) 31-34]

Also, he is the author of the book:

Henri Lebesgue: Lecons sur les constructions geometriques au college 
de France en 1940-1941. Paris : Gauthier - Villars, 1950
In pp. 148- 49 he describes the construction of the r. heptakaidecagon
(and in p. 145 of the r. pentagon).

I haven't seen HL's paper/book, only brief descriptions of his 
constructions (in the book) published in a Greek periodical.

Here is the construction of the r. heptakaidecagon:

Let (O) be a circle of center O.

                   A_6      |        A_4

   1. Draw the diameters XA perpendicular to YZ

   2. Quadrisect the radius OZ by B.

   3. Draw CB perpendicular to BA (C lies on OX)

   4. Draw the semicircle (C, CB) intersecting XA at D, D*

   5. Draw the semicircle (D, DB) intersecting XA at E,E*

   6. Draw the semicircle (D*, D*B) intersecting XA at F,F*

   7. Draw the semicircle of diameter AE*, intersecting OY at K

   8. Draw the semicircle (F, FK) intersecting XA at G, H.

   9. Draw the perpendiculars from G, H, intersecting the (O) at A_4, A_6.

Now, A:=A_1, A_4, A_6 are vertices of the r. 17-gon.
(We have arc(A_4A_6) = 4Pi/17. We bisect it to find A_5, and therefore
the  r.17-gon's side).

It is a memorizable construction: 5 semicircles and 4 perpendiculars.

The question is: Is it a construction of Andre-Marie Ampere (see paper above)
or H. Lebesque's himself ? (My source calls it as Lebesgue's.)

Antreas P. Hatzipolakis

Construction of a regular pentagon:

Let  N,S,E,W  be the points of a circle C with center O in the four
compass directions, and let M be the midpoint of ON.

Let E' and E" be the points where the circle with center M through E
meets the line NS.

Finally, let P' and P" be the points where the circle with center E
through E' meets C and let Q' and Q" be the points where the circle with
center E through E" meets C.

Then E, P', P", Q' and Q" form a regular pentagon.

- Floor van Lamoen

Back to Contents

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. Math ®
© 1994- The Math Forum at NCTM. All rights reserved.