Associated Topics || Dr. Math Home || Search Dr. Math

### Construct Polygon Given One Side

```
Date: 12/03/2001 at 03:27:24
From: Dieter Woerner
Subject: Construct polygon given one side

Yesterday my son asked me to help him with what I thought was an easy
problem. But I was searched for a solution all night (just thinking)
and this morning looked on the Internet, and all I found was the other
side: given one circle, inscribe a pentagon.

```

```
Date: 12/03/2001 at 10:09:18
From: Doctor Rob
Subject: Re: Construct polygon given one side

Thanks for writing to Ask Dr. Math, Dieter.

You must be talking about a regular polygon, that is, one with all
sides equal and all interior angles of equal measure.

You also have to know the number n of sides the regular polygon has.
Then all the interior angles will have measure 180/n degrees, or

Some regular polygons can be exactly constructed using straightedge
and compass, and some cannot. The rest can be approximately
constructed using a protractor.

For polygons that can be constructed with straightedge and compass,
see the following web page from the Dr. Math Frequently Asked
Questions (FAQ):

Regular Polygons
http://www.mathforum.org/dr.math/faq/formulas/faq.regpoly.html

Let the side length of a regular n-gon inscribed in a unit circle
be a, and the desired side length be A. Then draw a circle with
radius A/a, and use that to construct the regular n-gon inscribed
in it.

For odd numbers n, here is how to proceed. For n = 3 this is easy,
since you are just constructing an equilateral triangle with side A.
For n = 5 and 17, there are constructions on the page cited above.
The construction of a regular 257-gon is more complicated, and the
regular 65537-gon is MUCH more complicated, but both are known. That
covers all prime number values of n that are possible with fewer than
a million digits (and maybe all).

For composite n that are products of some of the numbers 3, 5, 17,
257, and 65537, construct the regular polygons with the same number
of sides as the prime factors of n, inscribed in the same circle and
sharing a common vertex. Then the shortest distance between any two
of the vertices will be the length of the side of the regular n-gon
inscribed in the same circle. That will give you n = 15 = 3*5,
51= 3*17, 85 = 5*17, 255 = 3*5*17, and so on.

The odd numbers n for which the regular n-gon cannot be exactly
constructed with straightedge and compass include n = 7, 9, 11, 13,
19, 21, 23, 25, ....  In those cases, the best you can do is this.
Draw one side. Use a protractor to approximate the first angle. Draw
the next side. Continue this until you have drawn all n sides, at
which point the polygon should be closed. If your approximations are
close enough, you won't be able to tell that it actually isn't,
because the error will be smaller than the width of your pencil lines.

For even numbers n, write n = 2^e*m, where m is odd. If m = 1, then
e >= 2. Construct a square with side A, then bisect all the central
angles e-2 times. If m >= 3, construct a regular m-gon (exactly if
possible, otherwise approximately), and then bisect all the central
angles e times. The intersection of all the bisectors so generated and
the circle form the vertices of a regular n-gon.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search