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. Thanks in advance.
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 Pi/n radians. 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum