Constructing a Regular PentagonDate: 21 Feb 1995 21:18:47 -0500 From: Eric Robbins Subject: Re: Pentagon Greetings, from Bramalea Secondary School in Brampton, Ontario We are interested in knowing how to construct a regular pentagon using a compass and a straight edge. Thanks for your time and help! Eric Robbins Date: 23 Feb 1995 16:58:24 -0500 From: Dr. Ken Subject: Re: Pentagon Hello there. Annie Fetter from the Geometry Forum gave me this reference. It's in a book called "Geometry for the Classroom" by C. Herbert Clemens and Michael A. Clemens. We must start by designating a certain length as "one unit." After that we must determine the length of Sqrt{5} - 1 units (if you don't know how you'd do that, write back; that's a whole other construction, but it's easier). Start with a horizontal line, L. Pick a point p on L that will be the center of the pentagon. At a distance of Sqrt{5} - 1 units from p along L, mark point z. Construct a perpendicular to L at z. Next, construct a circle of radius 4 units about p. The distance from point y (the point where the circle intersects L at the right) to point x (the point where the circle intersects the perpendicular to L at the top) is the length of one side of the pentagon. Place the compass point on x and mark off this distance on the circle. Then place the compass point on the point you just marked, and mark off the distance again. Repeat this [a total of] five times, and then connect all of the points you have marked. This is the regular 5-gon, or pentagon. So there you have it. If you wanted to construct a regular pentagon inscribed in a given circle, you could find the radius of the circle, and call its length 4. Then bisect it twice, and you've got a segment of length 1, and you can proceed as directed. I hope you have fun with it! -Ken "Dr." Math From: Cedric Boyd Subject: Re: Needed construction to construct a pentagon. Date: Tue, 6 May 1997 Hi my name is Cedric Boyd form Overland Park, Kansas. I am in Geometery and I am doing a project making a dodecahedren. If you would please explain to me how to find the length of Sqrt {5} - 1 units A.S.A.P.. If you have trouble remembering, it was the article in Ask Dr.Math on February 21, 1995. Thank you. Sincerely, Cedric Boyd cedricb@usa.net Subject: Re: Needed construction to construct a pentagon. Date: Tue, 6 May 1997 Hi there- In my opinion, this is one of the cutest constructions out there. First we'll construct a segment of length Sqrt{5}, and then we'll subtract 1 from it (mark off 1 unit on that segment, and what's left has length Sqrt{5} - 1). Start with a segment AB whose length is 1, and construct a perpendicular line to that segment, passing through B. On this new line, mark off a distance of 1 from point B, and call that new point C. So now if you draw segment AC, you'll note that triangle ABC is a 45-45-90 degree triangle, so its hypotenuse has length Sqrt{2}. Now do the same kind of thing you just did - construct a segment CD of length 1, perpendicular to AC. Then ACD is a right triangle whose hypotenuse (by the Pythagorean theorem) has length Sqrt{3}. I bet by now you see where this is going. You'll get the square root of 4, then the square root of 5. Note that this procedure can be used to get the square root of _any_ integer. So now that you've got a segment whose length is the Sqrt{5}, mark of a distance of 1 on that segment, and what's left will have length Sqrt{5} - 1. -Ken Williams |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/