Constructing a Regular Pentagon

Date: 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