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Inscribing a Regular Pentagon within a Circle


Date: 04/15/99 at 09:38:14
From: Michael Huang
Subject: Inscribe a regular pentagon within a circle

I'm trying to inscribe a regular pentagon within a circle with only 
the help of a compass and a strightedge. So far, I can construct it 
because I found the steps, but I can't manage to see what each step 
stands for and how you can justify that the pentagon is a regular 
pentagon.

These are the steps:

Start with a circle of radius 1.  
Construct perpendicular diameters AB and CD. 
Bisect the radius OB at M. 
With center M and radius MC, draw an arc intersecting radius OA at N. 

I claim that CN is the side of the regular pentagon inscribed in the 
circle. 

Hope you can explain this. Thanks.


Date: 04/15/99 at 14:02:03
From: Doctor Rob
Subject: Re: Inscribe a regular pentagon within a circle

Thanks for writing to Ask Dr. Math!

OM = 1/2. OC = 1. Using the Pythagorean Theorem, MC = sqrt(5)/2.  
Then MN = sqrt(5)/2, so ON = (sqrt[5]-1)/2. OC = 1. Using the 
Pythagorean Theorem again, CN = sqrt(10-2*sqrt[5])/2.  

Now according to the formulas on the following Web page

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

the side of a regular pentagon inscribed in a unit circle has to be
2*sin(36 degrees). Thus you are reduced to convincing yourself that
sin(36 degrees) = sqrt(10-2*sqrt[5])/4, or, equivalently, that
cos(36 degrees) = (1+sqrt[5])/4. You can do that by considering the
following diagram:

                                   R
                                 _,o
                              _,' / \
                           _,' 36/36 \
                   x+y _,-'     /     \
                    _,'        /       \
                _,-'         y/         \y
             _,'             /           \
          _,'               /             \
       _,'36     y      108/72     x     72\
     o'-------------------o-----------------o
   P                     S                   Q

All three triangles are isosceles, because they each have two angles
equal, so making SQ = x and QR = y, the above labels are correct for 
the lengths of sides. Using the fact that triangles PQR and RSQ are 
similar, since they have equal angles of 36, 72, and 72 degrees, you 
get

   PQ/QR = RS/SQ,
   (x+y)/y = y/x,
   1 + (y/x) = (y/x)^2,
   y/x = (1+sqrt[5])/2  (by the Quadratic Formula, discarding the
                         negative root as extraneous),
   2*(y/x)^2 = 3 + sqrt[5].

Then applying the Law of Cosines to RSQ,

   x^2 = y^2 + y^2 - 2*y^2*cos(36 degrees),
   cos(36 degrees) = (2*y^2-x^2)/(2*y^2),
                   = (2*[y/x]^2-1)/(2*[y/x]^2),
                   = (2+sqrt[5])/(3+sqrt[5]),
                   = (1+sqrt[5])/4,
   sin(36 degrees) = sqrt[1-cos^2(36 degrees)],
                   = sqrt[16-(1+sqrt[5])^2]/4,
                   = sqrt(10-2*sqrt[5])/4,
   CN = 2*sin(36 degrees).

Thus CN is the length of the side of a regular pentagon inscribed in a
unit circle.

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

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