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Regular Pentagon Construction Proof


Date: 11/26/2001 at 11:02:22
From: Joe Boehm
Subject: Geometry

What is the proof of the construction of a regular pentagon?


Date: 11/27/2001 at 10:45:43
From: Doctor Jubal
Subject: Re: Geometry

Hi Joe,

There is no one construction of a regular pentagon. There are many, 
many ways to do it, so you can't talk about proving that a 
construction works without giving the steps of the construction.

There's a nicely illustrated construction of a regular pentagon by Dr. 
Floor on the Dr. Math FAQ page at

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

and here's why it works.

Let's say that the radius of the circle centered at O through N,E,S,W 
is unity. M is the midpoint of ON, so OM = 1/2, and by the Pythagorean 
theorem, ME is sqrt(5)/2, so the radius of the circle at M through E 
is sqrt(5)/2.

Since this circle's radius is sqrt(5)/2, and OM = 1/2, then 
OE' = (sqrt(5) - 1)/2, and by the Pythagorean theorem, 
EE' = EP' = EP'' = sqrt(10 - 2 sqrt(5))/2.

Similarly, OE'' = (sqrt(5)+1)/2, so by the Pythagorean theorem, 
EE'' = EQ' = EQ'' sqrt(10 + 2 sqrt(5))/2.

And these are the values they need to be in a regular pentagon. The 
law of cosines (a trigonometric extension of the Pythagorean theorem), 
states that in any triangle, c^2 = a^2 + b^2 - 2ab cos C, where C is 
the angle between sides a and b.

Applied to triangle OEP' or OEP'', using EP' or EP'' as side c,

  (5 - sqrt(5))/2 = 1 + 1 - 2 cos EOP'

  (1 - sqrt(5))/2 = - 2 cos EOP'

  cos EOP' = (sqrt(5) - 1)/4

Consulting a table of trigonometric values shows that this corresponds 
to 

  EOP' = EOP'' = 2pi/5

Applied to triangle OEQ' or OEQ'', using EQ' or EQ'' as side c,

  (5 + sqrt(5))/2 = 1 + 1 - 2 cos EOQ'

  (1 + sqrt(5))/2 = - 2 cos EOQ'

  cos EOQ' = -(1+sqrt(5))/4

Which corresponds to

  EOQ' = EOQ'' = 4pi/5

So the angles EOP', EOP'', P'OQ', P''OQ'', and Q'OQ'' are all equal to 
2pi/5, and the segments OE, OP', OP'', OQ', OQ'' are equal to 1. By 
SAS congruency, EOP', EOP'', P'OQ', P''OQ'', and Q'OQ'' are congruent 
triangles. By congruency of corresponding parts, the sides of the 
pentagon are congruent. Also, since the interior angles of the 
pentagon are each the sums of corresponding angles in the congruent 
triangles, they are congruent. The pentagon has congruent sides and 
congruent angles, and thus it is regular.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

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

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