Regular Pentagon Construction ProofDate: 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/ |
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