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-1)/2. OC = 1. Using the Pythagorean Theorem again, CN = sqrt(10-2*sqrt)/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)/4, or, equivalently, that cos(36 degrees) = (1+sqrt)/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)/2 (by the Quadratic Formula, discarding the negative root as extraneous), 2*(y/x)^2 = 3 + sqrt. 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)/(3+sqrt), = (1+sqrt)/4, sin(36 degrees) = sqrt[1-cos^2(36 degrees)], = sqrt[16-(1+sqrt)^2]/4, = sqrt(10-2*sqrt)/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/
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