Date: 03/10/98 at 04:47:49 From: Melanie Subject: trisecting angles not divisible by 3 Hi. I have been faced with this rather annoying problem: "The measure in degrees of a given angle is 180/n, where n is a positive integer not divisible by 3. Prove that the angle CAN be trisected using Euclidean means." There isn't any 'rule' for trisecting angles. The only way I can think of is to construct a 60 degree angle to trisect a straight angle, and then to bisect this to obtain a right angle. How do you sugest I go about this question? Is there any relevance about the divisible by 3 stuff? (I suppose this relates to 60, which is a multiple of 3?) Please help, I so don't understand! Thank you, Melanie.
Date: 03/12/98 at 14:56:00 From: Doctor Wilkinson Subject: Re: trisecting angles not divisible by 3 This is a very nice little problem. At first glance, it looks very difficult or even impossible, but it turns out to be quite easy. Since n is not divisible by 3, it must be of the form 3q + 1 or 3q - 1, where q is a positive integer. Suppose that n = 3q + 1 Divide both sides by 3n, and you get 1/3 = q/n + 1/(3n) or 1/(3n) = 1/3 - q/n and, multiplying by 180, 180/(3n) = 180/3 - q(180/n) Now, you're given the angle 180/n. Draw a circle with its center at the vertex of the angle. Suppose the angle is AOB, where O is the center of the circle and going the short way from A to B around the circle is counterclockwise. Now set the compass to the radius of the circle, put the point of the compass at A and mark a point C on the circle, again going counterclockwise. This makes AOC 60 degrees, or 180/3. Now set the compass point on A and the other arm at B, and step clockwise from C for q steps. You will now be at a point X, with AOX equal to 180/3 - q(180/n) = 180/(3n), so you have trisected the given angle. I'll let you work out how to do the case when n = 3q - 1. -Doctor Wilkinson, The Math Forum http://mathforum.org/dr.math/
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