Trisecting an AngleDate: 11/21/96 at 12:45:13 From: Tracy J Henson Subject: Trisecting an angel I've recently been pondering the possibility of trisecting an angle. I wasn't sure if it was possible and found a FAQ submitted to you that stated it was indeed impossible. Has it been proved to be impossible or is it that no one has proved it possible? The reason I ask is that I think (but am not sure) that I have found a way to trisect an angle. I am a seventh grade math teacher, so I am not completely ignorant of mathematical proofs. I would appreciate some information on who would be the best person to contact with my "proof". I hope you believe me because I think this actually works. Date: 11/21/96 at 14:47:32 From: Doctor Tom Subject: Re: Trisecting an angel Hi Tracy, There is, in fact, a proof that a trisection is impossible. Many people think they have found trisections, but they either don't understand exactly what the problem is or their method is flawed. The problem requires the construction of the trisection using an (unmarked) straightedge and a compass. If, for example, you were allowed to make 2 marks on the straight-edge, there is a trisection. The straight-edge can only connect points already constructed, or use arbitrary points, and the compass can only be used similarly. An exact trisection in a finite number of steps is also required since it's easy to do a trisection that doesn't require too much accuracy. As a limiting case, it might be possible to perfectly trisect an angle in an infinite number of steps. The proof that it's impossible basically considers the arithmetic form of the points that can be obtained using a compass and straight-edge. In a sense, they are "quadratic" devices - the new points generated are solutions of quadratic equations of previously constructed points. So you can clearly get things like square roots, fourth roots, eighth roots, and so on (and it's a bit more complicated than that - what you can get are field extensions of degree 2 over whatever you begin with). But you CANNOT solve irreducible cubic equations this way - such solutions require extensions of degree 3, and no combination of multiples of 2 gives a multiple of 3 (to put it in an inexact, but hopefully clear, way). You can, beginning from nothing, construct a 60 degree angle. So if you can trisect anything, you can trisect the 60 degree angle which you produce. So if you have a trisection, you can construct, from scratch, a 20 degree angle. Hence you can construct the sine and cosine of 20 degrees. But it's easy to show that the sine and cosine of 20 degrees are roots of an irreducible cubic equation over the rational numbers. This is a contradiction, so a trisection is impossible. To get the details, read about fields and field extensions in an abstract algebra book. Also, to find out more about trisecting angles (or not being able to do so!), do a search under "trisecting an angle" at: http://mathforum.org/mathgrepform.html or take a look at this site: http://daisy.uwaterloo.ca/~alopez-o/math-faq/ So the bottom line is, there's probably something wrong with your trisection, not because I've seen it, but because I've studied field extensions and irreducibility. (And I'm not sure about trisecting "angels" as in your title - you'll need to talk to a Bible expert about that :^) -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 11/21/96 at 22:36:25 From: Tracy J Henson Subject: Re: Trisecting an angel Thanks for your swift response. I thought I found all the typos. I surely would never attempt to trisect or even bisect an angel. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/