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Relationship between constructibility of reg. polygons and cot(pi/N)

  1. Sep 9, 2014 #1
    Full title: Relationship between the constructibility of regular polygons and the reducability of trigonometric functions into expressions of square roots.

    I stumbled upon this after I derived the formula for the area of a triangle given it's side length x as a trigonometry exercise. ## A = \frac{1}{2}\sin{(60°)}x^2 = \frac{\sqrt{3}}{4}x^2 ##. I challenged myself to find the area formulas for every N-gon until I found a pattern. I did square, pentagon, and hexagon using basic trigonometry, and reducing the trigonometric functions into square root expressions using Wolfram.

    Things started to get weird when I got to heptagon or the 7-gon, though. Wolfram wouldn't reduce it, and my area equation was made ugly by the presence of a trig function. (I spent a good few hours researching how to find exact trig values by hand using the half-angle formula et al, before I realized it was impossible)

    Then I tried to find a general angle formula, but I couldn't write the later trigonometric functions in terms of sine. Eventually I figured out cot() is the only one that works (but I don't know why). This general pattern popped out at me:

    $$ A = \frac{N}{4}\cot{\frac{\pi}{N}} $$

    After I got to the 11-gon I started noticing another pattern: Whether or not the function ##\cot(\frac{\pi}{N})##, where N is the number of sides, is reducible has the same truth value as whether or not a given polygon is constructible. (I was amazed by this). Also, the relative complexity of each square-root expression is correlated with the relative complexity of the construction of that polygon. (As first seen with the long and convoluted 17-gon)

    The wiki page for the 17-gon mentions that its area and constructibility was determined by Gauss 200 years ago, while he was going through this thought process, sqeeee!). It also claimed he had a method of determining whether cot(pi/N) was in general reducible, based on some property of "fermat" primes.


    I had two other non-trivial observations:

    1. The square root expression of cot(pi/N) has a similarity with cot(pi/(2N)). Something to do with the half angle formula, maybe.

    2. If cot(pi/N) cannot be reduced, then cot(pi/(2N)) cannot be reduced either.

    I also found the entry in the OEIS of the integer sequence of non-constructible polygons:

    It mentions a generating function, but I don't know what the "totient function, phi" is yet.

    I can research what the totient function is on my own, but...


    I have four challenging questions that I think this board will be interested in:

    1. Does there exist a general algorithm for determining the exact value of ##\cot(\frac{\pi}{N})## in terms of square roots if such an expression exists?

    2. Likewise, is there a general algorithm to derive the steps of constructing a regular polygon with a number of sides ##N##?

    3. Does there exist a linear-time algorithm for converting a square-root expression into a set of polygon construction steps?

    Beyond a certain degree of self-interference and complexity, a system can be used as an analogue to a Turing machine (i.e. it can be used for arbitrary computation). If a system is beyond that point, it makes certain statements about that system undecidable, such as if it will enter into an infinite-regress or not.

    4. If you had infinite time, infinite paper, and an unmarked ruler and compass - could you do the same set of problems as a Turing machine?


    Here's my observations on the possible answers:
    * I suspect questions 1 and 2 have the same truth value.
    * I suspect question 3 is true (there is a linear time algorithm for converting between a square-root expression and a set of instruction for polygon construction)
    * I suspect question 4 is the negation of questions 1 and 2.

    NOTE: I won't be disappointed with answers like "One could write a whole original book on this topic, nobody knows yet." I just want to encourage discussion on an interesting problem. I'm about to leave for class, so I won't be back for a while.
    Last edited: Sep 9, 2014
  2. jcsd
  3. Sep 21, 2014 #2

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