Conference rusure::math

    If you let the length of OA be 1 unit, you can use the Pythagorean
    relation and "definition of a circle" to find these other lengths:
    
         OA = OX = OB = OY = 1    ! each is a radius (center at O)
         OC = CB = 1/2            ! each is half of OB = 1
         CA = sqrt(5) / 2         ! Pyth. CA^2 = OC^2 + OA^2
         CA = CD                  ! each is a radius (center at C)
         OD = (sqrt(5) - 1) / 2   ! OC + OD = CD
    
         AD^2 = OD^2 + OA^2       ! Pyth.
              = (5 - 2sqrt(5) + 1)/4 + 1
              = (10 - 2sqrt(5))/4 = (5 - sqrt(5))/2
    
         So AD = 1.17... which looks about right. (-:
    
    The claim is that AD is the length of the side of the pentagon
    that the circle (center O radius OA) circumscribes.  It's back to
    work now so I'll finish this later.
         
    Dan

A nice way to make a pentagon is with a length of wide ribbon, like
that with which you might wrap a present.

Tie a simple overhand knot in the middle of the ribbon and gently
pull it taut.  See the pentagon ?!

/Eric

    The length of the side of your construct is easily shown to be
    
    			sqrt(2.5 - sqrt(1.25)) 	radii
    
    The side of the pentagon is also
    
    			2.0 * sin(pi/5)
    
    H_FLOATING calculations verify that these two numbers are equal
    to 30+ decimal places, so I don't think it is just an approximate
    construction.
    
    I wonder if someone managed to get the first formula as the side
    of the pentagon from a different direction, and then found this
    construct as a way of geometrically evaluating this expression.
    
    /Bevin

    It is possible to show "symbolically" that the side computed
    as 2 sin(pi/5) and sqrt(5/2 - sqrt(5/4)) are equal.
    
    But as I said before, that will be in a later reply. (-:
    
    Dan

    Okay, here is the later reply.
    
    e^ix = cos x + i sin x
    
    e^(n ix) = (e^ix)^n = (cos x + i sin x)^n = ... [binomial expansion]
    e^(n ix) = cos nx + i sin nx
    
    If you equate the last two, use i^2 = -1, i^3 = -i, i^4 = 1, and
    substitute 1 - (sin x)^2 for (cos x)^2 or 1 - (cos x)^2 for (sin x)^2,
    you get a nice formula for cos nx as a polynomial of cos x, or sin nx
    as a polynomial of sin x.  For n=5 you get:
    
    sin 5x = 5 sin x - 20 (sin x)^3 + 16 (sin x)^5
    
    Now, let x take on the five values:
    
     x   =     0,     pi/5,     2 pi/5,     3 pi/5,    4 pi/5
    5x   =     0,       pi,       2 pi,       3 pi,      4 pi
    sin 5x =   0,        0,          0,          0,         0
    
    So for these five values of x,
    
              16 (sin x)^5 - 20 (sin x)^3 + 5 (sin x) = 0
    
    Or 16y^5 - 20y^3 + 5y = 0 where y = sin x
    
    This fifth degree equation can be solved:  factoring out y leaves a
    quadratic equation in y^2.  When you take the five roots and match
    them up with the corresponding values of x [i.e., the root y = sin x = 0
    goes with x = 0, etc.] then you see that the equality stated in .3
    between the computed length in .1 and 2 sin pi/5 is indeed true.
    
    This is the outline, you just have to fill in the details. (-:
    
    Dan
    

    If I recall, the most direct route to this is to observe that
    cos(2pi/5) = (sqrt(5) - 1)/4. Construct a 1, 2, sqrt(5) triangle,
    where 1 = radius of circle, using ruler and compasses, the construction
    can easily be done.
    
    Notice the value - similar to the Fibonacci ratio (sqrt(5) + 1)/2.
    
    Anybody know a construction for a regular heptagon? What regular
    n-gons cannot be constructed?

Re: .1, .5

As I said in .0, I'm not a math person.  The solution presented may be 
(probably is) valid, but to me the portion of it expounded in .5 is 
neither simple nor clear.  If it turns out that solutions of this type 
are the only possible ones for the problem, then I guess that's that.

Re: .6

Richard, the same drafting book from which I learned this pentagon 
construction contains a different construction for a general n-gon.  I
don't recall at present if it involves trial and error, but I'll check 
it out.

   The numbers n for which the regular n-gon is constructible with straight
edge and compass are those for which, for any odd prime divisor, p, of n:

     (i) p is Fermat (i.e. of the form 2^(2^a) + 1)

     (ii) p^2 does not divide n.

   So, for small numbers: 3,4,5,6,8 *are* constructible
   7 and 9 are not.

& in particular 17 *is* constructible (but don't ask me how)

Regards
Andrew

Re .9
	I thought the regular n-gon is constructible with straight
	edge and compass iff

		     e                                              e
		n = 2  p  p  ... p , e >= 0, m >= 0 (m=0 means n = 2 )
		        1  2      m

	where the p  are distinct Fermat primes.
	           i

	Thus the (2^32+1)-gon is not constructible because although
	2^(2^5)+1 is a Fermat number, it is not a Fermat prime:
	2^31 + 1 = 4294967297 = 641 x 6700417.

Re .7

Here's an alternate way to get the expansion of sin(5x).

We know that:
	sin(a+b) = cos(a) sin(b) + sin(a) cos(b)
	cos(a+b) = cos(a) cos(b) - sin(a) sin(b)

So,
	sin 2x = cos x sin x + sin x cos x = 2 sin x cos x
	cos 2x = (cos x)^2 - (sin x)^2 = 1 - 2 (sin x)^2
	sin 3x = cos 2x sin x + sin 2x cos x
	       = (1 - 2 (sin x)^2) sin x + 2 sin x cos x cos x
	       = sin x - 2 (sin x)^3 + 2 sin x (1 - (sin x)^2)
	       = 3 sin x - 4 (sin x)^3
	cos 3x = cos 2x cos x - sin 2x sin x
	       = (1 - 2 (sin x)^2) cos x - 2 sin x cos x sin x
	       = cos x ( 1 - 4 (sin x)^2 )
	sin 4x = ... = cos x ( 4 sin x - 8 (sin x)^3 )
	cos 4x = ... = 8 (sin x)^4 - 8 (sin x)^2 + 1
	sin 5x = ... = 16 (sin x)^5 - 20 (sin x)^3 + 5 sin x

Notice that in general, we have a polynomial in (sin x)^2, and a possible
factor of (cos x), depending on whether we're expanding a sin or cos, and
on whether we're expanding an odd or even multiple of x.

    Actually, .9 and .10 are saying the same thing:
    
    .9:  n is such that
    
         for any odd prime divisor, p, of n:
    
         (i) p is Fermat (i.e. of the form 2^(2^a) + 1)
    
         (ii) p^2 does not divide n.
    
    .10:  n is such that
    
              e                                             e
         n = 2 p  p  ... p , e >= 0, m >= 0 (m=0 means n = 2 )
                1  2      m
    
         where the p  are distinct Fermat primes.
                    i
    
    In either case, the only odd primes dividing n are Fermat primes,
    and any given Fermat prime dividing n does so only once.  Both
    descriptions even allow n = 1 and n = 2, although I suppose that
    those n could be permitted as "trivial" n-gons.
    
    Dan

    
>>  Re .7                but to me the portion of it expounded in .5 is
>>    neither simple nor clear.
    
    You already had a diagram, so I computed the edge length it implied,
    then someone gave an expression for what that edge length should be,
    so I figured, well, show they are the same.  So maybe it was a little
    messy. (-:
    
    But for something that is clear, I guess I would have approached it
    in the other direction, by first trying to find the easiest edge or
    diagonal length or angle to express symbolically.  Then I would go
    off to the side of the page and construct that.  Finally I would
    mark that off in the circle.  This would be easiest to explain,
    but the construction would look totally artificial.
    
    At least solving for the length in the given construction was okay, 
    so I will think about an easier way to compute the sine of 36 degrees.
    But the easiest first approach is always the "methodical" one, and
    simplifying it usually waits for someone's clever insight.  Maybe
    starting with .11 will clear things up.
    
    Dan

There is/was a discussion of construction of n-gons in the Usenet sci.math
newsgroup.  Apparently, some poor graduate student spent 10 years of his life
constructing the 65537-gon. (supposedly the faculty tried to get rid of him)

-Mike

    A usenet posting by a Paul Callahan, concerning
    the construction of the regular 65537-gon, states:
    
          "65537-gon" -- The fact that this is constructable
          with straight-edge and compass is interesting.  The actual
          steps are UNinteresting.  (Incidentally, I know of a
          one-step construction using ONLY a compass which
          produces an excellent approximation to a 65537-gon.)
    
    Dan

     What finer compliment can one give to Mr. Gauss than to say
     that if he were both alive today and a DECkie, that he would
     be an avid noter in CLT::MATH?

     This will sketch Gauss's solution to the construction of the
     regular p-gon for a Fermat prime p, by showing how it works
     for the particular case of p = 17 = 2^(2^k) + 1 for k=2.

     My reference is the book

       100 Great Problems of Elementary Mathematics
       Their History and Solution
       Heinrich Dorrie
       translated by David Antin
       Dover Publications, Inc., New York, copyright 1965.

     Let w be the "first" complex 17th root of one, i.e.,

                w = cos (2 pi / 17) + i sin (2 pi / 17)

     The roots of the polynomial equation z^17 = 1 are w^0 = 1,
     w^1 = 1, w^2, w^3, ..., up and including w^16.  z^17 - 1
     factors into (z - 1) and (z^16 + z^15 + ... + z^2 + z + 1).
     The root w^0 = 1 is the root to z - 1 = 0, and the sixteen
     non-zero powers of w are the roots of z^16 + ... + 1 = 0.

     Note that plugging w itself in for z in the latter equations
     show that the sum w^1 + w^2 + ... + w^16 = -1.

     Gauss's idea was to group the roots (the powers of w) into
     so-called "periods" [well, that's what this book calls them,
     even if Gauss didn't].  Each period is a sum of some of
     these roots in which each successive term is the g-th power
     of the preceding term, and the g-th power of the last term
     is the first term [it says "(hence the name period)"].  The
     exponent g is a primitive root of the prime p.  You can use
     g = 3 for Fermat primes p > 3, see note 208.       

     The first step will split the sum w^1 + w^2 + ... + w^16 =
     -1 into two periods, each a sum of eight of the terms.  The
     value of each sum will be solved for exactly.  Each step
     will split a period into a sum of two periods of half as
     many terms each, and solve for those sums.  I suppose that
     it is somewhere in here that p having to be a Fermat prime
     comes in, because the p-1 terms, being a power of two, you
     can keep halving the periods like that.

     Anyway, g, the primitive root of p, is such that the roots
     w^1, ..., w^(p-1) can be rewritten as

          z0 = w
          z1 = w^g
          z2 = w^(g^2)
          z3 = w^(g^3)
            ...
          z[p-2] = w^(g^(p-2))

     [Note that w^(qp + r) = w^r, because w^p = 1.]

     For p = 17, g = 3, rewrite the powers of 3, i.e., 3^0, 3^1,
     3^2, 3^3, ..., modulo 17 [see remark above about w^(qp + r)] as
     1, 3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4, 12, 2, 6, so
     that z0 = w, z1 = w^3, z2 = w^9, z3 = w^10, ..., z15 = w^6.

     The first period is z0 + z1 + ... + z[p-2] which is a
     permutation of w^1 + w^2 + ... + w^(p-1) which we have already
     seen is equal to -1.

     The next two periods are
        X = z0 + z2 + z4 + z6 + z8 + z10 + z12 + z14
          = w^1 + w^9 + w^13 + w^15 + w^16 + w^8 + w^4 + w^2
     and
        x = z1 + z3 + z5 + z7 + z9 + z11 + z13 + z15

     Compute the sum X + x; it is the sum of all of the powers of
     w from 1 to 16, i.e., X + x = -1.

     Compute the product Xx.  This will be a product of a sum of
     eight terms by a sum of eight terms.  Apparently the [sum of]
     sixty-four terms of the product reduce to four copies of the
     sum w1 + ... + w16 = -1 -- I don't know that for sure -- but
     the book says the product ends up as -4.

     The equation (t - X)(t - x) = 0 or t^2 -(X + x)t + Xx = 0
     can be written down once we know X + x and Xx; it is
     t^2 + t - 4 = 0, which has roots t1 = (-1 + sqrt(17))/2 and
     t2 = (-1 - sqrt(17))/2.

     Oh, gee, which is X and which is x?  Well, they draw a
     diagram and show that the terms in X are terms with
     positive real part and the terms in x have negative real
     parts.  The book notes that if r+s=17 then w^r, w^s have
     equal real parts.  If you are lazy you can compute X and x
     numerically from w = cos (2 pi /17) + i sin (2 pi /17).  Anyway,
     it turns out that X = (-1 + sqrt(17))/2, x = (-1 - sqrt(17))/2.

     The four four-term periods are
        U = z0 + z4 + z8 + z12 = w^1 + w^13 + w^16 + w^4
        u = z2 + z6 + z10 + z14
        V = z1 + z5 + z9 + z13
        v = z3 + z7 + z11 + z15

     and so U + u = X, V + v = x.  Computing Uu and Vv it turns
     out that Uu = -1 and Vv = -1, so that U and u are the roots
     of t^2 - Xt - 1 = 0 and V and v are the roots of t^2 - xt - 1 = 0.
     Again, the big letters were chosen that way because it turns
     out that Re(U) > Re(u) [where Re(a+bi) = a] and Re(V) > Re(v),
     so these roots are
        U = (X + sqrt(X^2 + 4))/2, u = (X - sqrt(X^2 + 4))/2
        V = (x + sqrt(x^2 + 4))/2, v = (x - sqrt(x^2 + 4))/2

     Of the two-membered periods one needs only
        F = z0 + z8 = w + w^16, f = z4 + z12 = w^13 + w^4
     when F + f = U, and Ff = V [multiply it out in terms of
     powers of w and compare to V represented that way].  The
     quadratic equation with roots F and f is t^2 - Ut + V = 0.

     So, to construct the repular "heptadecagon" (17-gon) using
     compass and straight-edge [the heptadecagon being inscribed
     into the circle in the complex plane centered at the origin,
     and with radius one]

         1) Construct X and x
         2) Construct U and V
         3) Construct F and f
         4) F = w + w^16, f = w^13 + w^4 are sums of complex
            conjugates.  Therefore F = 2Re(w), and the
            perpendicular bisector of the segment from the origin
            to F cuts the circle at w^1 and w^16.  Of the
            perpendicular bisector or the segment from the origin
            to f cuts the circle at w^4 and w^13.  From these you
            can mark off the rest of the vertices.

     Dan

     P.S. You do have to know how to "construct" lengths a
     rational multiple of a given segment, or the product of two
     lengths, or the square root of a length.

    I applied the technique of .-1 to the case of p=5.
    
    I solved all the way for w = the "first" complex fifth root of 1.
    = cos (2 pi / 5) + i sin (2 pi / 5)
    
        -1 + sqrt(5)     sqrt(10 + 2sqrt(5))
    =   ------------ + i -------------------
             4                    4
    
    To get w^2 from this, change those two of the three plus signs that are
    above a dashed line into minus signs.  w^3 will be the complex
    conjugate of w^2.  w^4 will be the complex conjugate of w.  w^5, of
    course, will be 1.  The only trick in simplifying products if you
    multiply these out yourself is to sometimes convert A sqrt(B) into
    sqrt(A^2 B) when both A and B are complicated.
    
    The length of a side this gives for a heptagon is
    sqrt(10 - 2sqrt(5))/2.
    
    Dan

Re: construction of n-gons.  The same book from which I got the 
construction detailed in .0, provides a construction for any n-gon.  
This construction involves trial, however.  I'm not sure that it 
qualifies as a legitimate construction; but if it does, then the theory 
is valid for *any* n-gon.

I can't draw a figure with character-cell graphics that is good enough 
to illustrate, so I'll try to be detailed in the verbiage.

To construct a regular polygon of n sides, given the length of a side:

1.  Lay off side AB equal to the given length, and with A as center and 
    AB as radius, draw a semicircle with C as the opposite end of the
    diameter (BAC).

2.  Divide the semicircle into the required number of parts by trial.
    Label the points thus located as D, E, etc., beginning at C and 
    proceeding toward B.

3.  Draw line AE, forming a second side of the polygon.

4.  Bisect AB and AE.  The intersection of the bisectors, O, is the 
    center of a circumscribing circle.  Draw the circle.

5.  Draw lines from A through the remaining points F, etc., on the
    semicircle to intersect the circle.  Connect these intersections
    by straight lines.

5a. (Alternate, not given in the book)  Having found the circle, you can
    step off the length of a side as many times as required to locate 
    the remaining vertices of the polygon.

Comments?

- Dick

    Re .-1
    
>>    2.  Divide the semicircle into the required number of parts by trial.

    This is not a "classical" construction, i.e., compass and
    straightedge only.  Using the classical methods you can approximate
    the required angles to within an arbitrary precision, enough to
    get a good looking result on paper.  But it is not a "theoretically
    perfect" regular n-gon.  Using extra tools, like a ruler (a ruled
    straightedge, i.e. with lengths marked off) and protracter [sp?] (with
    angles marked) you can construct more figures than with the
    classical stuff alone.
    
    Dan