Conference rusure::math

My intuition is that the inradius of a cyclic pentagon is _not_ uniquely
determined by the lengths of its sides and the order in which they occur.

It seems to me that there might be at least one extra degree of freedom.

Informal approach:  cut a straw into five segments, not all equal.  Draw a
circle of the "right" size and fit the segments into a pentagon inscribed in
the circle.  Now move them around and see if you can shift the angles
while keeping the segments inscribed and adjacent.  If I were a betting man,
my bet would be there's room to wiggle.

    Perhaps you would get better results if you lit candles and chanted out
    "Fire, Walk with me"    :-)
    
      John

This looked too hard to me, but since .1 raises doubts about uniqueness, I'll
take a shot at it.

Pick a side and call the length s1.  Now consider the right triangle formed by
half the side, the radius from the end point and the perpendicular bisector of
the side (a theorem in elementary geometry shows that this will pass through
the center).  Call the interior angle of this triangle at the center a1.  Then
s1 and a1 are related by

	a1 = arcsin ( s1 / ( 2 * r ) )

There are five such equations in six unknowns: a1 ... a5 and r.

There is also a sixth equation relating the ai

	2 * ( a1 + a2 + a3 + a4 + a5 ) = 2 * pi

This makes six equations in six unknowns.  There is no room for wiggle here, but
see a few comments on non-uniqueness below.

If I actually needed to find r, I would substitute the first five into the
sixth, giving one equation with the one unknown r.  Then I could solve 
for r numerically.

I presume that .0 is looking for a closed solution.  I suspect there is one, 
using either the formulas in .0 or the side-side-side trig relations, but I'll
leave that to the more algebraically inclined.

Now for the non-uniqueness:

In principle, the sixth equation should really equate to n * 2 * pi.  The case
n=1 is a pentagon. Reply .2 may have been referring to the fact that n=2 is
a pentagram.  But .0 explicitly asked about a pentagon, and we may take that
to exclude n > 1.

It is also true that arcsin is not uniquely defined.  Most of the cases I 
have found lead to five sided figures which are not strictly pentagons.  I can
draw one case where r is close to s/2.  There are two values of r which will
solve the six equations, corresponding to two pentagons, one of which encloses 
the center of the circle and one of which does not.  Still no wiggle, but not
a uniquely defined radius.

    If we make the further assumption that the pentagon in question is
    *simple* (i.e., no edges intersect except at a vertex, and all vertexes
    [it is *too* a proper plural] have exactly two edges meeting at them),
    then it is pretty easy to see that the solution is unique.  Multiple
    solutions for "r" are spurious.

    Imagine that an oracle has presented you with a set of sticks of the
    specified length, numbered in clockwise order, and (more oracularly)
    a circle of the correct radius.  Your job is to place the sticks on
    the circle to form the pentagon.

    You take the first stick (which, of course, could be any one of them)
    and place the "counterclockwise" end of it on some point on the
    circle's circumference.  This choice of starting point is the last
    free choice you have.  There are only two points that the other end
    can be placed on -- only one of which is clockwise of the first point
    (the two points may coincide if the the stick is twice the radius).

    You now take the next stick and place it one end on the just placed
    end of the first.  Once again you have only one choice of where to
    place the second end.

    This continues through the five sticks until you get to the last stick
    whose end-point will just line up with the starting point -- assuming
    that your oracle was right.

    If she was wrong and the circle is too large, then your actions would
    be no less constrained but the final point will come short of the
    initial point.  If the circle is too small, then the final point will
    overshoot.  Only one radius will work.

    There is *at most* a single solution.

    Any additional solutions will be spurious -- the same as other
    solutions, negative radii or whatever.

    The most interesting property to the analytic solution in .3 is that
    the radius *does not depend on the order of the sides*.  This is not
    obvious from my geometric argument.  Anyone with a geometric argument
    for this property.

					    Topher

RE: .4 (me)
    
    Never mind -- I thought of one myself in the lunch line.
    
    Imagine that we have the sticks in one order within the proper size
    circle as in .4.  We want to move the sticks around into another order
    in such a way so that the size of the circle doesn't need to change.
    The question is -- can we always do this?
    
    Let the segement of the circle formed the two radii from the center of
    the circle to an edges endpoints be called a 1-wedge.  A 2-wedge
    consists of two adjacent 1-wedges.  We can take any 2-wedge and reverse
    the order of its two edgeds by "flipping it over"  -- the circle will
    be unchanged by this.  By using a "bubble sort" we can use a series of
    such 2-wedge flips to put the edges in any desired order.  QED.
    
    Of course, we can switch any two edges by using either a 2-wedge flip
    or a 3-wedge flip, allowing us to use a the more efficient "exchange
    sort" rather than "bubble sort" but mathematics doesn't care about
    efficiency.
    
    Note that the arguments and formula in .3, .4 and here do not depend
    on the figure being a pentagon -- they apply to any simple n-gon.
    (with the exception that exchanging two edgess without affecting the
    order of the rest may need two i-wedge flips instead of just 1 if n>5).
    
    						Topher

re: .4

	Thanks for a good definition of a 'simple' pentagon.

>    You take the first stick (which, of course, could be any one of them)
>    and place the "counterclockwise" end of it on some point on the
>    circle's circumference.  This choice of starting point is the last
>    free choice you have.  There are only two points that the other end
>    can be placed on -- only one of which is clockwise of the first point
>    (the two points may coincide if the the stick is twice the radius).

OK, if a stick is sufficiently far from twice the radius, then choosing the 
counterclockwise point will lead to a non-simple pentagon.

But suppose that the stick length is very nearly twice the radius (and the
other stick lengths are large compared to this difference).  Now the clockwise
and counterclockwise choices will be very close, and neither will lead to 
a non-simple pentagon.  So we have another free choice.

I have not got a good geometric argument for why you can finish the pentagon
in both cases, but the algebraic argument is still good: there is still a
solution radius for the equation.


>    The most interesting property to the analytic solution in .3 is that
>    the radius *does not depend on the order of the sides*.  This is not
>    obvious from my geometric argument.  Anyone with a geometric argument
>    for this property.

Imagine that you have created a solution using your method.  You now have 5
isosceles triangles with all the double sides equal.  If you put them 
together with their unique angles touching, the unique angles will add up to
2 * pi and you will have your inscribed pentagon.  Now put them together in 
some other order.  You will still have an inscribed pentagon, with the same 
radius and the unique angles will still add up to 2 * pi.  The pentagon will,
in general, be different, but the radius of the circle is determined.

This works for any inscribed n-gon.

RE: rearranging the 5 sticks.

The `interior angle' argument (angles add up to 2*pi) applies if the simple
pentagon contains the center of the circle.  If it doesn't, a slightly
different argument is needed.

P.S.  I'm not convinced by the arguments that the radius is unique.  For a
counterexample to exist, I think it's necessary that for one radius, the
circle's center is inside the pentagon, and for another radius, the radius
is outside the pentagon.

>The `interior angle' argument (angles add up to 2*pi) applies if the simple
>pentagon contains the center of the circle.  If it doesn't, a slightly
>different argument is needed.

Good point.  The different argument is the same in mathematical form, but 
one cannot visualize it as I did, just pushing rigid wedges together.

>P.S.  I'm not convinced by the arguments that the radius is unique.  For a
>counterexample to exist, I think it's necessary that for one radius, the
>circle's center is inside the pentagon, and for another radius, the radius
>is outside the pentagon.

Not clear to me which arguments you find unconvincing.  In any case, you 
are correct that for the one counterexample I have found, the center of the
circle is inside and outside the pentagon for the two different radii.  Both
pentagons are still simple, by the earlier definition.  So it still looks like
a valid counterexample to me.

	What I want to do is solve the problem in a way that cleans up all
these issues of pentagon/pentagram, location of centre, simple polygons,
etc.   I also want to bring in one factor which has not been mentioned yet, the
polygon inequality (generalized triangle inequality).

	I'd like to return to Wally's formulation of the problem.   That:

	sin(a_i) = s_i / (2*r)		(for i = 1,..,n)
	
is clearly the case.   (s_i is the side length, a_i the half-angle, remember).
The tricky stuff comes in deciding what equality the a_i should satisfy.
I can restrict a_i to lie in the interval (0, pi/2].

	Suppose that we start with a vertex somewhere on Topher's oracular
circle.  As earlier replies have discussed, the order in which we take the
sides is irrelevant.   However, the direction in which the edges are measured
off is not irrelevant.   The most general solution will be when:

	sum (i) �a_i = k * pi

where the contribution of a_i is +ve or -ve depending on whether vertex i+1 is
to widdershins or to deasil (alias: to counter-clockwise or to clockwise) of
vertex i, if take the shorter route.

	So let's define x_i = �1, for all i, with x_n = 1, wlog.   Suppose
that all the x_i are known and fixed, and that so is k.

	a_n = k*pi - sum(i<n) x_i * a_i 

	Then we can banish a_n from the equation:

	sin(a_n) = s_n / (2 * r)

and simplify out the expression using standard trig identities for sin(a+b)
and cos(a+b), so that we are only dealing with a polynomial in sin(a_i) and
cos(a_i), for i < n.   We then express cos(a_i) = sqrt(1 - sin�(a_i)) and in 
turn sin(a_i) = s_i / (2 * r), and this gives us a necessary condition on r, 
for the given choice of the x_i and k.   Call this the Big Equation.   It
contains r and all the s_i, no trigonometric stuff but lots of square roots.
I'll pursue its solution perhaps at some future date.

	What will make it sufficient?   Well, r >= s_i/2 for all i, since
otherwise the circle is not big enough to contain some of the required chords.
That is all we need.

	Now, how many (sufficiently large) r can satisfy the Big Equation?
Let's look at:

	f(r) = sum(i) x_i * a_i(r)

(ie, our circle is no longer necessarily oracular.)

	df/dr = sum(i) -(x_i * s_i) / (r * sqrt(4r�-(s_i)�))

	Clearly, if all the x_i = 1, then there is no point at which
df/dr is >= 0, and thus f(r) is monotonic strictly decreasing.   As r becomes
very large, it asymptotes to s/r, where s is the semi-perimeter, which tends
to zero.   So (still assuming that x_i = 1 for all i) the intermediate value
theorem tells us that the number of solutions is:

	F = floor( f(s_m/2)/pi )

where s_m = max(i) s_i.   Wlog, let's take m = n (ie s_i =< s_n for all i).
If we have a solution, then one will be the "simple" one, with the centre of
the circle in the polygon.    If (for n = 5) we have two solutions, then the 
second will be the "pentagram", with the centre of the cirle in the interior 
*pentagon*.

	For n = 5, F =< 5/2, so no third solutions is possible.   Note that if
sum(i<n) a_i(s_m/2) < 1/2, then there are no solutions at all.   It would be 
nice to have a tidier expression for the conditions here.

	Now, if x_i are not all = 1, things are more complicated because f is
not necessarily monotonic.   f(r) goes to 0, as r goes to infinity, as
before, and we can compute F (with x_i now unrestrained) to give us a lower
bound on the number of r which exist.   However, I haven't had a chance to
check on the behaviour of df/dr, to give an upper bound on the number of
solutions.

	Note that Topher's concept of "simplicity" corresponds (using
convexity arguments) to one of:

	(i) k = 1, & all x_i = 1
	(ii) k = 0, & all x_i (i<n) = -1

	So he removes *most* of the tricky cases, but I allege still has to
handle the non-monotonic f for case (ii).

	There's still work to be done.

Regards,
Andrew.

	The Big Equation can be transformed into a polynomial equation
in r, although this is sufficiently messy for even n=4 that it looks like a 
excuse for my long-postponed baptism by MAPLE.   I'll be back when I've run
it.   For n=3, it converts into a quadratic (where l_i are the side lengths)

	r = prod(i)(l_i) / sqrt( sum(i<>j)(l_i�*l_j�) - sum(i)l_i^4 )

the denominator of which is a new form (for me) of my beloved Hero's equation.

	However, it's worth pointing out that r is not likely to be 
be particularly clean, for general polygons.   For instance, if n=7, and
all the edge lengths are the same, then the polygon is known (Gauss) not to be
constructible, and r can be a root of an irreducible sixtic equation.   Or, we
could imagine that one of the edges doubles back on the others, and this
enables r to be constructible, since we are essentially just building the
regular pentagon.   This folding property (for larger n) means that there are
basically an enormous number of possible irreducible polynomial which we know
must factor into the Big Equation.   I don't know if there is some natural to
code Topher's "simplicity" into the problem representation, which will exclude
these pathological solutions.

	The advantage of tackling the Big Equation however, is that it
gets rid of the essentially irrelevant distinction between cases where the
circumcentre lies inside or outside the pentagon, which marred -.1.

	One thing I find helpful in dealing with this problem is to imagine
n being very large, and all the edges being the same length.   It is then
clear that there are only a finite number of possible answers (corresponding
to different values of x_i and k), and moreover the relationships between r
and K becomes clear.

	The basic questions that I want to answer are:

	(1) How many possible r are there for a convex pentagon?
	(2) What are they?   (In terms of radicals, or ultra-radicals)
	(3) How does the situation change if the pentagon is not convex?
	(4) What solutions are constructible, in the Gauss/Galois sense?

Regards,
Andrew.

	re .10

>>	However, it's worth pointing out that r is not likely to be
>>be particularly clean, for general polygons.   For instance, if n=7, and
>>all the edge lengths are the same, then the polygon is known (Gauss) not to be
>>constructible, and r can be a root of an irreducible sixtic equation.

	The sixtic equation (is it really called sixtic?) for the
	side of a regular 7-gon is not irreducible.  x^6 + x^5 +
	x^4 + x^3 + x^2 + x + 1 factors into a product of two
	cubics, which can then be solved.  The roots as you say are
	not constructible though.

	Dan

RE: .9,.10 (Andrew)
    
    I don't want to take credit where credit is not due.  The concept of a
    "simple polygon" is a standard one in geometry.  I tailored the
    definition a little for the application, but informally a simple
    polygon is one which:
    
    	1) Is non-self interesecting.
    	2) No adjacent triple of vertexes are co-linear.
    
    I ignored the second requirement since it was implicit in the polygon
    being inscribed in a circle.  If anyone finds it usefull, keep in mind
    that being inscribed in a circle means that "simplicity" is equivalent
    to "convexity".
    
    					Topher

    FWIW, I considered the five side lengths of a,a,a,a,b.
    The `big equation' simplifies to:
    
    	  2        6       4      2
    	(B  - 16) R  + 20 R  - 8 R  + 1
    
    where B = b/a, and R = r/a.  This is a cubic in R^2.
    
    For what real values of B does it have more than one real root?
    
    Does this happen when B < 4 ?  (i.e., the longest side can't be
    larger than the sum of the other 4 sides).

	I've just got this MAPLE thing running, and I haven't much of
a clue how to do anything.   How does one substitute variables, for
instance?   The Tschirnhaus transformation would have been damn handy in
dealing with Peter's poly.

	So far as I can tell in my primitive state, the cubic that Peter
generated, has one or three real roots depending as B� is greater than or 
less than 32/27.   Ie, the smaller that B� is, the more choice there is.   
I can believe that.

	So for instance, if b = a (ie all sides the same) then we get
three values of R�:

	1/3

	1/2 + sqrt(5)/10

	1/2 - sqrt(5)/10

each of these is positive, and so gives us a positive r.   For each of these,
2r > a, so these all generate proper solutions.   This is quite a surprise!

	But if I take B� = 2, then I only get one real solution.

	We're on the way!

Regards,
Andrew.

	Of course, a moment's thought tells us that the result I quoted
for b=a in the previous reply is *not* surprising.   The three roots
correspond to equilateral triangle (double back), regular pentagon &
regular pentagram respectively.

	I'm off for fish & chips

Regards,
Andrew.

	re .11
        
>>		re .10
>>
>>	>>	However, it's worth pointing out that r is not likely to be
>>	>>be particularly clean, for general polygons.   For instance, if n=7, and
>>	>>all the edge lengths are the same, then the polygon is known (Gauss) not to be
>>	>>constructible, and r can be a root of an irreducible sixtic equation.
>>
>>		The sixtic equation (is it really called sixtic?) for the
>>		side of a regular 7-gon is not irreducible.  x^6 + x^5 +
>>		x^4 + x^3 + x^2 + x + 1 factors into a product of two
>>		cubics, which can then be solved.  The roots as you say are
>>		not constructible though.

        Oops.  I retract what I said in .10.  I was quoting a
        result from memory and quoted the wrong result. :-(
        
        It's not that x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 is
        reducible over the rationals, it may or may not be,
        though I doubt it.  The result I wanted to quote is that
        its zeroes, the seventh roots of unity, are solveable in
        radicals.  To see this note that if w is a zero of the
        polynomial, and a = w + 1/w, then a is a zero of the
        cubic polynomial y^3 + y^2 - 2y + 1.  This is solveable
        in radicals and then w, 1/w = (a +/- sqrt(a^2 - 4)) / 2
        is solveable, too.
        
        Hmmmm.  I wonder what the least n is such that e^(2 pi i / n)
        is not expressiblein radicals, if any.  But that's
        another topic. :-)
        
	Dan

1.)	f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1  
is irreducible over the rationals.

	Let x = y+1.   f(x) is irreducible iff f(y+1) is irreducible.

	f(y+1) = y^6 + 6y^5 + 15y^4 + 20y^3 + 15y^2 + 6y + 1
		     +  y^5 +  5y^4 + 10y^3 + 10y^2 + 5y + 1
		            +   y^4 +  4y^3 +  6y^2 + 4y + 1
				    +   y^3 +  3y^2 + 3y + 1
				    	    +   y^2 + 2y + 1
						    +  y + 1
						         + 1

	f(y+1) = y^6 + 7y^5 + 21y^4 + 35y^3 + 35y^2 + 21y + 7 

	Now, by Eisenstein's criterion, with p=7, this is irreducible.
	(The requirements are:
		p must divide every coefficient *except* the leading coeff.
		p^2 must not divide the trailing coefficient.)


2.)	Yes, I have no problems with the solubility *route*, except
you made a slight slip in your cubic, might I humbly suggest.   But I
question also the * motivation* in remark 3 below.   The cubic should be:

	y^3 + y^2 - 2y - 1.
                       ^

	That this works, we can see by expanding the cubic:

	w^3 * [(w + 1/w)^3 + (w + 1/w)^2 - 2(w + 1/w) + 1]
	
=	w^6       + 3w^4 	+ 3w^2     + 1
	    + w^5        + 2w^3        + w
		  - 2w^4        - 2w^2
                         -  w^3 

= f(w) = 0.   Since w <> 0, a is a root of the cubic.


3.)	You look for the smallest n such that e^(2*pi*i/n) is not expressible
by radicals.   I don't understand your question.

	An extension L:K is radical if L = K(a_1,...,a_m), where for each
a_j, there is an integer n(j) such that (a_j)^n(j) lies in K(a_1,...,a_(j-1)).
So a radical is some such a_j.

	Now here, z = e^(2*pi*n) is either going to lie in Q anyway, or it
can be immediately derived by extending Q by directly adding the radical z
itself!

	Please tell me if I've misunderstood you.

Regards,
Andrew.

>>	1.)	f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1  
>>	is irreducible over the rationals.
>>
>>		Let x = y+1.   f(x) is irreducible iff f(y+1) is irreducible.

        Clever proof!
        
>>	The cubic should be:
>>
>>		y^3 + y^2 - 2y - 1.
>>	                       ^

        Well, the next to last step in my derivation was
        
        	y^3 + y^2 + y = 3y + 1
        
        so I suppose you're right. :-)

>>	3.)	You look for the smallest n such that e^(2*pi*i/n) is not expressible
>>	by radicals.   I don't understand your question.

        You mean just saying it is 1^(1/n) is enough to make it
        "official"?  That doesn't seem fair to me. :-)  It also
        uses an n-th root although the polynomial is of degree
        n-1.  I.e. I would expect a solveable zero of a degree
        six polynomial to require only square, cube, and fifth
        roots (most likely nested).
        
        You understood the question, I just didn't think that
        that was the answer.
        
        Dan

>        You mean just saying it is 1^(1/n) is enough to make it
>        "official"?  That doesn't seem fair to me. :-)  It also
>        uses an n-th root although the polynomial is of degree
>        n-1.  I.e. I would expect a solveable zero of a degree
>        six polynomial to require only square, cube, and fifth
>        roots (most likely nested).

	The point is this:

	w = 1^(1/n) is a radical

=>	Q(w):Q has an abelian Galois group, G.

	This is always true, even if x^n-1 is reducible.   This is because
every root of the minimum polynomial, what ever it is, will be a zero of x^n-1,
and so will be a power of w.   Thus each Q-automorphism of Q(w) is of the form
w -> w^k.   Two such Q-automorphisms obviously commute.

	Then from this,	a fortiori, G is *ALWAYS* solvable.   S5 doesn't get
a look in.  We're working entirely with abelian groups.

=>	we can construct a chain of normal subgroups from I to G, each quotient
being prime cyclic.

=>	there exists a corresponding chain of extensions from Q(w) -> Q,
the degree of each of which is prime, p_i

	Dan asks, I think: "When is *every* p_i < n?"

	Well, how big is G?   

	say: n = prod(j) (q_j)^(a_j) where q_j are prime.
then E(n) = prod(j) (q_j - 1)*(q_j)^(a_j - 1) is the number of primitive
nth roots of 1.   (This is just the number of integers in {1,...,n} which are
coprime to n.)

	So from this, it's obvious that p_i < n.

	Dan's case with a 7th root, yields E(n) = 6 = 2.3, so it is clear that
a cubic and then a quadratic extension will build w.   Or a quadratic followed
by a cubic, equivalently.

	Also, Gauss' constructibility argument falls out of this.   If you
accept that constructibility with ruler and compasses is equivalent to only
permitting quadratic extensions, then our requirement becomes that:

	E(n) is a power of 2.

ie:	q_j > 2 => a_j = 1, & q_j-1 a power of 2.

	q_j = 2^x + 1 cannot be prime unless x is itself a power of 2, so
Bob's one's uncle, one has the Fermat numbers popping out.

Regards,
Andrew.

	I overlooked a more general allegation that Dan made.

>            I.e. I would expect a solveable zero of a degree
>        six polynomial to require only square, cube, and fifth
>        roots (most likely nested).

	Yes, this is correct.   A poly of degree n will have a splitting
field of degree dividing n!.   If it is solvable, then a chain of extension
fields each of prime degree exists, and such a prime will divide n!, so
is less than n.

Regards,
Andrew.

	Thanks Andrew.  You answered my questions.

	Dan

I think the problem of calculating the circumradius of a (convex) pentagon 
is intractable, in that its solution involves solving a polynomial equation
of degree >=5. 

The following approach works for triangles:

$maple
# (File 3sides.maple)
# Given the triangle {a,b,c} find the radius of the circumscribed circle.
# ============================================================
# Let A=angle at the center of the circle of the isoceles triangle (r,r,a),
# and B,C, resp. be angles similarly defined. Then sin(A) = a/(2r), etc.
# Since the 3 angles add to 2Pi, sin(A+B+C) = 0. 
# (In MAPLE, '"' means the immediately previous result.)
expand(sin(A+B+C)=0);
# Convert cosines to sines:
subs(	cos(A)=(1-sin(A)^2)^(1/2),
	cos(B)=(1-sin(B)^2)^(1/2),
	cos(C)=(1-sin(C)^2)^(1/2), ");
# Convert the sines to functions of the sides and radius:
subs(	sin(A)=a/(2*r),
	sin(B)=b/(2*r),
	sin(C)=c/(2*r), ");
solve(",r);
# A pair of solutions, one negative, are produced by the last operation.
quit;

Extending the approach to 4 sides exhausts the capacity of the 8Meg 3100 I 
am using. Extending the approach to 5 sides causes an error message to be 
generated which gives a clue to the impossibility of machine symbolic 
analysis. So it appears that the only method of solution is numerical
approximation, which can be obtained by iterating values of r in the
equation produced by the substitution steps outlined above, or by other
methods. 

If anyone else comes up with a brilliant insight, I would be willing to 
try it.

Lynn Yarbrough 

	The solution for n=4 splits into two solutions for s.   One of
them is the one given by Peter in the base note, and the second is the same
with A -> -A.

	So, if you take one of these, and make the substitution
D -> D*sqrt(4*r^2-E^2) + E*sqrt(4*r^2-D^2) (or something like that) 
	then square the new equation twice after rearranging, to get rid
of the nasty surds, you should have something which after getting rid of
any r=0 cases, will factor() into some smaller polys, which will then
crack.

	Must dash,
Andrew.

	I tried the strategy suggested in -.1, and it gives me a 
polynomial of degree 15 in r^2, A^2, B^2, C^2, D^2, E^2.   (ie, it would
be of degree 30, but every term is of entirely even degree in each of
the six indeterminates.   degree 15 is actually pretty good.   The
polynomial is only 239 blocks long as text!

	A basic sanity check suggests that it is correct, but I could do
more (eg try E=0, or A+B+C+D).   I gave the beast to MAPLE to factorize,
and MAPLE came back after an hour or so, but the polynomial was unreduced.

	I still haven't got any MAPLE documentation.   Does my experience
mean that the poly is *sure* to be irreducible, or is MAPLE only limited
in its guesses.

	Alternatively, can we come up with a plausible argument for why
the thing should be degree 15?

Regards,
Andrew.