[net.math] pizza

jce (06/21/82)

I went to get a pizza the other night, and, since I had
six friends over, I asked that he cut it in seven pieces.
The pizzaman said he couldn't do it exactly with just
straight-edge and compass (it was a greek pizzeria), and
I asked him why.  He started talking about automorphisms
and algebraic fields and other things I don't understand -
can anybody explain this in plain english?  Is there a general
rule?

djj (06/21/82)

Regarding the seven slices of pizza --

Could it be related to the fact that the prime factorization of
360 is 2 x 2 x 2 x 5 x 9 and the fact that 7 is prime and
not among the factors?

Dave Johnson
BTL - Piscataway
(...!pyuxcc!djj)

p.s. I realize that's not a very geometric approach, but it makes
     sense to me!

thomas (06/22/82)

There is a general rule for ruler and compass constructions, which can
be stated as follows:
	When the solution x of the problem is real and can be found
	by rational operations and (not necessarily real) square roots
	from the given numbers (line segment lengths) a, b, ..., the
	number (segment length) x can be constructed using ruler and
	compass.

In the case of dividing a circle into n equal parts, the condition is that
the number of divisors of n (phi(n)) is a power of 2.  Now, n can be written
in the form (2^m)(p1^m1)(p2^m2)...(pl^ml), where the p's are prime.  Then
phi(n) = (2^(m-1))(pq^(m1-1))...(pl^(ml-1))(p1-1)...(pl-1).  Thus, m1 thru
ml must all be 1, and (pi-1) must be a power of 2.  Now, it turns out that
all primes which are one greater than a power of 2 are of the form
2^(2^r)+1 for some integer r.  (Note, that not all numbers of this form
are prime, however, for example, 2^2^5 is divisible by 641.)  So, a
circle may be divided into n pieces if n is a product of some power of 2
and primes of this sort (to a most the first power).  The first few
primes in this sequence are 3, 5, 17, 257, 65537.  So, if you and 16
other friends were sharing a pizza, you could divide it into equal
pieces using only a straightedge and a compass.  The result is due to
Gauss, my presentation is a paraphrase of the presentation in Modern
Algebra, Vol 1, van der Waerden, pp183-187.

=Spencer

G:shallit (06/22/82)

Why you would want to explain it in plain English is beyond me.
Mathematical notation is so much more precise and clear for this
kind of thing.

So, after starting thus, here goes an attempt at an explanation.  It
has absolutely nothing to do with the prime factorization of 360, as
was suggested.  (How could it?  360 is an artificial number, decided
by man, not nature.  If we decided on 420 degrees in a circle, would
this make it possible to divide a circle into 7 parts?  Of course not.
But I digress.)

Gauss showed that the circle can be divided into n parts if and
only if n is 1) a Fermat prime (i. e. of the form 1+2*(2*k) ) or
(2) a product of such primes or (3) the product of a power of
2 and a set of Fermat primes.

His reasoning was approximately as follows:  A quantity can be
constructed by straightedge and compass if it can be expressed
by a finite number of applications of + - * / sqrt.

There are polynomials all of whose roots can be expressed in this
way; among such, there is one of minimal degree with a given
quantity as a root.  The degree of these polynomials is a power of 2.

The roots of x^n - 1 = 0 lie on a circle , spaced equally around
the circumference.  In order to divide the circle we must be able
to construct these points.  We can assume n is a prime or a power
of a prime, since if we can construct the roots of x^p - 1 = 0 and
x^q - 1 = 0 we can construct the roots of x^pq - 1 = 0.

We can certainly bisect an arc with straightedge and compass.  Hence
we can divide a circle into 2, 4, 8 , ... parts.  If we could
divide a circle into k parts (k odd) we could divide it into
(2^j)*k parts.

Look at x^p - 1 = 0, where p is prime.  Divide this by x-1 to
get an irreducible (unfactorable) equation of degree p-1.  Since
it is irreducible, it is of minimal degree and so p-1 must
be a power of 2.  The only numbers p-1 that are powers of 2 are
such that p = 1+2^2^n, a Fermat prime.  The only ones known
are 3,5, 17, 257, and 65537.

Hope this helps a little.

bj (06/24/82)

People have done wonderfull jobs showing that you can not divide
a pizza into 7 equal wedges, but that does not solve the original
problem - can you divide a pizza into seven pieces of equal size
and any shape (some people don't like crust anyway)?

The answer to that question is yes. Construct a line whose length
is  sqrt ( 1 / 7r )  where r is the radius of the pizza. Take a
circle with this radius from the center of the pizza and give it to
somebody. Then divide the ring into six pieces for everyone else.

jce (06/24/82)

yale-com!bj's solution to the pizza problem is brilliant, because
it wasn't even what I had in mind when I posted the original query.
Unfortunately, it was a little off - the correct radius for the
inner circle is r / sqrt(7).

wesw (06/25/82)

  What about a rectangular pizza?

grunwald (06/27/82)

#R:floyd:-29500:uiucdcs:10300002:000:427
uiucdcs!grunwald    Jun 26 23:18:00 1982

  why is it not possible to construct a septagon within the pizza, with the
longest radius of the septagon equal to the radius of the pizza?  Constructing
the septagon involves calculating the length of the outer edge and then setting
the compass to that length. Then, start anywhere on the edge of the Zah and
go around using the compass to measure off the distance on the perimeter.

Is there some non obvious catch to this?