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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

17.0. "Pizza time!" by RANI::LEICHTERJ () Sun Jan 29 1984 23:36

  29-JAN-1984 20:35           RANI::LEICHTERJ   


A neat problem that wandered around Usenet about a year ago; if no one
proposes a good solution, in a couple of days I'll give the most elegant
one I know, due to BJ Herbison (now, though not at the time, at DEC - on
ULTRA, I think):

It's easy to divide a (circular) pizza into 6 equal pieces using only a
compass and straightedge (since 360/6=60 degrees is a constructible angle).
How would you divide the same pizza into 7 equal pieces?  Note that, as in
classical geometry, what is sought is an EXACT solution, not a series of
approximations (which would be easy).
							-- Jerry
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17.1HARE::STANMon Jan 30 1984 15:1815
Well, you can't do it by dividing the circumference into 7 equal parts
and cutting to the center because that would mean you could construct
a regular heptagon.  Galois proved that the only regular n-gons that
can be constructed with straight-edge and compasses are of the form

			     k
			n = 2  X F

where F is a Fermat prime. (A Fermat prime is a prime of the form

			  m
			 2
			2   + 1 ,

e.g. 3, 5, 17, 257, etc.)
17.2HARE::STANMon Jan 30 1984 19:1822
I assume that when you want the pizza divided up into 7 equal parts that
you mean 7 parts that have the same area (not 7 congruent parts).

Anyhow, here's a solution:

Let us assume that the radius of the pizza is R.  We will cut a circular
piece of radius r out of the center that has an area equal to 1/7 of the
area of the pizza.  The remaining annulus (with area 6/7 of the pizza)
can then easily be divided into 6 equal pieces by inscribing a regular
hexagon in the circle and joining their vertices to the center of the
circle.

To find r, we note that pi r^2 = 1/7 pi R^2, so that R/r=sqrt(7).
But sqrt(7) is quadratic and can therefore be constructed with straight-
edge and compasses.  So we are done.

(An easy way to construct sqrt(7) is to draw a circle with diameter 8.
 Let P be a point on a diameter AB that divides the diameter into two pieces
 with lengths 1 and 7.  Erect a perpendicular at P until it hits the circle
 again at Q.  Then PQ=sqrt(7) because ABQ is a right triangle with altitude
 PQ and an altitude is the mean proportion between the segments it forms
 on the hypotenuse.)
17.3ULTRA::HERBISONMon Jan 30 1984 22:5615
Right.

The initial source of this problem was someone who wanted a clearer proof
that you could not 7-sect an angle with straight edge and compass (apparently
he had called a pizza place and asked for a pizza cut in 7 pieces and was
given a proof of why it was impossible over the phone!).  I saw a loop
hole in the way the question was asked and came up with the same solution
you did.

A refignment (suggested by the person at Bell Labs who submitted the initial
problem) is to divide the pizza in to 7 rings rather than one circle and
six truncated wedges.  If you do this than the solution can be generalized
to divide a pizza into any number of pieces.  However, this only works if
you party consists of exactly one person who likes eating the crust.
						B.J.
17.4HARE::STANSat Feb 04 1984 15:1714
A correction on 17.1:

A regular n-gon is constructible with straight-edge and compasses
if and only if n is of the form

		 k
		2  F  F  ... F
		    1  2      r

where each F  (if any) denotes a Fermat prime.
            i

My previous statement was incorrect.  For example, a regular
15-gon is constructible since 15=3 X 5, the product of two Fermat primes.
17.5Gauss had a solution?CACHE::MARSHALLThu Jun 26 1986 12:365
    Hi,
    	Didn't Gauss develop a constuction that would divide the
    circumference of a circle into 7 equal segments (or was it 17)?
    
    steve M	(beware the fractal dragon)