Pizza Theorem:
If a circular pizza is divided into 8, 12, 16, ... slices by making cuts
at equal angles from an arbitrary point, then the sums of the areas of
alternate slices are equal.
Pizza Lemma: [Upton 1967], [Barr 1969, problem 9], [Nelson 2004]
If two chords in a circle intersect at right angles, then the sum of the
squares of the lengths of the four segments formed is a constant
(the square of the length of the diameter).
4r^2 = a^2 + b^2 + c^2 + d^2
References:
Stephen Barr;
2nd Miscellany of Puzzles,
New York, 1969
(Dover Publ. Mathematical Brain Benders, 2nd Miscellany of Puzzles, 1982)
- problem 9: The Pot on the Crosspieces.
George Berzsenyi;
The Pizza Theorem,Part I (equality of off-center slices),
Quantum, Jan/Feb94, p29 (Math Investigations)
George Berzsenyi;
The Pizza Theorem,Part II (including the Calzone Theorem),
Quantum, Mar/Apr94, p29 (Math Investigations)
Larry Carter, John Duncan, Stan Wagon;
Problem 1457: The Center of a Sliced Pizza,
Mathematics Magazine. 67:4 (1994) problem : Carter, Duncan, Wagon
Mathematics Magazine. 68:4 (1995) 312-315 solution: Deiermann, Mabry
- a: For a point P inside a circle draw three chords throuh P making six
60 degree angles at P and form two regions by coloring the six
"pizza slices" alternating black and white. Prove that the region
containing the center has the larger area.
Larry Carter, Stan Wagon;
Proof without Words: Fair Allocation of a Pizza,
Mathematics Magazine. 67 (1994) 267
Hirschhorn, M. D. Hirschhorn, J. K. Hirschhorn, A. D. Hirschhorn, P. M. Hirschhorn;
The pizza theorem,
Austral. Math. Soc. Gaz., 26 (1999), 120-121.
http://www.maths.unsw.edu.au/~mikeh/
Rob Johnson;
http://www.whim.org/nebula/math/images/perpchord.gif
I came up with a pretty short proof based on the fact that when you have
two intersecting chords, opposing arcs on the circle sum to twice the
angle between the chords. Since the chords are perpindicular, we have
that the opposing arcs must be supplementary. Sliding these arcs and
chords together, we see that the chords form a right triangle with the
diameter of the circle as the hypotenuse.
Roger B. Nelsen;
Four Squares with Constant Area
Mathematics Magazine. 77:2 (April 2004) 135
A proof without words of the following statement: If two chords in a circle
intersect at right angles, then the sum of the squares of the lengths of the
four segments formed is a constant (the square of the length of the diameter).
C.E.M. Pearce;
More on the Pizza Theorem,
Austral. Math. Soc. Gazette 27: 4-5 (2000).
Stanley Rabinowitz;
Problem 1325,
Crux Mathematicorum 14 (1988) 77 problem : Stanley Rabinowitz
Crux Mathematicorum 15 (1989) 120-122 solution1: J"org H"arterich
Crux Mathematicorum 15 (1989) 120-122 solution2: Shiko Iwata
L. J. Upton;
Divisors of a Circle, Problem 660,
Mathematics Magazine. 40 (May 1967) problem : L. J. Upton
Mathematics Magazine. 41 (Jan-Feb 1968) 46 solution: Michael Goldberg
chephip:
La pizza Partage dŽune pizza,
http://chephip.free.fr/ie/pb116.html
http://chephip.free.fr/ie/sol116.html
--
http://www.mathematik.uni-bielefeld.de/~sillke/
mailto:Torsten.Sillke@uni-bielefeld.de