| Am I misinterpreting this? Let Q be a point on the circle. (Yes, it's
a degenerate case, but the theorem should be true for points
arbitrarily close to this.) Let P equal Q and n=4. Of four rays
extending from Q at right angles, two project outside the circle
immediately, giving an area of 0 in that quadrant. The other two form
chords of length 2p and 2q. p^2+q^2=r^2, where r is the radius of the
circle.
The area of the first segment is r^2*arcsin(p/r)-p*sqrt(r^2-p^2), which
can be found by integrating or by finding the portion of the circle
formed by the angle formed by rays from the center of the circle to the
chord's intersections with the circle and then subtracting the
triangular part. The area of the second segment is
r^2*arcsin(sqrt(r^2-p^2)/r)-p*sqrt(r^2-p^2). Numerically evaluating
the sum of these two areas for variables values shows they do not add
up to half the area of the circle.
What's wrong?
-- edp
Public key fingerprint: 8e ad 63 61 ba 0c 26 86 32 0a 7d 28 db e7 6f 75
To get PGP, FTP /pub/unix/security/crypt/pgp23A.zip from ftp.funet.fi.
For FTP access, mail "help" message to DECWRL::FTPmail or open Upsar::Gateways.
|
| .0 says n>4. The counterexample for n=4 doesn't require sqrt's and arcsins:
The two rays that extend into the circle intersect it at the ends of a
diameter. The area enclosed between the rays includes the half circle on
the far side of that diameter, together with a nonzero amount of triangle,
so clearly more than half the area.
I still find the theorem hard to believe...
|
| re .3
> What is the pizza theorem? I can guess as well as anybody, but would
> like a formal statement just to be sure.
.0 contains an implied formal statement of the Pizza theorem (which I
must admit I hadn't heard of but that's not saying much).
Reverse engineering from and elaborating on .0 ...
Let Q be any point inside a circle (center O) and create n rays emanating from
Q at equal angles (2pi/n). Let the rays meet the circle in the points V[1],...,
V[n].
Let P be any point inside the circle and join P to each of V[1],...,V[n]. This
divides the circle into n regions. Colour the regions alternately white and
black.
Hypothesis: The black area equals the white area i.e. this is a fair division
of the pizza (except that one person got all the burnt bits).
Considering the five different cases for three points (O, Q and P) which may or
may not be pairwise disjoint we have...
I Q = O = P
This corresponds to a robot/laser cutter and is obviously fair for all even n
(including n=2) and obviously unfair for all odd n.
II Q = O, P <> Q
Call this the Small Pizza Theorem. Proof below.
III Q <> O, P = Q
Call this the Regular Pizza Theorem. Superficial comments below.
As claimed in .1 this is obviously false for n = 4.
IV Q <> O, P = O
V Q <> O, P <> O, Q <> P
Stan doesn't distinguish between these last two.
Combine them and call this the Family Pizza Theorem. Superficial comments below.
What about more than two colours i.e. dividing for more than two people? What
about the cases for n that are not covered i.e. do the theorems definitely not
apply for other n?
Small Pizza Theorem
-------------------
Note that V[1],...,V[n] form the vertices of a regular polygon. Define a crust
as a region bounded by the chord between two adjacent vertices and the arc
between two adjacent vertices.
Clearly each crust is congruent and for even n each person gets the same number
of crusts so we can ignore the crusts and concentrate on the division of the
polygon into (non-congruent) triangles i.e. this is a theorem about regular
polygons as claimed in .0. Note that each triangle has one side of the regular
polygon as base so the length of this base is the same for all triangles.
It is easy to show that the theorem is true for n such that 4 | n (including
n = 4) by observing that opposite pairs of sides of the polygon are parallel
(true for all even n) and that the triangles having those sides as base are of
the same colour (true where 4 | n).
The area of the pair of triangles = � � base � (h1 + h2) where h1 is
the height of one triangle and h2 the height of the other. Now whereas it is
difficult to evaluate h1 and h2 separately, h1+h2 is a constant because the
altitude for one triangle is also an altitude for the other (bases parallel)
and thus h1+h2 is just the distance between the parallel bases. Thus the black
triangle area is just n/4 such constant areas and the white the same. This
establishes the result.
The above argument needs to be modified slightly if the point P lies within one
of the crusts. In that case for one pair of triangles the area is
� � base � (h1 - h2) but h1 - h2 is still the constant distance between opposite
parallel sides.
{In fact for a unit radius pizza h1+h2 = 2 cos (pi/n) and base = 2 sin (pi/n)
hence the combined area of a pair of opposite triangles is 2 cos (pi/n) sin
(pi/n) = sin (2pi/n). Thus the black triangle area is n/4 � sin (2pi/n) which
as a sanity check tends to pi/2 as n -> oo.}
One final observation on this theorem. It obviously does not apply for n = 2.
Regular Pizza Theorem
---------------------
The vertices do not form a regular polygon. The crusts are not congruent. It is
not obvious to me that the theorem is true. (Of the three theorems this is the
only one that accords with my impression of how a pizza is actually cut i.e.
with chords. When P <> Q the points V[i], Q, V[j] are not in general collinear
but that needn't constrain us.)
Family Pizza Theorem
--------------------
Q <> O, P = O is an easy special case to visualise because the slices are just
sectors of the circle (hence area proportional to angle of sector or length of
circumference). Still not obvious to me though...
|