| 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 |
(Stan has been writing a monograph on properties of regular polygons, and recently claims to have come up with a proof of the following interesting result:) From: DECPA::"72717.3515@CompuServe.COM" "Stanley Rabinowitz" 16-MAR-1994 18:15:25.75 Subj: Duality After months of study, I have finally found the elusive duality between the Pizza Theorem and my regular polygon theorems. It is the following generalization of the Pizza Theorem: Let Q be any point inside the circle (center O) and create n rays emanating out of Q at equal angles with each other, where n>4 is a multiple of 4. [Do not cut along these rays.] These rays produce n points on the circumference of the circle. Let P be any point inside the circle and make n cuts from P to those n points. This divides the pizza into n slices. Color the slices alternately white and black. Then the black area equals the white area. If P=Q, then we get the Pizza Theorem. If Q=O, then we get a theorem about regular polygons. For a general P, we get a new theorem! - stan -
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1859.1 | RUSURE::EDP | Always mount a scratch monkey. | Fri Mar 18 1994 12:50 | 27 | |
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
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| 1859.2 | Stan outlawed that case. | WIBBIN::NOYCE | DEC 21064-200DX5 : 130 SPECint @ $36K | Fri Mar 18 1994 13:26 | 8 |
.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... | |||||
| 1859.3 | DIR/TITL=PIZZA isn't too helpful | VMSDEV::HALLYB | Fish have no concept of fire | Fri Mar 18 1994 16:58 | 4 |
What is the pizza theorem? I can guess as well as anybody, but would
like a formal statement just to be sure.
John
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| 1859.4 | AUSSIE::GARSON | Hotel Garson: No Vacancies | Sat Mar 19 1994 21:44 | 105 | |