| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1160.1 | Rye, please | AKQJ10::YARBROUGH | I prefer Pi | Mon Dec 11 1989 13:01 | 5 |
| This is a two-dimensional special case of the Ham-and-cheese Sandwich
theorem: For any ham and cheese sandwich there exists a plane cut that
simultaneously halves the bread, the ham, and the cheese.
Lynn
|
| 1160.2 | I am no analyst... | HERON::BUCHANAN | Andrew @vbo DTN 828-5805 | Tue Dec 12 1989 17:56 | 35 |
| 1160.3 | Not Really a Proof; but- | WONDER::COYLE | Only 48.8% of my former self! | Tue Dec 12 1989 18:49 | 21 |
| If we look at the point that is the center of gravity for any triangle,
we also have a point through widh any line drawn on the plane of
the triangle will divide that triangle in half.
Now take any two triangles on the same plane. Determine the center
of gravity of both traiangles. The line that is defined by these
two points is the line that will divide both triangles in half.
In the special case of two overlapping triangles everything works
the same except when the centers of gravity are the same. In that
case any line drawn through that point will meet the criteria of
the division.
The center of gravity can be found by getting the average of all
the X,Y coordinates within the triangle. Since the triangle is
continuos within its boundaries the average will be within its
boundary and will have equal weight/area in opposite directions.
-Joe
|
| 1160.4 | not so simple, alas... | HERON::BUCHANAN | Andrew @vbo DTN 828-5805 | Tue Dec 12 1989 20:17 | 17 |
| 1160.5 | average vs average' | HERON::BUCHANAN | combinatorial bomb disposal squad | Thu Dec 14 1989 12:22 | 10 |
| The error in .3 is essentially a confusion between the concepts
of mean and median. Now the idea of mean generalizes naturally from
one dimension to two or many, becoming the centre of gravity, and something
free of the co-ordinate frame in which the data is described.
The median however does not generalize cleanly, even to two
dimensions, as we have seen.
Regards,
Andrew.
|
| 1160.6 | | AKQJ10::YARBROUGH | I prefer Pi | Thu Dec 14 1989 16:38 | 14 |
| I find it easier to think of the problem in terms of *moments*. The reason
the C.G. does not correspond to half the area is that some parts of the
area are at greater distances from the C.G. than others, therefore
contribute more leverage.
One way of approaching the problem is to consider the set of all bisectors
of the [possibly disjoint] area consisting of the two triangles, as a
function of their angle with the horizontal. Now let the angle vary
continuously, so that the bisector sweeps through the triangles. Show that
the fraction of area contributed by each triangle varies *continously*, and
you have it: when one is split 50-50, so will be the other. The trick is in
proving that there are no discontinuities.
Lynn
|