Conference rusure::math

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

>         <<< Note 1160.0 by HPSTEK::XIA "In my beginning is my end." >>>
>                   -< A simple(?) problem in plane geometry >-
>
>    Given two triangles on the plane, show that there exists at least one
>    line that divides each triangle into two parts of equal areas.

	Say that we are dealing with two non-overlapping triangles.   A
sketch of the proof follows.

	Consider one of the triangles, T2, :-).   For any angle, �,
there will be exactly one straight line of gradient t=tan� which
bisects T2.   If we give the line a direction, we can talk of the left
hand side, and the right hand side into which the plane and T2 have been
divided.

	So if we have such a line y*cos� = x*sin� + c, then we can regard
c as a function of �.   The trick now is to show that c is a continuous
function of �.   Note also that c(�) has period 2*pi with �.

	Assume for the moment that we have done this trick.   Now look at
how the line cuts the other triangle, T91.   Say a = a(�,c) is the area in the
lhs of the plane.   Then there exists �' such that a(�',c(�')) = 0, since
T91 is convex.   Now consider �" = �'+pi.   a(�",c(�")) = A, where A is
the area of T91.   Since c is a continuous function of �, and a is a
continuous function of � & c, a is a continuous function of �.   The
intermediate value theorem tells us that we can find ߰, with 
a(߰,c(߰))=A/2, as required.

	Now, about that trick...   The entire theorem is presumably true
for any two non-intersecting convex shapes, but for the triangles, it would
simply be adequate to find a formula for c in terms of the coordinates
of the vertices of T2, and �.   The with which I cannot be bothered (sic).

Apologies for syntactic japes,
Andrew.

    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
    
    

>    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.

	I don't think that this is true.   For instance, suppose I take
an equilateral triangle, centred on the origin, with a vertex at (0,1).
Then the other vertices fall on (�sqrt(3)/2, -1/2).   Now, the horizontal
line which bisects the triangle will be y = 1 - 3sqrt(2)/4.   This is
less than zero.

	By symmetry, then, the three area bisectors orthogonal to the three
altitudes will not meet at a single point.   (If they did, then this
would *have* to be the centre of symmetry, but we have seen that this
point is spurned by at least one area bisector.)

Regards,
Andrew.

	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.

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