Conference rusure::math

    I doubt if square vs circle makes any difference -- I'm pretty sure I
    could map any solution for one into a solution for another.  I would
    guess that this *has* been solved (perhaps with a proof that there is
    no such disection) or that would have been the open problem in the
    article.

				    Topher

What annoys me about this sort of problem is that it is either very simple
or very difficult, depending on what you call a dissection: i.e. are we
talking about using a pair of scissors, or cantorian set theory? The
simplistic approach never sees that a problem exists: one slice with the
scissors *removes* {x=.5} from the figure, Gordian-knot style. And for most
(if not all) practical applications this suffices. 

If we take the sharp scissors approach, the problem of squaring the circle
by dissection can be resolved by a simple argument. Note that any path cut
by scissors in a plane figure creates two boundaries of equal length on
either side of the cut, and if the curvature on one side is positive, that
on the other is its negative, and that the total curvature of the figure
(i.e. the path integral of arc length times curvature) is *invariant*. But
the unit circle starts out with a total curvature of 2*Pi, and the square
ends up with zero, so the dissection (into a finite number of pieces) is
impossible. "You can't get there from here."

I am not at all convinced that the set theorists have an answer to the 
question, *what happens to the curvature of the boundary?*. Allegedly they 
do, but not, to my mind, with a finite number of pieces.

RE: .2 (Lynn)

    Lynn, it rather depends on the definition of "piece".  One way of
    looking at it is that an infinite -- even uncountably infinite --
    number of pieces are allowed, but they are constrained to belong to
    a finite number of groups (potentially disconnected pieces) which are
    contrained to move rigidly with respect to each other.

    As for your curvature argument, how do you define the curvature of a
    sharp corner?  Isn't it, by the conventional definition of curvature,
    undefined (lazy-eight infinite)?

				Topher

>As for your curvature argument, how do you define the curvature of a
>sharp corner?  Isn't it, by the conventional definition of curvature,
>undefined (lazy-eight infinite)?

Right; that's why the (arc length)*curvature is critical. For a corner, the 
arc length is zero, so the function can be integrated piecewise along the 
path and the (countable) discontinuities can be ignored. Each cut adds
another 0-2 corners to the figure. 

    OK.  As long as "sharp scissors disection" excludes even connected
    fractal boundaries, your point is well taken.  Can this be extended
    with a similar simple argument to exclude finite circle-to/from-square
    disections with:

	1) Fractal but connected pieces?
	2) Non-fractal but disconnected pieces?

				Topher

    While I don't have an answer to such a problem, it probably rests
    on the "arithmetic" view of topology and analysis that's explained in
    books like Rudin.  Consider the result that the Hilbert curve passes thru
    "every" point of a square.  Any finite step in the construction only
    touches a finite number of points, yet in the limit it covers the
    square as a continuum (not just the rational points if I recall.)

    - Jim

    It seems that it is impossible to bisect a circle in the general sense.
    I believe that this also means that you cannot bisect a square, but
    I have not yet demonstrated the necessary equivalence.

    Here is how I would state the general circle bisection problem:

        Take the set of points representing a circle and its interior, C.
        Find a set of points S1 which is a "bisection set" (to be defined). 
        Let S2 = C - S1 (i.e., S2 is all points in the "circle" which are
        not in the set S1).  For S1 to qualify as a bisection set, there
        must be a rigid transformation T12 which maps the set S1 onto the
        set S2 (note that rigid transformations are, by definition, 1-1).

    On the plane, a rigid tranformation consists of a combination of a
    rotation around some point and a translation.  We don't really
    have to get any more formal than that for this purpose.

    Because of the symmetries provided by the circle, we can restrict our
    attention for T12, to rigid transformations which can be factored into
    a rotation around the center of C (0 to 2pi radians), followed by
    a translation in the positive x direction (0 to +oo).  (T21, the
    inverse transformation, is then restricted to transformations which
    can be factored into a translation in the negative x direction,
    followed by a rotation around the original center of C).

    We can immediately reject any transformation which includes a
    displacement of 0, i.e., a pure rotation.  The reason is that such a
    pure rotation maps the point at the center of the circle to itself.
    Thus if that center point is in S1, it will be mapped to a point in S1,
    not S2.  And if it is a point in S2, it will not be mapped to by any
    point in S1.  No transformation whose translation is 0 can be used.

    Now let's look at the effect of T12 on the entire circle (I'll use
    "circle" from now on to refer to the circle and its interior), without
    regard as to whether any point is a point in S1 or in S2.  At this
    point we need a figure.  Rather than an inadequate attempt at terminal
    art, I'll describe the figure (which is simple) and you draw it.

    First draw a circle, somewhat left of center of the area you are
    drawing in.  This represents C, untransformed.  Now draw a second
    circle of (roughly) the same size, with its center to the right of the
    first circle and overlapping it.  The result is like a two set Venn
    diagram (in a very real sense, it *is* a Venn diagram but the
    correspondences are a bit more literal than is usual).  This second
    circle represents the image of C as a result of the transformation T12,
    i.e., T12(C).  To help remember that there is a rotation involved
    (which is disguised by the rotational symmetries of the circle) draw a
    dot at 6 o'clock on the perimeter of the first (leftmost) circle and
    one somewhere else, say 4 o'clock, on the second.  The 4 o'clock dot is
    the image of the 6 o'clock dot.

    The figure has three (interior) regions.  The leftmost region consists
    of those points only in the first circle, label that "A".  The
    rightmost region consists of those points which are only in the second
    circel, label that "B". The middle region consists of those points
    which are in both circles, label that region AB.

    Consider the region A (which does not include the boundary which
    separate it from AB -- i.e., it is partially open).  A point in that
    region which was in set S2 would not be mapped into by any point in
    S1, since T12 "moves" all the points in the circle away from region
    A.  Therefore, all the points in region A must be in S1.

    Similarly consider a point which is mapped into region B by T12.  If
    that point is in S1, it will not be mapped into a point in S2, since
    it is mapped off of the circle entirely.  Therefore, all the points
    of the region of the circle which is mapped onto region B, must be
    in S2.

    Look now at the outer perimeter of region A.  If the translation
    part of the transformation is greater than 0, then the length of
    that segment of the outer perimeter will be greater than pi, i.e.,
    greater than half the circumfrence of the circle.  Similarly, since
    rigid tranformations are length preserving, the length of the segment
    of the circles perimeter which maps onto the outer perimeter of region
    B will also be greater than half the circumfrence of the circle.

    Thus, over half the circles circumference must be in S1, and over half
    the circles circumfrence must be in S2, i.e., not in S1.  But this
    is contradictory.  Therefore no such transformation can exist.  QED.

    The argument can be extended to the case where we include mirror
    reflections among the permissible components of a rigid transformation. 
    The latter half of the argument, where I showed that a transformation
    with a non-zero translation cannot be used is unchanged by this. A
    transformation which consists only of a reflection maps a diameter of
    the circle onto itself, so the argument against pure rotation can be
    applied also to pure reflection.  A reflection plus a rotation, still
    maps the center point onto itself, so the argument against pure
    rotation can also be applied to rotation composed with reflection.
    That covers all the bases, QED once again.

					    Topher