Conference rusure::math

I think this is a well-known result of Guggenheimer.  I'll see
if I can go dig up the reference.  I remember, though, that
additional hypothesis is necessary.  The curve must have a
continuously varying curvature.

A more simple problem:

Given a closed, bounded, 2-d shape, whose area is exactly 1, prove you can
always "locate" it on the 2-d plane, so it does not "hide" any integer grid
point (m,n) behind it.

Hint: Consider a rectangle with integer length & width containing the shape.
      Then, look at all the 1*1 squares inside...

...imagine the rectangle made from transparent glass and the shape inside from
blue glass. If you break apart all 1x1 squares and put them one on top of the
other, you look through and see some transparent, some blue. NOT ALL 1x1 IS
BLUE, since the original shape had area < 1, so take a needle, and pass it
through the glass (!!!) on a transparent point. Now re-arrange the 1x1 squares
to their original locations. The needle holes will form an integer grid, not
touching the original shape surface...

By now I am completely lost in this discussion. Nitsan is babbling; in .2
the area discussed is exactly 1, while by .4 it has shrunk to <1. BACK TO
THE ORIGINAL PROBLEM, GUYS! I asked about the dimensions of the square
imbedded in a regular pentagon. That shouldn't be too difficult. - Lynn

re .4	Now, now, Lynn.  The problem posed in .2 is interesting, and I'm
	glad to have seen the clever proof in .4.

	But back to the original question...

I think the side of a square inscribed in a regular pentagon with unit sides is:

	   cos(18) ( 1 + 2 sin(18) )
	--------------------------------, where the angles are in degrees.
	     3
	4 sin (18) - 2 sin(18) - cos(18)

> Does there always exist, for any simple closed curve in the plane,
> a square whose corners lie on the curve?

For a smooth, continuous, closed curve, the answer is yes.  The proof is
simple and clever, but a bit difficult to explain.  It's effectively the
same as the proof used for the following problem.

Is it always possible to place a square table on a bumpy or wavy floor so
that all four legs of the table touch the floor?

For all sufficiently smooth simple closed curves in the plane,
the existence of an inscribed square was proved by S'nirelman
and Jerrard.  For the case in which the curve is the boundary
of a convex region, elementary proofs were given by Emch, Zindler,
and Christensen.  The general case (an arbitrary simple closed curve
in the plane), has not been solved yet.

			References
			----------
Victor Klee, Some Unsolved Problems in Plane Geometry. Mathematics
	Magazine. 52(1979)131-145.
L. S'nirelman, On certain geometrical properties of closed curves
	(Russian). Uspehi Mat. Nauk., 10(1944)34-44.
R. Jerrard, Inscribed squares in plane curves. Transactions of the
	American Mathematical Society. 98(1961)234-241.
A. Emch, Some properties of closed convex curves in the plane.
	Amer. J. Math. 35(1913)19-28.
K. Zindler, Uber konvexe Gebilde, Monatsh. Math., 31(1921)25-57.
C. Christensen, A square inscribed in a convex figure (Danish).
	Mat. Tidsskr, B 1950(1950)22-26.

Voila?  You call that a proof? Define "rotate".  It is not at all obvious
that after rotating 90% a different three legs are touching the floor - why
can't it be the same three?

/Bevin

No, because the final situation could be

	D    A

	c    B

It is not clear that, in the process of revolving the chair, c gets any
closer to the floor, hence you may not be able to get it to touch down where
D was to start with.

/Bevin

The argument is correct IFF all four legs are coplanar. Then it is not possible
for one leg to remain in the air. - Lynn Yarbrough

Except that during the rotation the four legs do not stay in the plane they
started in.

/Bevin

If three of the legs do, then so does the fourth. That's the point of the
rotation argument. (If it makes you feel better, leave the table fixed and
rotate the floor.)

Let me see if I can help here.  Consider the ends of the four legs of the
table as four vertices of a square.  I assume the "floor" has enough room
to turn the square and that it can be represented by a continuous function
of x and y.  This means it is smooth and does not have funny places where it
goes under itself. 

Label the vertices of the square:

	A B

	C D.

Put vertex A on the floor at any point.  Rotate the square about a line through
A parallel to line BC.  This will bring vertex D to the floor.  Since the axis
of rotation passed through A, A has not moved; it is still on the floor. 
Rotate the square about line AD, to bring vertex B to the floor.  A and D are
still on the floor.  We now have three vertices on the floor.  It should be
noted that the point at which A first touched the floor is arbitrary, so it is
always possible to have three vertices on the floor.  In particular, I selected
the three points to be on the floor. 

Draw a "line" on the floor from A to B.  Start moving vertex A along this line.
Allow the square to rotate so that B and D remain on the floor.  Where on the
floor they are does not matter now.  If A reaches the place where B was without
C having touched the floor at some point, move vertex B to the location where
D was.  Consider the old and current orientations of the square, respectively:

	A B	C' A'
	    and
	C D	D' B'.

B and D were at the same points as A' and B' are now, respectively.  If
all four points touched the floor at some point, we are done; stop.
Originally, A was on the floor and C was off.  Now, C' must be off the floor
(Otherwise it would have touched the floor at some point, so all four points
would have been on the floor).  Consider the square held vertically, with
vertices B and D on the floor as they are above, but A and C in the air.
Rotate the square through line BD until either A or C touches the floor.  It
should be clear that A does not touch the floor last; otherwise it would not be
possible to get the square in the above orientation with A, B, and D on the
floor.  Now consider the new orientation.  Again, hold the square vertically
with A' and B' has shown above, but C' and D' in the air.  Rotate the square
through line A'B' until C' or D' touches the floor.  Point D' must not touch the
floor last, otherwise it would not be possible to get the square in the
above orientation with A', B', and D' on the floor.  But by symmetry, C' must
not touch the floor last, because A'B'D'C' is the same as BDCA.  Therefore,
C' and D' touch the floor at the same time.  All four points are on the floor.

This is not formal, but I hope I have broken things up into units fundamental
enough to convince everyone.

				-- edp

Re .16

Of course the four legs are coplanar, but the plane they start in is NOT
necessarily the plane they finish in.  Hence my statement that they do not
stay in the same plane - I meant the plane they start in.


Re .17

Much better, but still I find difficulty.   The place I dislike is the "by
symmetry".

You see

	A  B         C'  A'
              became 
	C  D         D'  B'

But you only moved A -> A'=B, and B -> B'=C.   You did not guarantee that
either D' = C, or that C' = A,  hence the two positions are not the same,
hence the symmetry argument breaks done.


Let me see if I can summarize what is taking place here - after placing the
square with three points in the surface (a trivial operation as .17 points
out) the remaining point is either above or below the surface (or in it,
in which case we are done).

Given two such configurations (even if only by naming the points differently)
it is possible to drag the square from the first to the second keeping two
points in the plane.  It is not obvious that it is possible to do this keeping
the third corner also in the surface such that it ends up on the remaining
point.

Assuming it is possible to do this to the third corner, the fourth corner
either (a) stays on the same side of the surface, or (b) passes thru the
surface on its way to the other side.  In case (b), we have a configuration
with four legs on the floor as it passes thru the surface.  In case (a) we
have gotten no where.

Note that the following cross section shows a case where rotating the square
around two of the corners has the third corner pass thru the surface THRICE
- hence the two corners do not uniquely - or even finitely, determine the
third...
                               .path of corner as it is rotated
                             .
                            .
		surface----.+
                           .|
			    . very steep but slight slope
			    |.
			    |  .
                            -----.---
                                                                        
/Bevin

PS: Sorry to be so "dense" - but this seems to be an area where it is easy
to misunderstand.

Re .18:

Okay, let's try again.  First, I will change my labels a little, to make things
clearer.  A, B, C, and D are vertices of the square.  a, b, c, and d are the
points on the surface where the square was first placed.  (Note that c is not
really on the surface, so I will not use it as a point, just a general region.)
So the initial configuration has A at a, B at b, and so on.  Now A is moved to
b.  Clearly, it is possible for at least B to remain on the surface while A is
dragged along.  Now imagine a circle with a radius whose length is equal to the
length of a side of the square.  The circle has its center at B and all of its
radii are perpendicular to line AB.

Whoops!  I was going to show that the circle must always pass through the
surface, so D could be placed at the point of intersection.  Then the path of
D as A and B are moved about would be the path described by that point of
intersection.  But consider this:  There is a steep hill in the surface:


	/-\
	| |
	| |
	| |
_______/   \_____

This hill is the same when seen from the front as when seen from the side.  If
point A is on the bottom of the hill and point B is on top, the circle
described above will be entirely above the surface.  This means it is not
always possible to put a third point of the square on the surface when two
points have already been positioned.  It seems Bevin has been correct.

Can anyone:

	a)	find something wrong with this or
	b)	find a way to show that A and B can always be dragged about
		without encountering such a feature in the surface or
	c)	prove the theorem in another way?


				-- edp