From: snewman@Xenon.Stanford.EDU (Steven Newman)
Subject: re: The Worm and the Blanket Problem
Organization: Computer Science Department, Stanford University.
Date: Sat, 12 Dec 1992 19:48:15 GMT

Here is a new upper bound for the worm-and-blanket problem.  The
blanket consists of a rectangle with width 2/3 and height 1/3,
semicircular caps on the left and right ends, and a "bump" on the top
side consisting of the upper half of a circle with radius 1/3 and
centered at the center of the rectangle.  More precisely, if the
rectangle is centered at (0,0), the blanket consists of the union
of the following objects:

  A rectangle extending from x=-1/3 to x=+1/3 and y=-1/6 to x=+1/6

  A circle with radius 1/6 and center (-1/3, 0)

  A circle with radius 1/6 and center (+1/3, 0)

  The upper half of a circle with radius 1/3 and center (0,0).

Here is a rough diagram of the shape; points A and B are the centers
of the two end-caps and point C is the center of the rectangle and
the upper semicircle.


               +++++
             +++++++++
           +++++++++++++
         +++++++++++++++++
        +++++++++++++++++++
     +++++++++++++++++++++++++
  +++++++++++++++++++++++++++++++
 +++++++++++++++++++++++++++++++++
 +++++++++++++++++++++++++++++++++
+++++++++++++++++++++++++++++++++++
+++++A+++++++++++C+++++++++++B+++++
+++++++++++++++++++++++++++++++++++
 +++++++++++++++++++++++++++++++++
 +++++++++++++++++++++++++++++++++
   +++++++++++++++++++++++++++++
     +++++++++++++++++++++++++

The area of this figure is 2/9 + pi/36 + (pi/27 - sqrt(3)/36) or
rougly 0.3777.

To see that this blanket will cover any worm, identify the points
1/6 and 5/6 along the length of the worm and call them P and Q.
Position the blanket so that C lies halfway between P and Q, and
the points A, B, C, P, and Q are all collinear.  Since P and Q
can be no more than 2/3 units apart, they must lie on A and B or
somewhere in-between.  Hence the [0,1/6] and [5/6,1] segments
of the worm will lie within the rectangle or the endcaps.  It
remains only to show that the [1/6,5/6] segment of the worm also
lies under the blanket.

It is easy to see that the [1/6,5/6] segment of the worm can't
extend to the left of point A or the to the right of point B,
so it must lie entirely inside the rectangle unless it sticks
out above or below.  The rectangle has a height of 1/3, so the
[1/6,5/6] segment can't stick out on both sides.  Assume it
sticks out on top or not at all (if it sticks out on the bottom
then we rotate by 180 degrees).

Let R be any point on the [1/6,5/6] segment of the worm; then
PR + QR <= 2/3.  I claim that R must lie within distance 1/3 of
(0,0); this proves that R is inside the blanket, since if it
isn't in the rectangle then it must be above the x axis and hence
lie inside the semicircular cap.

I suspect there is a trivial proof that shows RC <= 1/3 given
PQ + QR <= 2/3 and C is the midpoint of P and Q, but the only
proof I've come up with so far uses hairy calculus (I had to
prove that the distance from R as you slide along the line from
P to Q is a convex function).  Geometry buffs may know a better
one.

  - Steve Newman (snewman@cs.stanford.edu)
Newsgroups: rec.puzzles