Re 10cm worm to lie in a shape that encloses the maximum area.

        SPOILER

        Since the worm is free to curl anyway it wants we can
        assume that it is free to curl in a straight line. The blanket
        therefore has to be long enough to cover the streached worm
        ie 10CM.

        Now supposing the worm bends in the middle such that the  ends
        move together along the original line. When the angle that
        the worm bends reaches the 60 degree mark the worm could be moved
        such that the bend is now at one of its end points, now as the
        worm continues to bend the width of the blanket gets smaller
        again. Hence the maximum width of the blanket is the height
        of 5cm sided equilateral triangle = 4.33cm.

        Using this equilateral triangle as a starting point and extending
        the base 2.5cm on each side to make the straight worm condition.

        Now start folding the work at the ends, using 1cm increments you
        and bending the worm ends up 90 degrees you will find that you
        will form a new triangle that joins the apex of the old triangle
        to the ends of the straight line. This triangle marks the
        minimum area of blanket required.

        The area is therefore 4.33 x 10/2 = 21.65cm2

        This is a very complex puzzle to think of and I commend the author.
        Things appear simple till you start flipping things over and stuff.
        I am prettty sure this is the minimum area case though. The height
        and width I know are right, and the shape seems to fit all the test
        cases I can think of.

I think the important shapes to consider are the C shape, and the pi/3
V shape, and the maximal shapes in between. The latter are difficult
to analyze, but if we assume the result to be a straight edge on the
blanket, then the area of the final pentagon is (50+12.5*sqrt(3))/3 ~ 23.88

If someone wants to analyze the intermediate shapes, I think they go
as follows. A vertical segment of length w, followed by a tilted
segment, followed by a horizontal segment of length 2*x followed
by a tilted segment followed by a vertical segment of length w. When
x is zero, so is w and we get the V shape. When x is 10/6, w is 10/3
and we get the C shape. For any given x, w is chosen so as to maximize
the distance between the horizontal segment and the line joining the
endpoints, subject to the constraints of the length of the worm, and
the distance between the second bend and the far endpoint. The latter
should be no smaller than 2.5*sqrt(3) or the 'triangle' can be rotated.
I assume the worm is symmetrical.

Both remarks are correct. Unfortunately, everyone seems to be using different
interpretations of the problem. One poster assumed that the blanket must be
rectangular, but most don't; another said "I assume the worm is symmetrical",
although this is certainly not implied in the original statement. Clearly we
have to assume that the blanket can be moved to follow the worm, if the answer
is to be bounded, but what sort of movements is unclear: translations only?
direct congruences? all congruences (can the blanket be turned over)? Is the
blanket required to be convex?

A recent summary of the status of this class of problem can be found in
sections D18 (convex case) and G6 (non-convex case) of [1]. (Readers of
sci.math will no doubt be bored to death by my recommending this book
*yet again*.) The metaphor used there is "minimal comfortable living
quarters for a unit worm", rather than anything to do with blankets, which
makes rather better sense! Formally: "find the convex set K of least area
that contains a congruent copy of every continuous rectifiable curve of
length 1". Different constraints on K, the congruences, and the worm are
all mentioned.

For the case of (convex sets, all congruences, all unit worms) the cover
with least area known is apparently the following truncated rhombus

     (-1/2+sqrt(3)/4,1/4)  ---  (1/2-sqrt(3)/4,1/4)
                         /     \
                       /         \
              (-1/2,0) \         / (1/2,0)
                         \     /
                           \ /
                      (0,-1/(2*sqrt(3))

(not to scale!) with area 0.286084... [2], but this is probably not minimal.
If only translations are allowed, the equilateral triangle with height 1
(area 0.577350...) is known to be minimal. Multiply all areas by 100 cm^2
for a 10 cm worm, of course.

[1] Hallard T. Croft, Kenneth J. Falconer, & Richard K. Guy
    "Unsolved Problems in Geometry"
    Springer-Verlag (1991)   xv+198 pp
    ISBN 0-387-97506-3

[2] J. Gerriets & G. Poole
    "Minimum covers for arcs of constant length"
    Bull. Amer. Math. Soc. 79 (1973) 462-463

A----C
 \  |
  \ |
   \|
    B

The area would be maximised if the above situation occurs, i.e. AC=BC=5cm.
This would make the area 12.5 cm^2 I cannot think of any shape that would
require a greater area.

I calculated this for 1 bend in the worm (i.e. point c) and varied the
angles of C through 0-180 and found that I was right in thinking 90 degrees
needed the max area. Any other situation in which the vertices of the
bended worm have an orthogonal overlap would effectively make the worm
shorter, so this is the greatest area (I think)

The problem is to find the SMALLEST BLANKET that can be guaranteed to cover
the whole worm, whatever position its in.  So, for example, the above blanket
won't work because there's no way it will cover the worm if the worm is
stretched out (so 10cm long).

I REALLY would like a solution to this problem. ....

He poses the same problem, wrapped up in English prose for sugar, and
(here's the spoiler)....  

(gee, hope that pagebreak worked)
.... it's not solved yet, if I recall correctly. Anyway, it's all in
the book.  

C.