Constructing rectangles by folding a unit sqaure
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In this problem session we continued exploring the problems of origami
discussed in the last session.  In particular we tried to settle the
following open problems from last week.

Open Problem 1  -- Is it possible to fold a square sheet of paper so as to
construct a rectangle which cannot be inscribed within the square?

Open Problem 2 -- Can you characterize the complete set of rectangles which
can be made by folding a $1 \times 1$ square of paper?

Joe Mitchell proved that for a rectangle to be inscribed in a unit square,
if its length is $L$, its breadth cannot be greater than $sqrt{2} - x$.
The proof follows from the fact that the four corners of a maximal
rectangle can be restricted in only two ways:
1. The four corners of the rectangle lie on the four sides of the square.
2. Two opposite sides of the rectangle lie on two opposite sides of the
        square.

Saurabh Sethia answered the open problem 1 in affirmative by demonstrating
a way to construct a rectangle which cannot be inscribed within the sqaure.
The way to do it is shown in the following figure.  In the figure the
unit square is folded along line $AB$.  The rectangle constructed is
$ABCD$ shown by dashed lines.  Elimentary trignometry shows that
$l(AB)$ is $sqrt{1 + x^2}$ and $l(BC)$ is $(0.5 - x/2) sqrt{1 + x^2}$.

For the same length the maximal rectangle that can be inscribed in
the unit square without folding will have a breadth of
$sqrt{2} - sqrt{1 + x^2}$.  Thus we have:

I. Length = $sqrt{1 + x^2}
II. Breadth without folding = $sqrt{2} - sqrt{1 + x^2}$.
III. Breadth of rect $ABCD$ = $(0.5 - x/2) sqrt{1 + x^2}$.

It can be verified that for values of $x$ less than 0.268,
(III) is greater than (II).

Open problem 2 is still open.