TANGOES
Rex Games, Inc
2001 California Street
San Francisco CA
94109
(415)931-8200
(c) 1991

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All angles are intended to be mutilpes of 45 degrees. In particular, 
each edge will be parallel to one of 4 lines.
The lengths of the sides are as follows [followed by area]:
1) 1, 1, sqrt(2)   [1/2]
2) same as #1 
3) 1, 1, 1, 1  [1]
4) sqrt(2), 1, sqrt(2), 1  [1]
5) sqrt(2), sqrt(2), 2  [1]
6) 2, 2, 2sqrt(2) [2]
7) same as #6

Total area is 8  in these units.

Typical problem and solution: Form the pieces into two squares.

Solution: We must have 2 squares of area 4, hence  length 2 on a side.
Note that the lengths on a side can only add to  2  if the horizontal
lengths are rational; in particular, _all pieces must be rotated by 45
degrees_ from this initial position (followed by rotations which are
multiples of 90 degrees)

So for example, the square #3 is aligned with the big square. The medium
triangle #5 must have diagonal parallel to the square's edge; with #3 in place,
this requires the diagonal _be_ and edge, so #3 and #5 meet in the center.
This leaves a triangular hole with area 1.5, which then must be filled
with a small triangle #1. The remaining space has area 1-1/2, so must include
the other half-integral area (#2), and then the piece (#4) of area 1 must 
take up the remaining space. Trying the several edges of length sqrt(2) 
against that edge of #5 leaves only this case (up to reflection and
rotation of course):

+-----------+
| \   5   / |
| 1  \  /   |
+-----+  4  +
|     |   / |
|  3  | / 2 |
+-----+-----+

Each of #6 and #7 must be half the other square. This is the arrangement
in which the pieces are shipped.

(Actually one ought to _prove_ that no big triangle #6, #7 is in the same
square as the square piece #3. If it were, it would have to be the half
above the main diagonal, leaving room only for #1 and #2. But placing
#7 similarly in the other square nd aligning #5 along an edge of that
square leaves room only for an area-1 piece _in the shape of #5_, rather
than #4. Thus this is impossible.)

So goes a typical proof: given a desired shape to make, scale it so that
it has area 8, then calculate the lengths of various edges, each of
which would typically have to be of  the form  a sqrt(2) + b  (a,b integral);
there are only finitely many unions of edges with such an edge length total.