Torsten Sillke, Apr, 7, 1993

Martin Gardner [*] says, there is a unique dissection of the square into
five congruent parts.               ^^^^^^
                                    Proof?

[*] MG5SA: The Unexpected Hanging and Other Mathematical Diversions
    MG5SA: Simon & Schuster (1968)
    MG5SA.15.5 Dissecting a Square

Question A:
===========
For which k does there exist a dissection of the n-dimensional hypercube into
k conguent connected parts which are not boxes?
In other words:
 For which k do there exist subsets A_1, ..., A_n of the unit hypercube s.t.
   i) all A_i open, connected
  ii) the A_i are pairwise disjoined and congruent
 iii) the closure of the union of the A_i is the hypercube

If 1<k<=n, there is such a dissection. If in addition k is a prime, there
are infinitely many dissections.

Question B:
===========
For which k does there exist a dissection of the n-dimensional hypercube into
k conguent parts which are not generalized boxes?
(A generalized hyperbox is a product M = M1 * M2 * .. * Mn with the Mi
subsets of R.)

Question C:
===========
For which k does there exist a dissection of the n-dimensional hypercube into
k conguent polyhypercubes (need not be connected) which are not generalized
boxes?

Examples for C:
  The complement of four cubes in the 2^3 is congruent (rotation) to
  itself. This isn't true for the 2^4 any longer.

  The only trisection in congruent parts of the 3^3 is the trivial one.
  One has only to concern polycubes which are subsets of the 3*3*2 box.
  Some further reductions are possible: It must be possible for the
  polycube to match at least four corner cubes.

  The only tiling of the 5*5 square with congruent pentominoes (unconnected)
  is the I5 tiling.

  The only tiling of the 7*7 square with congruent heptominoes (unconnected)
  is the I7 tiling.

  Some tiling of the 9*9 square with congruent nonaminoes (unconnected):

  {0..8}*{0}=I9, {0..2}*{0..2}, {0,3,6}*{0..2}, {0,3,6}*{0,3,6}, {0..2}*{0,1,5}

  All solutions found so far are generalized rectanges: Set1*Set2.
  The {0..2}*{0,1,5} has non-trivial solutions:

   6 6 3 0 0 6 3 3 0
   6 6 3 0 0 6 3 3 0
   6 6 3 0 0 6 3 3 0
   7 7 4 1 1 7 4 4 1
   7 7 4 1 1 7 4 4 1
   7 7 4 1 1 7 4 4 1
   8 8 5 2 2 8 5 5 2
   8 8 5 2 2 8 5 5 2
   8 8 5 2 2 8 5 5 2

   6 6 4 4 4 6 0 0 0
   6 6 3 1 1 6 3 3 1
   6 6 3 1 1 6 3 3 1
   7 7 3 1 1 7 3 3 1
   7 7 4 4 4 7 0 0 0
   7 7 4 4 4 7 0 0 0
   8 8 5 2 2 8 5 5 2
   8 8 5 2 2 8 5 5 2
   8 8 5 2 2 8 5 5 2

   6 4 4 4 6 6 0 0 0
   6 1 1 3 6 6 1 3 3
   6 1 1 3 6 6 1 3 3
   7 1 1 3 7 7 1 3 3
   7 4 4 4 7 7 0 0 0
   7 4 4 4 7 7 0 0 0
   8 8 5 2 2 8 5 5 2
   8 8 5 2 2 8 5 5 2
   8 8 5 2 2 8 5 5 2

   6 6 3 3 3 6 0 0 0
   6 6 4 4 4 6 0 0 0
   6 6 4 4 4 6 1 1 1
   7 7 5 5 5 7 2 2 2
   7 7 3 3 3 7 2 2 2
   7 7 3 3 3 7 0 0 0
   8 8 4 4 4 8 1 1 1
   8 8 5 5 5 8 1 1 1
   8 8 5 5 5 8 2 2 2


   There is a dissection of the square into 15 congruent pieces (Karl Scherer).


Refs:

- Polyominoes of Order 3 do not exists, JoCT A 61 (1992) 130-136

- Dissection of a triangle in 3 similar pieces (Scherer), Quantum 4:5 (1994) 26-27

---------------------------------------------------------------------
Article 902 of sci.math.research:
Subject: Polygon division
Date: Sun, 2 May 1993

It is easy to divide an equilateral triangle into i congruent pieces
for i = 1, 2, 3, 4.  What about i = 5?

Now consider the problem of dividing a regular n-gon into i congruent
pieces.  Note that for n = 4, this is trivial; the same is true for
the circle, which could be viewed as a limiting case.

I am very interested in any answers to these questions; they seem to
elude any straightforward approach (exhaustive listings of the ways to
partition the polygon with curves is prohibitive).

-- Daniel Tunkelang (quixote@watson.ibm.com)