Two-Sided Hexa-Arc-Squares (rectangles)
60 pieces, total area = 360.
Since you can factorise 360 very well the list of constructions is endless:
rectangles, rings, replicas, jagged figures and nested squares.
has a large number of patterns for the 60 one-sided hexominoes. Those patterns can
be used with the arc-pieces, too. There are some differences, because some
pairs of mirror pieces in the hexomino set show up only once in the set of arc pieces and
some symmetric hexominoes have two corresponding pieces in the set ot arc pieces.
In the last rectangle these pieces are marked.
Now for the rectangles:
A: six 6x10 rectangles
B: six 5x12 rectangles
C: three 8x15 rectangles
D: two 9x20 rectangles
With these sets we can get the following rectangles: 5x72(B), 6x60(A),
8x45(C), 9x40(A), 10x36(A), 12x30(A), 15x24(C) and 18x20(A).
Brendan Owen has shown that ten 6x6 squares are impossible.
Rectangles with width 4 or 3 can't be filled, because the number of available border squares is too
small and pieces
which can contribute to opposite sides of a strip with width 4 divide the whole area in odd sized parts.
Two-Sided Hexa-Arc-Squares (replicas)
Four examples for replicas of n-ominoes with a similar hole are shown:
2 quadruplications of a 12-omino
3 pentuplications of the X-pentomino
4 pentuplications of the L-tetromino
6 quadruplications of the skew tetromino
The last problem is the hardest. The 6 quadruplications of the square tetromino are also possible and shown as
an example for rings. For the straight tetromino the problem cannot be solved due to the number of
available border squares. L and T-tetrominoes are open problems.
Two-Sided Hexa-Arc-Squares (rings)
Unlike rectangles with width 3 rings with width 3 can be constructed. We have some inner corners
and the pieces, which are useless for strips can now cover the border squares next to that point.
Nevertheless the two 18x18 - 12x12 rings shown in the title were hard to find.
Two-Sided Hexa-Arc-Squares (nested squares)
Two-Sided Hexa-Arc-Squares (jagged figures)
The jagged ring includes a free area of 369 square units. That's pretty much for a figure with two symmetry
The ten "houses" are an example for a figure with ten congruent copies.
39 pieces, area = 234.
Three rectangles are possible: 6x39, 9x26, 13x18.
110 pieces, area = 660.
72 pieces made up from dominoes area = 432
The 110 one-sided pieces do not pack a rectangle, because the number of balanced pieces under checkerboard
coloring is odd.
With the 72 domino pieces we can get a set of six congruent 12x6 rectangles or a set of nine
congruent 8x6 rectangles. From these sets different rectangles e.g. the 72x6 and the 54x8 can be made. Is it
possible to reach twelve 6x6 squares? Probably not, but it hasn't been proved yet.