Polyforms derived from the removal of half of one of the basic tiling shapes.

If take the set of tetrominoes and remove half a square (cut across the square) so that no original join is divided then we get the six truncated tetrominoes below.

This set is quite small and does not produce any interesting constructions. Removing a half square from the pentominoes gives us this set of 19 pieces. The only possible rectangles with the truncated pentominoes would be 4½ x 19 or 8½ x 9. Brendan Owen has shown that no solution exists for the 4½ x 19 and Patrick Hamlyn has found two solutions so far for the 8½ x 9.

An easier problem is to make the rectangle below where one of the pieces is used twice.

 

The removal of part of a polyomino by various types of diagoinal cuts or by adding squares to diagonally cut squares, dominoes or trominoes has been looked at in some detail. Most sets produced sre not very interesting but a few give some interesting figures. Click on the links (yellow rectangles) below to see samples. These sets are described in more detail below.

 

If we were to remove half a square with a single diagonal cut then we get 13 sliced tetrominoes of area 3½ unit squares. With this set we can make two 3x7 rectangles with one piece omitted or a 7x7 square with one piece used twice (or the second piece could be counted as a hole).

It is also possible to make a 'sliced rectangle' with the full set.

The one-sided set consists of the 25 pieces shown in the constructions below. No rectangle is possible unless we use one piece twice as in the figure at the right below.

A number of sliced rectangles can be made with this set.

If we allow an original join to be removed then we get one extra piece and this set of 14 pieces can make a square as well as a number of other symmetrical figures. This set has been referred to as the 3½-ominoes.

It is also possible to make a number of congruent pairs with the set.

We can also make four fold replicas of three of the pieces with the piece replicated used three times. The remaining pieces cannot be thus replicated as it is not possible to fill a 45º corner. It might also be possible to use the set to form a duplication and a triplication of one of the pieces but this is probably only possible for the piece shown at the right below. The piece omitted cannot be the piece replicated.

If we consider the one-sided set then we get the following 27 pieces which are shown forming the only two possible sliced rectangles. Also included are soem examples of the similar hole problem for this set.

Roel Huisman has some problems based on this and other polyforms.

There are 47 sliced pentominoes and 54 4½-ominoes with seven extra pieces. The sliced pentominoes can make a number of rectangles if one piece is used twice.

The 47 pieces can also make a number of 'sliced rectangles'.

It is also possible to make a similar hole construction with this set with one piece used twice.

The 54 4½-ominoes can make a 9x27 rectangle as shown here where the seven extra pieces are coloured paler than the rest.

An improvement on this would be to form three 9x9 squares as shown here.

Other figures formed with this set are shown below.

There are many variations on the 9x27 rectangle the first set here is by Roel Huisman

It is also possible to form two congruent shapes with this set. The second solution provides a new way of forming the 9x27 rectangle.

There are many other possible symmetrical figures to be made with this set some of which are shown below.

Roel Huisman has found sets of six and nine congruent shapes with this set.

If we consider the one-sided set of sliced pentominoes then there are 92 pieces which can make two 9x23 rectangles.

For the one-sided 4½-ominoes there are 106 pieces which can make a sliced rectangle.

If we omit the one piece with a hole then we get 210 5½-ominoes which Roel Huisman has packed into the series of rectangles shown here.

Roel Huisman has found the 36 pieces obtained by removing half a domino from a pentomino. Roel calls them domslicedpentominoes and a large numbed of constructions are possible with the set as can be seen here.

If we allow pentominoes to be cut so that an internal join is removed then we get an extra two pieces (shown lighter in the first diagram below). This set cannot make any rectangle.

Roel has also looked at constructions made with two sets of domsliced tetrominoes.

We could also look at the set of one-sided domslicedtetrominoes which gives the same number of pieces but ensures that for each pair of pieces both of the mirror pair is used. Rectangles are certainly possible as shown here where the left hand pair of rectangles give solutions to the 5x12 and 6x10 rectangles and the right hand figure shows a 4x15 rectangle. The first four constructions below based on an 8x8 square are all made of two congruent parts.

While no rectangle can be made with a single set of pieces, a variety of symmetrical figures can be made with a single set of pieces. Also the one sided set can make a number of sliced rectangles.

There are many other possible variations of adding/removing half squares/dominoes from the polyominoes. The following table gives some numbers of pieces for various sets. (Peter Esser has an excellent solver which works with all these sets and it can be downloaded from http://members.tripod.de/polyforms/psolver.zip.

If we remove a half triangle from pentiamonds ensuring that no existing join is cut then we get the following 14 truncated pentiamonds which have a total area equal to each of the figures shown.

Brendan Owen has used his solver to prove that the figure at the right has no solutions and has found a unique solution to the figure at the left. Below are a number of other solved and unsolved problems for this set. Owen has found 205 solutions for the figure at the top left.

Brendan Owen has produced solutions to all but one the above. Try to guess which one before looking at the solutions here.

If we remove a half triangle from the hexiamonds we get the following set of 40 pieces with a total area of 220 triangles.

Each of the figures below has an an area of 220 triangles and might be possible to make with these pieces.

Alternatively we can remove half a rhombus (by the long diagonal) from the polyiamonds to form rhombsliced polyiamonds. The diagram below shows the three rhombsliced tetriamonds; the seven rhombsliced pentiamonds and the twenty rhombsliced hexiamonds. This last set has a total area of 100 triangles and might have made some interesting figures such as a side 10 triangle or a number of parallelograms. Unfortunately the set has three pieces with an unbalanced colouring (a excess of two triangles each) which means that none of these figures can be made. Brendan Owen has placed 18 of them into a dodecagon as shown below.

Any figure which might be made with the full set must have a colour excess of 2 or 6. Patrick Hamlyn has produced this symmetric figure which has an excess of two.

If we look at the one-sided set we get 38 pieces. This solution is also by Patrick.

Finally pictures of the 9 truncated tetracubes -

and the 36 sliced tetracubes -

The number of pieces in the set depends on whether to just allow single cuts which produce the pieces at the left or to allow for halving cubes diagonally which will produce extra pieces as at the right (these are only some of the possible extra pieces). Peter Esser has done some work on these and similar pieces - see his Clipped and Extended Polycubes.