Pentominoes in 4 squares
Alexandre Muņiz
We
found this puzzle on
facebook of Puzzle Fun
Start
with the 12 pentominoes.
Make a single diagonal cut in two of the pentominoes.
Pack the resulting 14 pieces and 4 monominoes into four 4×4 squares.
We got a wonderful document of Helmut Postl.
Since the
Ipentomino is the only one which does not fit into a 4×4square, it has to
be split all the time. There seems to be no way to cut the Ipentomino twice,
so the second cut has to be made on a different pentomino. For all given
solutions, this is the F or the T.
Remark: Only holes of polyomino shape are considered. There are many
more possibilities if the holes may have diagonal borders. (This is an other
nice problem)
Two squares
contain 3 whole pentominos, and two squares contain 2 whole pentominoes.
1. Each of the four squares contain
exactly one monomino hole.
In the sequel, a thick black closed line is meant to surround a symmetric
shape which can be flipped over to give new solutions.


2. One square contains two
monomino holes.
In the following 4 solutions the third square contains two holes, and
the last square contains no hole.
Here, the I and F are cut.
Three squares contain 3 whole pentominos, and
the last square contains one whole pentomino.
In the following 3 (resp.6) solutions, the whole pentomino in the
last square is the F.
Here, the I and T are cut.
In the following solutions, the whole pentomino in the last square is
the T.
Here, the I and F are cut. Since the first three squares are always built in
the same manner, only the last square is drawn for the solutions
The last two
solutions have a speciality: The diagonal cut may not only be made
under 45°, but under various angles. (The additional drawing of the grey
triangles shall indicate some of the possible cuts.)
First case: 0° < φ ≤ 45°.
Second case: 45° ≤ φ ≤ arctan(2) = 63,43494882°.
So there is an infinity of solutions."
Aad
van de Wetering found an solution:
Helmut generalized the solution by varying the angle of the cuts.