Pentominoes in 4 squares
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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 I-pentomino is the only one which does not fit into a 4×4-square, it has to be split all the time. There seems to be no way to cut the I-pentomino 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
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.