The task, of course, is to place all of the pentominoes into the tray. Furthermore, when completed, a perfect checkboard pattern should result.
It was quite fun to build these pieces, but an experienced LEGO builder will realize that constructing the pieces so that the cubes of the pieces are strictly alternating black and white is not as easy as it would immediately seem. Even if you don't have enough pieces to build a complete set, I would challenge one to build just a single piece (say the 'cross or X' shaped piece). How to make the black and white cubes join to form the pieces as a whole? The question is left as an exercise for the reader.
There are many, many variations on this puzzle, some involving more than six pieces. A wonderful webpage at IBM is a good starting place to investigate this type of puzzle further.
Maybe it's easier seen that read, so below are some pictures of a set of polycubes I have built. The set consists of all monocubes, duocubes (there is only one of each), tricubes, tetracubes, and pentacubes, plus a single long hexacube (the '1x1x6' straight piece). Each piece is unique and 'reflections' are also built (meaning some are symmetric to each other). The first picture shows all of the white polycubes laid out.
I also constructed a LEGO box of dimensions 4-by-6-by-8 cubes. The pieces can be arranged so that they fit perfectly into this black box with no spaces inside. That, of course, is the puzzle.
The solution is not unique... there are many ways to fit all of the pieces inside the box, and, in fact, if a certain strategy is used, the task is not terribly difficult.
Note that 4x6x8 yields a total volume of 192 cubes. This number is factorable in many ways, so one might wonder whether the pieces can be fit into different sized boxes. They can. I have, for example, built a red box of dimensions 3x8x8 (also equals to 192 volume) and placed all the pieces within. I have done a 4x4x12 configuration as well, but have not constructed a box yet.
Now, each of the four 'levels' of the tower rotates about a central axis. Furthermore, one space where a (red) section should go is left blank. This allows one to slide rooms up or down into the empty space (and moving the space to a new location).
The object of the puzzle is to arrange the sections so that each column is all of one color and when stacked from bottom to top the sections are: Arch, Window, Window, Window with Roof (except for the red column which is missing one Window section, so it is just: Arch, Window, Roof).
The pictures below show the various interactions with the puzzle, and finally a solved tower.
To solve, the user must configure the tiles so that a continuous, connected path of white is formed. The solution is shown in the last photo below.
This puzzle has 48 tiles (a 7-by-7 board with on space empty).
The thumbnail picture to the right shows the final construction laid out (click on picture to enlarge). Other pictures are viewable with these links:
Note that 6-cubed equals 5-cubed plus 4-cubed plus 3-cubed (216 = 125 + 64 + 27). The picture to the right (which may be clicked to enlarge) shows three cubes made of LEGO bricks so that their proportions are 5, 4, and 3 (the black 3x3x3 cube is a little hard to see). Obviously, if each cube were made of individual unit cube pieces, one could disassemble the three cubes and reform a single larger cube of size 6x6x6 (it would be made of 216 little unit cubes).
But what is the fewest number of pieces that can be used to create both a set of three cubes as pictured as well as a single 6x6x6 cube?
As Brian stated in his email to me, "For example, you can remove a 3x3x3 section from the larger cube intact as one piece, and two 4x2x4 pieces to make the other cube, but then you're left with a mess to try and assemble the 5x5x5 cube using the least pieces."
After a couple of hours of building and rebuilding, I managed to fashion 9 pieces which build the three cubes pictured at right and will also form a single 6x6x6 cube.
Can you do the same? Can you do better than 9 pieces? I'm still trying to improve my result (though, I think it may be impossible sticking to polycube shaped pieces). Similar questions can be posed with other cube sizes (9-cubed equals 8-cubed plus 6-cubed plus 1-cubed, for example).
If you need a hints, I have a couple of pictures of the large cube constructed of the white, red, and black pieces: Angle #1, Angle #2.
If you have tired of this topic already, you may just want to look at my blurry picture of all the pieces lying about (now it should be fairly easy to build the 6-cube).
If you have not yet tired of this topic, you should be aware that geometry dissections are a topic in mathematics that has been active for many, many years. Both amateur and professional mathematicians have made many contributions in dissections of 2, 3, and more-dimensional shapes.
This example uses pieces that are formed from smaller cubes (or blocks), but, in general, dissections of various shapes can result in very peculiarly shaped pieces. One web-resource on this topic is MathWorld. Another is The Geometry Junkyard.
UPDATE: Shortly after posting the above paragraphs, a certain Don Reble correspoded with me and alerted me to Martin Gardener's book "Knotted Doughtnuts and Other Mathermatical Entertainments" (1986) which details a solution to the above dissection problem by E. H. Wheeler. That solution required only 8 polycubic pieces.
Don provided me with a slightly altered solution as the one as printed in Gardener's book was slighted flawed.
Anyway, using red, white, and blue LEGO bricks I constructed the pieces and assembled the three cubes as well as the single large (6x6x6) cube.
I also have a picture of the 8 pieces without assembly. If you are unable to reproduce the pieces from that picture, below is an ASCII representation of the assembled 6x6x6 cube, layer by layer, with each of the 8 pieces enumerated:
112222 112222 662222 662222 666668 666668 112222 112222 662222 662222 666668 666668 332222 332222 332222 666222 666668 666668 332222 332222 332222 777222 777555 777555 333334 333555 333555 777555 777555 777555 333334 333555 333555 777555 777555 777555
Thanks, Don, for the additional solution.