Dissection of the Rhombic Triacontahedron
It
is a wonderful fact that the five-colored
rhombic triacontahedron can be dissected into twenty three-colored
parallelepipeds
in a color consistent manner. This is probably my favorite paper model;
it is very simple and deserves to be well known.You should definitely drop
your mouse and make a paper model of this
right now! (As it will get a lot of handling, I recommend a glue-and-tabs
approach rather than the quicker tape-on-the-inside method.) This figure
shows a set which I constructed some twenty years ago. The twenty pieces
at left can be assembled into the hollow shell at right with red face touching
red face, yellow face touching yellow face, etc. There are several solutions.
As the figure shows, there are ten "pointy" rhombic parallelepipeds (top left) and ten "flat" rhombic parallelepipeds (bottom left). The faces of all the parallelepipeds are the same "golden rhombus" shape as the face of the rhombic triacontahedron, i.e., the lengths of the diagonals are in the proportion of the golden ratio. It follows that the face angles are approximately 63 degrees 26 minutes and 116 degrees 34 minutes.
Each parallelepiped is colored with just three colors (selected from the five colors of the rhombic triacontahedron) and is assembled so any two opposite faces are always the same color. There are exactly ten ways to choose three items from a set of five, namely: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345. Each of these ten color choices is used for one of the ten pointy blocks and for one of the ten flat ones. (Note that when you start to make a block, you may assemble faces of its three colors about a vertex in either clockwise or counterclockwise order. By the rule that opposite faces are the same color, the opposite vertex will end up with the opposite cyclic order, so every order will appear equally.)
Start by cutting out 120 sturdy rhombi, all the same size: 24 of each of five colors. Be pretty careful about the face shape, as a 60 degree angle would not do! Use two rhombi of color #1, two of color #2, and two of color #3 to make a pointy parallelepiped, being sure two faces of the same color are never adjacent. Then use the same color combination to make a flat parallelepiped. (To get the pointy shape, put three acute angles together at a vertex. To get the flat shape, put three obtuse angles together.) Next make two blocks each having two faces of each of colors #1, #2, and #4. Continue until all ten color combinations listed above are used in both a pointy and a flat parallelepiped.
For the outer shell, assemble 30 more rhombi into a rhombic triacontahedron, and divide along one of the decagonal "equators" (orthogonal to a 5-fold axis). These rhombi should be larger by two or three cardboard thicknesses, in order for the parallelepipeds to fit inside it. In the model illustrated, I glued additional rhombi to the inner surface, so no tabs can be seen.
Making the shell with six rhombi of each of five colors (as illustrated above) makes it fairly easy to assemble the puzzle because the shell serves as a guide when placing new pieces inside it. For a more difficult puzzle, one could use 30 clear plastic pieces to make the shell (although I haven't actually done this).
This construction was first described by Gerhard Kowalewski in his book Der Keplersche Korper und andere Bauspiele. (I learned German originally just so I could read this book, but now it is available in English translation.) Although Coxeter mentions it in Regular Polytopes, he doesn't describe the coloring aspect, and the model is not widely known.
After mastering this, you will be interested in the
color-matched dissection of the rhombic enneacontahedron.