Treebeard's Stumper Answer

Tricky Tetrahedra
Here's a tricky visualization stumper. Imagine a perfect tetrahedron. That's a triangular pyramid where every side including the base is an identical triangle. Now imagine lots of tetrahedra and start to color their sides. If you have four colors to work with, then how many different pyramids can you paint so that a duplicate cannot be found by rotating or turning the pyramids? For help, there's a nifty way to make a tetrahedron from a small envelope at http://tremor.nmt.edu/tetra.html. How about coloring a cube with six colors?
Four different ways of visualizing a tetrahedron: as a wire frame, or an animation (by Rüdiger Appel), or a colored model, or a paper plan ready to fold up. How many unique colorings are possible, with just four colors to work with, if every face has a single color?
It's hard to visualize coloring the faces of a triangular pyramid because of the strong symmetry. With two colors, each can cover 1, 2, or 3 sides. There are 6 ways of picking 2 different colors from the 4 available, so that makes 3 x 6 = 18 colorings. Similarly,
That makes a total of 36 unique colorings. There are details and colored pictures below.
with 1 color: 1 way x 4 color choices = 4 with 2 colors: 3 ways x 6 color choices = 18 with 3 colors: 3 ways x 4 color choices = 12 with 4 colors: 2 ways x 1 color choice = 2 Notes:
Here's how I figure it. The colors are RGBY (Red, Green, Blue, and Yellow), and the flag on the pyramid shows the color of the hidden side. What makes this such a hard visuaization stumper is that each of these tricky tetrahedra is identical when rotated due to the overwhelming symmetry. Try to visualize how these are all really the same!
With one color: This is easy. Each face is the same color, and there are four colors available, so there's 1 x 4 = 4 possible colorings.
With two colors: With two colors, each can cover 1, 2, or 3 sides. There are 6 ways of picking 2 different colors from the 4 available, so that makes 3 x 6 = 18 colorings.The colorings are:
RG (RGGG RRGG RRRG) RB (RBBB RRBB RRRB) RY (RYYY RRYY RRRY) GB (GBBB GGBB GBBB) GY (GYYY GGYY GGGY) BY (BYYY BBYY BBBY)
With three colors: With three colors, each can cover 2 sides in turn for 3 combinations. There are 4 ways of picking 3 different colors from the 4 available, so that makes 3 x 4 = 12 colorings.The colorings are:
RGB (RRGB RGGB RGBB) RGY (RRGY RGGY RGYY) GBY (GGBY GBBY GBYY) RBY (RRBY RBBY RBYY)
With four colors: This is the tricky one since there are two mirror images that cannot be matched by rotating, even though there's only one way of picking 4 colors from the 4 available. That makes 2 x 1 = 2 colorings.That's
4 + 18 + 12 + 2 = 36 colorings total with 4 available colors.Coloring a cube with six colors is much trickier than this! Symmetry and mirror images abound. I haven't had time to work it out, but Graybear sent this analysis in betwen figuring grades at school:
When I say 'triples adjacent', I mean that three sides of the same color are arranged like the 1, 2, and 3 on a standard die. 'Triples inline', would imply the 1, 2, and 6 would be the same color, for example. 'Doubles adjacent', likewise, would be two adjacent sides of the same color (e.g. 1 and 2), while 'doubles opposite would mean the 1 and 6 are the same color. You can probably deduce the rest.I've found two other answers on the Web at Ken Duisenberg's Puzzle of the Week that don't agree with Graybear's: either 2226 or 1866. Graybear is giving the total numbers for all six colors, so I divided his results by the number of ways to select n colors out of 6 to find how many ways there are of coloring the cube with just 1,2,3,...,6 colors, using each color at least once in each case. These divisors come from Pascal's Triangle:
I. One color: 6 possibilities II. Two colors A. 5 (sides of one color) + 1 (side of another color): 30 poss. B. 4 + 2 1) doubles adjacent: 30 poss. 2) doubles opposite: 30 poss. C. 3 + 3 1) triples adjacent: 15 poss. 2) triples inline: 15 poss. III. Three colors A. 4 + 1 + 1 1) singles adjacent: 60 poss. 2) singles opposite: 60 poss. B. 3 + 2 + 1 1) triples adjacent: 120 poss. 2) triples inline a) doubles adjacent: 120 poss. b) doubles opposite: 120 poss. C. 2 + 2 + 2 1) three doubles adjacent: 40 poss. ? 2) two doubles adj, one opp: 60 poss. 3) three doubles opposite: 20 poss. IV. Four colors A. 3 + 1 + 1 + 1 1) triples adjacent: 120 poss. 2) triples inline: 180 poss. B. 2 + 2 + 1 + 1 1) both doubles adj, singles opp: 90 poss. 2) both doubles adj, singles adj: 180 poss. 3) one double opp, one double adj: 90 poss. 4) both doubles opposite: 90 poss. V. Five colors: 450 poss. VI. Six colors: 30 poss. TOTAL: 6 + 120 + 600 + 750 + 450 + 30 = 1956
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 This row > 1 6 15 20 15 6 1So using each of n colors (n=1,2,3,4,5,6) at least once, in how many ways can you color a cube, such that a duplicate cannot be found simply by rotating the cube? Here are the candidates:
number of colors 1 2 3 4 5 6 total Graybear 1 8 30 50 75 30 1956 Graybear (2) 1 8 30 56 75 30 2046 Al Zimmermann & Gerd Baron 1 8 30 68 75 30 2226 Joseph DeVincentis 1 8 30 56 45 30 1866 ways to select that
many colors from 66 15 20 15 6 1 Graybear sent this update (07nov01):
I found 90 more combinations for four colors. They fall into the category of 1 double opposite, 1 double adjacent, singles adjacent, so there is a total of 180 in that category instead of 90. This brings my 4color total of 840 (56 x 15) in line with one of the other solutions. I'm now standing by my total of 2046 solutions.The disagreements are with 4 and 5 colors. This should make it much easier to think about and settle.
What about the other Platonic Solids: octahedrons, dodecahedrons, and icosahedrons? And there's many other polyhedra to color if we relax the requirement that all faces be congruent. Is there a general solution for how many unique ways there are to color a nfaced polyhedron with ncolors?
Here are a few Web links for more research:
 I found this puzzle as #281 in H.E. Dudney's Amusements in Mathematics (Dover, 1958). This book has a place of honor that moves back and forth between my bathroom and my study. Remarkable questions and interesting answers! It's in the good company of Sam Loyd, The Moscow Puzzles, Martin Gardner, and my old paperback collections of Isaac Asimov's essays. Other books come and go, but I keep finding new challenges in these. This puzzle is also discussed at Ken Duisenberg's Puzzle of the Week (Dec. 3, 1999).
 George W. Hart's Encyclopedia of Polyhedra and G. Korthals Altes' Paper Models of Polyhedra are proof that geometric polyhedra can become a passion equal to any. You can spend days at these sites.
 You can make a tetrahedron from a small envelope by following the directions at http://tremor.nmt.edu/tetra.html. You can also make polyhedra with straws and string. Tarquin Books sells wonderful books that you cut up and fold and glue together to make many mathematical models, including intricate polyhedra. The Zome System is another great tool for exploring polyhedra, but it's not cheap. There are many interesting JAVA animations on the Web, including here, here, and here. Ernst Haeckel, M.C. Escher, and Arthur Silverman find art in polyhedra. There are many more polyhedra links here, here, and especially here. The wizard Dan Winter takes polyhedra into another dimension, though his remarkable webpage does not seem to be available right now.
 My FourColor Christmas (15 Dec 2000) stumper is another tricky coloring puzzle.
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