Advanced Math/Two Advanced Questions
Good morning, sir. I'm hoping you can help me out as I've finally given up. For your information, I have no more than a high school education (Class of 02). I am an autodidact by nature and I focus heavily on mathematic studies. All of this is for my own understanding to better myself. As stated in your directions, I'm not looking for scrupulous detail, but rather, general concepts or references that I can use for guidance. The request is listed in each question.
(This question is hidden due to revealing answers to publically available IQ tests)
Question 1: Intersecting Polyhedra
(All I'm hoping to find are resources that can lead me to solving this type of problem if you cannot answer this readily.) When given two interpenetrating polyhedra in R3, how can I determine the number of bounded regions generated by the enclosed faces of the polyhedra?
For example, two mutually interpenetrating cubes, one of which is rotated 45 degrees about a line through the center of opposite faces, where the center points of the cubes are (0,0,0). The answer is obviously 9 bounded regions (for any rotation where MOD 90 <> 0), but how do I go about proving this or solving for more advanced combinations (e.g. cube and tetrahedron) where I'm allowed to know any information requested about the polyhedra (rotation, coordinates, planes, etc?
Question 2: Inequivalency (colors vs. diagonals)
Given a cube with diagonals across each of it's six faces, there are a limited number of distinct, inequivalent patterns that can be generated. (The answer is eight distinct patterns)
If the diagonals are drawn using any combination of two colors, how many rotationally distinct patterns are possible?
Unfortunately, I have been unable to find a way to safely use Burnside's Theorem to solve this. Using brute force, I was able to come up with eight diagonal patterns with the following orders: 24, 12, 6, 4, 2, 1 where they vary 1, 2, 4, 6, 12, and 24 ways respectively. The eight patterns use only one color.
Unless you can guide me towards determining the cycle index of the cube with the diagonals (i.e. proving 8 is correct), how can I determine the total number of distint colorings of the eight distinct diagonal patterns using two colors?
Here is some additional info about the cube to assist, if needed:
Cycle:(x1^6 + 3*x1^2*x2^2 + 6*x1^2*x4 + 6*x2^3 + 8*x3^2)/24
C(n):(n^6 + 3*n^4 + 12*n^3 + 8*n^2)/24
C(2) = 10
Cycle: (x1^8 + 9*x2^4 + 6*x4^2 + 8*x1^2*x3^2)/24
Please let me know if anything is unclear.
Thank you for your help.
Your first question is much too vague to provide some succinct formula or algorithm, but I will try to address it as best I can.
First, you have to be clear on what sorts of polyhedra you consider. Are the polyhedra convex
? Are they simply connected
? If you allow non-convex polyhedra (like a "star" shape) or non-simply-connected ones (like a donut-shape, squared off to be a polyhedron), then the question may become more complicated.
It's also not clear if your polygons can be stretched to be irregular, or if they are required to be fixed or regular in some way (which limits you to finitely many combinations).
This is a generalization of a problem that can be easily solved in the plane -- given a (regular or not) polygon with n sides (or n-gon), can you draw another n-gon that creates the maximum number of bounded regions? How many regions is it? Here we have to assume that the polygons are simply connected and bounded, but this makes the answer easy. In this case, there are 2n+1 regions possible -- just line them up and rotate them partway.
This is the method you want to generalize, but there are a few problems:
First, there are not regular polyhedra of all sizes. The "most" regular type of polyhedra is the set of Platonic solids
. The next is the set of Archimedean solids
. There are only finitely many such polygons.
However, there are still possibilities for doing what you ask, and I would invite you to consider the following:
Given any polyhedron, construct a polyhedron which is its dual
. For each face of the original polyhedron, draw . Not only is this an interesting construction, you can even phrase the question in the mathematical language of graph theory (see here
Given this construction, do you get exactly one region per vertex? But how many vertices are there? (You know how many there are on your original polyhedron, but that does not tell you how many there are on the dual!) And is there always just one big internal region? Why or why not?
That should give you plenty to think about for question 1, which was a bit too vague to not leave you with an open-ended answer.
For the second question, you don't need to crunch all the numbers computing cycle indices. I'm not sure you'd want to go so deep into the "machinery" to answer this question. It won't get you to the answer more efficiently or directly.
The faces of the cube are being acted on by symmetries of the cube. Each one has colored diagonals somehow, say red or blue. There are 2^12 colorings of a cube's diagonals (there are 6 sides, so 12 diagonals), not accounting for any symmetry. Now you have to consider which ones are fixed by each type of rotation. Following the example on Wikipedia
is not too hard:
The identity leaves all 2^12 unchanged of course.
What about 90° rotations (through the center of a face)?
Well, the four faces that change positions must be the same, only 2 choices for the diagonals of a single face which gets repeated. The other two faces must be all red or all blue, since their diagonals change positions. 2 choices each for those. So in total, only 8 configurations here are unchanged by a 90°. There are six of these rotations.
What about the 180degree rotations? There, the faces that change position must match up in opposing pairs, so there are 16 ways of doing that. The other two faces could be anything, because 180° doesn't affect them. So 4 ways of choosing each of those. Total here: 64. There are only three of these rotations.
And the 120 degree rotations along the big diagonals of the cube? There, the axis goes through two vertices, and the three faces touching one vertex all match (and likewise the other). So there are 16 configurations here, and there are eight such rotations.
And the 180 degree rotations through edges? (This is through an axis that goes through the centers of opposing edges.) This pairs up faces that must have the same colorings, so there are 4 ways for each pair of similarly-colored faces, for a total of 64 configurations. There are six of these rotations.
The full size of the group is (just like in the example on Wikipedia) 24.
So what you get is:
(1/24) ( 2^12 + 6×8 + 3×64 + 8×16 + 6×64 ) = 202
You should go through this computation yourself and make sure you agree that all these numbers seem right. One, because it would help you get things straight, and two, I may have made an error (although the answer coming out to an integer helps me feel like that is less likely).