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Little cube question

July 20, 2008 pm31 7:04 pm

There was a cube dissection problem the other day, all about conditional probability, inspired by this post on Math Notations. The puzzle here was hard. The one at Math Notations was classic.

Here’s something a bit easier. It challenges you to represent some quantities algebraically (if it’s very very easy for you, you might want to discuss without posting a spoiler, at least right away)

A cube has 8 vertices (corners), 12 edges, and 6 faces (flat sides).

If we cut it down the middle three times, each time parallel to a pair of faces, we get a 2x2x2 cube, and each little cube (cubelet is what I like to call them, and now there are 8 of them) is a corner.

If we cut each edge into thirds, parallel to a pair of faces, we get a 3x3x3 cube, 27 cubelets in all, with 8 corners, 12 edges, 6 cubelets that are on faces but not edges, and one center cubelet that we can’t see from outside. This is like a rubik’s cube (except for the center)

Now, the question: if we go wild slicing, and we get a cube that is b x b x b,

  1. how many cubelets will there be?
  2. how many corners?
  3. how many edges?
  4. how many face cubelets?
  5. how many center cubelets?
from → Math, mathematics, Puzzles
4 Comments leave one →
  1. dr rick permalink
    July 22, 2008 am31 2:15 am 2:15 am

    I suspect your audience here is divided into “can’t do” and “too easy to post on” :). Although it’s relatively easy to miss the simple form for #5 (the obvious route being do #1-#2-#3-#4, which is not obviously pretty).

    Reply
  2. jd2718 permalink*
    July 22, 2008 am31 2:27 am 2:27 am

    And I suspect you are right. Maybe start with a partial solution, or the answer, or just jump in full, and then look at those exponents…?

    Reply
  3. dr rick permalink
    July 22, 2008 pm31 7:00 pm 7:00 pm

    Well, what’s nice is that the equality that drops out of this (1-2-3-4=5, in the above) has a non-obvious and really beautiful proof… so maybe we should reformulate the question as that?

    Testing: n^3.

    Reply
  4. dr rick permalink
    July 22, 2008 pm31 7:08 pm 7:08 pm

    Well, I’ll give the key idea, and leave the rest as an exercise for another few hours… let’s call the numbers you’re asking for n_1, \ldots, n_5.

    Easy bits first: n_1 = n^3 and n_2=8.

    The key to the others: consider the “face pieces” on a single face. They form a square n-2 on each side (one in from the edges of the “whole face”. There are six faces and so six such squares, so the total number of pieces is n_4=6(n-2)^2. The other two quantities can be found similarly.

    The challenge now: prove, elegantly and in a single line (purely algebraically; of course, this analysis IS a geometric proof!), that n_1=n_2+n_3+n_4+n_5.

    Reply

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