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Painted Cube


Date: 06/21/2001 at 14:35:06
From: Ste 
Subject: Painted cube

We've been given the painted cube topic in preperation for gcse's 
next year. I've done loads of diagrams and tables, but just can't see 
or find a formula for the problem (it must be in algebra). Would you
please send me a formula with a brief explenation? 

Thanks, Dr. Math.
STE


Date: 06/21/2001 at 15:42:17
From: Doctor Greenie
Subject: Re: Painted cube

Hello, Ste -

You will learn a lot more about mathematics if you figure out the 
formulas on your own than if we give them to you. I will try to help 
you find a way to develop the formulas; if you are able to develop the 
formulas with the help I give you, then you will be able to provide 
explanations of the formulas without my help.

We have a large cube made up of smaller cubes, with each dimension of 
the large cube being equivalent to n of the smaller cubes. We paint 
the large cube, and then we want to know how many of the small cubes 
have paint on 0, 1, 2, 3, 4, 5, or 6 faces.

To do this, you want to visualize the large cube and determine where 
the small cubes are that have 0 faces painted, where the small cubes 
are that have 1 face painted, and so on.

Let's try to build equations (functions of n) that describe the 
numbers of small cubes with different numbers of faces painted. In the 
discussion that follows, I will use f0(n) to denote the function of n 
defining the number of small cubes with 0 faces painted, f1(n) to 
denote the function of n defining the number of small cubes with 1 
face painted, ..., and f6(n) be the function of n defining the number 
of small cubes with 6 faces painted.

If the "large" cube is a single small cube, then there is one small 
cube, and it is painted on all 6 faces. That is a special case, since 
in all the larger cubes there are no small cubes with 6 faces painted.  
In fact, in all the larger cubes there are no small cubes with more 
than three painted faces.

So we have (for n > 1)

    f6(n) = 0
    f5(n) = 0
    f4(n) = 0

Now let's visualize the picture to determine f3(n), f2(n), f1(n), and 
f0(n).

f3(n)...

The small cubes that have three faces painted are on the corners of 
the cube. How many corners are there on an n by n by n cube?  The 
answer will give you f3(n).

f2(n)...

The small cubes that have two faces painted are on the edges of the 
cube - except for the small corner cubes, which have three faces 
painted. How many edges does a cube have? And on an n by n by n cube, 
how many small cubes are there on each of those edges, not counting 
the small corner cubes? So how many small edge cubes (not counting 
corner cubes) does that make on all the edges together? The answer is 
f2(n).

f1(n)...

The small cubes that have one face painted are on the faces of the 
cube - except for the small edge cubes, which are painted on two 
faces. How many faces does a cube have? And on an n by n by n cube, 
how many small cubes are there on each of those faces, not counting 
the edge cubes? So how many small "face" cubes (not counting edge 
cubes) does that make on all the faces together?  The answer is f1(n).

f0(n)...

The small cubes that have no faces painted are on the interior of the 
large cube; these small cubes with no painted faces together form a 
smaller cube inside the large cube. If the large cube is n by n by n, 
what are the dimensions of the cube inside the large cube formed by 
all of the small cubes with no painted faces? So how many small cubes 
are there in that "interior" cube? The answer is f0(n).

Now, all together, f0(n), f1(n), ..., and f6(n) must account for all 
of the small cubes in the large cube. So you can perform a confidence 
check of your answers for f0(n), f1(n), f2(n), and f3(n) by seeing if  
together they add to the total number of small cubes in the large 
cube, which is n^3. If they do, then you probably did the reasoning 
correctly and have found the correct formulas for f0(n), f1(n), f2(n), 
and f3(n). If they don't, go back and check your work - at least one 
of the four formulas you found for f0(n), f1(n), f2(n), and f3(n) is 
incorrect.

See what you can do with this problem on your own now.  Write back if 
you would like more help with this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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