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Triangle Teachers Circle Chameleon Cubes Problem
The Chameleon Cube Problem
and Other Problems with Cubes
HAROLDB.REITER
UNIVERSITY OF NORTH CAROLINA
CHARLOTTE
Triangle Teachers’ Circle
Raleigh, NC
February 4, 2017
math2.uncc.edu/~hbreiter
Write me at hbreiter@uncc.edu for a copy of the 18 page set of problems
related to cubes.
1
Triangle Teachers Circle Chameleon Cubes Problem
1 Introduction
Averypopular contest topic in the United States is problems about arrange-
ments of cubes. Its popularity derives from the lovely mixture of combi-
natorics, spacial geometry, and number theory usually required for solving
such problems. This essay focuses on five problems, the painted cubes prob-
lems, the visible cubes problem, the chameleon cubes problems, the drilled
cube problem, and a problem about faces and vertices of a cubical die.
2 Painted Cubes Problem
Suppose all six faces of an a×b×c block of cubes are painted red. Then one
of the abc unit cubes is randomly selected and tossed like a die. Is it possible
that the probability of a red face showing up is exactly 2/7? This problem
clearly generalizes the one with a = b = c.
Solution: Yes. This solution is due to Anthony Chen, a fourth grader
as this is written. Since every face is equally likely, we are looking for values
of a ≤ b ≤ c such that 2ab+2bc+2ac 2
6abc =7.
But this is equivalent to
1 + 1 + 1 = 6,
a b c 7
which we can solve by inspection in exactly one way, a = 2,b = 3,c = 42.
For the second part, let p denote the probability that a painted face turns
up. Then
2ab+2bc+2ac =p.
6abc
That is ab+bc+ac = 3p. This is equivalent to saying 1 + 1 + 1 = 3p, and in the
abc 1 a b c
case that a = b = c, we have p = a, just as in the previous problem.
For a related problem, suppose all six faces of an a × b × c, a < b < c,
block of cubes are painted red. Then one of the abc unit cubes is randomly
selected and tossed like a die. The probability of a red face showing up is
exactly 2/9. What is the least possible volume of the block?
Solution: Therearelotsofblockswiththisproperty. Afeware(a,b,c) =
(2,7,42),(2,8,24),(2,9,18),(2,10,15). The last one has the least volume.
Without the a < b < c condition, the least volume would be obtained for the
(a,b,c) = (3,3,6) block.
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Triangle Teachers Circle Chameleon Cubes Problem
3 Visible Cubes Problem
Consider a large cube made from n × n × n unit cubes. Look at the large
cube from a corner so that you can see three faces. How many unit cubes
are in your line of vision?
Our objective here is to solve the problem in three different ways. To get
a better feel for the problem, let us build a table that shows, for small values
of n, the number of cubes visible from one corner.
n n3 number visible
1 1 1
2 8 7
3 27 19
4 64 37
Howdoesthetable continue? Make some guesses and then try to prove your
answer. We can use a technique called the method of finite differences. This
method fits a polynomial to a sequence of numbers. Let’s name the number
we’re looking for. Let G(n) denote the number of cubes visible from a corner
of the n × n × n cube. Notice that the sequence of first order differences
G(2)−G(1) = 6;G(3)−G(2) = 19−7 = 12;G(4)−G(3) = 37−19 = 18
has an interesting property. The differences are all multiples of 6. When
we explore such a sequence in which the sequence of successive differences is
eventually constant, we can build a polynomial that produces the sequence.
Since the second order differences are constant, we propose that G(n) is a
quadratic polynomial, G(n) = an2 + bn + c. From the table we can write
G(1) = 1,G(2) = 7 and G(3) = 19. So we can write
2
1 a+1b+c = 1
2
2 a+2b+c = 7
2
3 a+3b+c = 19
We can solve this without great difficulty to get G(n) = 3n2 − 3n + 1.
But what do these coefficients have to do with the problem? One way to see
this is to extend the chart by one more column that shows the cubes that
are not visible.
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Triangle Teachers Circle Chameleon Cubes Problem
n n3 number not visible number visible
1 1 0 1
2 8 1 7
3 27 8 19
4 64 27 37
Now you can see that we can count the number of visible cubes by
3 3
counting the invisible ones first. So, in general, G(n) = n − (n − 1) =
n3 − (n3 − 3n2 + 3n − 1) = 3n2 − 3n + 1. This doesn’t completely answer
the question however. What are the coefficients telling us? How does the
algebra help us reason geometrically? The answer is below.
Our third solution invokes the famous Inclusion/Exclusion Principle. We
can first estimate the number of visible cubes by noticing that we can see
2 2
three faces each with area n . But the number 3n clearly overcounts, because
all the cubes on each of the three visible edges are being counted twice. The
2
number 3n − 3n works fine except that the corner cube, the one nearest
us has been counted 3 times in 3n2 and then removed 3 times in −3n, so it
must be added back into the count. We get 3n2 −3n+1 as our final count.
Finally, we can say what the coefficients mean. The 3 in 3n2 is the number
of visible faces, the 3 in 3n is the number of visible edges, and the 1 is the
single corner cube that belongs to all three visible edges.
4 Chameleon Cubes Problems
The 3×3×3 Chameleon Cubes Problem.
You are given 27 unpainted cubes. Can you paint the faces with three
colors, red, white, and blue, so that when you’re done, you can assemble an
all red 3 × 3 × 3 cube, an all white 3 × 3 × 3 cube and an all blue 3 × 3 × 3
cube?
Solution: Yes, you can do this in essentially just one way.
Here is one solution. Since the total number of faces we can paint is
27 · 6 = 162, and since the all-red cube requires 6 · 9 = 54 red faces as do
the other two colors, we must be perfectly efficient in the following sense.
Each face we paint red must appear on the outside of the red cube. This
implies that there must be exactly one unit cube, the one not visible from the
outside, that has no red faces, and similarly exactly one that has no white
faces, and one that has no blue faces. These cubes must have exactly three
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