# The Problem

### A solid block of wood in the shape of a right-rectangular
prism measures a inches by b

inches by c inches. The block is dipped into red paint and is
then cut into one-inch cubes.

How many of the one-inch cubes will have no paint on them? Give
your answer in terms

of a, b, and c.

### (Source: Adapted from Mathematics Teaching in the
Middle School, Jan 1998)

# The Solution

### First, we need to consider what the total volume of the rectangular
prism will be, then we can determine how many cubes will not have
paint on them. Since we are given the dimensions of the cube to
be a inches by b inches by c inches, we know that the volume,
or total number of 1 inch cubes contained in the prism is equal
to abc.

### Next, we need to consider what each face of the prism will
look like. The prism will have 6 faces. The faces have the following
dimensions:

### a x b

### a x c

### b x c

### There are two faces for each of the above listed dimensions,
so the surface area of the prism is equal to the following:

### 2ab + 2ac + 2bc

### One might be tempted to think that this would be the number
of cubes that will be painted; however, this is an incorrect assumption.
Since all of the faces share edges, some of the cubes will be
painted on more than one side. The eight cubes located at the
verticies will be pained on three sides, and each cube along an
edge will be painted twice, once for each face it is a part of.

### Thus, we must now determine how many cubes will really be
pained. First we can consider the top and bottom of the prism.
We can use the two faces that measure axb for the top and bottom.
For the top we get the following number of cubes:

### 2a + 2(b - 2)

### The 2a comes from the length of the prism. This figure will
account for all of the cubes along the length of the top, all
four of the top verticies, and the a part of the side faces. We
need to subtract 2 from each width of b to account for the cubes
located at the vertices that we accounted for by counting a. Finally,
we can multiply this figure by 2 to give us the number of cubes
along the edge for the top and the bottom:

### 4a +4(b-2)

### Next, we can determine the number of cubes that will be pained
along the sides. Keep in mind that we already counted all of the
cubes that lie along the length of a, so we only need to account
for the height of the prism, but we must remember not to double
count the vertices. So, we know that we will have 4 edges of height
c to count, less the two points on each end of side c which lie
in the vertices that have been counted, giving the following:

### 4(c-2)

### We can combine these two figures to determine the number of
painted cubes along the edges of the prism:

### 4a +4b +4c -16

### Next, we must determine how many cubes that are not edges
have a face on the surface of the prism. We already stated that
the sides of the prism are ab, ac, and bc. But we took off one
cube on each side to account for the edges. This gives us new
dimensions for each side:

### (a-2)(b-2)

### (a-2)(c-2)

### (b-2)(c-2)

### As before, each of these rectangles appears twice on the prism,
so we have the following number of cubes that are painted only
on one face:

### 2(ab-2a-2b+4)

### 2(ac-2a-2c+4)

### 2(bc-2b-2c+4)

### By comining the above number of cubes we get:

### 2ab+2ac+2bc-4a-4b-4c+24

### Next we want to add this number to the number of cubes we
determined were located along the edges of the prism:

### (2ab+2ac+2bc-4a-4b-4c+24) +(4a+4b+4c-16) = 2ab+2ac+2bc+8

### Now we know how many cubes were indeed painted, so we must
subtract amount from the total number of cubes contained in the
rectangular prism:

### abc-(2ab+2ac+2bc+8)