Painted Cube Investigation

Investigation Rules

In this investigation I had to investigate the number of faces on a cube, which had had its outer surface painted red. I had to answer the question, 'How many of the small cubes will have no red faces, one red face, two red faces, and three faces?' From this, I hope to find a formula to work out the number of different faces on a cube sized 'n x n x n'.

The Rules and Solving the Problem

To solve this problem, I built different sized cubes (2 x 2 x 2, 3 x 3 x 3, 4 x 4 x 4) and made tables to help find the patterns.

The diagrams above justify the answers for 2 x 2 x 2, 3 x 3 x 3 and 4 x 4 x 4 cubes.

Check

To check if my formulas were correct, I drew a 5 x 5 x 5 cube and checked if my predictions were correct.

The table proves that the formulas are correct. All my formulae added together should be equal to the total number of small cubes in the large cube which is n3 . Here are the added formulae: n3 = (n – 2)3 + 6 (n – ...

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The diagrams above justify the answers for 2 x 2 x 2, 3 x 3 x 3 and 4 x 4 x 4 cubes.

Check

To check if my formulas were correct, I drew a 5 x 5 x 5 cube and checked if my predictions were correct.

(n - 2)(n - 2)(n - 2) + 6(n - 2)(n - 2) + 12(n - 2) + 8
(n2 - 4n + 4)(n - 2) + 6(n2 - 4n + 4) + 12n - 24 + 8
n3 - 4n2+ 4n - 2n2 + 8n - 8 + 6n2 - 24n + 24 +12n - 24 + 8
n3 + 6n2 - 6n2 + 24n - 24n + 32 - 32
n3

Exceptions

None of the formulae will work for the cube 1 x 1 x 1: This pattern does not work because there is more than one vertex on one cube and therefore there is not going to be 8 cubes with 3 painted sides.

Explaining the Rules

The formula for 0 painted faces is (n – 2)3 . It is n – 2 because the top and bottom ‘lines’ of a cube can’t have a cube that has no faces painted. The answer has to be cubed so that it can be multiplied by the length, height and width.

The formula for 1 painted face is 6(n – 2)2 . It is n – 2 because the top and bottom ‘lines’ of a cube can’t have a cube that has 1 face painted. The answer is squared so that the number of cubes with 1 face painted PER FACE is found. The answer is multiplied by 6 because there are 6 faces in a cube.

The formula for 2 painted faces is 12 (n – 2). The diagram shows us that

it is n – 2 because there are 2 corner cubes per each half of a face.

These 2 cubes cannot have 2 painted faces. The answer to n – 2 has to be

multiplied by 2 to get the corner cubes for the whole face and then multiplied by 6 as there are 6 faces in a cube. 6 x 2 = 12

The formula for 3 painted faces is n = 8 (unless it is a 1 x 1 x 1 cube). Only, a corner cube can have 3 of its faces painted. There are 4 corner cubes in half a cube. So when 4 is multiplied by 2 to get the answer for the number of corner cubes in the whole cube, it is 8. (The corner cubes are shown in white).

It is impossible for any cube to have 4 or 5 painted faces and for any cube bigger than a 1 x 1 x 1, it is impossible to have a cube with 6 painted faces.

Extending the Investigation

I extended the investigation to cuboids. I used measurements 2 x 3 x 4, 2 x 3 x 5 and 3 x 4 x 5.

Exceptions to the rule are cuboids with one or two dimensions of 1.

Check

To check I used a 3 x 4 x 6 cuboid.