# Find internal surface area of painted cube

Suppose that a wooden cube, whose edge is $3$ inch, is painted red, then cut into $27$ pieces of $1$ inch edge. Find total surface area of unpainted?

First of all, I have tried to draw the cube using MS Paint, below is given picture:

The surface area of cube is $6\cdot a^2$, where $a$ is the length of cube, when cube is painted, I'm trying to imagine it visually, how much surface would be unpainted?

If it's painted,does not it mean that all pages are painted?

EDIT:

A $3\times3\times3$ cube gets painted red. Then it gets split into $27$, $1\times1\times1$ cubes.

Find the number of painted faces:

On the $3$ inch cube, there are $9$ $1$-inch faces on each face, and there are $6$ such faces, so the total red 1-inch faces is: $$9\cdot 6=54$$ Find the total number of faces of all the cubes:

There are $6$ faces per cube and $27$ cubes, so: $$27\cdot 6=162.$$ Of these, we know $54$ are painted: $$162-54=108$$ There is a total of $108$ unpainted $1\times1$ squares each having an area of 1, so there are $108$ unpainted square inches.

EDIT:

I'd like to post also my approaches if the cube would be divided into two identical parts, actually if we have a cube with length $a$, divided into $2$ parts, then we get two cubes, which volume is $a^3/2$.

So length is the cubic root of $a^3/2$, or length of this two cubes is $a$ divided by cubic root of $2$.

Now we have length $3$, which means that it's length would be $3$ divided by cubic root of $2$, on each face with length $3$, we would have: $$\frac{9}{3/\sqrt[3]{2}}=3\cdot \sqrt[3]{2}$$ in total $4$ such cubes, so $2*4$ of such faces, total face would be $2*6=12$,so unpainted would be $12-8=4$ is it correct?

• I'm sorry but if you consider the shape just being cut into two pieces, you are actually "morphing" the cube in order to obtain two identical cubes. This, in reality is impossible because from a cube two identical sub-cubes can not be generated by one slice. – alandella Aug 10 '13 at 10:34
• so it means that we only can cut cube into several part,if it's volume is divided by number of cut and get cube with exact number of length?? – giorgi Aug 10 '13 at 10:35
• in your case we have divided by $27$,because volume of cube is $27$ – giorgi Aug 10 '13 at 10:38
• Yes, but this works because if you re-attach every sub-cube you get back to the original one, and this works even with random slices, as @ Blue pointed out. But in your case you can not re-attach two identical cubes to form the original cube, because you modified the shape of the slices. In the other case we didn't talked about dividing the volume because this would modify the shape of the sub cubes! – alandella Aug 10 '13 at 10:43

This answer requires some spatial examples, so I think it will be better if I generalize the case of a simple cube, which can be applied to all possible situations.

As you said, the total area covered by the paint in a cube with lenght $a$ is equal to: $$S_{\mathrm{ext}}=6\cdot a^2$$ But then the division into $27$ cubes with lenght $1/3$ of the original gives the new surface area, greater than the original because new slices prodce more internal surface; so now the total surface is expressed by: $$S_{\mathrm{tot}}=27\cdot 6\cdot (a/3)^2=\frac{27\cdot 6}{9}\cdot a^2=3\cdot 6\cdot a^2=3S_{\mathrm{ext}}$$ Now think about a Rubik's Cube, with only the cubes facing the outside are painted: since you know that the total amount of paint occupies the external surface $S_{\mathrm{ext}}$, then to know the surface not covered in paint you have to simply subtract to the total sliced surface $S_{\mathrm{tot}}$ the external one: $$S_{\mathrm{unpaint}}=S_{\mathrm{tot}}-S_{\mathrm{ext}}=3S_{\mathrm{ext}}-S_{\mathrm{ext}}=2S_{\mathrm{ext}}=12\cdot a^2$$ Now you have basically derived the sum of the internal surface of every little cube, because the sum of the external is nothing more than the total covered in paint, or $S_{\mathrm{ext}}$ (with the Rubik's Cube case, the only cubes coloured).

EDIT: If the number of slices is not defined, it is generalized to a single parameter $n\in \mathbb{N}$, then the new surface tends to be greater as $n$ gets larger.

Let's assume that one cut is repeated on every edge of the cube, so if an edge is divided in three (with two slices), then the number of cubes created is expressed as follows: $$N_{\mathrm{cubes}}=(n+1)^3$$ So if I "slice" a cube for every edge once ($n=1$), I get $N_{\mathrm{cubes}}=(2)^3=8$ little cubes.

The surface of each sub-cube is expressed by: $$S_{\mathrm{sub-cb}}=6\cdot\left(\frac{a}{n+1}\right)^2$$ Then the total surface is expressed as th single one multiplied by the total number of sub-cubes created $N_{\mathrm{cubes}}$: $$S_{\mathrm{tot}}=(n+1)^3\cdot 6\cdot\left(\frac{a}{n+1}\right)^2=(n+1)\cdot 6\cdot a^2=(n+1)\cdot S_{\mathrm{ext}}$$ Now, applying the same deduction as before, the surface not covered with paint is given by a simple subtraction between the total surface and the ponly painted one, which is the external one: $$S_{\mathrm{unpaint}}=S_{\mathrm{tot}}-S_{\mathrm{ext}}=(n+1)\cdot S_{\mathrm{ext}}-S_{\mathrm{ext}}=n\cdot S_{\mathrm{ext}}$$ Now remember that the parameter $n$ is the number of slices, and not the sub-cubes created; as your previous example, the slices were $2$, and so this number appeared inside the final result.

• what if figure is not cube?original or new figure,for example cube is cutted into $27$ rectangular solid or something like this – giorgi Aug 10 '13 at 9:06
• Well, if the cuts are not regular, the inital equation is different and so the reasoning behind its derivation. – alandella Aug 10 '13 at 9:09
• or for example instead of $27$,it is divided into half part?forumula would be same yes or – giorgi Aug 10 '13 at 9:10
• @ giorgi If it is divided into $n$ parts (so $2$, $3$ ...), I'd like to explain a more genealized case. Mind if I edit the answer to add this new case? – alandella Aug 10 '13 at 9:11
• ok i would be happy,thanks very much – giorgi Aug 10 '13 at 9:14

Consider cutting-up a rectangular solid ("shoebox"). We'll say that the top and bottom faces (each of area $R$) are painted red, the front and back faces (each of area $G$) are painted green, and the left and right faces (each of area $B$) are painted green.

As we slice the box, it helps to think of the pieces sticking together in place in the shape of the original box, rather than thinking that we suddenly have a lot of mini-boxes.

When you slice the shoebox parallel to a pair of faces, you create two unpainted surfaces (one on each side of cut) in the same shape as those painted faces. Those surfaces might get sub-divided by other cuts in other directions, but the total area of each surface will always match the area of a painted face.

Thus, a cut parallel to faces of a particular color creates two unpainted surfaces, each with the same area as one of those colored faces. Making $r$ cuts parallel to the red faces, $g$ cuts parallel to the green faces, and $b$ cuts parallel to the blue faces, creates a total unpainted surface area of $$\text{total unpainted area} \quad = \quad 2 r R + 2 g G + 2 b B \qquad (1)$$

Note. Formula $(1)$ holds no matter where the cuts are made (so long as they're made parallel to the faces). That is, the spacing between cuts is irrelevant.

In the case of a $3\times 3 \times 3$ cube, the faces have area $$R = G = B = 9$$ Cutting this into $27$ sub-cubes involves making $2$ cuts in each direction: $$r = g = b = 2$$ So, by $(1)$, $$\text{total unpainted area} \quad = \quad 2\cdot 2 \cdot 9 + 2\cdot 2\cdot 9 + 2 \cdot 2 \cdot 9 \quad = \quad 108$$

• Thanks for pointing this out, this not only helped the author but even me. It'll be interesting if the case wether a specific angle of slicing $\theta$ is generalized though. – alandella Aug 10 '13 at 10:29
• i have edited my question and my approaches for case of two cube,please see if it is correct – giorgi Aug 10 '13 at 10:32
• @Blue we can divide cube into $a$ part if volume is divided into $a$ and we get length ,from which i can take cubic root? – giorgi Aug 10 '13 at 10:44
• @giorgi: If you have a perfect clay cube with side-length $a$ (and volume $a^3$) and slice it into $n$ equal-volume pieces, then you can imagine re-shaping the smaller pieces of clay into perfect cubes of volume $a^3/n$ and side-length $a/\sqrt[3]{n}$. That's true. However, that's a very different situation than in the classic painted-cube problem, where $n$ is specifically chosen to have a whole-number cube root so that creating perfect sub-cubes is a simple matter of making evenly-spaced cuts in each direction; no "re-shaping" required. – Blue Aug 10 '13 at 11:16

I think this question is set for testing one's 3D figure visualization and counting ability.

There are two ways to count the total unpainted area

Method_1

*For the corner blocks,

_There are 8 in total, each has 3 faces unpainted (1 top/bottom + 2 internal faces)

_Sub-total unpainted area = (1 x 1) x 3 x 8 = 24 = p

*For the central block,

_There is only 1, it has 6 (all) faces unpainted

_Subtotal unpainted area = (1 x 1) x 6 = 6 = q

*For the centered block on one face,

_There are 6 in total, it has 5 faces unpainted (1 top/bottom + 4 internal faces)

_Subtotal unpainted area = (1 x 1) x 5 x 6 = 30 = r

*For the remaining not-yet counted (but of the same nature),

_There are 12 in total, it has 4 faces unpainted (1 top/bottom + 3 internal faces)

_Subtotal unpainted area = (1 x 1) x 4 x 12 = 48 = s

Total unpainted area = p + q + r + s = 108

Method_2

Total painted area = 3 x 3 x 6 = 54

If the original was cut before it has been painted, total original surface area = (1 x 1) 6 x 27 = 162

Total unpainted area = 162 - 54 = 108