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Geometry Project of the Month

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A cube with sides n units long is painted on all faces. It is then cut into cubes with sides 1 unit long.

Explain how many of these smaller cubes will have paint on:

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Annie says:

This month you guys made me work pretty hard! We got 64 submissions, 52 of which might be considered correct to a degree - most had the general forms for the equations right on.

Here's what I looked for: first, the general form of the equation. As I said, most folks got this okay. Second, some explanation of where those equations came from, preferably in relation to the cube itself, not just in relation to a pattern of numbers. Third, an example - how do the equations work? Thirty-two teams got at least 2 points from this process, and 11 got 3 or more.

Here's where the "more" came from: several folks, including the winners, explained that there are restrictions on n, and also provided some sort of checking mechanism to make sure their answers were correct.

I chose two solutions as the winners, as they fulfilled all of the criteria. Of the other nine solutions that scored a 3, I felt that three of them were more clear and thorough than the others, so we have three honorable mentions. I was torn for a while because so many people/teams got the solution correct, but there really were some that were better than others. However, this is by far the greatest number of correct solutions we've ever gotten for a Project of the Month, so kudos to all those who participated!

This problem is a good example of something for which is not hard to get the "right" answer, but it is hard to explain it well, and that's where the differences lay.

Here are the honorees this month:


November Project of the Month

Winners:

Honorable Mention:

The winners receive certificates and totally cool Geometry Forum t-shirts which say on the back "Winners, Project of the Month, 1995-96." Honorable mention recipients get certificates and some pretty cool keychains. Well worth the effort in either case.

All of the solutions, along with my comments, follow below. - Annie


Winners

From: highsted@rmplc.co.uk

From: Martina Smith
Grade: Year 10 (14 years old)
School: Highsted School Sittingbourne Kent UK

3 surfaces: 8
2 surfaces: 12(n-2)
1 surface: 6(n-2)^2
0 surfaces: (n-2)^3
so long as n is greater than 1.

because...

n       0 faces         1 face          2 faces         3 faces
2       0               0               0               8
3       1 (1^3)         6 (6x1)         12 (12x1)       8
4       8 (2^3)         24 (6x4, 6x2^2) 24 (12x2)       8
5       27 (3^3)        54 (6x9, 6x3^2) 36 (12x3)       8
This works because every cube has 8vertices/ corners 12 edges but the two cubes at the end are corners so 12x(n-2) 6 faces with a square of faces in the centre but the cubes at the edge are counted as edge so 6x(n-2)^2 a smaller cube in the middle which can not be painted which will be (n-2)^3.

CHECK

Does (n-2)^3+6(n-2)^2+12(n-2)+8=n^3?
(n-2)^3=(n-2)(n-2)(n-2)
       =(n-2)(n^2 -4n +4)
       =n^3 -4n^2+4n-2n^2+8n-8
       =n^3 -6n^2 +12n -8
6(n-2)^2=6(n^2-4n+4)
        = 6n^2 -24 n + 24
12(n-2) = 12n-24


n^3-6n^2+12n-8+6n^2-24n+24+12n-24+8= 

n^3   -6n^2+6n^2   +12n-24n+12n   -8 +8
=n^3

Comments:

Martina's solution is very clear and concise. She writes out the equations first thing, explains the restriction on n, and gives some examples and an explanation. She then checks her solution - not just for one cube, but in general. This is an excellent way of making sure everything works out! I really like this solution, as not only is it right, but it is very easy to follow.

Winners

From: p1g1

Below, please find our solution to the problem of the month. This team was made up of "Pistol" Pete Bodnar, "Andromeda" Andrea Bytnar, and "Wild" Bill Norberg. We are in Honors Geometry in Shaler Area High School.

Problem: A cube with sides n units long is painted on all faces. It is then cut into cubes with sides 1 unit long. Explain how many of these smaller cubes will have paint on:

a. 3 surfaces; b. 2 surfaces; c. 1 surface; and d. no surfaces.

Solution: Our group discovered two different methods to solve the Project of the Month, November 1995.

Procedure 1: The first formula we originated is:

[(# of sides of a cube)*(# of times the desired smaller cube is displayed on one face of the whole cube)] / (# of sides painted on desired cube).

Explanation: Once the number of times the smaller cubes (with 3, 2, or 1 painted sides) appears on one face of the original cube is found, we may now calculate the exact number of cubes for each.

When using an example with dimensions of 3*3*3, the formulas will be:

        a. For small cubes with paint on 3 sides:
             (6*4)/3 = 8 cubes

        b. For small cubes with paint on 2 sides:
             (6*4)/2 = 12 cubes

        c. For small cubes with paint on 1 side:
             (6*1)/1 = 6 cubes

        d. For small cubes with paint on no sides:
           Since the total number of 1*1*1 that can fit into the original cube
           is 27,
       and the total number of smaller cubes we have so far is 26 (a+b+c), we
       can see that 27-26 = 1 cube
There can only be 1 cube with paint on no sides.

Figure 1


Procedure #2: Using inductive reasoning, we can form generalizations about the smaller cubes. We know that the only smaller cubes that will generate cubes with paint on three sides are the corners. Since we know that there are always eight corners on a cube, there will always be eight cubes that generate cubes with three painted sides.

We also know that the smaller cubes along the edges between the corners will produce the cubes with paint on two sides. We can then formulate a general equation to find the exact number of cubes with paint on two sides:

[(Unit of Cube)-2]* # of edges on a cube= # of small cubes.

We subtract two because we can no longer include the corners. We only want the cubes in between them. Since there are two corners per edge, we subtract two. Of course, there are twelve edges on any cube, and the unit of the cube for this assignment is n.

Furthermore, we know that the cubes left on each surface of the original cube will generate the smaller cubes with paint on only one side. We thus form the equation:

[(Unit of Cube)- 2]^2* # of faces on a cube = # of smaller cubes.

Once again, we subtract two for the rows in which the corners are contained because we can no longer include them. Since the figure we desire is two dimensional, we square it.

Of course, the number of faces on a cube is six, and the unit of the cube is n.

Moreover, we can also calculate the number of cubes with no sides painted with an equation. We know that these cubes will be the remaining internal cubes after the external shell is husked. This equation is:

[(Unit of Cube)-2]^3= # of smaller cubes.

We cube this number because the figure we desire is a three-dimensional one. We subtract two for the same reasoning again, and the unit of the cube is n.

It is worthwhile to note that n cannot equal 1 and n must be greater than or equal to 2.

Let us now look at an example:

        Using 5*5*5:
        a. For small cubes with paint on 3 sides:
           Always 8 cubes.

        b. For small cubes with paint on 2 sides:
        (n-2)*12 = (5-2)*12 =  3*12 = 36 cubes.

        c. For small cubes with paint on 1 side:
        (n-2)^2*6 = (5-2)2* 6 = 32*6 = 9*6 = 54 cubes.

        d. For small cubes with paint on no sides:
        (n-2)^3 = 3^3 = 27 cubes.
It is also worthwhile to note that we may check this process. Since the measurements of the example are 5*5*5, the total number of 1*1*1 cubes is 125.
So:
   (# of cubes w/ 3 painted sides) + (# of cubes w/ 2 painted sides) +
   (# of cubes w/ 1 painted side) + (# of cubes w/ no painted sides)
should equal 125.  And.......

   (8) + (36) + (54) + (27) = 125

It does equal 125.  The equations for Procedure #2 are then:

   b. (n-2) * 12
   c. (n-2)2 * 6
   d. (n-2)3

   (Remember a is always 8)
Figures 2 and 3

Therefore these numbers explain for themselves how many of these cubes would have their respective numbers of painted sides.

Comments:

These folks provided the general solutions, an explanation, the restrictions on n, an example, and a check. The solution is a little hard to follow, though. There is a lot of writing that isn't necessary, and it would be nice to see the general equations clearly laid out somewhere. And though they found "two" ways of getting the answers, I would argue that the second one is much easier to use, and I would have left the first one out. But they covered it all, and were fairly thorough.

Honorable Mention

From: Ruth Carver

Tara Shephard, Katrina Myers, Courtney Piper
Mount St. Joseph Acdemy- Grade 10
Problem of the Month

Starting off, we drew cubes with dimensions of 3x3x3 units, 4x4x4 units, 5x5x5 units, and 6x6x6 units. We figured out that the area of a cube is equal to n cubed . We looked at each cube and figured out the numbers of smaller cubes that would have paint on 3, 2, 1, and 0 sides for each different volume. We had to do this in our heads. We tried to look for a pattern in the answers we came up with, but found none. The only thing we did discover is that the number of smaller circles painted on 3 sides would always be 8 because that are found at the corners of the larger cube.

A cube always has 8 corners, no matter what its dimensions. When we couldn't find a pattern for the other numbers, we decided to look at the way. We could find each answer. Knowing that a cube with paint on 3 sides would be on a corner, we realized that a cube with paint on 2 sides would lie on an intersection of 2 faces of the larger cube.

We counted the edges of the cube and knew we would have to multiply by 12 (edges). Looking at each individual edge, we noticed that 2 of the cubes on it, the corners, were already accounted for. therefor we let "n" equal the length (height and width) of the side and subtracted by 2. We came up with the equation 12(n-2). When we plugged in the dimensions of our 4 cubes, the answers matched the ones we had found! Therefore the # of cubes with paint on 2 faces always equal 12(n-2). {Ex. V=5x5x5 with 2 faces painted =12(5-2)=12(3)=36 cubes}

With this information, finding the rest was fairly easy. Cubes with 1 face painted would lie in the center of each face in the larger cube. We realized that the perimeter of the face changed because the cubes on the edges had been used already. Only the center was left. If the # of small cubes going across is x by the # of small cubes going down, the total # of cubes with paitn on 1 face is found. Another way of saying this is (n-2) squared. Then we multiplied the equation by 6 to include each face giving us 6(n-2)^2. {Ex. V= 5x5x5 -with 1 face painted=6(5-2)^2=6(3)^2=6(9)=54}

Again, the answers we got matched the ones from earlier. Finding the number of faces with paint on 0 faces was the easiest. Since all the cubes on the outside were covered, we imagined that they were lifted off and we were left with a cube with new dimensions and therefore a new volume. The new dimensions would be (n-2)(n-2)(n-2) or (n-2)^3. Since none of the smaller cubes on this cube are "within" the orginal one. We knew this was the answer. {Ex. V=5x5x5 with 0 faces are painted=(5-2)cubed= (3)cubed= 27.} All of the equations we had came up with worked out to the right answers. Finally, we looked for a pattern with in the equations. 3 faces was a constant 8 because a cube always has 8 corners. 2 faces is dealing with the perimeter of each individual face together, therefor, we simply found the answer by finding the lengthof each face and the formula for the area fo a square is n squared , so our equation applies. 0 faces entailed volume, and the area of a cubeis n cubed, so again my equation applies. We had fun when doing this project, and we were even more excited when we came to an answer.

Comments:

This group provided all the "necessary" elements. They also provided a nice narrative of how they figured out the answer. It is often helpful to read about how other people approach problems, including the problems they ran into and how they eventually worked their way out. I also like that they compared the answers that they found each way.

This solution is pretty hard to read, though. More paragraphs would have made a big difference, and the equations should have been stated somewhere clearly so that the reader knows where things are headed. You really have to read the whole solution to figure out what's going on. I also would have liked to see the table of values they worked out for the first part (where they couldn't find an obvious solution).

This was one step below the winners in that they didn't check the answers, either as a whole or totally up one of the examples. They also didn't say that n _couldn't_ be certain things. Small details, but important in a complete solution.

Honorable Mention

From: patience@muskox.alaska.edu

From: Chris Klint
Grade: 10
School: Juneau-Douglas High School

Cubes with 3 painted sides: There will be only 8 cubes with three sides painted because there are only 8 locations (the cube's vertices) where three sides meet.

Cubes with 2 painted sides: The equation 12(n-2) will solve for this value, representative of the 12 cube edges times the n side length, minus 2 to represent the corner cubes found at either end of any given edge.

Cubes with 1 painted side: The equation 6(n-2)squared will solve for this value, representative of the 6 sides for which the equation is repeated times the n side length minus 2 (representative of the four 1-unit edge cube rows on the edges of a side, paired on opposite edges) squared to represent a side with a square formed one unit inside the edge and containing all cubes with one painted side, as opposed to the edges which have two or three sides painted.

Cubes with no painted sides: The equation (n-2)cubed will solve for this, representative of a cube formed within the actual cube with a one-unit layer on the surface containing all painted cubes. The remaining cubes in the interior of the cube (and thus unpainted) are contained in this equation.

I thought I'd put these equations in practice here with cubes of various sizes:

Example 1: n=4

The first answer is obvious: only the 8 corners have 3 sides painted.

The equation 12(4-2) yields the number 24, a number confirmed by drawing a cube and counting. On each of the 12 edges, the 2 center cubes on the 4-unit edges have paint on 2 sides.

The equation 6(4-2)squared yields 24 yet again, with the 4 central cubes on each of the 6 sides being the only cubes painted on only one side.

The equation (4-2)cubed yields the nuber 8, with the few cubes unaccounted for here represented in this equation.

Example 2: n=120

Once again, for all its mass, this cube also is entitled to only 8 cubes with 3 sides painted.

12(120-2) yields the massive figure 1,416, a number I shall not to confirm by drawing and counting. When divided by the 12 edges present, though, this number yields a more modest 118, a number representative of the 120-unit edge minus the 2 corners at either end.

6(120-2)squared yields an even more astounding figure for surface area, 83,544-an astronomical number until you consider that one side of this giant cube has a surface area of 14,400 units. When multiplied by 6, this will come out to 86,400 units, an even greater number-proof that the edges have indeed been subtracted from this figure of the cubes with one side.

(120-2)cubed yields the incredible figure of 1,643,032, a figure made surprisingly believable by the numbers we have previously seen.

These proofs should be sufficient to prove these equations. If you have further questions or other numbers you would like plotted as a cube please contact me. Thanks for your time!

Comments:

Chris gave some really good, clear explanations along with his solution, and included two very good examples - one normal number, meaning one for which you really could just count the cubes, and then one really big number, for which you really wouldn't want to try to count the cubes. This is where having the general solution really comes in handy.

As with the other honorable mentions, no check and no restraints on n were the only real holes.

Honorable Mention

From: Vince Devlin

The following solution is submitted by:
Matt D'Emilio, Grade 10, Haverford High School

I used a table and induction to find a pattern for each part of the problem. My results and the reasons are:

3 surfaces: Always 8 because this will occur at every corner square and there are always 8 coner squares.

2 surfaces: There are always 12(x-2). the (x-2) removes the "outer" cubes and there are 12 locations on the cubes edges, not counting the corners.

1 surface: There are 6(x-2)(x-2). Again the (x-2) removes the "outer" cubes and the two (x-2)'s measures a "square" on each of six faces.

0 surfaces: There are (x-2)(x-2)(x-2). Once again, the (x-2) removes the "outer" cubes and what you have is a cube of squares inside the cube that is 2 cubes smaller in each dimension. Table:

3 surfaces
sides     2     3     4     5     
number    8     8     8     8

2 surfaces
sides     2     3     4     5     
number    0    12    24    36

1 surface
sides     2     3     4     5     
number    0     6    24    54

0 surfaces
sides     2     3     4     5     
number    0     1     8    27

Comments:

Chris does a nice clear job of presenting his solution and gives some good examples. This is another no-frills solution which works well.

Again, no check and no restraints on n. Otherwise right on the money.

- Annie

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23 January 1996