OtherSticksProblemIdeas

VMT Spring Fest ideas

Other ideas about the sticks problem or new problems

Add your group's ideas here:

• Team A: For now, we have a table of the number of cubes for an increase in N. Later, we are going to add the faces and sticks amount:

N Cubes
1; 1
2; 4
3; 10
4; 20
5; 35
6; 56
7; 84
8; 120
9; 165
10; 220
N; 1/6N^3+1/2N^2+1/3N

We found these values by first finding the number of cubes in each level. It was equivalent to the orignal squares problem, with a pyramid. Each level above it had the same pattern, except N (as defined in the original problem), was 1 smaller. This meant that any size of the 3D pyramid, with maximum height N, maximum width N, and the third dimension of maximum depth N, increasing in one sirection per dimension (analogous to the squares/sticks problem). We found that the bottom (largest) level would have 1/2N^2+1/2N, as we found in the original problem, and we used the first few values as coordinates, assuming that a function related N to the sum of the sequence 1/2N^2+1/2N. Then, we assumed that the equation for the partial sums would a be a degree higher than the starting sequence's degree. With this in mind, our equation would have the form aN^3+bN^2+cN+d, and we plugged in four values of N and the results they output (found by calculating partial sums by hand), coming up with this system: a+b+c+d=1 8a+4b+2c+d=4 27a+9b+3c+d=10 64a+16b+4c+d=20 We solved this system with a computer program (from online) that presumably uses Cramer's rule once for each of the variables.
Then, we moved on to faces as a function of N
What we found was that the faces was quite similar to the sticks-squares original problem. After finding, for N={1,2,3}, the number of faces, we found that it started with 6 faces, then 14, then 24...The sticks, original, was 4, then 10, then 18! The equation was really the same, except that, if you started at 0 for faces and sticks, you first add six for faces, but 4 for sticks, then 8 for faces, 6 for sticks, etc. Thus, since the equation for the sticks was N^2+3N, the linear degree only changes--N^2+5N!
N; Faces
1; 6
2; 14
3; 24
4; 36
5; 50
6; 66

• Team B:

To investigate the number of sticks in a flat faced pyramid with n levels with 1 block increase in length and width per level. Also, to find as many approaches and put them to use. We eventually found 4 different strategies and applied them, such as divide the problem up, finding a basic pattern, and use recursion to solve problems. We also found a formula, its origins, and how to use it.

f(n)=4n(n+1)+(n+1)^2+f(n-1) and f(0)=0.
We first determine the number of squares in each level of the pyramid, 1 cubes in first level, 4 cubes in second level, 9 cubes in third level, and so on. Then we divide each level into 3 parts, the top, the bottom and the middle. The top is the same as the bottom part. They are just a bunch of squares in square format. When divided top or bottom into vertical or horizontal, the equation for # of sticks is n(n+1). Times that by 2 and you get a top or bottom, and times it by 2 again to get the total for top and bottom. The result is 4n(n+1). Then there is the middle which are straight columns, and they are n+1 by n+1 on the side, so (n+1)^2. The equation for each level is 4n(n+1)+(n+1)^2, and use a recursive fuction and you can get the total sticks for the pyramid.

So in session 3, our team tried to understand Team C's formula and how it was derived. We found out we could use triangular numbers because we took the diamonds one stage at a time. We put a big square around each stage. We figured out that if you subtract the stage from the square, you will get a triangular number.

In session 4, we continued our progress on the diamond problem. We found that if we filled up the diamound with more squares and get an easier square with 2n+1 as the dimension. So the number of squares in the big square is (2n+1)^2. We then minus the squares that we added on which was at the 4 corners, which grow in the same pattern as the triangle number in the first session. We used the formula for # of squares from the first session and times it by 4 to calculate the 4 corners that we add on to make the big square. The final formula for the # of squares in the diamond is (2n+1)^2-n(n+1)/2*4. We tested it several times to check if it works.

We then move on to understand Team C's formula for summing up the total # of sticks in n-level diamond. We first tried to used the big square and then minus the extra corners, but the corners turns out to be to hard to calculate. Then we tried to simplify Team C's equation to help as find a lead, but we found out that their stick equation is wrong. We then decide to find out a whole new equation and tried to divide the sticks up into vertical and horizontal groups like we did before with all the other problems. The groups can be further divided into 2 equal parts. We found a pattern.

1st level: 1
2nd level: 1+3
3rd level: 1+3+5
4th level: 1+3+5+7
5th level: 1+3+5+7+9
nth level: (2*n)*n/2

We then found out that each of these can be by calculated by (2*n)*n/2 which simplified into n^2. n^2 can then be multiplied by 4 and get the total of sticks in a nth leveled diamond. The final equation is 4(n^2).

• Team C:

For the original problem:
sides:
N(N+3)
squares:
n(n-1)/2


Explanations for the above formulas can be found in the wiki for our previous session.

We also found formulas for a diamond-like arangement of the squares:
sides:
(n^2+(n-1)^2)*2+n*3-2
squares:
n^2+(n-1)^2

By "sides" we mean the three squares a side of the diamond is comprised of.

We decided that while an explicit formula to calculate the number of squares or sides is clearer for calculating, a recursive formula is easier when one is trying to determine how a particular series or pattern grows.<p>

Next, we did a hexagon made of triangles. n is the side length, again.
The number of sides is:
9n(n+1)-6n<p>

The number of triangles is:
6n^2<p>

Crescent Team 2

Patterns we noticed for cubes: 1.Number of cubes= N+previous number of cubes OR (1+2+3.....+N)

2. Number of sticks=4[ N^2 + (N+1)] E.g: pattern 2- 4(4+3)=28

3. Let number of cubes be Y. Number of sticks= 12Y-4(Y-1) E.g: pattern 2- 12(3)-4(3-1)= 28

4. Number of sticks=8N+previous number of sticks E.g: Pattern 4- 8(84)+52=124

5.number of cubes for : first pattern:1N second pattern:1.5N third pattern:2N fourth pattern:2.5N and so on

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