# User:Frivolous Consultant/sandbox

Comment: If this is not a neologism it is original research, which is not allowed on Wikipedia. Any subject on Wikipedia needs to be sourced to secondary coverage. If the author doesn't have a proper title for this article it is probably because there is no secondary coverage/independent analysis. Sionk (talk) 11:37, 8 February 2013 (UTC)

# Tessellation conglomerate

(open to a better name)

A tessellation conglomerate is mainly a mathematical puzzle that is a tessellation of one shape that forms another, specific shape. A person is expected to find the maximum number of the specific shapes within.

## Squares in a square

(the picture and some of this section is from "Square pyramidal number")

A 5 by 5 square grid, with three of its 55 squares highlighted.

This most common tessellation conglomerate involves finding the total number of squares in a square grid. An example of a 5 by 5 square can be used to visualize this. The first thing to be done is to figure out the number of the easily seen small squares. 25 of them would be found. To continue, the counting would need to be made to include the 2 by 2 squares in addition to the 1 by 1 squares. This also applies to the 3 by 3 squares and all increasing dimensions until the conclusion of 5 by 5 is reached. Since the squares don't have to be the same size, it is more difficult to determine the amount of squares. A way to figure this out is by substituting n for one of the sides of the square.

$n^2 + (n-1)^2 + (n-2)^2 + (n-3)^2 + \ldots + 1^2 = \frac{n(n+1)(2n+1)}{6} = \sum_{k=1}^nk^2$ [1]

That is, the solution to the puzzle is given by the square pyramidal numbers.

(there was a part in Square pyramidal numbers about finding the number of rectangles, but I looked into it and don't think it works for this)

### Finding squares in rectangles

It is only a matter of time until this same sort of challenge comes up but utilizing a rectangle. When this happens, the following formula can be used to overcome this problem. As would be expected, l is the length and w is the width, the width, for these purposes, being the shortest edge.

$\sum_{k=0}^{w-1}(l-k)(w-k)$

This can be thought of as multiplying the length and the width then adding the product of one less of each amount, continuing until there are no squares left on one side. For these purposes, the width is the shortest edge.

examples: 4×3 rectangle

4×3 + 3×2 + 2×1....................or....................(4-0)(3-0) + (4-1)(3-1) + (4-2)(3-2) = 21

12+6+2

21

### Non-rectangular conglomerates

Some configurations of squares could be in a shape other than a polygon. They could be any shape with connecting squares, even with holes on the inside. One way to account for this is to find all the squares in one, full rectangular area then the other one that uses some of the same squares, and subtract the amount of area and their squares where the rectangles overlap. This method, however, does not work in all cases. When it doesn't work, this formula can be used:

$\sum_{k=1}^{n-1}k(n-k)$ [2] (The formula isn't technically in there, it's a condensed version, and I cut off the zeros because they're useless in this case.)

In this case, n is the number of squares going across a rectangle or square's side that is up against another rectangle or square. For this situation, the number of squares in each joining rectangle is determined without leaving any spaces and with no overlap. This amount is added to the number of squares from using the formula.

#### Imperfect conditions

There are times when the formula for irregular shapes doesn't work. It only works when the area where the squares come together is at least one square longer than either side of the rectangle or squares. Any shorter and problems are run into. To compensate for this, the following formula can be used:

$n(n+1)/2$ [3] or $\sum_{k=1}^nk$

For this, n is the number of rows of squares missing in the arrangement from a particular side of where the squares come together. The answer reached by this is meant to be subtracted from the answer of the preceding formula. In situations where the number of rows missing is drastic, this formula can be rendered invalid.

## Cubes in a cube

Adding a third dimension further complicates the task of finding all the cubes. The number of cubes in a perfect cube can be determined using the formula $[n(n+2)/2]^2.$ [4]

### Cubes in rectangular prisms

Just like with squares, a perfect cube isn't always guaranteed. To account for this problem, a second formula is used with the width still being the shortest side.

$\sum_{k=0}^{w-1}(l-k)(w-k)(h-k)$

examples: 5 by 3 by 4 rectangular prism

5×3×4 + 4×2×3 + 3×1×2..............or..............(5-0)(3-0)(4-0) + (5-1)(3-1)(4-1) + (5-2)(3-2)(4-2) = 90

60+24+6

90

### Irregular shapes

The cubes in this type of tessellation conglomerate also can be in a shape other than a rectangular prism. A slice of where rectangular prisms come together is used just like where the squares come together in the section above. As long as the slice is a perfect square, this can be used:

$\sum_{k=1}^{n-1}k^2(n-k)$

Just as before, n is one of the sides of the slice which, since it's a perfect square, is both of it's sides. Added to the amount of cubes in the adjacent shapes, the total number of cubes can be found.

#### Formula exceptions

This employs the same idea as finding the number of squares in a square. For the number of square slices missing, they can be thought of as one of the sides of a tessellation conglomerate square with that amount being subtracted from the total amount. When the slice of squares doesn't end up being in a square, the same can be applied, but the square's formula with the summation $\sum$ needs to be used instead. The minimum number of cubes needed on the side is determined slightly differently from the squares; it's one less than the shortest side of the joining slice.

## Triangles in a triangle

When triangles are tessellated, the number of the small triangles are harder to determine. The number of component triangles are $n^2$ [5] (This part isn't as relevant as the rest, but just as unknown. I won't be too upset if you think this part should go.)

Triangle tessellation conglomerates have special circumstances. Since a triangle can't have the number of triangles increased on one side without increasing the same amount on the other two sides, the formula is more simplified. Triangles have two formulas based on if the number of triangles on one side is even or odd.

$evens: n(n+2)(2n+1)/8$ [6] $odds: [n(n+2)(2n+1)-1]/8$

As long as it's in the shape of a triangle, these formulas work for any size.

## Hexagons in a hexagon

When in a perfect hexagon, a simple formula can be used to determine the number of component hexagons. The variable, in this case, refers to one side of the hexagon.

$n^3-(n-1)^3$ [7] (I don't get this because, although my formula works and the numbers are all at the reference, their formula is wrong. Or am I just looking at it the wrong way?) (Also, the same as Triangles in a triangle)

Although hexagons can't form as smooth of figures, it is still possible to find the full amount hexagons. For a perfect hexagon, the formula is the cube of one side or: $n^3$ [8]