1. ## Help with Structures

Hey. I’m having trouble with this work I’ve been set about structures. I understand some of it, which I’ll explain later, but I really need your help with it!

My task is to investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes. For example, 2 by 2 by 2 cube, 3 by 3 by 3 cube etc etc etc…

I desperately need help on getting a formula to work this out. I know already, as my teacher gave us these results, that a 1 by 1 by 1 cube has the following:

8 Three-Joints ( as do all cubes)
0 Four Joints
0 Five Joints
0 Six Joints

and a 2 by 2 by 2 has:

8 three joints
12 four joints
2 five joints
1 six joints

I really need a formula in order to calculate how many joints a 3 by 3 by 3 has, and with that formula to use it on lets say a 55 by 55 by 55 cube.

All help would be appreciated!

2. Sorry, I was vague and confusing, here's the task given:
---
Rigid Structures are constructed using unit rods.

An example is shown below:

* Pic opf a 2 by 2 by 2 cube*

The individual rods in the structures are held together using different types of joints. (Just think of making cubes using matchsticks and playdough*.

A three joint would be on the vertices of each cube no matter what the size and so therefore there'll be 8 three-joints.
--

3. Hello, carlito1!

This is a killer of a problem . . .

1-by-1-by-1 has:
8 three-joints, 0 four-joints, 0 five-joints, 0 six-joints . . . I agree!
But a 2-by-2 by-2 has:
8 three-joints, 12 four-joints, 6 five-joints, 1 six-joints

As you pointed out, every cube has 8 vertices . . . hence, eight 3-joints.

A 4-joint occurs on each of the twelve edges of the cube.

A 5-joint occurs in the center of each of the six faces of the cube.

And there is a 6-joint in the very center of the cube.

Note that: $8 + 12 + 6 + 1 \,= \,27 \,=\,3^3$
The total number of joints is a cube, namely: $(n+1)^3$

I just counted up a 3-by-3-by-3:
8 three-joints, 24 four-joints, 24 five-joints, 8 six-joints

Of course, the cube has 8 verticies = 8 three-joints.
There are 2 four-joints on each of the 12 edges: 24 four-joints.
There are 4 five-joints on each of the 6 faces: 24 five-joints.
There are 8 six-joints inside the cube.
. . [Think of the six-joints forming a 2-by-2-by-2 cube.]
And: $8 + 24 + 24 + 8 \,=\,64\,=\,4^3$

I think I've got the pattern for a cube of side $n$.

There will alway be 8 vertices: $8\text{ three-joints}$

There will be $n-1$ four-joints on each of the 12 edges: $12(n-1)\text{ four-joints}$

There will be $(n-1)^2$ five-joints on each of the 6 faces: $6(n-1)^2\text{ five-joints}$

There will be $(n-1)^3$ six-joints inside the cube: $(n-1)^3\text{ six-joints}$

And this checks out:
$8 + 12(n-1) + 6(n-1)^2 + (n-1)^3 \;=$ $n^3 + 3n^2 + 3n + 1 \;= \;(n+1)^3$

4. Thanks for your help. Can you help me count up 4 by 4 by 4 up to 12 by 12 by 12 ***pretty please*** and show me how many joints are in there like you previously have done.

Also:

What does this equation denote?

5. Hello, carlito1!

Can you help me count up 4 by 4 by 4 up to 12 by 12 by 12 ?
I gave you the formulas . . . just plug in $n = 4,5,6,...,12$

An $n$-cube has:

Three-joints: $8$
Four-joints: $12(n-1)$
Five-joints: $6(n-1)^2$
Six-joints: $(n-1)^3$

Also: $8 + 12(n-1) + 6(n-1)^2 + (n-1)^3\;=\;(n+1)^3$
What does this equation denote?
It denotes the total number of joints in an n-cube.
It's a good way to check your arithmetic.

For example: a $7$-cube has

. . . . . . . . $8$ three-joints
$12(6) = \;72$ four-joints
$6(6^2) = 216$ five-joints
. . $6^3 = 216$ six-joints

And: $8 + 72 + 216 + 216 \:=\:512\:=\:8^3$

6. Thanks so much for your help. I had problems with it and didn't understand it that fully until you explained. I'm very greatful!