Hello rainer

I'm not sure that I agree with your working so far - that is, if I'm understanding your requirements.

Take a look at the diagram I've attached. Originally Posted by

**rainer** Every corner of the cube must have a line connecting it to every other corner. I am interested in the sum of the length of these lines.

In one cube the sum, if I'm not mistaken, is:

$\displaystyle 12+12\sqrt{2}+4\sqrt{3}$

I agree with this. This sum can be thought of as:(1) The sum of the lengths of the lines lying in the plane $\displaystyle A_1B_1C_1D_1\;(=4+2\sqrt2)$; PLUS

The additional lengths created when the plane $\displaystyle A_2B_2C_2D_2$ is added; which in turn is comprised of: (2) The sum of the lengths of the lines in the plane $\displaystyle A_2B_2C_2D_2\;(=4+2\sqrt2)$ (the same as (1)); PLUS

(3) The sum of the distances of each of the points in plane #1 from each of the points in plane #2.

This last sum, (3), can be found by taking the sum of the distances of one of the points in plane #1 - say, $\displaystyle A_1$ - from each of the points in plane #2, and multiplying by 4 (since there are four points in plane #1, each indistinguishable from each other). This sum (3), then, is $\displaystyle 4(1+2\sqrt2+\sqrt3)$.

Thus, when we add the totals (1), (2) and (3) together, we get:$\displaystyle \underbrace{4+2\sqrt2}_{\color{blue}\text{(1)}}+\u nderbrace{4+2\sqrt2}_{\color{blue}\text{(2)}}+\und erbrace{4(1+2\sqrt2+\sqrt3)}_{\color{blue}\text{(3 )}}=12+12\sqrt2+4\sqrt3$

Now, when I place another cube right smack alongside the first cube, there are now 12 planets connected by lines of force.

The sum of these lines, if I'm not mistaken, is:

$\displaystyle 12+12\sqrt{2}+4\sqrt{3}+8+10\sqrt{2}+4\sqrt{3}+8\s qrt{5}+4\sqrt{6}$

If I understand you correctly, I think you are mistaken here.

When we add four more 'planets' in plane #3, we get, in addition to the distances above:(4) The sum of the lengths of the lines in plane #4; this is, of course, the same as (1) and (2); namely $\displaystyle 4+ 2\sqrt2$. PLUS

The sum of the all the distances of $\displaystyle A_1, B_1, ..., C_2, D_2$ from each of the points $\displaystyle A_3, ..., D_3$; which comprises:(5) the distances of the points in plane #2, which is the same as (3) above; namely $\displaystyle 4(1+2\sqrt2+\sqrt3)$; and

(6) the distances of the points in plane #1, which is (taking the distances from $\displaystyle A_1$ and multiplying by 4) $\displaystyle 4(2+2\sqrt5+\sqrt6)$.

So I reckon that the total for the 12 planets is:$\displaystyle \underbrace{4+2\sqrt2}_{\color{blue}\text{(1)}}+\u nderbrace{4+2\sqrt2}_{\color{blue}\text{(2)}}+\und erbrace{4(1+2\sqrt2+\sqrt3)}_{\color{blue}\text{(3 )}}+\underbrace{4+2\sqrt2}_{\color{blue}\text{(4)} }+\underbrace{4(1+2\sqrt2+\sqrt3)}_{\color{blue}\t ext{(5)}}$ $\displaystyle +\underbrace{4(2+2\sqrt5+\sqrt6)}_{\color{blue}\te xt{(6)}}$$\displaystyle =12+2\sqrt2+4\sqrt3+\color{red}16\color{black}+10\ sqrt2+4\sqrt3+8\sqrt5+4\sqrt6$

To see the pattern being created here, break the expression above into three components:(1) for plane #1

(2) + (3) for plane #2

(4) + (5) + (6) for plane #3

and notice that (1), (2) and (4) are the same; (3) and (5) are the same; (6) is the new expression formed from the distances of the points in the first plane from the last. So we can re-arrange these as: $\displaystyle 3\times$ (1) + $\displaystyle 2\times$ (3) + $\displaystyle 1\times$ (6)

i.e.$\displaystyle 3\times(4+2\sqrt2) + 2\times\Big(4(1+2\sqrt2+\sqrt3)\Big)+1\times\Big(4 (2+2\sqrt5+\sqrt6)\Big)$

So when the next new plane is added comprising points $\displaystyle A_4, ..., D_4$, there will be additional totals equal to:(4) + (5) + (6) for the points in planes #2, #3 and #4; PLUS

(7) the distances from plane #1 to plane #4; namely $\displaystyle 4(3+2\sqrt{10}+\sqrt{11})$.

So this, when re-arranged, is equivalent to:$\displaystyle 4\times$ (1) + $\displaystyle 3\times$ (3) + $\displaystyle 2\times$ (6) + $\displaystyle 1\times$ (7)

$\displaystyle =4\times(4+2\sqrt2) + 3\times\Big(4(1+2\sqrt2+\sqrt3)\Big)$ $\displaystyle +2\times\Big(4(2+2\sqrt5+\sqrt6)\Big)+1\times\Big( 4(3+2\sqrt{10}+\sqrt{11})\Big)$

... and so on. I hope you can see the pattern emerging. I realise that I'm still quite a way short of developing a formula for the total with $\displaystyle n$ planes, but at least you should be able to see how to continue - if you still want to, that is!

Lastly:
AND: In this scheme of things, the cubes line up in a straight line. Would the sum of lengths be the same if the cubes were arranged differently (but always so that the xth cube is added in such a way that four of its corners line up exactly with the four corners of one of the other cubes)?

Thanks

No, I'm sure that the sum *won't* be the same if the cubes are not arranged in a straight line.

Grandad