[Guest]

(when I use 'n' this is what i'm using)

For the FIRST pyramid: n=9

For the SECOND pyramid: n=10

Each pyramid contains a square base consisting of n^2 blocks.

Remember that depending on whether 'n' is odd or even the pyramids change slightly.

Take each pyramid and coat it in paint. (so that only the outside faces get painted)

Now, disassemble each pyramid and count how many of the blocks have...

NO Faces painted: ?

ONE Face painted: ?

TWO Faces painted: ?

THREE Faces painted: ?

FOUR Faces painted: ?

FIVE Faces painted: ?

and

The TOTAL number of blocks: ?

Do this for each pyramid.

Now determine formulas to give you the number of painted faces and total number of blocks for each pyramid.

- The formula MUST work for ANY square based pyramid with an 'n' of at least 5 (I choose 5 because it's the smallest pyramid with at least 1 block with no painted faces, I'm not sure if the formulas would get wacky below n=5 but this is just in case they do!)

** I'm posting this before I've actually worked out all the formulas how I would like them to be.

** And how I would LIKE for them to be is without using sigma notation (or prove it's impossible to come up with a formula without using sigma notation)

** I would prefer them to be in terms of 'n' or you could use a combination with f0, f1, f2, ... , f5 with f0 being the number of blocks with NO faces painted, f1 being the number of blocks with ONE face painted, etc.

Have Fun!

- K4D

(if you didn't check out my cube version of this you can see it Here!)

[Gmaster479]

Well I can do total # of blocks

1st Pyramid

81+49+25+9+1=165

100+64+36+16+4=220

165+220=385

[bushindo]

I'll tackle the total number of blocks first

First of all, this is a modified version of the sum of square problems. The sum of squared natural numbers up to n, S_n, is

S_n = 1² + 2² + ... + n² = n³/3 + n²/2 + n/6

For n = even, the total number of blocks S_e,n is

S_e,n = 2² + 4² + ... + n²

= 4( 1² + 2² + ... + (n/2) ² )

= 4( S_n/2 )

= 4( n³/24 + n²/8 + n/12 )

For n = odd, the total number of blocks, S_o,n, is quite easy if we use the above two equations

S_o,n = S_n - S_e,n-1

S_o,n = n³/3 + n²/2 + n/6 - 4( (n-1)³/24 + (n-1)²/8 + (n-1)/12 )

Now let's tackle the pyramids,

Lets assume n = even, just so that n/2 is a nice number too.

f5 = 0

f4 = 4

f3 = (n-2)*4 + 4( n/2 -1 )

f2 = 4( (n-4)/2 (n/2 -1 ) )

f1 = (n-2)²

f0 = S_e,n - ( f1 + f2 + f3 + f4 + f5)

The case for odd n should be easy to derive as a slight modification of the above. There should be a single set of equation that can accommodate any n regardless of whether it is odd or even, but I'm not feeling ambitious right now.

[Prof. Templeton]

I assume that the bottom gets painted.

Zero painted faces = 35

One painted face = 49

Two painted faces = 36

Three painted faces = 40

Four painted faces = 4

Five painted faces = 1

Total 165

I'll try to put together some formulas tomorrow if no one else does. Some of them look pretty straight-forward.

Edited April 29, 2009 by Prof. Templeton

[Guest]

(I think)

0 faces = 56

1 face = 64

2 faces = 48

3 faces = 48

4 faces = 4

5 faces = 0

total 220

[HoustonHokie]

if N is the total number of blocks,

N(n) = (1/6)n³ + (1/2)n³ + (1/3)n

works for both odd and even values of n

the number of blocks with 0 painted faces = N(n-2)

Basically, it's the pyramid completely contained inside the outer pyramid.

Edited April 29, 2009 by HoustonHokie

[HoustonHokie]

N₁(n) = (n - 2)² for all n > 2

They are the interior blocks on the bottom layer.

N₂(n) = N(n) - N₀(n) - N₁(n) - N₃(n) - N₄(n) - N₅(n) for all n > 2

Basically, everything that's left over after doing all the others.

N₃(n) = {[(n - 1) / 2 - 1] + (n - 2)} * 4 for odd n > 1

N₃(n) = [n / 2 - 1 + (n - 2)] * 4 for even n > 2

These are the blocks on the four corners of the interior layers and the outer blocks that are not corners on the bottom layer.

N₄(n) = 4 for all n > 2

They are the blocks in the four corners of the bottom layer.

N₅(n) = 1 for odd n > 1

N₅(n) = 0 for even n

It's the solitary block on top of the pyramid.

Edited April 29, 2009 by HoustonHokie

[Guest]

Also:

Total = n(n+1)(n+2)/6