Pyramid

"Magic" Pyramid

Many of you may be familiar with the concept of Magic Squares and for those of you who aren't, you're welcome to follow that link and check them out.

Their basic premise, the thing that makes them "magical" is that the arrangement of numbers produce equal values when added vertically, horizontally, and sometimes diagonally. I've seen a number of expansions on this basic object, making them more complex by increasing the size (or order) of the square, and by increasing the dimensions (magic cubes). I've also seen some number puzzles floating around on the internet, where you're asked to arrange the numbers around the perimeter of a triangle to make all sides equal. Once again, expansion has come. This time with the help of Smojack, who worked with me to nail down the formulas. I give you: the "Magic" Pyramid.

In order to begin construction of our pyramid, we need to break it down into component parts, to make it easier to work with. On any given pyramid, we will have 4 faces, one of which is displayed above with its component parts labeled:

  • Tip - Comprised of 2 sections. 3 appear on each face for a total of 12 on the pyramid.
  • Side - Comprised of 3 sections. Also 3 per face and 12 per pyramid.
  • Center - The Center of each face, 1 section, 1 per face, 4 per pyramid.

Now, you may have seen something like this before. If you'll note the example on the left, adding up each line produces identical results, that is: 2+6+1 = 1+5+3 = 3+4+2 = 9. With our pyramid, we're seeking similar results, but for twelve faces, each comprised of seven parts. What we've found is that unlike its smaller cousin, where equality is not an option, it is exactly what is needed for the pyramid. In order to complete its construction, each of the tips are made equal, as are each of the sides. That does leave the Centers however, each being a single piece, cannot possibly be equal to one another. I'll leave it to you to determine what this means. Does this mean the pyramids are not "magical," or are the centers the product of the pyramids, much like the sigils which were drawn over the magic squares?

Bringing Equality to All Sides

This is where the math begins. For right now, I don't have anything that makes nice Sigmas quickly, I'll revise this later if it still seems important.

We're going to make 3 series of numbers, one for each of the component pieces of the side. To make it a little easier to read later on, I'm going to name them. I'm not a math teacher, I'm sure there's a real convention for doing this, but I don't know what it is.

  • 1{} = {a, a+1... a+5}
  • 2{} = {a+6, a+7... a+11}
  • 3{} = {b+7, b+8... b+12}
  • 4{} = {b, b+1... b+5}
  • 5{} = {c+10, c+8...c}
  • 6{} = {c+11, c+9...c+1}
  • 7() = (b+6) intentionally unused

Where (b > a+11) or (b+12 < a) and where ((c > a+11) or (c+11 < a)) and ((c > b+12) or (c+11 < b)) In plain English: Where you don't use the same number twice.

  • 1{} + 3{} + 5{} = all equal results (S)
  • 2{} + 4{} + 6{} = all equal results (S)

You now have 12 equal sets of 3 - The Walls. You've also got b+6 - The First Center Piece.

Topping off the Tips

The tips, much like the sides all need to be made equal for our chosen method to work. With only 2 parts per tip instead of 3, this becomes a much simpler task.

  • 8{} = {d, d+1,... d+11}
  • 9{} = {e+11, e+10,...e}

Again, avoid overlapping numbers. 8{} + 9{} = all equal results (T)

Get it Together and Bringing it Back to Me.

Well that's it, then, isn't it? If we have 12 equal sides and 12 equal end pieces, we have all 12 sides of our pyramid being perfectly equal. The sum of each face should be (S)+2(T), and the perimeter should be 3(S)+3(T).

An Ounce of Example is worth a Pound of Formula.

  • a = 1
  • b = 25
  • c = 13
  • d = 38
  • e = 50
  • 1{} = {1, 2, 3, 4, 5, 6}
  • 2{} = {7, 8, 9, 10, 11, 12}
  • 3{} = {32, 33, 34, 35, 36, 37}
  • 4{} = {25, 26, 27, 28, 29, 30}
  • 5{} = {23, 21, 19, 17, 15, 13}
  • 6{} = {24, 22, 20, 18, 16, 14}
  • 7() = (31) intentionally unused

1+32+23 = 2+33+21 = 7+25+24 = 8+26+22 = 56

  • 8{} = {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49}
  • 9{} = {62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51}

38+62 = 39+61 = 49+51 = 100

Therefore Each edge = (S)+2(T), 56+2(100) = 256 (In binary 0001 0000 0000! Magic!)

Now that the sets have been created, there's nothing left but to put them onto our Pyramid. With 4 sides and 16 pieces per side, that's 64 pieces (In binary 0100 0000! Magic!) We know what all our sides and tips are, and we've got one center already (b+6 = 31), and another where we skipped over 50 between 8{} and 9{}. Just go ahead and fill in the balance, 63, 64 and your magic pyramid is complete.

The perimeter of all four faces is equal to 468, and the sum of all 12 sides is 256. This can be easily changed by changing the ranges of our a-e, either by transposing them or by shifting them up to 3 in either direction, or apart to change the center numbers.

I'm certain there are other ways to do this, but this is the one that we've developed. Feel free to play with it, have fun, come up with your own unique center combinations and permutate them into whatever your heart desires.

If you do devise a different way to create these pyramids, please, let us know. We'd love to hear from you and post your credited work.