Magic Square

[BMAD]

Classic magic number square problem but hopefully with a slight twist.

If all the rows, columns, and diagonals add up to 33 using any number you desire. Is the center uniquely determined?

[gavinksong]

Here's a simple magic square that adds up to 15:

4 9 2
3 5 7
8 1 6

We can add (33 - 15) / 3 = 6 to each cell to raise the total up to 33, turning our original square into:
10 15 8
9 11 13
14 7 12

We also have the option of doubling each cell and then incrementing each cell by one to raise the sum to 33, and that gives us:
9 19 5
7 11 15
17 3 13

which I honestly thought would give us a different center, but I guess not...

Are we allowed to have negative numbers or fractions?

Edited July 29, 2013 by gavinksong

[BMAD]

Here's a simple magic square that adds up to 15:

4 9 2

3 5 7

8 1 6

We can add (33 - 15) / 3 = 6 to each cell to raise the total up to 33, turning our original square into:

10 15 8

9 11 13

14 7 12

We also have the option of doubling each cell and then incrementing each cell by one to raise the sum to 33, and that gives us:

9 19 5

7 11 15

17 3 13

which I honestly thought would give us a different center, but I guess not...

Are we allowed to have negative numbers or fractions?

yes. any type of real number (no complex)

[Rainman]

Yes. Let's label the squares as follows:

ABC

DEF

GHI

Now if we add them row by row we get 3*33 = (A+B+C)+(D+E+F)+(G+H+I). However, if we add the four lines that go through the center, we get 4*33 = (A+E+I)+(B+E+H)+(C+E+G)+(D+E+F) = (A+B+C+D+E+F+G+H+I)+3*E. Subtract the first equation from the second to get 33=3*E, which fixes E at 11.