More about Magical properties of Magic square
Magic squares of any order can be constructed with a set of numbers, so that its sum in rows,
columns and main diagonals will be same. They have many curious mathematical properties.
Such a magic square may be assumed that its small squares are uniformly filled with weights
Instead of numbers, (The magnitude of weight is exactly equal to the corresponding number).
When this magic square is pivoted at its CG, it is found that it is in an unstable equilibrium.
Under this condition, net moment about any symmetrical axis will be equal to zero. That is,
the clock wise moment will be equal to the anti-clock wise moment.
This can be verified with a typical example, a 3 x 3 magic square filled with natural numbers from
1 to 9.
x
2 | 3½ | 3½ | 6 |
9 | 2½ | 2½ | 1 |
4 | 1½ | 1½ | 8 |
y
With respect to the vertical axis xy , both the clock wise and anti-clock wise moment are equal and
is equal to 7½ X + 15 Y, where
X and Y are the perpendicular distance from the axis xy to the CG of the component weight.
The equidistant rows and columns with respect to symmetrical axis are interchangeable, where as
odd row or column, if any,
will be common, The interchangeable columns and rows are having not only equal sum but also the
sum of their of squares will also be equal.
2+7+6= 4+3+8 = 15 and
22 + 72 + 62 = 42 + 32 + 82 = 89
Since the middle column is common, its value is equally shared by both the interchangeable
columns. Considering this equal shares,
We have,
5½ + 11½ + 5½ = 9½ + 3½ + 9½ = 22½
The sum of their squares again equals
2(5½)2 + (11½)2 = 2 (9½)2 + (3½)2 = 192.75
For the interchangeable rows,
6½+ 9½ + 6½ = 8½ + 5½ + 8½
2(6½)2 + (9½)2 = 2 (8½)2 + (5½)2 = 174.75
For a main diagonal (ascending), by cancelling out the moment due to weights in the common
diagonal, we have,
7(d/2) + 2d + 9(d/2) 1(d/2) + 8d + 3(d/2) = 10 d
Even though they are interchangeable sets ,they are not equidistant, and hence the sum of their
squares will not be same.
For any system under equilibrium, the moments with respect to any pair of vertical or horizontal axes
passing through edges will be equal.
Since the perpendicular distances between the reference axis and the weights in a row/column are
x,2x,3x , we can derive
For vertical axis ,right edge; 2x+14x + 18x =34x
9x+10x+3x = 22x or 2(34x) +22x = 90 x
4x+6x+24x = 34x
For vertical axis,left edge; 6x+14x+6x= 26x
x+10x+27 x= 38X or 2(26x) + 38x= 90 x
8x +6x+ 12 x = 26x
As the set of numbers is interchangeable and equidistant, we have the additional square relation
2(34)2 + (22)2 = 2 (26)2 + (38)2 = 2796
The horizontal top and bottom axes give,
2(32x) + 26x = 2(28x) + 34x = 90x
2(32)2 + (26)2 = 2(28)2 + (34)2 = 2724
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