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A $3\times3$ grid contains altogether six squares that are formed by its nine entries: there are five squares whose sides are parallel to the sides of the grid (four small ones and a big one), and there is one square that is tilted by $45$ degrees:

  a b c             a b    b c                     a    c          b
  d e f  contains   d e    e f    d e    e f                    d     f
  g h i                           g h    h i       g    i          h

Determine all possible ways of arranging the nine numbers $1,2,3,\ldots,9$ in a $3\times3$ grid, such that the four numbers in the corners of each of the six squares add up to the same sum.

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  • $\begingroup$ So... this is asking for a 3x3 magic square where a+b+d+e = b+c+e+f = d+e+g+h = e+f+h+i = a+c+g+i = b+d+f+h? $\endgroup$ – Bulldogg6404 Feb 19 '15 at 10:30
  • $\begingroup$ I wonder if there are any solutions for larger grids? The number of squares increases quite rapidly, but there might at least be a solution for a 4×4 grid.) $\endgroup$ – squeamish ossifrage Feb 19 '15 at 11:49
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Summing the big square and the tilted square, we get for the common sum $S$
$$ 2S = (a+c+g+i)+(b+d+f+h) = 45-e.$$ Summing the entries in the four small squares $$4S = (a+b+d+e)+(b+c+e+f)+(d+e+g+h)+(e+f+h+i).$$ Summing the big square, twice the tilted square and four times entry $e$
$$3S+4e = (a+c+g+i)+2(b+d+f+h)+4e.$$ This shows $4S=3S+4e$ and $S=4e$. Then $2S+e=45$ and $S-4e=0$ shows $e=5$ and $S=20$.

One of the small squares uses $e=5$ and the number $1$. Then the other two entries are distinct and have sum $14$. Then the other two entries can only be $6$ and $8$. Then $1$ must be in a corner:

   1 6 c
   8 5 f
   g h i

Since 1 can be in any of the four corners and 6 and 8 can switch places, there are eight symmetric solutions.

One of the small squares uses $e=5$ and the number $7$. Then the other two entries are distinct and have sum $8$. Then the other two entries can only be $2$ and $6$. Then $7$ must be in a corner:

   1 6 7
   8 5 2
   g h i

The tilted sqaure has $20=b+d+f+h=6+8+2+h$ and $h=4$. Then $g$=3$ and $i=9$.

   1 6 7
   8 5 2
   3 4 9

Alltogether there are eight solutions. The other seven solutions come from rotating my solution by 90, 180, 270 degree and reflecting it on the middle line.

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Solutions:

(a, b, c, d, e, f, g, h, i) = (1, 6, 7, 8, 5, 2, 3, 4, 9)
(a, b, c, d, e, f, g, h, i) = (1, 8, 3, 6, 5, 4, 7, 2, 9)
(a, b, c, d, e, f, g, h, i) = (3, 4, 9, 8, 5, 2, 1, 6, 7)
(a, b, c, d, e, f, g, h, i) = (3, 8, 1, 4, 5, 6, 9, 2, 7)
(a, b, c, d, e, f, g, h, i) = (7, 2, 9, 6, 5, 4, 1, 8, 3)
(a, b, c, d, e, f, g, h, i) = (7, 6, 1, 2, 5, 8, 9, 4, 3)
(a, b, c, d, e, f, g, h, i) = (9, 2, 7, 4, 5, 6, 3, 8, 1)
(a, b, c, d, e, f, g, h, i) = (9, 4, 3, 2, 5, 8, 7, 6, 1)

How?

proof-by-exhaustion

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  • 2
    $\begingroup$ If I am not mistaken those 8 answers are actual the same if you treat mirrored and rotated answers as the same. $\endgroup$ – Ivo Beckers Feb 19 '15 at 12:03
  • $\begingroup$ @IvoBeckers Indeed they are, you can see the 5 in the middle, each solution has a mirror image, and the rotations are observable by noticing that two numbers always keep they their place in terms of "first four" and "last four" $\endgroup$ – dmg Feb 19 '15 at 13:22

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