## Partially Magic Squares

1. Place the numbers 1-9 into a 3x3 grid, such that the four sides and the two diagonals each have the same sum (greater than 15.) What is the number in the center of the grid?
2. Disregarding rotations and reflections, find all partially magic squares which use the numbers 1-9 and in which at least six lines of three numbers have the same sum (other than 15).

Source: Original.

Solutions:
1. Disregarding rotations and reflections, there is only one solution. It was found by Bill Hewson, Joseph DeVincentis, Robert Kelnhofer, Bob Odineal and Joan McCormick:
```7  2  9
5  3  1
6  4  8```
2. Joseph DeVincentis has a very thorough solution. He solved the problem with 15 as another possible sum. I later asked only for solutions which don't sum to 15 and he said that drops the number of solutions to 12.
```I found a total of 38 solutions to this part, including some solutions
using every possible set of 6 or 7 lines adding to a common sum that
doesn't cause an immediate contradiction.  Every digit appears in the
middle in some solution, but only 12, 15, and 18 seem to be allowed as
common sums.

The possible sets of magic lines are limited, as follows:
If all three rows add up to the common sum, then either
all three columns do, or at most one column does, since the
sum of the rows and the sum of the columns is the same.

A. One possibility is that all the rows and columns add up to
the common sum, but neither diagonal does (or, alternately,
one diagonal does).  Several forms of this can be found as
wraparound-rotations of the standard 3x3 magic square:

Standard magic square:
8 3 4
1 5 9
6 7 2

Wrapped H or V:
6 7 2   1 5 9   3 4 8   4 8 3
8 3 4   6 7 2   5 9 1   9 1 5
1 5 9   8 3 4   7 2 6   2 6 7

Wrapped H and V (in these cases, one diagonal adds to 15):
5 9 1   7 2 6   2 6 7   9 1 5
7 2 6   3 4 8   4 8 3   2 6 7
3 4 8   5 9 1   9 1 5   4 8 3

Note that this gives one square for each center digit.
This includes all squares of this type.
Proof:
Since the sum of all the digits is 45, and the three rows use
each digit once, the sum of each row and column must be 15.
There are only two pairs of numbers that can be in the same row
with 9: 1 and 5, or 2 and 4.  Likewise, 1 can only be in a row
with 9 and 5, or 8 and 6.  So, one row (or column, but ignoring
rotations, put this into a row) has to have 1, 9, and 5 in it
in some order.  The column with 9 must have 2 and 4, and the
column with 1 must have 6 and 8.  So, the column with 5 has
7 and 3 in it.  Now, if 8 is in the same row with 2, the other
number in that row must be 5, so this cannot be.  8 must be in
the row with 4 (and 3), and 2 must be in the row with 6 (and 7).
Thus, the rows and columns of any such square must be the same
as the rows and columns of the standard 3x3 magic square, but
possible out of order, permuted, etc.  Also note that to keep the
sums consistent, having the entries in a row permuted requires
that the columns are rearranged in the order of that permutation,
so all the rows will have their entries permuted the same way, so
we can just consider the reorderings of the rows and columns.
Ignoring reflections, there are just 3 ways to arrange the 3 rows
and 3 ways to arrange the columns.  Any of these row orders can
go with any of the column orders, and this gives the 9 squares above.

B. Since you can't have just one orthogonal line not add to the
common sum, in all other cases, both diagonals add to the
common sum, as do 4 of the other lines.  Ignoring rotations and
reflections, five possible cases result from this:

B1: Two opposite edge lines don't add to the common sum
B3: One edge line and one center line, in the same direction, don't
B4: One edge line and one center line, in different directions, don't
B5: The two center lines don't add to the common sum.

There are solutions for every one of these cases!

In cases B1 and B3, three lines in the same direction add to the
common sum, and those three lines contain all the numbers and thus
add to 45, so the common sum can only be 15.

In cases B1 and B2, all four lines through the center number add up
to the common sum.  If we add these up, using the terminology of
part 1, 4M = sum(1..9) + 3X.  The only answers that work are X = 1, 5, 9,
which make M respectively 12, 15, 18.  It happens that you can pair up the
remaining numbers to give four sums of M through the center in each of
these cases, so there's no problem yet, although the reasoning above
limits us to X=5 in case B1.

Case B1. Let's make the outer columns the ones that don't add up.
We know the common sum is 15 and the center number is 5. Now all the
other numbers pair up into groups that must go on opposite sides of 5.
1-9 must go somewhere.  Suppose it goes across the middle.  Then 8-2
must go into the top and bottom rows.  8 is too large to go together
with any of the other numbers greater than 5, so it goes with 3 and 4.
similarly, 2 goes with 6 and 7.  This is all we needed for this case.
However, we have some freedom left. We can arrange those numbers in
the top row in any order. The numbers in the bottom row are then
forced into place. The numbers in the middle row are still free to
be swapped.  However, swapping these and the upper corners just flips
the thing over.  To avoid reflections, keep 8-3-4 on top, and 1-5-9
in that order in the middle.  If the top is in 8-3-4 order, we get the
standard magic square; for the 5 other arrangements, we get these
solutions:

8 4 3   3 8 4   3 4 8   4 8 3   4 3 8
1 5 9   1 5 9   1 5 9   1 5 9   1 5 9
7 6 2   6 2 7   2 6 7   7 2 6   2 7 6

Now, suppose 9-1 is not in the middle.  Then 9 can only go with 2 and 4
to make a solution; 1 goes with 6 and 8.  3 and 7 are left across the
middle.  Again we have the freedom of putting the top digits in any order.
This time, put 6-1-8 in the top, and 7-5-3 across the middle in that order.
If the top is in the order 6-1-8, we get the standard magic square, and
for the other 5 arrangements we get these solutions:

6 8 1   1 6 8   1 8 6   8 6 1   8 1 6
7 5 3   7 5 3   7 5 3   7 5 3   7 5 3
9 2 4   2 4 9   4 2 9   9 4 2   4 9 2

Since 9-1 either has to be in the middle or not in the middle, these are
all the solutions for this case.

Case B2. Make the right column and bottom row the ones that don't add up.
With this combination of rows, there will be no rotations and only one
reflection of each answer to worry about.  The center number is 1, 5, or 9,
corresponding to common sums of 12, 15, and 18, and the remaining numbers
in each case have to exist in specific pairs on opposite sides of the
center.  There are many ways to arrange the pairs for each center number,
and some of them work:

If 1 is in the center, the sum of all six rows is 72, and this sum contains
every number at least once, but the upper left number is included three
times, the center number (1) 4 times, and the other numbers at the top or
left twice each.  After subtracting the center, this leaves us with:
68 = sum(2..9) + sum(2..9) + upper left - (3 numbers in lower right)
sum(2..9) = 44, so this says 3 numbers in lower right = 20 + upper left.
The largest possible sum of 3 numbers in the lower right is 24, so the
number in the upper left is 2, 3, or 4.

If it's 2, 9 goes opposite it, and the other numbers at the top are a pair
that adds to 10, 3-7 or 4-6, and the other numbers on the left are the
other of these pairs.  Out of these four numbers, one pair must go along
the diagonal with 1, and so the pair has to add to 11, and that can be
only 7 and 4.  This leaves just one way fill in the square, ignoring
reflections:

2 3 7
6 1 5
4 8 9

If 3 is in the upper left, 8 is opposite it, and the other pairs at the
top and left add to 9: 2-7 and 4-5.  Again 7-4 go on the other diagonal,
and we get:

3 2 7
5 1 6
4 9 8

If 4 is in the upper left, 7 is opposite it, and the other pairs at top
and left add to 8: 3-5 and 2-6.  5-6 goes on the other diagonal:

4 3 5
2 1 9
6 8 7

If 9 is in the middle, all the same calculations work as their
tens-complements, (that is, swap each number with 10 minus itself)
and we get the tens-complements of these three answers:

8 7 3   7 8 3   6 7 5
4 9 5   5 9 4   8 9 1
6 2 1   6 1 2   4 2 3

If 5 is in the middle, the common sum is 15, and since 2 rows add to 15,
the remaining three numbers in the other row add to 15, so we have only
the standard magic square in this case.

Case B3. Let the middle and right columns be the ones that don't add up.
We know the common sum is 15.  It looks like 2, 4, 6, or 8 could go in the
middle instead of 5, and still have three pairs of numbers to go opposite
them; in the case of 4, these are 2-9, 3-8, and 5-6; for 2, they are
6-7, 5-8, and 4-9; for 6 and 8, take the tens-complements.  There are two
other squares that have three lines going through them, the upper left and
lower left corners; these must also come from the set 2, 4, 5, 6, 8.
The numbers 1, 3, 7, and 9 must go into four of the other six spaces.
For each center number besides 5, two of these four do not appear in the
three lines through the center digit which add to the common sum, and
therefore must go in the middle column.  If 5 is in the center, the
leftover pair of numbers that goes in the center column with it cause
that column to add to 15 also, so we end up with the standard magic square.

For 2 in the middle, the only way to get 15 in the left column is to
put 4, 5, 6 there, and 1 and 3 need to go in the middle column so,
ignoring reflections, we want the top corners to add to 14 and the bottom
corners to add to 12.  This happens for these two cases:

6 1 8   5 1 9
4 2 9   6 2 7
5 3 7   4 3 8

For 4 in the middle, if we don't have either 8 or 9 on the left,
the left column will add up to less than 15; both of them is too much, so
we have one or the other of them.  But there's no pair available to match
with 9 to add to 15 so we must have 8-2-5 on the left in some order.
Then 1 and 7 go in the middle column, so we need, say, top corners adding
to 14, bottom corners to 8.  This yields two solutions:

5 1 9   8 1 6
8 4 3   2 4 9
2 7 6   5 7 3

For 6 or 8 in the middle, we get four more solutions that are the
tens-complements of these:

4 9 2   5 9 1   5 9 1   2 9 4
6 8 1   4 8 3   2 6 7   8 6 1
5 7 3   6 7 2   8 3 4   5 3 7

Case B4: Let the middle column and bottom row be the ones that don't
add up.  Take the sum of the two diagonals and the middle row, and
subtract the left and right columns.  The result is the common sum,
and it must equal three times the center number.  Also note that
three lines go through the center number, each of which must add to
three times the center number; this only works for 4, 5, and 6.
If 5 is in the center, the last line through the center also adds
to 15, so all the columns and 2 rows add to 15, so the last row does
as well, and we only have the standard magic square.

For 4 in the center, 8 and 9 must go in the middle column. If 9 goes
on top, 1 and 2 must go beside it; if 8 is on top, 1 and 3 must go
beside it.  These entries force all the other numbers, ignoring
reflections, as follows:

1 9 2   1 8 3
5 4 3   6 4 2
6 8 7   5 9 7

For 6 in the center, we get the tens-complements of the above answers:

9 1 8   9 2 7
5 6 7   4 6 8
4 2 3   5 1 3

Case B5. In part 1 I considered the case of having the two center
lines not add to the common sum.  If we remove the restriction that
the common sum be greater than 15, we have the five cases in the
table from part 1, but X=1 and X=9 have no way to achieve the
specified sums.  X=7 gives the tens-complement of the square from
part 1:

1 8 3
9 7 5
2 6 4

X=5 results only in the standard magic square.  The diagonals must
contain 2 of the following 4 sets of numbers:  1-5-9, 2-5-8, 3-5-7, 4-5-6.
If 1-5-9 is in a diagonal, then no matter what you put on the
other diagonal, one outer edge row will need at least a 10 to add
to 15.  If you put 2-5-8 on a diagonal, 4-5-6 must go on the other
to avoid needing a 10; then the standard 3x3 magic square is forced.
If you put 3-5-7 and 4-5-6 on the diagonals, repeats of 4 and 6 are
forced within the square.

Summary of part 2:
Type  Number of solutions
A      8
B1    10
B2     6
B3     8
B4     4
B5     2
total  38

Joseph DeVincentis```

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