Write one number in each red circle so that the sum of the numbers along each blue, diagonal line is always the same number $x.$
Find the maximum value of $x$ .
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Side Thoughts: I got thinking about the set of solutions, all possible ways to fill in the grid such that all diagonals have a common sum of zero. Any linear combination of solutions is another solution, so the solution set is a vector space. I am pretty sure there are 16 degrees of freedom - 16 linearly independent solutions that form a basis for the vector space.
I thought of the 16 basis solutions as each associated with a 1x1 diamond in the grid. Placing {1, -1, 1, -1} around any such diamond (or any larger tilted rectangle, for that matter) and leaving all other entries as 0 is a solution. The remaining question is whether these 16 'diamond solutions' are linearly independent. It is not too hard to show they are. This is most clear for the 12 diamonds that include a circle on the boundary, but the logic can be extended to cover the 4 remaining diamonds in the center of the grid.
The value of each corner is equal to the sum x. Take the biggest blue line. It goes through 2 corners and 4 other circles. The sum of those 4 other circles are x - x - x = -x. Take the 4 diagonals that go through those 4 circles (so not the main diagonal). All the points covering those 4 diagonals have a total sum of 4x. Now look at those points, exept for the first 4. They go through exactly 4 diagonals, without crossing, so their sum is also 4x. We can observe that the sum of our 4 chosen points is 0. Now -x=0 so that x=0.
Indices or labels or a drawing would be useful here.
This proof does answer the original question.
I'm going to take issue with the original question, noting the difference between "Write one number in each red circle" and "In each red circle, write a number". The way this problem is written implies that each red circle contains
the same
number. If that's a condition of the problem, the solution is very simple. Only if we allow the numbers to vary is there any challenge involved. Wieter's solution solves the more general problem.
It took me a moment to get it. So if you start with the long NW-SE diagonal, then the 4 circles other than the corners lie on four different NE-SW diagonals. Now look at the other circles on those diagonals, and observe that they constitute the circles of four NW-SE diagonals.
Now this does not account for a complete solution. But given the problem as stated, imagine we had filled in numbers and gotten our maximum x. We could just scale every number in the red dots by a factor $k>1$ to get a new maximum of $kx>x$ for all positive x (if for some reason x were negative to begin with multiply every number by -1). So by contradiction there cannot be a maximum x that is not zero. Thus if there is in fact a maximum for x its only feasible value would be 0, which of course can be achieved by just pluggin in 0 in all red dots.
That's how I approached the problem. It shows that 0 is the only possible maximum solution. Showing it's the only possible solution is a different issue, that Wieter ably addresses.
Imagine the circles were each filled in with black or white, such that no two adjacent circles are the same color. Then the sum of all the numbers in the black circles can be shown to equal both 5x and 6x, so it follows that x=0.
-06 | -01 | +11 | -04 | +03 | -25 |
+02 | -15 | -09 | -17 | +25 | -02 |
+05 | +06 | +23 | +12 | +21 | -14 |
+08 | -08 | -11 | -12 | -05 | +30 |
00 | -20 | -07 | +07 | -13 | +04 |
+20 | +01 | +13 | -16 | -03 | +24 |
==> this could also be another solution... nothing mentioned in the question to lead to a single definite solution.
You have misunderstood
Claim: The only (and hence maximum) value for $x$ is 0 for all square grids with an even number of red circles.
Proof: Suppose the size of the square grid is $2n$ . If $n = 1$ then each corner must have value $x$ as they lie on their own blue line however the longer diagonal blue lines have sum $2x$ . So we must have $x = 2x$ and so $x = 0$ .
Suppose $n > 1$ is the smallest value for which we can fill in the grid with $x \neq 0$ . Now consider the central $2(n - 1)$ square and do the following: For each red circle on the boundary of this square, add to its value the value of the neighbouring red circles which lie outside this square. This gives a square of size $2(n-1)$ with the same value for $x$ . This contradicts the minimality of $n$ and so $x = 0$ for all values of $n$ .
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Since 6x = 5x, the only possible value of x is zero. This is the same as Peter Byers' solution, and Wieter Jacobs looked at the exact same entries.