SOLUTION: A grid below contains one large square and four small squares with circles on each corner of the smaller squares. Show that, up to rotation and reflection, there is only one way to

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Question 1166308: A grid below contains one large square and four small squares with circles on each corner of the smaller squares. Show that, up to rotation and reflection, there is only one way to fill the empty circles with the numbers 1 to 9 so that the sums of the numbers at the vertices of all five squares are the same.
Link to figure/diagram:
https://imgur.com/xrRbfsu
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


I can only get you started on this, which is to say I don't have a complete solution, but perhaps you will be able to make something work from what I have here.

In the first place, the sum of the digits from 1 to 9 is 45: , so the average is 45 divided by 9 is 5. So if you are going to have 5 equal sums using 4 of the digits, it stands to reason that that sum must be 4 times 5 or 20. Further, my intuitive reaction is that the digit 5 should be in the center.

There are twelve ways to create a sum equal to 20 using four of the digits 1 through 9, specifically: 

    9821   8741  *7652
    9731   8732   7643
    9641  *8651
    9632   8642
   *9542  *8543

Note that there are exactly four of these combinations that contain a 5. Assuming the 5 needs to be in the center, the four marked combinations need to be the combinations that go around the smaller squares, and since two of these combinations contain an 8, the 8 must be in the center of one of the sides rather than in a corner so that the 8 can be part of two different small squares. By the same token, the 9 and the 7 have to be in outside corner spots.

Given all of that, the following is a solution that withstands both rotations and reflections. But you are on your own from here to prove that the solution is unique discounting rotations or reflections. 

         9  2  7

         4  5  6

         3  8  1

Also, 9542 and 7652 share the digits 52 and therefore must be adjacent. Likewise 8651 and 8543 share 85, 9542 and 8543 share 54, and 8651 and 7652 share 65, creating three additional forced adjacencies.

Really, everything flows from the assumption that the 5 is in the center. If you can demonstrate that fact as a necessity, you have it solved.

John

My calculator said it, I believe it, that settles it


I > Ø
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