Ken's POTW


A Few Final Sums
  1. Can you place values at the vertices of a cube (not necesseraly different) so you end up with a regular die (faces 1 to 6, with 1 opposite 6, 2 opposite 5, 3 opposite 4), when you take the sum over the vertices of each face? [Also, in a regular die, faces 1-2-3 are clockwise around one of the vertices. -KD] Can you find a way to end up with a non-regular die (still faces 1 to 6, but not all opposite sides give 7 when added)? [In how many ways can it be done with non-negative values? Unique non-negative values? -KD]

  2. Revisiting the Olympic Rings puzzle, here are some more linking rings:
    1. Consider three circles, all intersecting one another, with their centers at the corners of an equilateral triangle (and radii shorter than the side of the triangle.) There are seven regions defined by these crossing circles. Place the numbers 1 to 7 in these regions, such that each circle has the same total within it.
    2. Consider four circles, all intersecting one another, with their centers at the corners of a square (and radii shorter than the side of the square.) There are 13 regions defined by these crossing circles. Place the numbers 1 to 13 in these regions, such that each circle has the same total within it.
    3. Linking squares: Place the numbers 1 to 9 into a 3x3 grid, such that each corner block of four squares, and the sum of the four independent corner squares, is the same. (5 equal sums.)
Source: 1. Bert Sevenhant. 2. Original
Solutions to problem 1 were received from Nick Baxter and Kirk Bresniker. Both found different solutions to model a regular die.

Nick's comments:
For any die with this sort of numbering, compute the total of all faces. Since each vertex is on 3 faces, the total would be 3 times the sum of the vertices. Since any two opposite faces include each vertex just once, the sum of all the face values is exactly 3 times the sum of ANY pair of opposite faces. Since the total must be 21, we see that the opposite faces must always be 7.

There are 10 possible numberings to achieve this cube.
(In each below, the front=1, right=2, bottom=3, top=4, left=5, back=6.) 
  3---1
 /|  /|
0---0 |
| 2-|-0
|/  |/
0---1

  4---0
 /|  /|
0---0 |
| 1-|-1
|/  |/
0---1

  2---2
 /|  /|
0---0 |
| 2-|-0
|/  |/
1---0

  3---1
 /|  /|
0---0 |
| 1-|-1
|/  |/
1---0

  4---0
 /|  /|
0---0 |
| 0-|-2
|/  |/
1---0

  2---1
 /|  /|
0---1 |
| 3-|-0
|/  |/
0---0

  3---0
 /|  /|
0---1 |
| 2-|-1
|/  |/
0---0

  1---2
 /|  /|
1---0 |
| 3-|-0
|/  |/
0---0

  2---1
 /|  /|
1---0 |
| 2-|-1
|/  |/
0---0

  3---0
 /|  /|
1---0 |
| 1-|-2
|/  |/
0---0
Nick Baxter answered the remaining problems:
Problem 1 with unique values: 

   .1---.2
   /|  /|
 .3---.4|
  |.4-|-1.3
  |/  |/
3.2---1.1
Problem 2: 
2.1.  the circles can be (1,3,4,6)(1,2,4,7)(2,3,4,5)

2.2.  the circles can be (1,3,4,6,7,8,11)(1,2,3,6,7,9,12)
                        (1,2,4,5,7,8,13)(2,3,4,5,7,9,10)
        the configuration can be determined uniquely from this information,
        so I don't need to draw!!

2.3.  the sum must be 20 and the center must be 5;
        turn a magic square 45 degrees:
        9 4 3
        2 5 8
        7 6 1
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