## Task 5 - 340635

In the schema of Figure a, one can enter the numbers 0, 1, 2, 3, 4, 5, 6 in place of the letters such that the four "side sums" are equal:

$$
a+b+c=c+d+e=e+f+g=g+h+a
$$

An example, here with the value 14 for the four "side sums," is shown in Figure b.

(a) Provide three such entries, one with the value 2 of the four "side sums," one with the value 13 of the four "side sums," and one with the value 17 of the four "side sums"!

(b) An entry with the value 18 for the four "side sums" is also possible; however, it is not possible if the schema is to be formed with domino tiles as shown in Figure c.

Show that this is true; explain the difference!

(c) Fritz Schlaumeier looks at the completed schema for the "side sum" of 14, thinks for a long time, and then says:

"The four numbers for \( b, d, f, \) and \( h \) are in a very special relationship to each other. This relationship also holds for any filling with a different 'side sum'."

Give such a relationship and show that Fritz is right!
a)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0300.jpg?height=203&width=208&top_left_y=344&top_left_x=684)
b)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0300.jpg?height=203&width=215&top_left_y=344&top_left_x=949)
c)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0300.jpg?height=203&width=205&top_left_y=344&top_left_x=1211)
(a) The figure shows an example for entries with the "side sums" 2, 13, and 17.
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0300.jpg?height=200&width=448&top_left_y=665&top_left_x=676)

| 5 | 6 | 6 |
| :---: | :---: | :---: |
| 6 |  | 5 |
| 6 | 5 | 6 |

(b) The difference can be explained as follows:

For the scheme from the figure in part (a), an entry with the "side sum" 18 is possible by entering a 6 in all eight fields.

However, since a domino set contains only one (6;6) tile, such an entry is not possible with the tiles of a domino set.

The only way to represent 18 as the sum of three addends, which can only be the numbers 0, 1, 2, 3, 4, 5, 6, is $18 = 6 + 6 + 6$. Therefore, there is no other way to achieve the "side sum" 18 than with a 6 in all fields.

For this, more than one (6;6) tile would be required.

(c) A relationship between the numbers for $b, d, f, h$ is $b+f=d+h$.

Proof:

Due to the equality of all four "side sums," for example, the equations $a+b+c=c+d+e$ and $e+f+g=g+h+a$ also hold, each containing an edge number on both sides.

Therefore, the equations $a+b=d+e$ and $e+f=a+h$ also hold.

From these, it follows that $(a+b)+(e+f)=(d+e)+(h+a)$. Since the same summands, namely $a$ and $e$, appear on both sides of this equation, it follows that $b+f=d+h$.