Problem 5. Six boys and six girls stood in a circle, alternating. Each of them wrote a non-zero number in their notebook. It is known that each number written by a boy is equal to the sum of the numbers written by the adjacent girls, and each number written by a girl is equal to the product of the numbers written by the adjacent boys. What can the sum of all twelve numbers be?
Answer: 4.5

Solution. Let's choose any five people standing in a circle in a row, with the outermost being boys. Let their numbers be $x, x y, y, y z, z$. Then $y=x y+y z$. Since $y \neq 0$, we can divide by it, obtaining $x+z=1$. Thus, the sum of any two numbers said by boys standing three apart is 1.

Let the boys say the numbers $a_{1}, a_{2}, \ldots, a_{6}$ (in this order around the circle). Then

$$
1=a_{1}+a_{3}=a_{2}+a_{4}=\ldots=a_{5}+a_{1}=a_{6}+a_{2} .
$$

We can notice that

$$
6=\left(a_{1}+a_{3}\right)+\left(a_{2}+a_{4}\right)+\ldots+\left(a_{5}+a_{1}\right)+\left(a_{6}+a_{2}\right)=2\left(a_{1}+a_{2}+\ldots+a_{6}\right)
$$

from which the sum of all the boys' numbers is 3.

On the other hand, the sum of the boys' numbers is twice the sum of all the girls' numbers, since each girl's number is a summand for two boys. Therefore, the sum of the girls' numbers is 1.5. The total sum is then 4.5.

It remains to note that such a situation is indeed possible: all boys can write the number $\frac{1}{2}$, and all girls can write the number $\frac{1}{4}$.

## Criteria

The highest applicable criterion is used:

7 p. Any complete solution to the problem.

7 p. It is proven that the sum can only be 4.5, but no valid example of a permissible arrangement of numbers is provided.

1 p. It is stated that the sum of the girls' numbers is half the sum of the boys' numbers.

3 p. It is proven that the sum of the numbers of boys standing "three apart" is 1.

0 p. Correct answer.

1 p. Correct example of number arrangement.