Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be equal to the same value k. Find the value of k. Provide at least one set of four numbers that satisfy the condition. Describe all possible sets of such numbers and determine how many there are.
Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:

$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2}+x_{3}}{x_{1}+x_{4}}=k
\end{aligned}
$$

Let $A=x_{1}+x_{2}, B=x_{3}+x_{4}$. Then from (2) we get $A=k B, B= k A, A B \neq 0$, from which $(k^{2}-1) A=0, k= \pm 1$. The same values will be obtained by analyzing relations (3) and (4).

If $k=1$, then equations (2)-(4) take the form $x_{1}+x_{2}=x_{3}+x_{4}, x_{1}+x_{3}= x_{2}+x_{4}, x_{1}+x_{4}=x_{2}+x_{3}$, from which we find the general solution $x_{1}=x_{2}=x_{3}= x_{4}=C \neq 0$. This case does not meet the condition (all numbers turned out to be equal).

If $k=-1$, then each of equations (2)-(4) takes the form $x_{1}+x_{2}+x_{3}+ x_{4}=0$, and the general solution is

$x_{1}=A, x_{2}=B, \quad x_{3}=C, \quad x_{4}=-A-B-C, \quad(A+B)(A+C)(B+C) \neq 0$.

Answer: $k=-1$.

$x_{1}=A, \quad x_{2}=B, \quad x_{3}=C, \quad x_{4}=-A-B-C, \quad(A+B)(A+C)(B+C) \neq 0$.

The set of tuples $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ is infinite.