Problem 1043

If $x = a$ and $y = f$, for all $z, t \in \{b, c, d, e\}$ we have $\frac{x}{y} = \frac{a}{f} \le \frac{z}{t}$, where the equality holds for $z = a$ and $t = f$. Since $z \in \{b, c, d, e\}$, it follows that $z \ge b$, hence $a \ge b$. Therefore, $a = b$. Similarly, we infer that $e = f$.

We now choose $x = c$, $y = d$. Then $\frac{c}{d} = \frac{z}{t}$, where $z, t \in \{a, f\}$. But $c \le d$, hence $c = d$, or $\frac{c}{d} = \frac{a}{f}$.

If $c = d$, the 6-tuple $(a, a, c, c, f, f)$ is obviously a solution. Indeed, if $x \ne y$, we choose $z = x$ and $t = y$, and if $x = y$, we choose $z = t \ne x$.

If $c \ne d$, we have $\frac{c}{d} = \frac{a}{f}$, thus $d = \frac{cf}{a} \ge f$. It follows that $a = c$ and $d = f$. Similarly, it is easy to prove that the 6-tuple $(a, a, a, f, f, f)$ is a solution.

Thus, the 6-tuple which satisfy the required condition are $(a, a, b, b, c, c)$, with $a \le b \le c$, $(a, a, a, b, b, b)$, with $a \le b$, and all their permutations.