74th NMO Selection Tests for JBMO

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.