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Algebra Difficulty 5.3 AIME, harder Prove it Saudi Arabia

Let SS be a set of positive real numbers with five elements such that for any distinct a,b,ca, b, c in SS, the number ab+bc+caa b + b c + c a is rational. Prove that for any aa and bb in SS, ab\frac{a}{b} is a rational number.

Solution

Let a,b,ca, b, c be three distinct elements in SS.
If we denote by Si\mathcal{S}_i, the set of subsets of SS of ii elements, i=2,3i=2,3, we notice that
∑{x,y}∈S2xy=13∑{x,y,z}∈S3(xy+yz+zx)∈Q. \sum_{\{x, y\} \in \mathcal{S}_2} x y = \frac{1}{3} \sum_{\{x, y, z\} \in \mathcal{S}_3} (x y + y z + z x) \in \mathbb{Q} .
If we denote by Sx,i\mathcal{S}_{x, i}, the set of subsets of S∖{x}S \setminus \{x\} of ii elements, i=2,3i=2,3, for x∈Sx \in S, we notice that
ac=∑{x,y}∈S2xy−12(∑{x,y,z}∈Sa,3(xy+yz+zx)+∑{x,y,z}∈Sc,3(xy+yz+zx)). a c = \sum_{\{x, y\} \in \mathcal{S}_2} x y - \frac{1}{2} \left( \sum_{\{x, y, z\} \in \mathcal{S}_{a, 3}} (x y + y z + z x) + \sum_{\{x, y, z\} \in \mathcal{S}_{c, 3}} (x y + y z + z x) \right) .
Hence ac∈Qa c \in \mathbb{Q}. Similarly bc∈Qb c \in \mathbb{Q}. We deduce that ab=acbc∈Q\frac{a}{b} = \frac{a c}{b c} \in \mathbb{Q}.

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