20th Turkish Mathematical Olympiad

The answer is $\frac{1}{20^{2012}}$. Let us treat more general case when $P$ is the set of all $n$ tuples. If $A=B=P$ then $f(A,B)=\frac{1}{20^n}$. We prove that $f(A,B)\le \frac{1}{20^n}$ by induction over $n$.

$n=1$. Suppose that $A=\{1,2,\dots ,a+c\}$, $B=\{20-b-c+1,\dots ,20\}$. Then $|A\cap B|=c$, $|A|=a+c$, $|B|=b+c$ and $f(A,B)=\frac{c}{(a+c)(b+c)}=\frac{1}{20+ab/c}\le \frac{1}{20}$. Suppose that the statement is correct for $n-1$. Let $A={\bigcup }_{i=1}^{20}{A}_i$, where elements of the set $A_i$ are obtained from elements of $A$ having last entry $i$ by removing this last entry. By definitions $|A|={\sum }_{i=1}^{20}|A_i|$ and $A_1\subset A_2\subset \cdots \subset A_{20}$. Let $B={\bigcup }_{i=1}^{20}{B}_i$, where elements of the set $B_i$ are obtained from elements of $B$ having last entry $i$ by removing this last entry. By definitions $|B|={\sum }_{i=1}^{20}|B_i|$ and $B_1\supset B_2\supset \cdots \supset B_{20}$. Now $$|A\cap B|=\sum _{i=1}^{20}|A_i\cap B_i|\le \frac{1}{20^{n-1}}\sum _{i=1}^{20}|A_i|\cdot |B_i|\le \frac{1}{20^{n-1}}\cdot \frac{1}{20}\left(\sum _{i=1}^{20}|A_i|\right)\left(\sum _{i=1}^{20}|B_i|\right)=\frac{1}{20^n}|A|\cdot |B|$$ (The first inequality is valid due to inductive hypothesis, the second inequality is the Chebyshev's rearrangement inequality). Done.