Problem

Source: Romania TST 1 2012, Problem 5

Tags: arithmetic sequence, algebra proposed, algebra



Let $p$ and $q$ be two given positive integers. A set of $p+q$ real numbers $a_1<a_2<\cdots <a_{p+q}$ is said to be balanced iff $a_1,\ldots,a_p$ were an arithmetic progression with common difference $q$ and $a_p,\ldots,a_{p+q}$ where an arithmetic progression with common difference $p$. Find the maximum possible number of balanced sets, so that any two of them have nonempty intersection. Comment: The intended problem also had "$p$ and $q$ are coprime" in the hypothesis. A typo when the problems where written made it appear like that in the exam (as if it were the only typo in the olympiad). Fortunately, the problem can be solved even if we didn't suppose that and it can be further generalized: we may suppose that a balanced set has $m+n$ reals $a_1<\cdots <a_{m+n-1}$ so that $a_1,\ldots,a_m$ is an arithmetic progression with common difference $p$ and $a_m,\ldots,a_{m+n-1}$ is an arithmetic progression with common difference $q$.