2009 USAMO Problems/Problem 6

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Problem

Let $s_1, s_2, s_3, \ldots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \cdots.$ Suppose that $t_1, t_2, t_3, \ldots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$. Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.

Solution

Suppose the $s_i$ can be represented as $\frac{a_i}{b_i}$ for every $i$, and suppose $t_i$ can be represented as $\frac{c_i}{d_i}$. Let's start with only the first two terms in the two sequences, $s_1$ and $s_2$ for sequence $s$ and $t_1$ and $t_2$ for sequence $t$. Then by the conditions of the problem, we have $(s_2 - s_1)(t_2 - t_1)$ is an integer, or $(\frac{a_2}{b_2} - \frac{a_1}{b_1})(\frac{c_2}{d_2} - \frac{c_1}{d_1})$ is an integer. Now we can set $r = \frac{b_1 b_2}{d_1 d_2}$, because the least common denominator of $s_2 - s_1$ is $b_1 b_2$ and of $t_2 - t_1$ is $d_1 d_2$, and multiplying or dividing appropriately by $\frac{b_1 b_2}{d_1 d_2}$ will always give an integer.

Now suppose we kept adding $s_i$ and $t_i$ until we get to $s_m = \frac{a_m}{b_m}$ in sequence $s$ and $t_m = \frac{c_m}{d_m}$ in sequence $t$ so that $(t_m - t_i)(s_m - s_i)$ is an integer for all $i$ with $1 \le i < m$, where $m$ is a positive integer. At this point, we will have $r$ = $\frac{\prod_{n=1}^{m}b_n}{\prod_{n=1}^{m}d_n}$, because these are the least common denominators of the two sequences up to $m$. As we keep adding $s_i$ and $t_i$, $r$ will always have value $\frac{\prod_{n=1}^{m}b_n}{\prod_{n=1}^{m}d_n}$, and we are done.

See Also

2009 USAMO (ProblemsResources)
Preceded by
Problem 5
Followed by
Last question
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All USAMO Problems and Solutions

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