Problem 2098

Solution. The answer is $n \ge 13$. First, we show that any $n \ge 13$ satisfies the desired condition. Suppose for the sake of contradiction that $a_1 \le a_2 \le \dots \le a_n$ are integers such that $\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n)$ and no three are the side lengths of an acute triangle. We conclude that
$$a_{i+2}^2 \ge a_i^2 + a_{i+1}^2 \qquad (2)$$
for all $i \le n-2$. Letting $\{F_n\}$ be the Fibonacci numbers, defined by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \ge 2$, repeated application of (2) and the ordering of the $\{a_i\}$ implies that
$$a_i^2 \ge F_i \cdot a_1^2 \qquad (3)$$
for all $i \le n$. Noting that $F_{12} = 12^2$, an easy induction shows that $F_n > n^2$ for $n > 12$. Hence, if $n \ge 13$, (3) implies $a_n^2 > n^2 \cdot a_1^2$, a contradiction. This shows that any $n \ge 13$ satisfies the condition of the problem.

On the other hand, for any $n < 13$, we may take $a_i = \sqrt{F_i}$ for $1 \le i \le n$, so that
$$\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n)$$
holds because $F_n \le n^2$ for $n \le 12$. Further, for $i < j$, we have $F_i + F_j \le F_{j+1}$, which shows that for $i < j < k$, we have $a_k^2 \ge a_i^2 + a_j^2$. Hence, $\{a_i, a_j, a_k\}$ are not the side lengths of an acute triangle. Therefore, all $n < 13$ do not satisfy the conditions of the problem, and the answer is $n \ge 13$.