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Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 8.7

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#1 h Thursday at 11:59 PM • 1 Y

Y by kiyoras_2001

Find the smallest real number $a$ such that for any positive integer number $n > 2$ and any arrangement of the numbers from 1 to $n$ on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than $a$ .

Proposed by Mykhailo Shtandenko

This post has been edited 2 times. Last edited by mshtand1, Yesterday at 12:00 AM

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#2 h Yesterday at 8:45 PM

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Let show, that $a=2$

For $n=2k$ place numbers in next order : $1,k+1,2,k+2,...,k,2k$ then all ratios are $\frac{2k}{1}, \frac{k+i}{i}, \frac{k+i-1}{i}$ with $1 \leq i \leq k$ and minimal ratio is $\frac{2k-1}{k}$ can be very close to $2$

Now let prove, that always there is such pair
For $n=2k$ first place numbers $(k, k+1,k+2,...,2k)$ Easy to see, that should be at least two pair of adjacent numbers from this subset, but then their is ratio $\leq \frac{2k-1}{k}<2$
For $n=2k+1$ first place numbers $(k+1,k+2,...,2k+1)$ and there is at least one pair of adjacent numbers, but their ratio is $\leq \frac{2k+1}{k+1}<2$

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KAME06

#3 h Today at 1:07 AM

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Suppose that $a<2$ . Consider $n=2k+3, k \ge 0$ and the next arrangement on a circle: $(x_1, x_2, ...., x_n)=(k+2, 1, k+3, 2, k+4, 3, ..., 2k+2, k+1, 2k+3)$ . Notice that, for even $i$ , $2x_i =x_i+x_i \le x_{i-1}=x_i+(k+1)$ and $2x_i =x_i+x_i < x_{i+1}=x_i+(k+2)$ , so every ratio of $(x_j, x_{j+1}), 0<j<n$ is greater or equal than $2$ .
Now, consider the ratio $\frac{2k+3}{k+2}$ . See that $\lim_{k \rightarrow \infty} \frac{2k+3}{k+2} = 2$ (or see that $y=2$ is an asymptote of $\frac{2k+3}{k+2}$ and that $\frac{2k+3}{k+2}$ grows for $k \ge 0$ ).
So, for a $k$ large enough, any ratio of $(x_1, x_2, ...., x_n)=(k+2, 1, k+3, 2, k+4, 3, ..., 2k+2, k+1, 2k+3)$ is going to be greater than $a$ . Contradiction.
That implies that $a \ge 2$ . If $a=2$ , we have these cases:
-If $n=2k+1, k > 0$ , suppose that there exist an arrangement of numbers where any ratio of a pair of adjacent numbers is greater or equal than $2$ . We notice that $k+1, k+2, ..., 2k+1$ can't be adjacent because any multiple of them is greater than $2(k+1)$ . So, there are $k+1$ spots in the circle that need to be filled with at least one number, but we just have $1, 2, ..., k$ numbers. Contradiction.
-If $n=2k$ , suppose that there exist an arrangement of numbers where any ratio of a pair of adjacent numbers is greater or equal than $2$ . We notice that $k+1, k+2, ..., 2k$ can't be adjacent because any multiple of them is greater than $2(k+1)$ . So, there are $k$ spots in the circle that need to be filled with at least one number. Consider the number $k$ . $k$ can't be adjacent with $k+1, k+2, ..., 2k-1$ , because any multiple of $k$ is greater than them. That implies that $k$ has to be adjacent with one number of $1, 2, ..., k-1$ , but in that case there's going to be no more than $k-1$ spots filled. Contradiction.
We conclude that $a=2$ is the answer.

This post has been edited 2 times. Last edited by KAME06, Today at 1:18 AM

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