5. Jure drew a regular 9-sided polygon. He wanted to arrange the numbers from 1 to 9 at its vertices such that the sum of the numbers in any three consecutive vertices would not exceed a natural number $n$. For what smallest number $n$ would he have succeeded?

## 53rd Mathematical Competition for High School Students in Slovenia

Selection Competition, April 1, 2009

## Solutions to the Problems

Each problem is worth 7 points. All mathematically and logically correct solutions are equally valid. When evaluating each problem, reasonably consider the provided scoring guide. A contestant should not receive more than 3 points for any problem if the partial solution does not clearly lead to the final solution of the problem.
IV/5. We will show that $n=16$. As the figure shows, the numbers from 1 to 9 can be written in the vertices of a regular 9-gon such that the sum of any three consecutive numbers is at most 16.

We will prove that for $n<16$, it is not possible to arrange the numbers from 1 to 9 such that the sum of the numbers in any three consecutive vertices is at most $n$. Suppose this is possible. If we sum all the sums of the numbers from three consecutive vertices,

![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-18.jpg?height=320&width=308&top_left_y=1876&top_left_x=1528)
each number will be counted three times, as it appears in three such sums. Therefore, the resulting number is $3 \cdot(1+2+\ldots+9)=135$. On the other hand, each sum of the numbers from three consecutive vertices is at most $n$, and there are 9 such sums, so the total sum is at most $9 n$. It follows that $9 n \geq 135$ or $n \geq 15$.

We will also show that the value of $n$ cannot be 15. In this case, the upper bound in the above estimate would be an equality, meaning that all sums of the numbers from three consecutive vertices are equal to 15. If we denote the numbers in four consecutive vertices by $a_{1}, a_{2}, a_{3}$, and $a_{4}$, this means that $a_{1}+a_{2}+a_{3}=15$ and $a_{2}+a_{3}+a_{4}=15$, from which it follows that $a_{1}=a_{4}$, which is not possible, as we are arranging 9 different numbers. Therefore, $n$ must be at least 16.
Example of an arrangement when $n=16$ ..... 2 points
(Check if the written arrangement for $n=16$ meets the conditions of the problem.
There are multiple correct arrangements, only one of which is given in the figure.)
The sum of all sums of the numbers from three consecutive vertices is 135 ..... 1 point
The sum of all sums of the numbers from three consecutive vertices is at most $9 n$ ..... 1 point
Conclusion $n \geq 15$ ..... 1 point
If $n=15$, all sums of the numbers from three consecutive vertices are equal ..... 1 point
Conclusion that this is not possible 1 point

[^0]:    The record of a solution with 15 correctly solved problems

    2 points

    Conclusion that from 13 points and at least 16 correct answers, it follows that there are at least 12 incorrect ones ..... 2 points

    Determination that in this case, the total number of problems is 28 .............................................. 2 points

    Conclusion that the maximum number of correctly solved problems is 15 . . . . . . . . . . . . . . . . . . . . . . 1 point

    (If the case with 16 correct answers is considered, and the cases with 17, 18, ..., 24 correct answers are not, award a maximum of 4 points.)