# Numbers from 1 to 10 put in a circle

Write the numbers from $$1$$ to $$10$$ in a circle. Consider all the groups of three consecutive numbers and their sums. Show that we can find a group with sum at least $$18$$.

Approach 1 ( not usefull ): The arithmetic mean of all sums of the groups is $$\frac{3(1+2+3+\dots+10)}{10} = \frac{165}{10} = 16.5$$ so, at least one group has sum $$17$$. But that doesn't help.

Approach 2: Suppose that there exist a configuration in which all groups have sums less that $$18$$. Therefore, $$10$$ and $$9$$, $$10$$ and $$8$$, $$8$$ and $$9$$, $$10$$ and $$7$$ can not be in the same group.

From the first three observations between $$8$$, $$9$$ and $$10$$ there are group of numbers of size, two, two and three.

Here I am stuck. I suppose looking at where 7 can be placed and with lots of case work you can get to a contradiction. Is this a good approach ? Do you have a better alternative ?

Also feel free to change the tags, I do not know which tags are appropriate

• I don’t like statistics, but I love the probabilistic method. It would have been so nice if your first approach had worked out... Mar 15, 2020 at 12:20
• Approach 1: If $a_1+a_2+a_3=a_2+a_3+a_4$ then $a_1=a_4$. So to avoid $18$s, the sums must alternate $16$ and $17$... Mar 15, 2020 at 12:53

Let us say the numbers in clockwise direction are $$a_{i},i=1,2,...,10$$ with $$a_{1}=1$$.
\begin{aligned} \sum_{i=2}^{10}{a_{i}}&=2+3+...+10\\ &=3\times 18 \end{aligned}
At least one of $$(a_2+a_3+a_4)$$, $$(a_5+a_6+a_7)$$, $$(a_8+a_9+a_{10})$$ is greater than or equal to $$18$$