Exercise 11. The integers $1,2, \ldots, 20$ have been arranged around a circle, in some order. For each of these integers $k$, Matthieu counts how many integers are less than $k$ among the 9 integers that follow $k$ when traversing the circle clockwise; he counts $A(k)$. He also counts how many integers are less than $k$ among the 9 integers that follow $k$ when traversing the circle counterclockwise; he counts $B(k)$. Matthieu then notes that $A(k)=B(k)$ for all $k$. What is the number diametrically opposite to 11 on the circle?
Solution to Exercise 11 From now on, to simplify the terms used, when we talk about a number "before" (or "after") another number, it means that the first number is among the 9 numbers preceding (or following) the second number, looking in the clockwise direction.

In the following, we denote $o(a)$ as the number diametrically opposite to the integer $a$ on the circle.

Let's start with some useful observations.

- First, the only number that does not appear before or after $a$ is the number $a$ itself.
- Furthermore, if two numbers $a$ and $b$ are opposite each other on the circle, then $a$ is before and $b$ is after (or $a$ after and $b$ before) any number on the circle other than $a$ and $b$.

After these general observations, let's look at which number can be the opposite of each number on the circle.

Since $A(2)=B(2)$, there are as many numbers less than 2 before and after 2. Since 1 is the only integer less than 2 among the integers on the circle, we deduce that $A(2)=B(2)=0$ and $o(2)=1$.

We can repeat this same reasoning step by step to show that the numbers $2k$ and $2k-1$ are diametrically opposite on the circle.

Indeed, if we already know that $o(2i)=2i-1$ for all $1 \leq i < k$, we can prove that $2k$ and $2k-1$ are also diametrically opposite.

To do this, suppose by contradiction that $o(2k) \neq 2k-1$. Then $2k-1$ is either before or after $2k$. Considering the pairs of opposite numbers $(2i-1, 2i)$ for $1 \leq i \leq k-1$, we get $k-1$ numbers before $2k$ and $k-1$ numbers after $2k$ on the circle. Considering the number $2k-1$, which is either before or after $2k$ on the circle, and since all other numbers on the circle are greater than $2k$, we get $A(2k)=k-1$ and $B(2k)=k$ or $A(2k)=k$ and $B(2k)=k-1$, which is impossible. Thus, $o(2k)=2k-1$ as claimed.

Finally, we have shown that the numbers $2k$ and $2k-1$ are always diametrically opposite on the circle for $1 \leq k \leq 10$. In particular, the number diametrically opposite to 11 on the circle is 12.

Remark: We could have provided a very similar solution by noting with the same reasoning that the number diametrically opposite to 19 on the circle is necessarily 20, and then descending step by step to 11 instead of ascending to 11.