Exercise 7. A set of $n$ non-zero and distinct real numbers is said to be regular if, when these numbers are written on the board in ascending order, the difference between two adjacent numbers is always the same, regardless of the two adjacent numbers chosen. For example, the set $\{4,18,-3,11\}$ is regular because if the numbers are written in ascending order, $18-11=11-4=4-(-3)$. A set $A$ of non-zero and distinct real numbers is said to be super-regular if it is regular and the set formed by the inverses of the numbers in set $A$ is also regular.

What is the largest integer $n$ such that there exists a super-regular set of $n$ non-zero and distinct real numbers?
## Solution to Exercise 7

Answer: $n=4$

In this exercise, we aim to demonstrate that 4 is the largest integer satisfying the property stated in the problem. The solution necessarily consists of two parts. First, we show that if a super-regular set has size $k$, then $k \leqslant 4$; this step is called the analysis. Second, we show that there indeed exists a super-regular set of size 4; this step is called the construction or synthesis.

In the following, we call the common difference of a regular set the difference between the largest number and the second largest number in that set.

## Analysis:

Step 1: We start by showing that there does not exist a super-regular set of size 3 where all elements have the same sign.

By contradiction, we assume a super-regular set $A=\{a, b, c\}$ such that $a < b < c$ and all elements are strictly positive. Let $r = b - a > 0$ be the common difference of $A$. Then $c - b = r$, so $c = b + r$ and $b = a + r$.

We define $B=\left\{\frac{1}{a}, \frac{1}{a+r}, \frac{1}{a+2r}\right\}$ as the set composed of the inverses of the elements of $A$. We verify that $\frac{1}{a+2r} < \frac{1}{a+r} < \frac{1}{a}$. The common difference of $B$ is given by:

$$
\frac{1}{a} - \frac{1}{a+r} = \frac{1}{a+2r} - \frac{1}{a+r}
$$

After simplification, this gives the relation $2a + 3r = 0$. Taking $a = -1$, we obtain the set $A = \{-3, -1, 1, 3\}$, which we verify is regular with a common difference of 2, and the set $B = \left\{-\frac{1}{3}, -1, 1, \frac{1}{3}\right\}$ composed of the inverses of the elements of $A$ is also regular with a common difference of $\frac{2}{3}$. Therefore, the set $A$ is super-regular, providing an example of a super-regular set of size 4.

Remark: The reasoning presented above may seem difficult to intuit if one does not already know the value of $n$. Here, a strategy to tackle the problem is to look for a super-regular set of size 3, and then a super-regular set of size 4. Generally, looking for small configurations that satisfy the properties of a problem is a good way to approach it and make conjectures.

Comment from the graders: Very few students had the time to tackle the problem and score points: only about ten students scored at least one point, and no one solved it entirely. It was relevant to consider the case where all the real numbers in the set have the same sign, as long as one kept in mind that this was not the only case. Note that many students confuse inverse and opposite: the opposite of 2 is -2, and its inverse is $\frac{1}{2}$.

High school exercises