
Problem 1. Find the least positive integer number $n(n \geq 3)$ with the following property: for any colouring of $n$ different points $A_{1}, A_{2}, \ldots, A_{n}$ on a line and such that $A_{1} A_{2}=A_{2} A_{3}=\ldots=$ $A_{n-1} A_{n}$ in two colours, there are three points $A_{i}, A_{j}, A_{2 j-i}(1 \leq i<$ $2 j-i \leq n$ ) which have the same colour.


Solution: Assume the two colours are white and black. Consider 8 points coloured as follows: $A_{1}, A_{2}, A_{5}, A_{6}$ (white), $A_{3}, A_{4}, A_{7}, A_{8}$ (black). Obviously no three points $A_{i}, A_{j}, A_{2 j-i}(1 \leq i<2 j-i \leq n)$ have the same colour and therefore $n \geq 9$.

If we can show that $n=9$ has the required property, we will be done. Suppose there are 9 points coloured black or white and no three points $A_{i}, A_{j}, A_{2 j-i}(1 \leq i<2 j-i \leq n)$ have the same colour.

First assume that for $i=3, i=4$ or $i=5$ points $A_{i}$ and $A_{i+2}$ have the same colour (say white). Then the points $A_{i-2}, A_{i+1}, A_{i+4}$
should be black (note that $i-2 \geq 1$ and $i+4 \leq 9$ ), which is a contradiction.

Suppose now that for $i=3,4,5$ the points $A_{i}$ and $A_{i+2}$ have different colours. Without loss of generality assume $A_{5}$ is a white point. Then $A_{3}$ and $A_{7}$ are black. Because of the symmetry we may suppose that $A_{4}$ is white and $A_{6}$ is black. Consiquently $A_{8}$ is white, $A_{2}$ is black $(2+8=2 \cdot 5)$ and $A_{9}$ is white $(7+9=2 \cdot 8)$. Therefore $A_{1}$ should be both white $(1+3=2 \cdot 2)$ and black $(1+9=2 \cdot 5)$, which is again a contradiction.

Consequently the assumption is not true and so $n=9$.
