
Problem 11.3. For any natural number $n \geq 3$ denote by $m(n)$ the maximum number of points which can be placed inside or on the outline of a regular $n$-gon with side 1 in a way that the distance between any two of them is greater than 1 . Find all $n$ for which $m(n)=n-1$.


Solution: We prove first that $m(n)=n-1$ for $n=4,5,6$. Let $n$ be one of the above numbers and let $B_{1}, B_{2}, \cdots, B_{n}$ be points satisfying the conditions of the problem for the regular $n$-gon $A_{1} A_{2} \cdots A_{n}$ of side 1 and centre $O$. It is obvious that $O B_{i} \leq 1$ and therefore no three points $O, B_{i}, B_{j}, 1 \leq i \neq j \leq n$ lie on a line. Furthermore at least one of the angles $O B_{i} B_{j}$ is less than $\frac{2 \pi}{n} \leq 90^{\circ}$ and it follows from the Cosine Theorem that $B_{i} B_{j}^{2} \leq O B_{i}^{2}+O B_{j}^{2}-$ $2 O B_{i} O B_{j} \cos \frac{2 \pi}{n}=B_{i}^{\prime} B_{j}^{\prime 2}$ where $B_{i}^{\prime}$ and $B_{j}^{\prime}$ are points on the segments $O A_{1}$ and $O A_{2}$ such that $O B_{i}^{\prime}=O B_{i}, B_{j}^{\prime}=O B_{j}$. When $n=4,5,6$ the greatest side in $\triangle O A_{1} A_{2}$ is $A_{1} A_{2}=1$ and therefore $B_{i} B_{j} \leq B_{i}^{\prime} B_{j}^{\prime} \leq 1$, which is a contradiction. Thus $m(n) \leq n-1$ if $n=4,5,6$. It is easily seen that for these $n$ there exist $n-1$ points on the outline of a regular $n$-gon $A_{1} A_{2} \cdots A_{n}$ with the required property. Therefore $m(n)=n-1$ if $n=4,5,6$. We shall prove now that if $n \geq 7$ then $m(n) \geq n$. Let $B_{i} \in A_{i} A_{i+1}$ be points such that $A_{i} B_{i}=2^{i-1} \varepsilon, 1 \leq i \leq n-1$ where $0(1-\varepsilon)^{2}+4 \varepsilon^{2}+2 \varepsilon(1-\varepsilon)^{2}=1+3 \varepsilon^{2}>1$. Similarly $B_{i} B_{i+1}>1$ when $1 \leq i \leq n-2$. Further, it is clear that $B_{1} B_{n-1}>A_{1} A_{n}=1$. Since $O A_{1}=O A_{2}=\cdots=O A_{n}>1$, $A_{i} A_{j}>1$ when $|i-j| \geq 2$ and $B_{i} \rightarrow A_{i}$ when $\varepsilon \rightarrow 0$, it follows that we can make $\varepsilon$ so small that $O B_{i}>1$ when $1 \leq i \leq n-1$ and $B_{i} B_{j}>1$ when $|i-j| \geq 2$. Then the points $B_{1}, B_{2}, \cdots, B_{n-1}, O$ satisfy the conditions of the problem and so $m(n) \geq n$ when $n \geq 7$. Since it is obvious that $m(3)=1$, we come to the conclusion that $m(n)=n-1$ only when $n=4,5,6$.

## Spring mathematics tournament -1998
