Estonian Mathematical Olympiad

Denote the vertices of the polygon as $V_1,\dots ,V_n$ and let the real numbers in these vertices be $v_1,v_2,\dots ,v_n$, respectively. Moreover, denote $V_{n+1}=V_1$ and $v_{n+1}=v_1$. Define the value of the side $V_i{V}_{i+1}$ of the polygon to be $s_i=v_{i+1}-v_i$.

Observe that $v_1,\dots ,v_n$ satisfy the conditions of the problem if and only if the corresponding differences $s_1,\dots ,s_n$ satisfy $\max \{|s_1|,\dots ,|s_n|\}\le 1$ and $s_1+\cdots +s_n=0$. Therefore the problem can be reformulated as finding the least non-negative real number $C$ such that, for arbitrary real numbers $s_1,\dots ,s_n$, assumptions $\max \{|s_1|,\dots ,|s_n|\}\le 1$ and $s_1+\cdots +s_n=0$ would imply $\min \{|s_1|,\dots ,|s_n|\}\le C$.

Clearly $\min \{|s_1|,\dots ,|s_n|\}\le 1$ for any choice of $s_1,\dots ,s_n$ satisfying the assumptions. If $n=2k$ for some $k\in \mathbb{N}$ then choosing $s_1=\cdots =s_k=1$ and $s_{k+1}=\cdots =s_{2k}=-1$ would imply $\min \{|s_1|,\dots ,|s_n|\}=1$. Hence $C=1$.

Let now be $n=2k+1$ for some $k\in \mathbb{N}$. Taking $s_1=s_2=\cdots =s_{k+1}=\frac{k}{k+1}$ and $s_{k+2}=\cdots =s_{2k+1}=-1$ establishes $\min \{|s_1|,\dots ,|s_n|\}=\frac{k}{k+1}$. We show that $\min \{|s_1|,\dots ,|s_n|\}\le \frac{k}{k+1}$ whenever $s_1,\dots ,s_n$ satisfy the assumptions. To this end, suppose the contrary, i.e. $\min \{|s_1|,\dots ,|s_n|\}>\frac{k}{k+1}$. Reorder $s_1,\dots ,s_n$ as $d_1,\dots ,d_n$ so that $d_1\ge d_2\ge \cdots \ge d_n$. Note that there exists $l\in \{1,\dots ,n\}$ such that $d_l>\frac{k}{k+1}$ and $d_{l+1}<-\frac{k}{k+1}$.

If $l\ge k+1$ then $$0=s_1+\cdots +s_n=d_1+\cdots +d_n\ge l\cdot d_l+(n-l)d_n>l\cdot \frac{k}{k+1}-(n-l)\ge (k+1)\cdot \frac{k}{k+1}-k=k-k=0,$$ contradiction. In the case $l<k+1$, the proof is analogous (one can consider the opposite differences $-d_n,-d_{n-1},\dots ,-d_1$). Hence $C=\frac{k}{k+1}=\frac{\lfloor \frac{n}{2}\rfloor }{\lfloor \frac{n}{2}\rfloor +1}$.