Czech-Polish-Slovak Match

By scaling, we can assume w.l.o.g. that the perimeter of $P$ has length $2$. Let $X_1,\dots ,X_n$ be the vertices of $P$, and let $x_i=|X_i{X}_{i+1}|$, where $X_{n+1}=X_1$; then ${\sum }_{i=1}^n{x}_i=2$. Since $P$ is a non-degenerate polygon, we have that $x_i<1$ for all $i=1,2,\dots ,n$. To prove the claim it is sufficient to partition the cyclic sequence $(x_1,\dots ,x_n)$ into three intervals such that the sum in each interval is strictly smaller than $1$; the endpoints of these intervals correspond to the selected vertices $A,B,C$ of $P$. To this end, we distinguish two cases: either there is some $p$ such that ${\sum }_{i=1}^p{x}_i=1$, or there is no such $p$.

In the first case, the perimeter of $P$ can be broken into two polygonal chains of length $1$ each, both with endpoints $X_1$ and $X_{p+1}$. Since $x_i<1$ for all $i$, both these chains consist of at least two segments. Since $n>4$, we infer that the four segments incident to $X_1$ and $X_{p+1}$, namely $X_n{X}_1$, $X_1{X}_2$, $X_p{X}_{p+1}$ and $X_{p+1}{X}_{p+2}$, are pairwise different, and they do not constitute the whole perimeter. Hence $x_n+x_1+x_p+x_{p+1}<2$, so either $x_n+x_1<1$ or $x_p+x_{p+1}<1$. In the former subcase we can take intervals $(x_n,x_1)$, $(x_2,\dots ,x_p)$ and $(x_{p+1},\dots ,x_{n-1})$, and in each of them the sum is clearly smaller than $1$. Symmetrically, in the latter subcase we can take intervals $(x_p,x_{p+1})$, $(x_{p+2},\dots ,x_n)$, and $(x_1,\dots ,x_{p-1})$.

We are left with the second case. Let $q$ be the largest index such that ${\sum }_{i=1}^q{x}_i<1$. Since $x_1<1$ and $x_n<1$, we have that $1\le q\le n-2$. By the choice of $q$ and the fact that the first case was not applicable, we have that ${\sum }_{i=1}^{q+1}{x}_i>1$, hence ${\sum }_{i=q+2}^n{x}_i=2-{\sum }_{i=1}^{q+1}{x}_i<1$. As $x_{q+1}<1$, we can take intervals $(x_1,\dots ,x_q)$, $(x_{q+1})$, and $(x_{q+2},\dots ,x_n)$, and in each of them the sum is strictly smaller than $1$. $\square$