Can the vertices of a regular $20$-gon be labeled with numbers $1, 2, ..., 20$ in such a way that each label is used exactly once and for every four consecutive vertices the sum of their labels is less than $43$?

First I tried to find some example. Then I got nothing , so decided to try something else. I wrote that sum of $1+2+3...20=210<43×5$ So it seems to have a solution. But I think something is disturbing this problem to have a solution. I tried to group all numbers into 4 parts . If I colour all verexes in 4 colours it will be 1,2,3,4,1,2,3,4....4 on 20-gon so if I choose 4 consicutive vertexes they all will be different colours. That means that there's a way to divide all 20 numbers into 4 groups of each colour or there's not , so we've to prove that it's impossible.

  • $\begingroup$ Let us say the points are $p_1,p_2,...,p_{20}$. Then the question demands $p_i+p_{i+1}+p_{i+2}+p_{i+3}\lt 43$ for $i=1,2,..20$ .Then we would have $4(p_1+p_2+..+p_{20})\lt 20×43$i.e $4×210\lt 860$. So I think, may be it's possible. $\endgroup$ Jun 22, 2020 at 7:11
  • $\begingroup$ yes, but we need to show that it's possible or prove that it's not $\endgroup$ Jun 22, 2020 at 7:23

1 Answer 1


No. Suppose the labels are $p_1,p_2,\dots,p_{20}$ in order (and consider indices modulo $20$). If it were possible, then $p_j+p_{j+1}+p_{j+2}+p_{j+3}\le42$ for all $1\le j\le 20$; summing these $20$ inequalities yields $4(p_1+p_2+\cdots+p_{20}) \le 840$. But $4(p_1+p_2+\cdots+p_{20}) = 4(1+2+\cdots+20) = 840$ as well, which would force $p_j+p_{j+1}+p_{j+2}+p_{j+3}=42$ exactly for all $1\le j\le 20$. But it's impossible for $p_1+p_2+p_3+p_4=42=p_2+p_3+p_4+p_5$ since $p_1$ and $p_5$ are distinct.

(One moral: when dealing with integers, always convert inequalities like $s<43$ to $s\le42$.)

  • $\begingroup$ thanks for answering $\endgroup$ Jun 22, 2020 at 8:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .