Some three consecutive numbers sum to at least $32$

Solution 1:

Suppose that the sums of sequences of three adjacent numbers in the circle are $s_1,s_2,\dots,s_{20}$. When you form the grand sum $s_1+s_2+\cdots+s_{20}$, in effect you’re adding up the numbers from $1$ through $20$ three times (why?), so you know the total. If all of the $s_k$ were less than $32$, what would the maximum possible total be?

Solution 2:

We can in fact prove that there is always three successive integers in this arrangement, whose sum is at least $33$. Though it is not clear if $33$ is the optimal lower bound.

Consider the circular arrangement of numbers starting from $1$ as follows. $$1 , a_1, a_2, \ldots, a_{19}$$ where $a_1,a_2,\ldots a_{19} \in \{2,3,4,\ldots,20\}$.

Note that $1 + a_1 + a_2 + \cdots a_{19} = 210$. Note that at least one of $a_1,a_4,a_7,a_{10},a_{13},a_{16},a_{19}$ must be $\leq 14$.

Say $a_1 \leq 14$, then we get that $1 + a_1 + a_2 + \cdots a_{19} \leq 1 + 14 + a_2 + \cdots a_{19}$. Let $s$ be the maximum possible sum of three consecutive elements. Then we have that $$(a_2 + a_3 + a_4) + (a_5 + a_6 + a_7) + \cdots +(a_{17} + a_{18} + a_{19}) \leq 6s$$

Hence, we get that $$210 = 1 + a_1 + a_2 + \cdots a_{19} \leq 1 + 14 + a_2 + \cdots a_{19} \leq 6s + 15$$ i.e. $$6s \geq 195 \implies s \geq 32.5.$$ Hence, $$s \geq 33.$$

The same argument works if $a_1 > 14$ and one of $a_4,a_7,a_{10},a_{13},a_{16},a_{19} \leq 14$. For instance, if $a_7 \leq 14$, then rearrange the sum as $$1 + (a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + a_7 + (a_8 + a_9 + a_{10}) + (a_{11} + a_{12} + a_{13}) + (a_{14} + a_{15} + a_{16}) + (a_{17} + a_{18} + a_{19}).$$