Maximum Neighbouring Sums

WLOG let $V_1 = 1$ . Observe that the 3 neighbouring sums $(V_2+V_3+V_4), (V_5+V_6+V_7), (V_8+V_9+V_{10})$ sum to $55 - V_1 = 54$ . Thus by pigeonhole principle, at least one is larger than or equal to $54/3 = 18$ .

We now only need to construct s.t. the maximum sum is 18. Arrange as follows:

$1,7,3,8,5,4,9,2,6,10.$

The resulting neighbouring sums are:

$11,18,16,17,18,15,17,18,17,18.$

Hence, we see that 18 is indeed the minimum value.

For all $i$ , we define $f(i)= V_{i-1} + V_i + V_{i+1}$ , where the indices are read modulo $10$ . Note that for all $\sigma$ , we have $\displaystyle \sum \limits_{i=1}^{10} f(i)= 3 \left( 1 + 2 + \cdots + 10 \right) = 165$ , since each of the numbers must appear thrice in the sum. It follows that $\max \left( f(i) \right ) \geq \dfrac{165}{10} = 16.5$ , i.e. $\max \left( f(i) \right) \geq 17$ .

Before we proceed, we prove the following claim.

Claim: $\forall \ i, f(i+1) \neq f(i)$

Proof: On the contrary, assume there exists an $i$ such that $f(i+1)= f(i)$ . This is equivalent to $V_i + V_{i+1} + V_{i+2} = V_{i-1} + V_i + V_{i+1}$ , which in turn implies $V_{i+2}= V_{i-1}$ , which is a contradiction since all $V_i$ 's are distinct. $\blacksquare$

Now, suppose there exists a permutation $\sigma$ for which $\max \left( f(i) \right) = 17$ . Note that more than $5$ $f(i)$ 's cannot be $17$ , since otherwise there would exist an $i$ such that $f(i)= f(i+1)$ . If less than $5$ $f(i)$ 's are equal to $17$ , we have $\displaystyle \sum \limits_{i=1}^{9} f(i) \leq 4 \times 17 + 6 \times 16 < 165$ , which is a contradiction. We thus conclude there exist precisely $5$ $i$ 's such that $f(i)= 17$ . Let $S$ denote the sum of the other $f(i)$ 's. Then, $S + 5 \times 17 = 165 \implies S= 80.$ But since the other $5$ $f(i)$ 's are atmost $16$ and $80= 16 \times 5$ , all the other $f(i)$ 's must be $16$ .

By the claim, it follows that the sequence $\left \{ f(i) \right \} _{i=1}^{10}$ alternates between $16$ and $17$ . In other words, if $f(i)= 17$ , $f(i+1)= 16$ , and vice-versa.

Let $\left \{ K_1, K_2, \cdots , K_6, K_7 \right \}$ be any seven consecutive vertices of the polygon where $f\left( K_1 \right) = 17$ , so $f\left( K_2 \right) = 16, f \left( K_3 \right) = 17, f \left( K_4 \right)= 16, f \left( K_5 \right) = 17, f\left( K_6 \right) = 16, f \left( K_7 \right)= 17.$ Then, note that $K_1 + K_2 + K_3 + K_4 + K_5 + K_6 = f\left( K_2 \right) + f \left( K_3 \right) = 33$ $K_2 + K_3 + K_4 + K_5 + K_6 + K_7 = f \left( K_3 \right) + f \left( K_6 \right) = 33.$ Combining them, we find out $K_1= K_7$ , a contradiction.

The minimum potential value of $\max \left( f(i) \right)$ , thus, is $18$ . A quick hack shows that this can indeed be attained for the permutation $\sigma = \{1, 6, 8, 4, 3, 10, 5, 2, 9, 7\}$ . We thus conclude our desired answer is $\boxed{18}$ .

The total of all 10 neighbouring sums is always 3*(1+2+...+10) = 165, because each number is included thrice. The highest neighboring sum is therefore at most ceiling(165/10) = 17.

Place each neighbouring sum in front of the physically centre of the three numbers it contains, like this:

$\begin{aligned} \text{Numbers} \qquad \qquad &amp; a &amp; &amp; b &amp; &amp; c &amp; &amp;d &amp; &amp;\ldots \\ \text{Neighbouring sums} &amp; &amp; &amp; a+b+c &amp; &amp; b+c+d &amp; &amp; &amp; &amp; \ldots\\ \end{aligned}$

Suppose two adjacent neighbouring sums are equal. Then a+b+c = b+c+d; therefore, a = d. Contradiction. Therefore the assumption is false: no two neighbouring sums are equal.

Suppose the highest neighbouring sum is 17. There must be at least 5 17s among the neighbouring sums. Because no two adjacent neighbouring sums are equal, they must alternate between 17 and 16.

$\begin{aligned} \text{Numbers} \qquad \qquad &amp; a &amp; &amp; b &amp; &amp; c &amp; &amp; d &amp; &amp; e &amp; &amp; f &amp; &amp; g &amp; &amp; \ldots \\ \text{Neighbouring sums} &amp;17 &amp; &amp; 16 &amp; &amp; 17 &amp; &amp; 16 &amp; &amp; 17 &amp; &amp; &amp; &amp; &amp; &amp; \ldots \\ \end{aligned}$ Then a + b + c + d + e + f = 17 + 16 = 33, and b + c + d + e + f + g = 16 + 17 = 33; therefore, a - g = 33 - 33 = 0, therefore a = g. Contradiction. Therefore, the assumption is false: the highest neighbouring sum cannot be 17.

The highest neighbouring sum can be 18, as shown here:

$\begin{aligned} \text{Numbers} \qquad &amp; 10 &amp; &amp; 5 &amp; &amp; 2 &amp; &amp; 9 &amp; &amp; 7 &amp; &amp; 1 &amp; &amp; 6 &amp; &amp; 8 &amp; &amp; 4 &amp; &amp; 3\\ \text{N'g sums} \qquad &amp; 18 &amp; &amp; 17 &amp; &amp; 16 &amp; &amp; 18 &amp; &amp; 17 &amp; &amp; 14 &amp; &amp; 15 &amp; &amp; 18 &amp; &amp; 15 &amp; &amp; 17\\ \end{aligned}$

Therefore, the answer is 18.

[Calvin - Slight edits to align the 'diagrams' as per latter posts]