The vertices of a regular 10-gon are labeled V_1, V_2, \ldots V_n, which is a permutation of \{ 1, 2, \ldots, 10\}. Define a neighboring sum to be the sum of 3 consecutive vertices V_i, V_{i+1} and V_{i+2} [where V_{11}=V_1, V_{12}=V_2]. For each permutation \sigma, let N_\sigma denote the maximum neighboring sum. As \sigma ranges over all permutations, what is the minimum value of N_\sigma?
To determine the minimum value of N_σ, we need to find the permutation σ that yields the smallest maximum neighboring sum.
First, let's consider the possible neighboring sums. Since we have a regular 10-gon, each vertex is adjacent to two other vertices. Therefore, there are 10 pairs of neighboring vertices, and each pair contributes one neighboring sum.
For simplicity, let's label the neighboring sums S_1, S_2, ..., S_10, corresponding to the pairs of vertices V_1V_2, V_2V_3, ..., V_9V_10, V_10V_1.
Now, let's think about the properties of these neighboring sums. Since each neighboring sum involves three consecutive vertices, we have the property that every vertex appears in exactly two neighboring sums. For example, V_1 appears in S_1 and S_10, V_2 appears in S_1 and S_2, and so on.
Notice that the sums of these neighboring sums must sum up to the sum of all the vertices. Since the sum of the vertices from 1 to 10 is 55, the sum of the neighboring sums must also be 55.
Now, for the minimum value of N_σ, we want to minimize the maximum neighboring sum. To achieve this, we want to distribute the numbers from 1 to 10 evenly among the neighboring sums. This means we want to minimize the difference between the largest and smallest neighboring sums.
Since the neighboring sums must sum up to 55, each neighboring sum should ideally be 55/10 = 5.5. However, since the neighboring sums involve integer vertices, we need to consider the rounding.
To minimize the difference, we want to distribute the numbers such that the largest neighboring sum is as close to 5.5 as possible without exceeding it. This means the largest neighboring sum should be either 5 or 6.
To achieve this distribution, we can start by assigning the vertices in increasing order to the neighboring sums until the sum reaches 55/2 = 27.5. At this point, we switch to placing the remaining vertices in decreasing order to distribute them evenly.
Let's illustrate this with an example: 1,2,3,4,5,5,4,3,2,1. Assigning these vertices to the neighboring sums, we get the following neighboring sums: 3+7+11 = 21, 2+6+10 = 18, 1+5+9 = 15, 9+5+1 = 15, 10+6+2 = 18, 11+7+3 = 21, 1+5+9 = 15, 2+6+10 = 18, 3+7+1 = 11, and 4+8+2 = 14.
The maximum neighboring sum in this case is 21, which is the largest we can achieve without exceeding 5.5.
Now, let's generalize this for any permutation σ. Since the differences between neighboring sums depend on the order of the permutation, the goal is to find the permutation σ that minimizes the maximum neighboring sum.
One approach to finding such a permutation is to use an algorithm called "greedy algorithm." The greedy algorithm starts by assigning the smallest available vertex to the smallest neighboring sum, then assigns the next smallest available vertex to the neighboring sum with the smallest current value, and so on. This ensures an even distribution of the vertices among the neighboring sums.
In this case, we can apply the greedy algorithm to distribute the vertices according to the approach described above: assign the vertices in increasing order until the sum reaches 27.5, then distribute the remaining vertices in decreasing order.
By applying the greedy algorithm, we find that the minimum value of N_σ is 21, which is achieved when using the permutation 1,2,3,4,5,5,4,3,2,1 or its rotations.
Therefore, the answer to the question is 21.