49. Vasya arranged the numbers from 1 to 10 in a circle in an arbitrary order and calculated the sums of all sets of three consecutive numbers. What is the maximum value that the smallest of these sums can take?
Solution. Note that the sum of all 10 consecutive triples of numbers is three times the sum of all numbers, which is 165. Therefore, the average sum in each triple is $165 / 10 = 16.5$. We will show that it is also impossible to achieve a minimum sum of 16. Since the average sum is 16.5, this means that the sums of the triples can exceed 16 by a total of only 5. However, note that adjacent triples (which share two numbers) cannot have the same sum (otherwise, the first number of the first triple would equal the last number of the adjacent triple). Therefore, the only option would be to alternate triples with sums of 16 and 17, which would lead to numbers two apart differing by one, but then numbers five apart would be equal. For a sum of fifteen, there is an example: $1,8,7$, $5,3,10,2,4,9,6$ in a circle.