To find the largest possible number of red integers in a circle of ten positive integers, we start by revisiting the condition for a number to be colored red: a number \( a_i \) is red if it is equal to the average of its two neighbors, that is,
\[ a_i = \frac{a_{i-1} + a_{i+1}}{2} \]
This can be rearranged to:
\[ 2a_i = a_{i-1} + a_{i+1} \]
or equivalently,
\[ a_{i-1} + a_{i+1} - 2a_i = 0 \]
This equation is reminiscent of a linear recurrence relation. Specifically, if we define the differences \( d_i = a_{i} - a_{i-1} \), we can express the condition for \( a_i \) to be red as:
\[ d_{i+1} - d_i = 0, \]
which indicates that the sequence of differences \( d_i \) must be constant.
Assuming that the differences are all equal, we can denote that constant difference by \( d \). Thus, we can express the integers sequentially as:
\[ a_1 = a_0 + d, \quad a_2 = a_0 + 2d, \quad \ldots, \quad a_{10} = a_0 + 9d. \]
This means that all integers are arithmetic sequences defined by initial value \( a_0 \) and common difference \( d \).
To maximize the number of red integers, we can illustrate that the structure of the sequence results in some \( a_i \) values being equal to the average of their neighbors. Explicitly, every third number can be made red under the uniform arithmetic condition where \( a_i = k \), \( a_{i+1} = k + d \), and \( a_{i+2} = k + 2d \) form an arithmetic sequence. This would imply:
\[ a_i = \frac{a_{i-1} + a_{i+1}}{2}. \]
By cycling through the elements of the circle, we recognize that if \( d \) is set to zero or the same value, many numbers would equal each other, leading to each number fulfilling the condition for being red.
Upon setting, say, all numbers to the same value \( n \) (making all differences zero), every integer satisfies the average condition:
\[ a_i = \frac{a_{i-1} + a_{i+1}}{2} = n = n. \]
All ten integers would be red.
Thus, the maximum possible number of red integers in this configuration ends up being:
\[ \boxed{10}. \]
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