2011 AIME II Problems/Problem 3
The average angle in an 18-gon is . In an arithmetic sequence the average is the same as the median, so the middle two terms of the sequence average to . Thus for some positive (the sequence is increasing and thus non-constant) integer , the middle two terms are and . Since the step is the last term of the sequence is , which must be less than , since the polygon is convex. This gives , so the only suitable positive integer is 1. The first term is then
You could also solve this problem with exterior angles. Exterior angles of any polygon add up to . Since there are exterior angles in an 18-gon, the average measure of an exterior angles is . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is . Since there are even number of exterior angles, the middle two must be and , and the difference between terms must be . Check to make sure the smallest exterior angle is greater than : . It is, so the greatest exterior angle is and the smallest interior angle is .
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