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2005 AMC 12A Problem 13

Problem 13 of 25IntermediateAlgebraCounting & Probability

In the five-sided star shown, the letters A,A, B,B, C,C, D,D, and EE are replaced by the numbers 3,3, 5,5, 6,6, 7,7, and 9,9, although not necessarily in that order. The sums of the numbers at the ends of the line segments AB,AB, BC,BC, CD,CD, DE,DE, and EAEA form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

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Solution

Every number appears as an endpoint of exactly two of the five segments, so the total of the five sums is 2(3+5+6+7+9)=60. 2(3 + 5 + 6 + 7 + 9) = 60. The middle term of a five-term arithmetic sequence is its mean, namely 60÷5=12.60 \div 5 = 12. Thus, the correct answer is D.

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Concepts: arithmetic sequence · double counting

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