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2005 AMC 10A Problem 17

Problem 17 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 this 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 this order. What is the middle term of the arithmetic sequence?

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Solution

Every number is an endpoint of two segments, so the five segment sums total 2(3+5+6+7+9)=60.2(3 + 5 + 6 + 7 + 9) = 60. The middle term of a five-term arithmetic sequence equals its mean, which is 605=12.\dfrac{60}{5} = 12. Thus, the correct answer is D.

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Concepts: arithmetic sequence · double counting · mean

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