2005 AMC 10A Problem 17

Below is the professionally curated solution for Problem 17 of the 2005 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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Concepts:arithmetic sequence double counting mean

Difficulty rating: 1660

17.

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

$9$

$10$

$11$

$12$

$13$

Solution:

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

Thus, the correct answer is D.

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