2005 AMC 12A Problem 13

Below is the professionally curated solution for Problem 13 of the 2005 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12A solutions, or check the answer key.

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Concepts:arithmetic sequence double counting

Difficulty rating: 1620

13.

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 that 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 that order. What is the middle term of the arithmetic sequence?

$9$

$10$

$11$

$12$

$13$

Solution:

Every number appears as an endpoint of exactly two of the five segments, so the total of the five sums is $2(3 + 5 + 6 + 7 + 9) = 60.$

The middle term of a five-term arithmetic sequence is its mean, namely $60 \div 5 = 12.$

Thus, the correct answer is D.

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