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2005 AMC 12A Problems/Problem 13

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Problem

The regular 5-point star ABCDE is drawn and in each vertex, there is a number. Each A,B,C,D, and E are chosen such that all 5 of them came from set \{3,5,6,7,9\}. Each letter is a different number (so one possible way is A = 3, B = 5, C = 6, D = 7, E = 9). Let AB be the sum of the numbers on A and B, and so forth. If AB, BC, CD, DE, and EA form an arithmetic sequence (not necessarily in increasing order), find the value of CD.

(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13

Solution

AB + BC + CD + DE + EA = 2(A+B+C+D+E). The sum A + B + C + D + E will always be 3 + 5 + 6 + 7 + 9 = 30, so the arithmetic sequence has a sum of 2 \cdot 30 = 60. Since CD is the middle term, it must be the average of the five numbers, of \frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}.

See also

2005 AMC 12A (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
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