2005 AMC 12A Problems/Problem 13
Problem
In the five-sided star shown, the letters , , , and 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 , , , , and form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?
Solutions
Solution 1
(i.e., each number is counted twice). The sum will always be , so the arithmetic sequence has a sum of . The middle term must be the average of the five numbers, which is .
Solution 2
Let the terms in the arithmetic sequence be , , , , and . We seek the middle term .
These five terms are , , , , and , in some order. The numbers , , , , and are equal to 3, 5, 6, 7, and 9, in some order, so Hence, the sum of the five terms is But adding all five numbers, we also get , so Dividing both sides by 5, we get , which is the middle term. The answer is (D).
Solution 3
Not too bad with some logic and the awesome guess and check. Let . Then let and . Our arithmetic sequence is so our answer is .
Solution by franzliszt
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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All AMC 12 Problems and Solutions |
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