2005 AMC 10A Problem 17

You do not need to find where each number goes. Add up all five sums instead.

1. Each letter is used twice. The segments go $A$ to $B$, $B$ to $C$, $C$ to $D$, $D$ to $E$, and $E$ back to $A$. Every letter is at the end of exactly two of them. For example, $A$ is on $AB$ and $EA$.
2. Total of the five sums. Since each number is counted twice,
$$\text{total} = 2 \times (3 + 5 + 6 + 7 + 9) = 2 \times 30 = 60.$$
3. Middle term of an arithmetic sequence. In five evenly spaced numbers, the middle one is the average. For example, $10, 11, 12, 13, 14$ has average $12$, the middle term. The terms below and above the middle balance out.
4. Compute. The average of the five sums is $\frac{60}{5} = 12$.

It is possible to do this: $A = 3$, $B = 9$, $C = 5$, $D = 6$, $E = 7$ gives sums $12, 14, 11, 13, 10$, which sort to $10, 11, 12, 13, 14$.

The answer is $\boxed{\textbf{(D)}\ 12}$.

Two ideas work together. First, when every number is used in exactly two sums, the sums together add to twice the numbers. Second, in an arithmetic sequence with an odd number of terms, the middle term equals the average. Neither idea needs the arrangement, which is why the problem says "not necessarily in this order."

The five segments $AB, BC, CD, DE, EA$ form a closed loop through the five letters, so each letter is an endpoint of exactly two segments. Adding up all five segment sums therefore counts every vertex value twice:
$$\text{(sum of the five terms)} = 2(3 + 5 + 6 + 7 + 9) = 2 \cdot 30 = 60.$$

In an arithmetic sequence with an odd number of terms, the terms are symmetric about the middle one, so the middle term equals the average of all terms:
$$\frac{60}{5} = 12.$$

(An assignment does exist, e.g. $A = 3, B = 9, C = 5, D = 6, E = 7$ gives sums $12, 14, 11, 13, 10$, the sequence $10, 11, 12, 13, 14$.)

The answer is $\boxed{\textbf{(D)}\ 12}$.