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2005 AMC 10A Problem 17

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2005 AMC 10A Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Sequences & Series · difficulty 3 of 5 · about 3 min · position P16-20

In the five-sided star shown, the letters AA , BB , CC , DD , and EE are replaced by the numbers 33 , 55 , 66 , 77 , and 99 , although not necessarily in this order. The sums of the numbers at the ends of the line segments AB‾\overline{AB} , BC‾\overline{BC} , CD‾\overline{CD} , DE‾\overline{DE} , and EA‾\overline{EA} form an arithmetic sequence, although not necessarily in this order. What is the middle term of the arithmetic sequence?

figure

  1. A)

    99

  2. B)

    1010

  3. C)

    1111

  4. D)

    1212

  5. E)

    1313

Answer

D

Key insight

Every vertex lies on exactly two segments, so the five sums total 2(3+5+6+7+9) = 60, and the middle term of five in arithmetic progression is their average, 12.

Solution

The five segments AB,BC,CD,DE,EAAB, 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:
(sum of the five terms)=2(3+5+6+7+9)=2⋅30=60.\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:
605=12.\frac{60}{5} = 12.

(An assignment does exist, e.g. A=3,B=9,C=5,D=6,E=7A = 3, B = 9, C = 5, D = 6, E = 7 gives sums 12,14,11,13,1012, 14, 11, 13, 10, the sequence 10,11,12,13,1410, 11, 12, 13, 14.)

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

Why this works

Two ideas combine: double counting (a quantity summed over edges can be re-summed over vertices) and the fact that the mean of an arithmetic sequence is its middle term. Neither requires knowing the arrangement, which is why the problem says "not necessarily in this order" twice: the order is irrelevant.

The trap

Trying to find the actual assignment of numbers to vertices by trial and error instead of using the double-counting shortcut.

Common mistakes

  • Trying to find the actual assignment of numbers to vertices by trial and error instead of using the double-counting shortcut.
  • Dividing the vertex total 3030 by 55 to get 66, forgetting that each segment sum involves two vertices.

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects