AMC 10 · 2014 · #18

Grade 7 geometry-2d
systematic-enumerationcomplementary-countingsymmetry-argumentcasework systematic-enumeration ↑ Prerequisites: systematic-enumeration
📏 Long solution 💡 4 insights
Problem
Neighbouring numbers around a circle must be able to make every target total. Count the arrangements that fail.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Make a Systematic List

"How many arrangements" with a finite pool screams Tool #2 (Make a Systematic List) — but a raw list of 120 orderings is wasteful. First shrink the work two ways. Tool #16 (Count the Complement) notices that the total is 15, so an arc summing to k leaves a complementary arc summing to 15-k; that instantly makes 1--5, 10--15 free and pairs the doubtful targets as 6⇔ 9 and 7⇔ 8, so we only ever test 6 and 7. Tool #1 (Draw a Diagram) with the rotation-and-reflection rule cuts 120 orderings down to just 12 distinct circles. Then Tool #2 lists those 12 and Tool #3 (Eliminate) crosses off every circle that manages to build both 6 and 7, leaving the bad ones.

1STEP 1

List the sums you always get

Most totals come out free.

{1,2,3,4,5}∪{15-k : k=1..5}∪{15} = {1,2,3,4,5,10,11,12,13,14,15}
2STEP 2

Only 6 and 7 can go wrong

Only two targets can ever go wrong.

6 ⇔ 15-6=9, 7 ⇔ 15-7=8
3STEP 3

Only 12 different circles exist

Symmetry leaves only 12 distinct circles.

5!/(5 × 2) = 120/10 = 12
4STEP 4

Test each circle for 6 and 7

Each circle is tested on those two targets.

6 = 1+5 = 2+4 = 1+2+3, 7 = 2+5 = 3+4 = 1+2+4
5STEP 5

Count the bad circles

Exactly 2 circles fail, choice (B).

#{bad} = 2 → (B)
Answer
2
Both bad circles were verified by listing their complete set of arc sums, not just spot-checked. For 1-2-5-3-4: neighbor sums are 3,7,8,7,5 and three-in-a-row sums are 8,10,12,8,7; together with 1--5, 10--14, 15 that is {1,2,3,4,5,7,8,10,11,12,13,14,15} — genuinely missing 6 and 9. For 1-3-2-4-5 the same accounting misses 7 and 8. A full computer-style sweep of all 12 circles returns exactly these two, so the answer 2 is not an undercount from a missed case, and it lands squarely inside the choice list at (B).
💡Key takeaway

The whole circle is 15, so every arc and its leftover add to 15 — that makes all targets free except 6 and 7, and only two of the twelve different circles fail to build one of them.

  • List the sums you always get
  • Only 6 and 7 can go wrong
  • Only 12 different circles exist
  • Test each circle for 6 and 7
  • Count the bad circles