AMC 10 · 2003 · #10

Grade 8 counting
combinations-basicrotation-isometryregular-pentagon systematic-enumerationsymmetry-argument ↑ Prerequisites: combinations-basic
📏 Long solution 💡 3 insights 📊 Diagram
Problem
A regular pentagon has five sides, each with one outward-pointing equilateral triangle position beside it. Glue on exactly two of those five triangles. Two finished figures count as the same when one can be turned or flipped onto the other. Count how many genuinely different figures come out.

Pick an answer.

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

The picture makes the five triangle positions look different, but the pentagon's own symmetry makes them interchangeable, so the honest way to count is to list every choice and then decide which listed choices are secretly the same figure. Tool #2 (Make a Systematic List) produces all ten two-side choices with nothing missed and nothing doubled. Tool #1 (Draw a Diagram) turns the pentagon into five numbered slots around a ring, which is the whole picture the problem depends on. Tool #5 (Look for a Pattern) finds the feature that never changes when the pentagon is turned or flipped — how far apart the two chosen sides are around the ring — and sorting by that feature is what collapses ten choices into a much smaller number of shapes. The final and most-skipped move is proving the surviving families really are different, not just different-looking.

1STEP 1

Number the five slots around the ring

Number the slots around a ring, so a figure is just a pair of slots.

slots {1,2,3,4,5} on a ring; figure = a pair of slots; turn: i ↦ i+1 (wrapping 5↦ 1)
2STEP 2

List all ten choices

There are 10 such pairs, but that counts choices, not shapes.

C(5, 2)=(5 · 4)/2=10 choices of two slots
3STEP 3

Sort by the gap, and watch the list collapse

Turning and flipping cannot change the gap, and on a five-ring only two gaps exist.

gap 1: {1,2},{2,3},{3,4},{4,5},{5,1} gap 2: {1,3},{2,4},{3,5},{4,1},{5,2} 5+5=10
4STEP 4

Show the two families really are different shapes

One gap makes the triangles touch and the other keeps them apart, so the two are genuinely different, choice (B).

gap 1 → triangles share a vertex; gap 2 → triangles share no point; answer 2 → (B)
Answer
2
The split is 5+5=10, so both families are non-empty and every one of the ten choices is accounted for exactly once — the count cannot be smaller than 2 unless the families merge, and the shared-vertex test showed they do not. Choice (A) 1 would require the touching figure to be congruent to the separated one. Choice (E) 5 counts choices with one slot pinned rather than shapes. An independent arithmetic check comes from the averaging rule for counting shapes: over the pentagon's 10 symmetries, count how many of the 10 pairs each one leaves unchanged — the do-nothing move fixes all 10, each of the 4 real turns fixes 0 (a turn shifts both slots, so no pair survives), and each of the 5 flips fixes exactly 2 pairs (the two pairs of slots the flip swaps). The average is (10+4 · 0+5 · 2)/10=20/10=2, agreeing with the direct count. Note this averaging is the correct rule; simply dividing 10 by 5 happens to give 2 here only because no pair is left unchanged by a real turn, and that is a fact about this pentagon, not a general shortcut.
💡Key takeaway

When a shape can be turned and flipped, count the features that survive those moves — here only whether the two triangles touch — instead of counting the positions you happened to label.

  • Number the five slots around the ring
  • List all ten choices
  • Sort by the gap, and watch the list collapse
  • Show the two families really are different shapes