AMC 10 · 2017 · #13

Grade 8 geometry-2d
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
combinations-basicsymmetry-argumentpermutations-basic burnsides-lemmacasework ↑ Prerequisites: combinations-basic
📏 Long solution 💡 3 insights 📊 Diagram
Problem
Six disks sit in a triangle: one on top, two in the middle row, three across the bottom. Paint three of them blue, two red, and one green. Two paintings count as the same if one becomes the other by rotating or flipping the whole triangle. Find how many genuinely different paintings there are.

Pick an answer.

(A)
6
(B)
8
(C)
9
(D)
12
(E)
15
How to solve
Strategy Make a Systematic List

Tool #2 (Make a Systematic List): a "how many different" question with fixed color counts starts by counting every painting as if positions were labeled — 6!/3! 2! 1!=60. Tool #17 (Visualize Spatial Relationships): mentally rotating and folding the triangle reveals its 6 symmetries and how each one shuffles the disks. Tool #16 (Change Focus): instead of chasing distinct paintings directly, I count, for each symmetry, how many paintings it leaves unchanged, then average — that average is the count of distinct paintings (Burnside's idea). Tool #7 (Identify Subproblems): the average splits into easy pieces — handle the identity, the two rotations, and the three reflections separately.

1STEP 1

Count every labeled painting

With the spots labeled there are 60.

6!/3! 2! 1!=720/(6 · 2 · 1)=60
2STEP 2

List the triangle's six symmetries

The triangle has six symmetries.

{identity, rot 120°, rot 240°, 3 reflections} → 6 symmetries
3STEP 3

Paintings fixed by each rotation

A rotation leaves no painting alone.

corners all one color and midpoints all one color → impossible; fixed=0
4STEP 4

Paintings fixed by each reflection

Each reflection leaves four alone.

2_which pair is blue × 2_blue/green on axis = 4 per reflection
5STEP 5

Average the fixed counts

Averaging the fixed counts gives 12.

(60+0+0+4+4+4)/6=72/6=12 → (D)
Answer
12
The pieces are consistent: 60 labeled paintings, an order-6 symmetry group, and fixed counts 60,0,0,4,4,4 summing to 72, a clean multiple of 6. If symmetry never collapsed anything you would expect roughly 60/6=10 paintings; the answer 12 is a bit higher because the rotations fix nothing (they merge fewer paintings), which nudges the count above 10. Twelve is among the choices, so (D) fits.
💡Key takeaway

Count all 60 labeled paintings, then for each of the triangle's 6 flips and turns count how many it leaves unchanged (60,0,0,4,4,4); the average 72 ÷ 6=12 is how many are truly different.

  • Count every labeled painting
  • List the triangle's six symmetries
  • Paintings fixed by each rotation
  • Paintings fixed by each reflection
  • Average the fixed counts

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