AMC 10 · 2017 · #18

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 3 of them blue, 2 red, and 1 green. Two paintings count as the same if one becomes the other by rotating or flipping the whole triangle. How many genuinely different paintings are there?

Pick an answer.

(A)
6
(B)
8
(C)
9
(D)
12
(E)
15

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Ignore symmetry: arranging the multiset {B,B,B,R,R,G} in six spots gives 6!3!⋅2!⋅1!\frac{6!}{3!·2!·1!} = 60 labeled paintings.

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

List the triangle's six symmetries

Its symmetries: the identity, 120° and 240° rotations, and 3 reflections through a corner and the opposite midpoint — 6 symmetries total.

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

Paintings fixed by each rotation

A rotation needs all 3 corners one color and all 3 midpoints one color, impossible from 3 blue, 2 red, 1 green — so each rotation fixes 0.

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

Paintings fixed by each reflection

Each reflection pairs off-axis disks (one color each) and fixes 2 axis disks; blue-pair/red-pair choice ×2 and axis blue/green ×2 give 4.

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

Average the fixed counts

Burnside: average the fixed counts (60, 0, 0, 4, 4, 4) over all 6 symmetries — 60+0+0+4+4+46\frac{60+0+0+4+4+4}{6} = 726\frac{72}{6} = 12 distinct paintings, choice (D).

(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 606\frac{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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