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2017 AMC 10B Problem 18

2017 AMC 10B Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Arrangements with Restrictions · difficulty 4 of 5 · about 6 min · position P16-20

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

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

Answer

Key insight

Fix the unique green disk on a corner or a midpoint; the 10 red placements then fall into 6 classes under the one remaining reflection.

Solution

The six disks form a triangle: three corner disks and three edge-midpoint disks. The symmetries are the three rotations and three reflections of the triangle; every symmetry sends corners to corners and midpoints to midpoints.

Use the unique green disk to break the symmetry. Up to rotation, green is either on a corner or on a midpoint.

Case 1: green on a corner. The remaining symmetry is the reflection through that corner and the opposite midpoint. Choose $2$ of the other $5$ disks to be red: $\binom52 = 10$ ways. Under the reflection, a placement is unchanged only if the two reds are a mirror pair: the two other corners, or the two other midpoints ( $2$ placements). The remaining $8$ placements pair off into $4$ classes. Distinct paintings: $2 + 4 = 6$ .

Case 2: green on a midpoint. The remaining symmetry is the reflection through that midpoint and the opposite corner. Again $\binom52 = 10$ red placements; the mirror pairs are the two other midpoints or the two other corners ( $2$ placements), and the other $8$ pair off into $4$ . Distinct paintings: $6$ .

Total: $6 + 6 = 12$ .

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

Why this works

A unique object (the green disk) lets you fix its position and kill most of the symmetry group; what survives is a single reflection, and counting orbits of one reflection is easy: fixed placements count once, the rest count in pairs. This is Burnside's lemma done by hand.

Alternative approach

Burnside directly: the identity fixes $\frac{6!}{3!\,2!\,1!} = 60$ colorings; each nontrivial rotation fixes $0$ (it would need a monochromatic $3$ -cycle of corners and of midpoints, impossible with $2$ reds and $1$ green); each of the $3$ reflections fixes $4$ (green on one of the $2$ fixed disks, reds on one of the $2$ swapped pairs). Orbits: $\frac{60 + 0 + 0 + 3 \cdot 4}{6} = 12$ .

The trap

Dividing the 60 raw colorings by 6 symmetries to get 10, which is wrong because some colorings are fixed by a reflection.

Common mistakes

Dividing the 60 raw colorings by 6 symmetries to get 10, which is wrong because some colorings are fixed by a reflection.
Treating "corner" and "midpoint" disks as interchangeable, or forgetting that the reflection fixes two disks (one corner and one midpoint), which changes which red pairs are self-symmetric.

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects