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2017 AMC 10B Problems/Problem 18

Revision as of 14:04, 30 August 2017 by Euler14 (talk | contribs) (→‎See Also)

Problem

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?

[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution

Solution 1

First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$. For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \boxed{\textbf{(D) } 12}$.

Solution 2

Denote the $6$ discs as in the first solution. Ignoring reflections or rotations, there are $\binom{6}{3} * \binom{3}{2} = 60$ colorings. Now we need to count the number of fixed points under possible transformations:

1. The identify transformation. Since this doesn't change anything, there are $60$ fixed points

2. Reflect about a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$. Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection

3. Rotate by $120^\circ$ counter clockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$, $4$, and $6$ will be the same. Similarly, the colors of circles $2$, $3$, and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation.

By Burnside's lemma, the total number of colorings is $(1*60+3*4+2*0)/(1+3+2) = \boxed{\textbf{(D) } 12}$.


Solution 3

Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as [asy] filldraw(circle((0,0),1),green); draw(circle((2,0),1)); draw(circle((4,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((3,sqrt(3)),1)); draw(circle((2,2sqrt(3)),1)); draw(circle((8,0),1)); filldraw(circle((10,0),1),green); draw(circle((12,0),1)); draw(circle((9,sqrt(3)),1)); draw(circle((11,sqrt(3)),1)); draw(circle((10,2sqrt(3)),1)); [/asy] Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are $\dbinom{5}{2}=10$ of these. However, due to reflection we must divide this by two. But, in two of these cases, the reflection is itself, so we must subtract these out before dividing by 2, and add them back afterwards, giving $\frac{10-2}{2}+2=6$ arrangement in this case.

Now, look at the second case. We again must pick two of the five remaining circles, and like in the first case, two of the reflections give the same arrangement. Thus, there are also $6$ arrangements in this case.

In total, we have $6+6=\boxed{\text{\bf(D) }12}$.

Solution by tdeng

Solution: Burnsides+Guessing, for if you are running out of time but want some free points

By burnsides, we quickly find the set $G$ of all group actions that can be applied to one coloring to obtain another. $G={s_1,s_2, s_3,120^\circ,240^\circ, e}$, where all $s_k$ are lines of reflection symmetry and $120^\circ,240^\circ$ are the rotations, and $e$ is the identity. Note that all $C(6,3)*C(3,2)=60$ colorings are fixed by the identity. Let's say there are a total of $K$ other fixed points (that let's say we don't have time to calculate) from the other group actions. Using Burnsides, we have $(60+K)/6=10+K/6$. At this point, we can clearly get rid of (A), (B), and (C). We have now narrowed down to two answers, and if you're lucky, you can guess $\boxed{\text{\bf(D) }12}$

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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