Difference between revisions of "2017 AMC 10B Problems/Problem 18"
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First we figure out the number of ways to put the <math>3</math> blue disks. Denote the spots to put the disks as <math>1-6</math> from left to right, top to bottom. The cases to put the blue disks are <math>(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)</math>. For each of those cases we can easily figure out the number of ways for each case, so the total amount is <math>2+2+3+3+1+1 = \boxed{\textbf{(D) } 12}</math>. | First we figure out the number of ways to put the <math>3</math> blue disks. Denote the spots to put the disks as <math>1-6</math> from left to right, top to bottom. The cases to put the blue disks are <math>(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)</math>. For each of those cases we can easily figure out the number of ways for each case, so the total amount is <math>2+2+3+3+1+1 = \boxed{\textbf{(D) } 12}</math>. | ||
Revision as of 19:57, 16 February 2017
Problem
In the figure below, 
of the 
disks are to be painted blue, 
are to be painted red, and 
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?
Solution
Solution 1
First we figure out the number of ways to put the blue disks. Denote the spots to put the disks as 
from left to right, top to bottom. The cases to put the blue disks are 
. For each of those cases we can easily figure out the number of ways for each case, so the total amount is 
.
Solution 2
Denote the discs as in the first solution. Ignoring reflections or rotations, there are 
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 
fixed points
2. Reflect about a line of symmetry. There are lines of reflections. Take the line of reflection going through the centers of circles and 
. Then, the colors of circles and must be the same, and the colors of circles 
and must be the same. This gives us fixed points per line of reflection
3. Rotate by 
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 , , and will be the same. Similarly, the colors of circles , , and will be the same. This is impossible, so this case gives us 
fixed points per rotation.
By Burnside's lemma, the total number of colorings is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
