Difference between revisions of "2017 AMC 12B Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | + | Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green ball is in a corner. This yields <math>6</math> possible arrangements for the <math>3</math> blue balls and <math>2</math> red balls in the remaining available slots. Now, consider the case that the green ball is on an edge. This yields <math>6</math> more possible arrangements for the <math>3</math> blue balls and <math>2</math> red balls in the remaining available slots. Thus, our answer is <math>6 + 6 = \boxed{\bold{(D)}\, 12}</math> | |
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| + | Solution by akaashp11 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2017|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:46, 19 February 2017
Problem 13
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?
![[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]](http://latex.artofproblemsolving.com/e/b/a/eba8bbf32167d960fbb94db72325cb3a2f89c234.png)
Solution
Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green ball is in a corner. This yields possible arrangements for the blue balls and red balls in the remaining available slots. Now, consider the case that the green ball is on an edge. This yields more possible arrangements for the blue balls and red balls in the remaining available slots. Thus, our answer is 
Solution by akaashp11
See Also
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
