Difference between revisions of "2011 AMC 12A Problems/Problem 16"
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== Solution == | == Solution == | ||
| − | + | We can do some casework when working our way around the pentagon from <math>A</math> to <math>E</math>. At each stage, there will be a makeshift diagram. | |
| + | |||
| + | 1.) For <math>A</math>, we can choose any of the 6 colors. | ||
| + | |||
| + | A : 6 | ||
| + | |||
| + | 2.) For <math>B</math>, we can either have the same color as <math>A</math>, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and <math>D</math> will be affected by both <math>A</math> an <math>B</math>. | ||
| + | |||
| + | A : 6 | ||
| + | B:1 B:5 | ||
| + | |||
| + | 3.) For <math>C</math>, we cannot have the same color as <math>A</math>. Also, we can have the same color as <math>B</math> (<math>E</math> will be affected), or any of the other 4 colors. Because <math>C</math> can't be the same as <math>A</math>, it can't be the same as <math>B</math> if <math>B</math> is the same as <math>A</math>, so it can be any of the 5 other colors. | ||
| + | |||
| + | A : 6 | ||
| + | B:1 B:5 | ||
| + | C:5 C:4 C:1 | ||
| + | |||
| + | 4.) <math>D</math> is affected by <math>A</math> and <math>B</math>. If they are the same, then <math>D</math> can be any of the other 5 colors. If they are different, then <math>D</math> can be any of the (6-2)=4 colors. | ||
| + | |||
| + | A : 6 | ||
| + | B:1 B:5 | ||
| + | C:5 C:4 C:1 | ||
| + | D:5 D:4 D:4 | ||
| + | |||
| + | 5.) <math>E</math> is affected by <math>B</math> and <math>C</math>. If they are the same, then <math>E</math> can be any of the other 5 colors. If they are different, then <math>E</math> can be any of the (6-2)=4 colors. | ||
| + | |||
| + | A : 6 | ||
| + | B:1 B:5 | ||
| + | C:5 C:4 C:1 | ||
| + | D:5 D:4 D:4 | ||
| + | E:4 E:4 E:5 | ||
| + | |||
| + | 6.) Now, we can multiply these three paths and add them: | ||
| + | <math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120</math> | ||
| + | |||
| + | 7.) Our answer is <math>C</math>! | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}} | {{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}} | ||
Revision as of 19:19, 17 February 2012
Problem
Each vertex of convex pentagon 
is to be assigned a color. There are 
colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
Solution
We can do some casework when working our way around the pentagon from 
to 
. At each stage, there will be a makeshift diagram.
1.) For , we can choose any of the 6 colors.
A : 6
2.) For 
, we can either have the same color as , or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and 
will be affected by both an .
A : 6 B:1 B:5
3.) For 
, we cannot have the same color as . Also, we can have the same color as ( will be affected), or any of the other 4 colors. Because can't be the same as , it can't be the same as if is the same as , so it can be any of the 5 other colors.
A : 6 B:1 B:5 C:5 C:4 C:1
4.) is affected by and . If they are the same, then can be any of the other 5 colors. If they are different, then can be any of the (6-2)=4 colors.
A : 6 B:1 B:5 C:5 C:4 C:1 D:5 D:4 D:4
5.) is affected by and . If they are the same, then can be any of the other 5 colors. If they are different, then can be any of the (6-2)=4 colors.
A : 6 B:1 B:5 C:5 C:4 C:1 D:5 D:4 D:4 E:4 E:4 E:5
6.) Now, we can multiply these three paths and add them:
7.) Our answer is !
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
