2011 AMC 10A Problems/Problem 22
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Problem 22
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 1
Let vertex be any vertex, then vertex be one of the diagonal vertices to , be one of the diagonal vertices to , and so on. We consider cases for this problem.
In the case that has the same color as , has a different color from and so has a different color from and . In this case, has choices, has choices (any color but the color of ), has choice, has choices, and has choices, resulting in a possible of combinations.
In the case that has a different color from and has a different color from , has choices, has choices, has choices (since and necessarily have different colors), has choices, and has choices, resulting in a possible of combinations.
In the case that has a different color from and has the same color as , has choices, has choices, has choices, has choice, and has choices, resulting in a possible of combinations.
Adding all those combinations up, we get .
Solution 2
First, notice that there can be cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.
Case : There are ways of assigning each vertex a different color.
Case : There are ways. After picking four colors, we can rotate our pentagon in ways to get different outcomes.
Case : There are ways of arranging the final case. We can pick colors for our pentagon. There are spots for the first pair of colors. Then, there are possible ways we can put the final pair in the last spaces. But because the two pairs are indistinguishable, we divide by .
Adding all the possibilities up, we get
~ZericHang
Solution 3
There are ways to assign a color to . WLOG, give vertex a color; we can multiply by at the end. Since vertices and cannot have the same color, there are ways to assign colors to vertex . Using this same logic, there are ways to assign a color to vertices , , and , giving a total of ways. However, vertex cannot be the same color as vertex . To use complementary counting, we need to find the amount of ways for and to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are ways, except we must subtract the number of ways for a triangle. Each time, there is less vertex, so times less ways to color. This process stops when there are only vertices left; in this case there are simply ways to color this figure. So in conclusion, there are ways.
Solution 4
This problem is a direct application of the chromatic polynomial of a graph, represented by , which returns a polynomial representing the number of ways to color a graph with colors such that no two adjacent vertices share the same color. In other words, it gives us a polynomial in terms of where we can plug in to get our answer.
Well, if we want to find for a graph , we should probably first draw the graph, right? We can first draw five vertices (without any edges) in the shape of a pentagon. After connecting the diagonals, we get a star. We attempt to shift the vertices around to simplify the graph, which we quickly realize is isomorphic (can be turned into) the five-cycle, or the pentagon.
Finding on a graph with a cycle of length at most 3 is straightforward -- we first pick a vertex v. It has 0 colored (visited) neighbors, so we can color it in k ways. We then move on to the vertices adjacent to v, etc, and at the end we multiply all these together. For example, the chromatic polynomial of a triangle is . However, cycles with length > 3 introduce ambiguity, and thus the above technique fails. Thus, we need to use the recursive formula , where represents removing e and represents contracting e, or collapsing the two endpoints of e into one. When we hit a graph where the longest cycle has length 3, we can use the first technique to quickly find .
After about four iterations of the algorithm, we get that the chromatic polynomial is . Plugging in 6 gives us .
~Helloworld1
Solution 5 (constructive counting)
In pentagon , fix any vertex . Now draw diagonal . There are six choices for vertex and choices for vertex
Now draw diagonal . Since cannot be the same color as vertex , we have choices for . Again, we have five choices for vertex (draw diagonal ).
Thus there are choices for vertices and and combinations for and .
To determine the final count, we consider two cases for the final combinations of and , which uniquely determines . Then, we multiply by since the choices of and are independent from these two cases.
Case : and are the same color. There are possible pairs (This is because and are not chosen from the same 5 colors. cannot be as they are on a diagonal, but can be. ~ primegn), and thus we have choices for . There are cases here.
Case : and are different. There are possible combinations and we have choices for (not the color of nor ). In this case we have cases.
Our final count is , which is answer .
~FIREDRAGONMATH16
Solution 6 (Casework)
WLOG, draw such that point is at the top, and write the letters in counterclockwise order. WLOG, fill in first. There are ways to do so. From here we proceed with casework on the color of :
- Case 1: is the same color as
In this case, there is option for , options for , options for , and options for . .
- Case 2: and are different colors
In this case, there are options for , options for , options for , but we notice we need to do casework on the color of .
- Case 2.1: and are the same color
There are options for , options for , option for , options for , and options for . .
- Case 2.2: and are different colors
There are options for , options for , options for , options for , and options for . .
Summing up the cases, .
~JH. L
Solution 7
Notice that a minimum of, and a maximum of colours can be used.
- Case : colours used.
This is straight forward and there are ways.
- Case : colours used.
If and have same colours, they have to be adjacent. Let be directly right of . consecutive points from clockwise have different colours. They can be coloured in ways. And can be chosen in ways, bringing the total to .
- Case : colours used.
Note that points can't have the same colour as at least of the lines joining them will be a diagonal. Then the arrangement is . The point with distinct colour can be chosen in ways and can be coloured in ways. The next two points clockwise can be coloured in ways and the next two in ways. The total is .
Finally, adding all cases, the answer is .
~IamBATMAN1313
Solution 8
Name the pentagon . Then and have to be of different colours.
can be coloured in ways, in ways (Without the one used for ), in ways (Without the one used for ), in ways (Without the one used for ).
Now we find the expected value of how many ways can be coloured.
If and are of the same colour, and are always different. This happens times and leaves options for .
If and are of different colours then there is a chance that and are of same colour and chance that they are different. The first one leaves ways to colour and the second one leaves .
Calculating all these, we find the expected value for the number of ways can be coloured is
Therefore total number or ways .
~IamBATMAN1313
Solution 9 (Elimination of answers)
First, we know that there are ways to assign a color to vertex . Then, there are ways to assign a color to vertex , because it doesn't depend on vertex . There are also ways to assign a color to vertex , because it can't be the same color as vertex . Notice that both vertices and have 5 or 4 choices, depending on whether , , and are the same color. This means that the we know that the amount of choices must be bigger than , which eliminates options and .
Then, notice that the amount of choices must be smaller than , because the expected amount of choices and have are less than 4.5. (There is a greater probability that , , and , are both different). This eliminates option , which leaves us with options and .
We know that the amount of choices must be a multiple of , so is eliminated, leaving us with option
~Jonathanzhou18
Video Solution
~IceMatrix
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
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Followed by Problem 23 | |
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