2021 Fall AMC 10B Problems/Problem 23

Problem

Each of the $5{ }$ sides and the $5{ }$ diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?

$(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1$

Solution 1

Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.

$\textbf{Case 1}$ : all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection.

$\textbf{Case 2}$ : 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.

$\textbf{Case 3}$ : 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are ${4\choose2}$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6\cdot 2 = 12$ ways to color the pentagon so that no such triangle has the same color for all of its sides.

These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}$

~KingRavi

Solution 2 (Ramsey's Theorem)

This problem is related to a special case of Ramsey's Theorem, R(3, 3) = 6. Suppose we color every edge of a $6$ vertex complete graph $(K_6)$ with $2$ colors, there must exist a $3$ vertex complete graph $(K_3)$ with all it's edges in the same color. That is, $K_6$ with edges in $2$ colors contains a monochromatic $K_3$ . For $K_5$ with edges in $2$ colors, a monochromatic $K_3$ does not always exist.

This is a problem about the probability of a monochromatic $K_3$ exist in a $5$ vertex complete graph $K_5$ with edges in $2$ colors.

Choose a vertex, it has $4$ edges.

$\textbf{Case 1}$ : $\quad$ When $3$ or more edges are the same color, there must exist a monochromatic $K_3$ . Suppose the color is red, as shown below.

There is only $1$ way to color all the edges in the same color. There is $\binom{4}{3} = 4$ ways to color $3$ edges in the same color. There are $2$ colors. The probability of $3$ or more edges the same color is $\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}$ . So the probability of containing a monochromatic $K_3$ is $\frac{5}{8}$ .

$\textbf{Case 2}$ : $\quad$ When $2$ edges are the same color, graphs that does not contain a monochromatic $K_3$ can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic $K_3$ .

There are $\binom{4}{2} = 6$ ways to choose $2$ edges with the same color. For the other $4$ vertices there are $\binom{4}{2} = 6$ edges among them, there are $2^6 = 64$ ways to color the edges. There are only $2$ cases without a monochromatic $K_3$ .

So the probability without monochromatic $K_3$ is $\frac{2}{64} = \frac{1}{32}$ .

The probability with monochromatic $K_3$ is $1 - \frac{1}{32} = \frac{31}{32}$ .

From case 1 and case 2, the probability with monochromatic $K_3$ is $\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}$

~isabelchen

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2021 Fall AMC 10B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2 (Ramsey's Theorem)

See Also