Question

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?

$\dfrac 23$

$\dfrac{105}{128}$

$\dfrac{125}{128}$

$\dfrac{253}{256}$

$1$

Solution:

Count the complement: colorings of the $10$ edges of $K_5$ with no monochromatic triangle.

At any vertex, if $3$ incident edges had the same color, then the edges among their other endpoints would all have to be the other color, making a monochromatic triangle. Thus each vertex has exactly $2$ red and $2$ blue incident edges.

So the red edges form a $2$ -regular graph on $5$ vertices, which must be a $5$ -cycle. The number of labeled $5$ -cycles is $\frac{(5-1)!}{2}=12$ .

There are $2^{10}=1024$ total colorings, so the desired probability is $1-\frac{12}{1024}=\frac{253}{256}.$

Thus, the answer is D .

2021 AMC 10B Fall Problem 23

Solution:

Problem 23 in Other Years