Question

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

$\dfrac{2}{7}$

$\dfrac{5}{42}$

$\dfrac{11}{14}$

$\dfrac{5}{7}$

$\dfrac{6}{7}$

Answer: D

Concepts:complementary counting stars and bars

Difficulty rating: 1650

Video solution:

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Written solution:

Count the complement: triangles with no two chosen vertices adjacent. Starting at a chosen vertex, let $x,y,z$ be the positive numbers of unchosen vertices in the three gaps. Then $x+y+z=5$ , which has $\binom42=6$ positive solutions.

There are $8$ choices for the starting vertex, and each triangle is counted from each of its $3$ vertices, so the complement contains $8\cdot6/3=16$ triangles. Out of $\binom83=56$ total triangles, the desired probability is $1-16/56=5/7.$

Thus, D is the correct answer.

2018 AMC 8 Problem 23

Answer: D

Difficulty rating: 1650

Video solution:

Written solution:

Problem 23 in Other Years