2018 AMC 8 Problems/Problem 23

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?

$[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]$

Solution 1

We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and also $4$ places to put the remaining point, so we have $8\cdot4$ choices. The total amount of choices is ${8 \choose 3} = 8\cdot7$ . Thus our answer is $\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}$

Solution 2 (Complementary)

We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is $\frac{8\cdot6-8}{8\cdot7}=\boxed{\textbf{(D) } \frac 57}$

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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2018 AMC 8 Problems/Problem 23

Solution 1

Solution 2 (Complementary)

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