2018 AMC 8 Problems/Problem 23

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?

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$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

Solutions

Solution 0

Choose side "lengths" $a,b,c$ for the triangle, where "length" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing $a=b$ if the triangle is isosceles: $a+b+c=5$ , where either [ $a\leq b$ and $a < c$ ] or [ $a=b=c$ (but this is impossible in an octagon)].

Options are: $a=0$ with $b,c$ in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and $a=1$ with { 1,3 ; 2,2} $5/7$ of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is $\boxed{\textbf{(D) } \frac 57}$

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 $4$ 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

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

$\frac{8\cdot6-8}{{8 \choose 3 }}$ $=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}$ .

Solution 3 (Stars and Bars)

Let $1$ point of the triangle be fixed at the top. Then, there are ${7 \choose 2} = 21$ ways to chose===Solution 1=== the other 2 points. There must be $3$ spaces in the points and $3$ points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and $2$ extra points (k-1) distributed so by the stars and bars formula, ${n+k-1 \choose k-1}$ , there are ${4 \choose 2} = 6$ ways to arrange the bars and stars. Thus, the probability is $\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}$ .

Simple Complementary Counting

POOOPY POOOPY POOP

Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=2678

~ pi_is_3.14

Video Solutions

https://www.youtube.com/watch?v=VNflxl7VpL0

https://youtu.be/YeYDixFXsvA

~savannahsolver

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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