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, starting from the shortest side, and going clockwise: $a+b+c=8$ , where $a\leq b,$ a \leq c $. Options are: 1,1,6 ; 1,2,5 ; 1,3,4 ; 1,4,3 ; 1,5,2 ; 2,2,4 ; 2,3,3.$ \boxed(5/7)$of these have a side with length 1, which corresponds to an edge of the octagon.

===Solution 1=== We will use constructive counting to solve this. There are$ (Error compiling LaTeX. Unknown error_msg)2 $cases: Either all$ 3 $points are adjacent, or exactly$ 2$points are adjacent.

If all$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)\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}Template:8 \choose 3$$ (Error compiling LaTeX. Unknown error_msg)=\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=== By rotational symmetry, choose an arbitrary point for one vertex. Then choose one of the 5 non-adjacent vertices, out of 7 possible. Sum the number of remaining non-adjacent vertices, for all 5 cases: 3+2+2+2+3=12, out of 6 possible for each. These are non-edge triangles, so the probability of edge triangles is$ (Error compiling LaTeX. Unknown error_msg)1-\frac{12}{7\cdot 6}= 5/7$

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

gg

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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