2018 AMC 8 Problems/Problem 23
Contents
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?
Solutions
Solution 0
Choose side lengths for the triangle, starting from the shortest side, and going clockwise:
, where
a \leq c
\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)23
2$points are adjacent.
If all$ (Error compiling LaTeX. Unknown error_msg)38
2
8
4
8\cdot4
{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}2
8
6
3
3
8
\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}
1
{7 \choose 2} = 21
3
3
2
{n+k-1 \choose k-1}
{4 \choose 2} = 6
\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
~savannahsolver
gg
See Also
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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