Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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<math>\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}</math> | <math>\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}</math> | ||
− | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
We will use constructive counting to solve this. There are <math>2</math> cases: Either all <math>3</math> points are adjacent, or exactly <math>2</math> points are adjacent. | We will use constructive counting to solve this. There are <math>2</math> cases: Either all <math>3</math> points are adjacent, or exactly <math>2</math> points are adjacent. | ||
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===Solution 3 (Stars and Bars)=== | ===Solution 3 (Stars and Bars)=== | ||
Let <math>1</math> point of the triangle be fixed at the top. Then, there are <math>{7 \choose 2} = 21</math> ways to chose the other 2 points. There must be <math>3</math> spaces in the points and <math>3</math> points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and <math>2</math> extra points (k-1) distributed so by the stars and bars formula, <math>{n+k-1 \choose k-1}</math>, there are <math>{4 \choose 2} = 6</math> ways to arrange the bars and stars. Thus, the probability is <math>\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}</math>. | Let <math>1</math> point of the triangle be fixed at the top. Then, there are <math>{7 \choose 2} = 21</math> ways to chose the other 2 points. There must be <math>3</math> spaces in the points and <math>3</math> points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and <math>2</math> extra points (k-1) distributed so by the stars and bars formula, <math>{n+k-1 \choose k-1}</math>, there are <math>{4 \choose 2} = 6</math> ways to arrange the bars and stars. Thus, the probability is <math>\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}</math>. | ||
− | + | ||
+ | == Video Solution == | ||
+ | https://youtu.be/5UojVH4Cqqs?t=2678 | ||
==Video Solution== | ==Video Solution== | ||
− | https://www.youtube.com/watch?v=VNflxl7VpL0 | + | https://www.youtube.com/watch?v=VNflxl7VpL0 |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2018|num-b=22|num-a=24}} | {{AMC8 box|year=2018|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:02, 26 October 2021
Contents
Problem
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 1
We will use constructive counting to solve this. There are cases: Either all points are adjacent, or exactly points are adjacent.
If all points are adjacent, then we have choices. If we have exactly adjacent points, then we will have places to put the adjacent points and also places to put the remaining point, so we have choices. The total amount of choices is . Thus our answer is
Solution 2
We can decide adjacent points with choices. The remaining point will have choices. However, we have counted the case with adjacent points twice, so we need to subtract this case once. The case with the adjacent points has arrangements, so our answer is
Solution 3 (Stars and Bars)
Let point of the triangle be fixed at the top. Then, there are ways to chose the other 2 points. There must be spaces in the points and points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and extra points (k-1) distributed so by the stars and bars formula, , there are ways to arrange the bars and stars. Thus, the probability is .
Video Solution
https://youtu.be/5UojVH4Cqqs?t=2678
Video Solution
https://www.youtube.com/watch?v=VNflxl7VpL0
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.