2018 AMC 8 Problems/Problem 23
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.
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 Solutions
https://youtu.be/5UojVH4Cqqs?t=2678
https://www.youtube.com/watch?v=VNflxl7VpL0
~savannahsolver
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.