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, 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 if the triangle is isosceles: , where either [ and ] or [ (but this is impossible in an octagon)].
Options are: with in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and with { 1,3 ; 2,2} of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is
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 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===Solution 1=== 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 by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=2678
~ pi_is_3.14
Video Solutions
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 |
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