Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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Choose side "lengths" <math>a,b,c</math> 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 <math>a=b</math> if the triangle is isosceles: <math>a+b+c=5</math>, where either [<math>a\leq b</math> and <math>a < c</math>] or [<math>a=b=c</math> (but this is impossible in an octagon)]. | Choose side "lengths" <math>a,b,c</math> 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 <math>a=b</math> if the triangle is isosceles: <math>a+b+c=5</math>, where either [<math>a\leq b</math> and <math>a < c</math>] or [<math>a=b=c</math> (but this is impossible in an octagon)]. | ||
Revision as of 14:30, 16 December 2023
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?
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Solutions
Solution 1
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
