Difference between revisions of "2018 AMC 8 Problems/Problem 23"
(→Solution) |
(→Solution 2 (Complementary)) |
||
Line 19: | Line 19: | ||
==Solution 2 (Complementary)== | ==Solution 2 (Complementary)== | ||
− | We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8 | + | We can decide <math>2</math> adjacent points with <math>8</math> choices. The remaining point will have <math>6</math> choices. However, we have counted the case with <math>3</math> adjacent points twice, so we need to subtract this case once. The case with the <math>3</math> adjacent points has <math>8</math> arrangements, so our answer is <math>\frac{8\cdot6-8}{8\cdot7}=\boxed{\textbf{(D) } \frac 57}</math> |
==See Also== | ==See Also== |
Revision as of 13:53, 15 July 2019
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?
Solution
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 (Complementary)
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
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.