Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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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>. | ||
+ | |||
+ | ===Simple Complementary Counting=== | ||
+ | By rotational symmetry, choose an arbitrary point for one vertex. Then choose one of the 5 non-adjacent vertices, out of 7 possible. Sum the number of remaining non-adjacent vertices, for all 5 cases: 3+2+2+2+3=12, out of 6 possible for each. These are non-edge triangles, so the probability of edge triangles is <math>1-\frac{12}{7\cdot 6}= 5/7</math> | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 15:10, 2 January 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?
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
.
Simple Complementary Counting
By rotational symmetry, choose an arbitrary point for one vertex. Then choose one of the 5 non-adjacent vertices, out of 7 possible. Sum the number of remaining non-adjacent vertices, for all 5 cases: 3+2+2+2+3=12, out of 6 possible for each. These are non-edge triangles, so the probability of edge triangles 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
gg
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.