Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 8 Problems/Problem 23"

(→‎Solution 2)
(→‎Solution 3)
Line 23: Line 23:
 
==Solution 3==
 
==Solution 3==
  
−
Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is <math>\frac{21 - 6\21) = </math>\frac{5/7}.
+
Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is (21 - 6)\21 = 5/7.
  
 
==See Also==
 
==See Also==

Revision as of 00:44, 12 November 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?

[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) { A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]

Solution 1

We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and also $4$ places to put the remaining point, so we have $8\cdot4$ choices. The total amount of choices is ${8 \choose 3} = 8\cdot7$. Thus our answer is $\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}$

Solution 2

We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is $\frac{8\cdot6-8}{{8 \choose 3 }}$$=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}$

Solution 3

Let 1 point of the triangle be fixed at the top. Then, there are 7C2 = 21 ways to chose the other 2 points. There must be 3 spaces in the points and 3 points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 things (triangle points) and 2 things (extra points) distributed so there are 4C2 = 6ways to arrange the bars and stars. Thus, the probability is (21 - 6)\21 = 5/7.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&oldid=111274"