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

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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?
 
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]
+
<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==
 
==Solution==

Revision as of 16:42, 24 November 2018

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

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 obviously 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 $8*4$ choices. The total amount of choices is ${8 \choose 3} = 8*7$. Thus our answer is $\frac{8+8*4}{8*7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}$

Solution 2 (Complementary)

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 $\frac{8*6-8}{8*7}=\boxed{\textbf{(D) } \frac 57}$

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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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