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Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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==Problem 23==
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LOOOOOOL
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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?
 
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<asy>
 
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size(3cm);
 
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pair A[];
 
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for (int i=0; i<9; ++i) {
 
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  A[i] = rotate(22.5+45*i)*(1,0);
 
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}
 
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filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black);
 
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for (int i=0; i<8; ++i) { dot(A[i]); }
 
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</asy>
 
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<math>\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 14:22, 24 November 2018

LOOOOOOL

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 (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

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