2018 AMC 8

Complete problem set with solutions and individual problem pages

Problem 23 Hard

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?

  • A.

    \frac 27

  • B.

    \frac {5}{42}

  • C.

    \frac {11}{14}

  • D.

    \frac 57

  • E.

    \frac 67

Answer:D

Solution 1

Choose side "lengths" a,b,c for the triangle, where "length" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing a=b if the triangle is isosceles: a+b+c=5, where either [a\leq b and a < c] or [a=b=c (but this is impossible in an octagon)].

Options are: a=0 with b,c in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and a=1 with { 1,3 ; 2,2}. \frac{5}{7} of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is \boxed{\textbf{(D) } \frac 57}

 

Solution 2

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 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}.