2018 AMC 8 Problems/Problem 23
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 0
Choose side "lengths" 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
if the triangle is isosceles:
, where either [
and
] or [
(but this is impossible in an octagon)].
Options are: with
in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and
with { 1,3 ; 2,2} $\frac{5/7}$ (Error compiling LaTeX. Unknown error_msg) of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is
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
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===Solution 1=== 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
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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