Difference between revisions of "2018 AMC 8 Problems/Problem 23"
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==Solutions== | ==Solutions== | ||
| + | ===Solution 0=== | ||
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| + | Choose side lengths <math>a,b,c</math> for the triangle, starting from the shortest side, and going clockwise: <math>a+b+c=8</math>, where <math>a\leq b, </math>a \leq c<math>. Options are: 1,1,6 ; 1,2,5 ; 1,3,4 ; 1,4,3 ; 1,5,2 ; 2,2,4 ; 2,3,3. </math>\boxed(5/7)<math> of these have a side with length 1, which corresponds to an edge of the octagon. | ||
| + | |||
===Solution 1=== | ===Solution 1=== | ||
| − | We will use constructive counting to solve this. There are <math>2< | + | We will use constructive counting to solve this. There are </math>2<math> cases: Either all </math>3<math> points are adjacent, or exactly </math>2<math> points are adjacent. |
| − | If all <math>3< | + | If all </math>3<math> points are adjacent, then we have </math>8<math> choices. If we have exactly </math>2<math> adjacent points, then we will have </math>8<math> places to put the adjacent points and also </math>4<math> places to put the remaining point, so we have </math>8\cdot4<math> choices. The total amount of choices is </math>{8 \choose 3} = 8\cdot7<math>. |
| − | Thus our answer is <math>\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}< | + | Thus our answer is </math>\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}<math> |
===Solution 2 === | ===Solution 2 === | ||
| − | We can decide <math>2< | + | We can decide </math>2<math> adjacent points with </math>8<math> choices. The remaining point will have </math>6<math> choices. However, we have counted the case with </math>3<math> adjacent points twice, so we need to subtract this case once. The case with the </math>3<math> adjacent points has </math>8<math> arrangements, so our answer is </math>\frac{8\cdot6-8}{{8 \choose 3 }}<math></math>=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}<math> |
===Solution 3 (Stars and Bars)=== | ===Solution 3 (Stars and Bars)=== | ||
| − | Let <math>1< | + | Let </math>1<math> point of the triangle be fixed at the top. Then, there are </math>{7 \choose 2} = 21<math> ways to chose===Solution 1=== the other 2 points. There must be </math>3<math> spaces in the points and </math>3<math> points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and </math>2<math> extra points (k-1) distributed so by the stars and bars formula, </math>{n+k-1 \choose k-1}<math>, there are </math>{4 \choose 2} = 6<math> ways to arrange the bars and stars. Thus, the probability is </math>\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}<math>. |
===Simple Complementary Counting=== | ===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 <math>1-\frac{12}{7\cdot 6}= 5/7 | + | 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 </math>1-\frac{12}{7\cdot 6}= 5/7$ |
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
Revision as of 16:51, 2 January 2023
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?
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Solutions
Solution 0
Choose side lengths 
for the triangle, starting from the shortest side, and going clockwise: 
, where 
a \leq c
\boxed(5/7)$of these have a side with length 1, which corresponds to an edge of the octagon.
===Solution 1===
We will use constructive counting to solve this. There are$ (Error compiling LaTeX. Unknown error_msg)2
3
2$points are adjacent.
If all$ (Error compiling LaTeX. Unknown error_msg)3
8
2
8
4
8\cdot4
{8 \choose 3} = 8\cdot7$.
Thus our answer is$ (Error compiling LaTeX. Unknown error_msg)\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}
2
8
6
3
3
8
\frac{8\cdot6-8}Template:8 \choose 3$$ (Error compiling LaTeX. Unknown error_msg)=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}
1
{7 \choose 2} = 21
3
3
2
{n+k-1 \choose k-1}
{4 \choose 2} = 6
\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}$.
===Simple Complementary Counting=== By rotational symmetry, choose an arbitrary point for one vertex. Then choose one of the 5 non-adjacent vertices, out of 7 possible. Sum the number of remaining non-adjacent vertices, for all 5 cases: 3+2+2+2+3=12, out of 6 possible for each. These are non-edge triangles, so the probability of edge triangles is$ (Error compiling LaTeX. Unknown error_msg)1-\frac{12}{7\cdot 6}= 5/7$
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=2678
~ pi_is_3.14
Video Solutions
https://www.youtube.com/watch?v=VNflxl7VpL0
~savannahsolver
gg
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. 
