Difference between revisions of "2005 AMC 8 Problems/Problem 21"

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<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24 </math>
 
<math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24 </math>
  
==Solution==
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==Solution 1==
The number of ways to choose three points to make a triangle is <math>_6 C _3 = 20</math>. However, two of these are a straight line so subtract <math>2</math> to get <math>\boxed{\texetbf{(C)}\ 18}</math>.
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The number of ways to choose three points to make a triangle is <math>\binom 63 = 20</math>. However, two* of these are a straight line so we subtract <math>2</math> to get <math>\boxed{\textbf{(C)}\ 18}</math>.
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*Note:  We are assuming that there are no degenerate triangles in this problem, and that is why we subtract two.
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==Video solution==
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https://www.youtube.com/watch?v=XQS-KVW1O6M  ~David
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=20|num-a=22}}
 
{{AMC8 box|year=2005|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 09:51, 17 June 2024

Problem

How many distinct triangles can be drawn using three of the dots below as vertices?

[asy]dot(origin^^(1,0)^^(2,0)^^(0,1)^^(1,1)^^(2,1));[/asy]

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution 1

The number of ways to choose three points to make a triangle is $\binom 63 = 20$. However, two* of these are a straight line so we subtract $2$ to get $\boxed{\textbf{(C)}\ 18}$.

  • Note: We are assuming that there are no degenerate triangles in this problem, and that is why we subtract two.

Video solution

https://www.youtube.com/watch?v=XQS-KVW1O6M ~David

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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