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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>
  
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==Solution==
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==Solution 1==
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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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==Solution 2==
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'''Case 1: One vertex is on the top 3 points'''
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Here, there is <math>\binom 31 =3</math> ways to choose the vertex on the top and <math>\binom 32 =3</math> ways the choose the <math>2</math> on the bottom, so there is <math>3 \cdot 3=9</math> triangles.
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'''Case 2: One vertex is on the bottom 3 points'''
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By symmetry, there is <math>9</math> triangles.
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Otherwise, the triangle is degenerate.
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Our final answer is <math>9+9=\boxed{\text{(C) 18}}</math> ~[[Ddk001]]
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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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:57, 29 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.

Solution 2

Case 1: One vertex is on the top 3 points

Here, there is $\binom 31 =3$ ways to choose the vertex on the top and $\binom 32 =3$ ways the choose the $2$ on the bottom, so there is $3 \cdot 3=9$ triangles.

Case 2: One vertex is on the bottom 3 points

By symmetry, there is $9$ triangles.

Otherwise, the triangle is degenerate.

Our final answer is $9+9=\boxed{\text{(C) 18}}$ ~Ddk001

Video solution

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

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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