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

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'''Case 1: One vertex is on the top 3 points'''
 
'''Case 1: One vertex is on the top 3 points'''
  
Here, there is <math>3 \choose 1</math> ways to choose the vertex on the top and <math>3 \choose 2</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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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.
  
 
'''Case 2: One vertex is on the bottom 3 points'''
 
'''Case 2: One vertex is on the bottom 3 points'''

Revision as of 17:18, 8 January 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 (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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