2005 AMC 8 Problems/Problem 21

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

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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2005 AMC 8 Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

Video solution

See Also