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

Revision as of 08:51, 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 *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 <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'''
-By symmetry, there is <math>9</math> triangles.
-Otherwise, the triangle is degenerate.
-Our final answer is <math>9+9=\boxed{\text{(C) 18}}</math> ~[[Ddk001]]
 ==Video solution==

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

Revision as of 08:51, 17 June 2024

Contents

Problem

Solution 1

Video solution

See Also