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== | ||
| + | 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. | ||
==Video solution== | ==Video solution== | ||
Revision as of 16:21, 8 January 2024
Contents
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]](http://latex.artofproblemsolving.com/5/a/2/5a2b71fd41fb8101c84f6e83e00a34af1e6c98ef.png)
Solution
The number of ways to choose three points to make a triangle is 
. However, two* of these are a straight line so we subtract 
to get 
.
- 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 (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
