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 | + | 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''' | ||
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Otherwise, the triangle is degenerate. | Otherwise, the triangle is degenerate. | ||
− | Our final answer is <math>9+9=\boxed{\text{(C) 18}}</math> ~Ddk001 | + | Our final answer is <math>9+9=\boxed{\text{(C) 18}}</math> ~[[Ddk001]] |
==Video solution== | ==Video solution== |
Latest revision as of 20:57, 29 June 2024
Contents
[hide]Problem
How many distinct triangles can be drawn using three of the dots below as vertices?
Solution 1
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.
Solution 2
Case 1: One vertex is on the top 3 points
Here, there is ways to choose the vertex on the top and ways the choose the on the bottom, so there is triangles.
Case 2: One vertex is on the bottom 3 points
By symmetry, there is triangles.
Otherwise, the triangle is degenerate.
Our final answer is ~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.