2003 AMC 12A Problems/Problem 7
- The following problem is from both the 2003 AMC 12A #7 and 2003 AMC 10A #7, so both problems redirect to this page.
Contents
Problem
How many non-congruent triangles with perimeter have integer side lengths?
Solution
By the triangle inequality, no side may have a length greater than the semiperimeter, which is .
Since all sides must be integers, the largest possible length of a side is . Therefore, all such triangles must have all sides of length , , or . Since , at least one side must have a length of . Thus, the remaining two sides have a combined length of . So, the remaining sides must be either and or and . Therefore, the number of triangles is .
Video Solution
https://www.youtube.com/watch?v=gII2tj5TZZY ~David
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 AMC 10 Problems and Solutions |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.