Difference between revisions of "2015 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
+ | Listing all possible triangle side lengths satisfying the constraints, we find the following: | ||
+ | <math>\text{2-3-4}\\ | ||
+ | \text{2-4-5}\\ | ||
+ | \text{2-5-6}\\ | ||
+ | \text{3-4-6}\\ | ||
+ | \text{3-5-6} | ||
+ | </math> | ||
+ | |||
+ | Thus the answer is <math>\fbox{\textbf{(C)}\; 5}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=11|num-b=9}} | {{AMC12 box|year=2015|ab=B|num-a=11|num-b=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:15, 4 March 2015
Problem
How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
Solution
Listing all possible triangle side lengths satisfying the constraints, we find the following:
Thus the answer is .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.