Difference between revisions of "2015 AMC 12B Problems/Problem 10"

(Problem)
(Solution)
Line 5: Line 5:
  
 
==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?

$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$

Solution

Listing all possible triangle side lengths satisfying the constraints, we find the following:

$\text{2-3-4}\\ \text{2-4-5}\\ \text{2-5-6}\\ \text{3-4-6}\\ \text{3-5-6}$

Thus the answer is $\fbox{\textbf{(C)}\; 5}$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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

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