Difference between revisions of "2003 AMC 12A Problems/Problem 7"

Line 1: Line 1:
 +
{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #7]] and [[2003 AMC 10A Problems|2003 AMC 10A #7]]}}
 
== Problem ==
 
== Problem ==
 
How many non-[[congruent (geometry) |congruent]] [[triangle]]s with [[perimeter]] <math>7</math> have [[integer]] side lengths?  
 
How many non-[[congruent (geometry) |congruent]] [[triangle]]s with [[perimeter]] <math>7</math> have [[integer]] side lengths?  

Revision as of 16:50, 31 July 2011

The following problem is from both the 2003 AMC 12A #7 and 2003 AMC 10A #7, so both problems redirect to this page.

Problem

How many non-congruent triangles with perimeter $7$ have integer side lengths?

$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$

Solution

By the triangle inequality, no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$.

Since all sides must be integers, the largest possible length of a side is $3$. Therefore, all such triangles must have all sides of length $1$, $2$, or $3$. Since $2+2+2=6<7$, at least one side must have a length of $3$. Thus, the remaining two sides have a combined length of $7-3=4$. So, the remaining sides must be either $3$ and $1$ or $2$ and $2$. Therefore, the number of triangles is $\boxed{\mathrm{(B)}\ 2}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS