# 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.

## 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 (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 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 byProblem 6 Followed byProblem 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.