2003 AMC 10A Problems/Problem 7

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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 one side may have a length greater than half the perimeter, 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 $2 \Rightarrow B$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 10 Problems and Solutions
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