2003 AMC 10A Problems/Problem 7: Difference between revisions

Revision as of 10:15, 15 January 2008

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

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

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 $2 \Rightarrow B$ .

2003 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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 == See Also ==
-*[[2003 AMC 10A Problems]]
+{{AMC10 box|year=2003|ab=A|num-b=6|num-a=8}}
-*[[2003 AMC 10A Problems/Problem 6|Previous Problem]]
-*[[2003 AMC 10A Problems/Problem 8|Next Problem]]
 [[Category:Introductory Geometry Problems]]