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| − | == Problem ==
| + | #REDIRECT[[2003 AMC 12A Problems/Problem 7]] |
| − | How many non-congruent triangles with perimeter <math>7</math> have integer side lengths?
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| − | <math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math>
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| − | == Solution ==
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| − | By the [[triangle inequality]], no one side may have a length greater than half the perimeter, which is <math>\frac{1}{2}\cdot7=3.5</math>
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| − | Since all sides must be integers, the largest possible length of a side is <math>3</math>
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| − | Therefore, all such triangles must have all sides of length <math>1</math>, <math>2</math>, or <math>3</math>.
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| − | Since <math>2+2+2=6<7</math>, at least one side must have a length of <math>3</math>
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| − | Thus, the remaining two sides have a combined length of <math>7-3=4</math>.
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| − | So, the remaining sides must be either <math>3</math> and <math>1</math> or <math>2</math> and <math>2</math>.
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| − | Therefore, the number of triangles is <math>2 \Rightarrow B</math>.
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| − | == See Also ==
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| − | {{AMC10 box|year=2003|ab=A|num-b=6|num-a=8}}
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| − | [[Category:Introductory Geometry Problems]]
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