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

Latest revision as of 16:49, 31 July 2011

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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?
-<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math>
-== Solution ==
-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>
-Since all sides must be integers, the largest possible length of a side is <math>3</math>
-Therefore, all such triangles must have all sides of length <math>1</math>, <math>2</math>, or <math>3</math>.
-Since <math>2+2+2=6<7</math>, at least one side must have a length of <math>3</math>
-Thus, the remaining two sides have a combined length of <math>7-3=4</math>.
-So, the remaining sides must be either <math>3</math> and <math>1</math> or <math>2</math> and <math>2</math>.
-Therefore, the number of triangles is <math>2 \Rightarrow B</math>.
-== See Also ==
-{{AMC10 box|year=2003|ab=A|num-b=6|num-a=8}}
-[[Category:Introductory Geometry Problems]]