Difference between revisions of "2003 AMC 12A Problems/Problem 7"

Revision as of 13:26, 18 November 2006

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 $2+2+2=6<7$ , atleast 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 $2 \Longrightarrow \mathrm{(B)}$ .

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 == Problem ==
-How many non-congruent triangles with perimeter <math>7</math> have integer side lengths?
+How many non-[[congruent (geometry) | congruent]] [[triangle]]s 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>
+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>
@@ Line 17: / Line 17: @@
 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>.
+Therefore, the number of triangles is <math>2 \Longrightarrow \mathrm{(B)}</math>.
 == See Also ==