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

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== Problem ==
 
== Problem ==
How many non-[[congruent (geometry) | congruent]] [[triangle]]s with [[perimeter]] <math>7</math> have [[integer]] side lengths?  
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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>
 
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5 </math>
  
 
== Solution ==
 
== 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>.  
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By the [[triangle inequality]], no side may have a length greater than the semiperimeter, 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>  
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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>\boxed{\mathrm{(B)}\ 2}</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>, atleast 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 \Longrightarrow \mathrm{(B)}</math>.  
 
  
 
== See Also ==
 
== See Also ==
*[[2003 AMC 12A Problems]]
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{{AMC10 box|year=2003|ab=A|num-b=6|num-a=8}}
*[[2003 AMC 12A/Problem 6|Previous Problem]]
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{{AMC12 box|year=2003|ab=A|num-b=6|num-a=8}}
*[[2003 AMC 12A/Problem 8|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 16:49, 31 July 2011

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 side may have a length greater than the semiperimeter, 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 $\boxed{\mathrm{(B)}\ 2}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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