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Difference between revisions of "2003 AMC 12A Problems/Problem 7"
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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>.  
 
By the [[triangle inequality]], no side may have a length greater than the semiperimeter, 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>. 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>.  
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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>.
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== Video Solution ==
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https://www.youtube.com/watch?v=gII2tj5TZZY  ~David
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 20:45, 19 July 2023

The following problem is from both the 2003 AMC 12A #7 and 2003 AMC 10A #7, so both problems redirect to this page.

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 $\boxed{\mathrm{(B)}\ 2}$.

Video Solution

https://www.youtube.com/watch?v=gII2tj5TZZY ~David

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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