Difference between revisions of "2003 AMC 12A Problems/Problem 7"
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| + | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #7]] and [[2003 AMC 10A Problems|2003 AMC 10A #7]]}} | ||
== Problem == | == Problem == | ||
| − | How many non-[[congruent (geometry) | congruent]] [[triangle]]s 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> | <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 | + | 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> | + | 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>. |
| − | + | == Video Solution == | |
| − | + | https://www.youtube.com/watch?v=gII2tj5TZZY ~David | |
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== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2003|ab=A|num-b=6|num-a=8}} | |
| − | + | {{AMC12 box|year=2003|ab=A|num-b=6|num-a=8}} | |
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{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.
Contents
Problem
How many non-congruent triangles with perimeter 
have integer side lengths?
Solution
By the triangle inequality, no side may have a length greater than the semiperimeter, which is 
.
Since all sides must be integers, the largest possible length of a side is 
. Therefore, all such triangles must have all sides of length 
, 
, or . Since 
, at least one side must have a length of . Thus, the remaining two sides have a combined length of 
. So, the remaining sides must be either and or and . Therefore, the number of triangles is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
