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? | + | 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>. |
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− | Therefore, all such triangles must have all sides of length <math>1</math>, <math>2</math>, or <math>3</math>. | ||
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− | Since <math>2+2+2=6<7</math>, | ||
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− | Thus, the remaining two sides have a combined length of <math>7-3=4</math>. | ||
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− | So, the remaining sides must be either <math>3</math> and <math>1</math> or <math>2</math> and <math>2</math>. | ||
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− | Therefore, the number of triangles is <math> | ||
== 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]] |
Revision as of 16:49, 31 July 2011
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 .
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 |