Difference between revisions of "2006 AMC 12B Problems/Problem 10"

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== Problem ==
 
== Problem ==
{{problem}}
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In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
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<math>
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\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47
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</math>
  
 
== Solution ==
 
== Solution ==
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<math>
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\text {(A) } 43
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</math>
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If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: x+15>3x, so 2x<15 and x<7.5. Now, since we want the greatest perimeter, we want the greatest integer x, which is 7. Then, the first side has length 3*7=21, the second side has length 7 and the third side has length 15, and so the perimeter is 21+7+15=43.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2006|ab=B|num-b=9|num-a=11}}

Revision as of 10:47, 26 February 2011

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Problem

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? $\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$

Solution

$\text {(A) } 43$ If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: x+15>3x, so 2x<15 and x<7.5. Now, since we want the greatest perimeter, we want the greatest integer x, which is 7. Then, the first side has length 3*7=21, the second side has length 7 and the third side has length 15, and so the perimeter is 21+7+15=43.

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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