Difference between revisions of "2006 AMC 12B Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | < | + | <cmath>\text {(A) } 43\\</cmath> |
| − | \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: <cmath>\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 \\</cmath> Now, since we want the greatest perimeter, we want the greatest integer x, and if <math>x<7.5</math> then <math>x=7</math>. Then, the first side has length <math>3*7=21</math>, the second side has length <math>7</math>, the third side has length <math>15</math>, and so the perimeter is <math>21+7+15=43</math>. |
| − | </ | ||
| − | 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 | ||
== 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 10:54, 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?
Solution
![\[\text {(A) } 43\\\]](http://latex.artofproblemsolving.com/a/7/c/a7ce9a1ea6bcd7ceb3afb6b446e390265954be91.png)
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: ![\[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 \\\]](http://latex.artofproblemsolving.com/8/6/f/86fb35d522abca48f82b4898918ec5877bb61226.png)
Now, since we want the greatest perimeter, we want the greatest integer x, and if 
then 
. Then, the first side has length 
, the second side has length 
, the third side has length 
, and so the perimeter is 
.
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| 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 | |
