Difference between revisions of "2006 AMC 12B Problems/Problem 10"
Armalite46 (talk | contribs) (→Solution) |
|||
Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
− | 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: <math></math>\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 | + | 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: <math></math>\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 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 \Rightarrow \boxed{\text {(A)}}</math>. |
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:18, 8 December 2013
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
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: $$ (Error compiling LaTeX. )\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.