Difference between revisions of "2006 AMC 10B Problems/Problem 10"
m (→Solution) |
m (→Solution) |
||
| Line 17: | Line 17: | ||
<math> x < \frac{15}{2} </math> | <math> x < \frac{15}{2} </math> | ||
| − | The | + | The greatest integer satisfying this inequality is <math>7</math>. |
| − | So the | + | So the greatest possible perimeter is <math> 7 + 3\cdot7 + 15 = 43 \Rightarrow A </math> |
== See Also == | == See Also == | ||
Revision as of 16:41, 2 June 2021
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
Let 
be the length of the first side.
The lengths of the sides are: , 
, and 
.
By the Triangle Inequality,
The greatest integer satisfying this inequality is 
.
So the greatest possible perimeter is 
See Also
| 2006 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
