Difference between revisions of "2006 AMC 10B Problems/Problem 10"
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The lengths of the sides are: <math>x</math>, <math>3x</math>, and <math>15</math>. | The lengths of the sides are: <math>x</math>, <math>3x</math>, and <math>15</math>. | ||
− | By the | + | By the [[Triangle Inequality]], |
<math> 3x < x + 15 </math> | <math> 3x < x + 15 </math> | ||
Line 17: | Line 17: | ||
<math> x < \frac{15}{2} </math> | <math> x < \frac{15}{2} </math> | ||
− | The largest integer satisfing this inequality is <math>7</math> | + | The largest integer satisfing this inequality is <math>7</math>. |
So the largest perimeter is <math> 7 + 3\cdot7 + 15 = 43 \Rightarrow A </math> | So the largest perimeter is <math> 7 + 3\cdot7 + 15 = 43 \Rightarrow A </math> | ||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=9|num-a=11}} | |
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:17, 4 July 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
Let be the length of the first side.
The lengths of the sides are: , , and .
By the Triangle Inequality,
The largest integer satisfing this inequality is .
So the largest 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 |
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