Difference between revisions of "2006 AMC 10B Problems/Problem 10"
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== Problem == | == 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 <math>15</math>. What is the greatest possible perimeter of the triangle? | ||
| + | |||
| + | <math> \textbf{(A) } 43\qquad \textbf{(B) } 44\qquad \textbf{(C) } 45\qquad \textbf{(D) } 46\qquad \textbf{(E) } 47 </math> | ||
| + | |||
== Solution == | == Solution == | ||
| + | Let <math>x</math> be the length of the first side. | ||
| + | |||
| + | The lengths of the sides are: <math>x</math>, <math>3x</math>, and <math>15</math>. | ||
| + | |||
| + | By the [[Triangle Inequality]], | ||
| + | |||
| + | <math> 3x < x + 15 </math> | ||
| + | |||
| + | <math> 2x < 15 </math> | ||
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| + | <math> x < \frac{15}{2} </math> | ||
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| + | The greatest integer satisfying this inequality is <math>7</math>. | ||
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| + | So the greatest possible perimeter is <math> 7 + 3\cdot7 + 15 =\boxed{\textbf{(A) } 43}</math> | ||
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== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2006|ab=B|num-b=9|num-a=11}} | |
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:57, 26 January 2022
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 
. 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 | ||
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