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Difference between revisions of "2006 AMC 10B Problems/Problem 10"

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<math> x < \frac{15}{2} </math>
 
<math> x < \frac{15}{2} </math>
  
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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 ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 13:36, 18 July 2006

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?

$\mathrm{(A) \ } 43\qquad \mathrm{(B) \ } 44\qquad \mathrm{(C) \ } 45\qquad \mathrm{(D) \ } 46\qquad \mathrm{(E) \ } 47$

Solution

Let $x$ be the length of the first side.

The lengths of the sides are: $x$, $3x$, and $15$.

By the Triangle Inequality,

$3x < x + 15$

$2x < 15$

$x < \frac{15}{2}$

The largest integer satisfing this inequality is $7$.

So the largest perimeter is $7 + 3\cdot7 + 15 = 43 \Rightarrow A$

See Also

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