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

 
 
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== Problem ==
 
== Problem ==
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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?
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<math> \mathrm{(A) \ } 43\qquad \mathrm{(B) \ } 44\qquad \mathrm{(C) \ } 45\qquad \mathrm{(D) \ } 46\qquad \mathrm{(E) \ } 47 </math>
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== Solution ==
 
== Solution ==
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Let <math>x</math> be the length of the first side.
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The lengths of the sides are: <math>x</math>, <math>3x</math>, and <math>15</math>.
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By the [[Triangle Inequality]],
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<math> 3x < x + 15 </math>
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<math> 2x < 15 </math>
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<math> x < \frac{15}{2} </math>
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The largest integer satisfing this inequality is <math>7</math>.
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So the largest perimeter is <math> 7 + 3\cdot7 + 15 = 43 \Rightarrow A </math>
 
== See Also ==
 
== See Also ==
*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=9|num-a=11}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:17, 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?

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

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

2006 AMC 10B (ProblemsAnswer KeyResources)
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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