# 2006 AMC 10B Problems/Problem 10

## 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 greatest integer satisfying this inequality is $7$.

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

## See Also

 2006 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 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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