# Difference between revisions of "2006 AMC 12B Problems/Problem 10"

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

## Solution $$\text {(A) } 43\\$$ If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: $$\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 \\$$ Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$. Then, the first side has length $3*7=21$, the second side has length $7$, the third side has length $15$, and so the perimeter is $21+7+15=43$.