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, so 2x<15 and x<7.5. Now, since we want the greatest perimeter, we want the greatest integer x, which is 7. Then, the first side has length 3*7=21, the second side has length 7 and the third side has length 15, and so the perimeter is 21+7+15=43.

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2006 AMC 12B Problems/Problem 10

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