2006 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2006 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

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Concepts:triangle inequality optimization

Difficulty rating: 1450

10.

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?

$43$

$44$

$45$

$46$

$47$

Solution:

Let the sides be $x,$ $3x,$ and $15.$ The triangle inequality requires $x + 3x \gt 15,$ so $x \geq 4,$ and $x + 15 \gt 3x,$ so $x \leq 7.$

The perimeter $4x + 15$ is largest when $x = 7,$ giving $7 + 21 + 15 = 43.$

Thus, the correct answer is A.

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