2006 AMC 12B Problem 10

Attempt Problem 10 of the 2006 AMC 12B below, then check your answer against the professionally curated solution 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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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

Let the sides be

x,

3x,

and

15,

then apply the triangle inequality

The binding condition is

x + 15 \gt 3x,

which caps how large

x

can be

Let the sides be

x,

3x,

and

15.

The triangle inequality requires

x + 3x \gt 15,

x \geq 4,

and

x + 15 \gt 3x,

x \leq 7.

The perimeter

4x + 15

is largest when

x = 7,

giving

7 + 21 + 15 = 43.

Thus, the correct answer is A.

Problem 10 in other years