2006 AMC 10B Problem 10

Call the sides $x$, $3x$, and $15$, with $x$ a positive integer. The perimeter $4x + 15$ grows with $x$, so we want the largest $x$ allowed by the triangle inequality.

For large $x$ the longest side is $3x$, and the sum of the other two must exceed it:
$$x + 15 > 3x \quad\Longrightarrow\quad 15 > 2x \quad\Longrightarrow\quad x < 7.5.$$
So $x \le 7$. With $x = 7$ the sides are $7, 21, 15$: check $7 + 15 = 22 > 21$, $7 + 21 > 15$, $15 + 21 > 7$, all fine.

The maximum perimeter is $7 + 21 + 15 = 43$.

The answer is $\boxed{\textbf{(A)}\ 43}$.

Only one of the three triangle inequalities can bind when maximizing: the one where the longest side is on the big side. Identify which side is longest for the values that matter, write that inequality, and take the largest integer below the bound. The inequality is strict, so $x = 7.5$ (which would give a degenerate triangle) is excluded even before integrality.

• Letting 15 be the longest side (x + 3x > 15) and pushing x too large, or using x + 15 >= 3x and taking x = 7.5.
• Forgetting the sides must be integers and reporting a non-integer perimeter, or answering $4 \cdot 7 = 28$ without adding $15$.