2006 AMC 10B Problem 10

This is a "greatest possible" question: find the largest triangle that works, then show a bigger one fails.

1. Name the sides. One side is $x$, another is $3x$, and the third is $15$, where $x$ is a whole number. The perimeter is $x + 3x + 15 = 4x + 15$, so a bigger $x$ gives a bigger perimeter. We want $x$ as large as possible.
2. The rule for a triangle: any two sides added together must be longer than the third side. When $x$ is large, $3x$ is the longest side, so the rule that can fail is
$$x + 15 > 3x.$$
3. Take $x$ from both sides: $15 > 2x$, so $x < 7.5$. The largest whole number allowed is $x = 7$.
4. $x = 7$ works. The sides are $7$, $21$, $15$. Check all three rules: $7 + 15 = 22 > 21$, $7 + 21 = 28 > 15$, and $15 + 21 = 36 > 7$. It is a real triangle.
5. $x = 8$ fails. The sides would be $8$, $24$, $15$, and $8 + 15 = 23$ is not more than $24$.

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

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

When you want a triangle as big as possible, only one of the three triangle rules actually stops you: the one that says the two shorter sides must add to more than the longest side. Figure out which side is longest, write that one rule, and take the largest whole number that fits. The rule uses "more than," not "at least," so the boundary value itself is not allowed.

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 $\boxed{\textbf{(A)}\ 43}$.