2017 AMC 10A Problem 10

Four lengths form a quadrilateral with positive area exactly when the longest one is strictly less than the sum of the other three (the same idea as the triangle inequality, one side more).

Let the fourth rod have length $d$.

If $d$ is the longest side, we need $d < 3 + 7 + 15 = 25$, so $d \le 24$.

Otherwise $15$ is the longest side, and we need $15 < 3 + 7 + d$, so $d > 5$, i.e. $d \ge 6$.

So $d$ can be any integer from $6$ to $24$: that is $24 - 6 + 1 = 19$ values. But the rods of length $7$ and $15$ are already in use, so they cannot be chosen again. That leaves $19 - 2 = 17$.

The answer is $\boxed{\textbf{(B)}\ 17}$.

The polygon inequality gives a two-sided bound on the unknown side: too long and it exceeds the rest, too short and the existing longest side exceeds the rest. Once you have an interval of integers, read the problem again for excluded values; "one each of every length" is the signal that $7$ and $15$ are gone.

• Forgetting that rods of length 7 and 15 are already on the table and answering 19.
• Using non-strict inequalities ($d \le 25$ or $d \ge 5$), which allow a degenerate quadrilateral of zero area.