2017 AMC 10A Problem 10

For three sticks to make a triangle, the longest must be shorter than the other two put together. Four sticks work the same way: the longest must be shorter than the other three put together. Otherwise the other three cannot reach around to close the shape.

Call the length of the fourth rod $d$.

1. $d$ cannot be too long. If $d$ is the longest side, it must be shorter than $3 + 7 + 15 = 25$. So $d$ is at most $24$.
2. $d$ cannot be too short. If $15$ is the longest side, it must be shorter than $3 + 7 + d = 10 + d$. So $d$ must be more than $5$, that is, at least $6$.
3. So $d$ can be any whole number from $6$ to $24$. That is $24 - 6 + 1 = 19$ lengths.
4. Joy has only one rod of each length, and the $7$ cm and $15$ cm rods are already on the table. Both are in the list from $6$ to $24$, so remove them: $19 - 2 = 17$.

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

The "longest side is less than the rest" rule traps the missing side from both ends: it cannot be longer than the others combined, and it cannot be so short that $15$ becomes too long. After you find the range, read the problem once more for numbers that are not allowed. "One each of every length" is the clue that $7$ and $15$ are gone.

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 $\boxed{\textbf{(B)}\ 17}$.