Question

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

$16$

$17$

$18$

$19$

$20$

Solution:

Note that no one side can be greater than or equal to the sum of the other side lengths.

Let $x$ be the length fourth rod. Then we have that $x \lt 3 + 7 + 15$ and $x + 3 + 7 \gt 15$ Simplifying, we know that $5\lt x\lt 25.$ Counting the number of integers in this range, we are left with $25 - 5 - 1 = 19$ values for $x.$

The rods with length $7$ and $15$ are already being used, however, so $x$ cannot equal these.

This leaves $19 - 2 = 17$ viable solutions for $x.$

Thus, B is the correct answer.

2017 AMC 10A Problem 10

Solution:

Problem 10 in Other Years