2017 AMC 10A Problems/Problem 10

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Problem

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?

$\text{(A) 16}\qquad\text{(B) 17}\qquad\text{(C) 18}\qquad\text{(D) 19}\qquad\text{(E) 20}$

Solution

The triangle inequality generalizes to all polygons, so you can just use $x < 3+7+15$ and $x+3+7>15$ to get $5<x<25$, so $x$ takes on 19 values. Then, of course, you subtract 2 because 7 and 15 are already used, to get $\boxed{17}$.