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# Difference between revisions of "2017 AMC 10A Problems/Problem 10"

## 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?

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

## Solution

The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15 yields $5. Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\boxed{\textbf{(B)}\ 17}$

## Video Solution

~savannahsolver

 2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions