2017 AMC 12A Problems/Problem 6

Problem

Joy has $30$ thin rods, one each of every integer length from $1 \text{ cm}$ through $30 \text{ cm}$ . She places the rods with lengths $3 \text{ cm}$ , $7 \text{ cm}$ , and $15 \text{cm}$ 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 quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7$ = 30. This means Joy can use the 19 possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $\boxed{\textbf{(B)}}$

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 12 Problems and Solutions

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2017 AMC 12A Problems/Problem 6

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