2017 AMC 12A Problem 6

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

Answer: B

Concepts:triangle inequality counting integers in a range

Difficulty rating: 1350

Small hint:

Four lengths form a quadrilateral exactly when the longest is less than the sum of the other three

Big hint:

This gives

5\lt n\lt 25

; then remove the rods already on the table

Solution:

Four lengths form a quadrilateral with positive area if and only if the longest is strictly less than the sum of the other three. With a fourth rod of length

n,

this requires

15\lt 3+7+n

and

n\lt 3+7+15,

so

5\lt n\lt 25.

The integers from

6

to

24

give

19

values, but the rods of length

7

and

15

are already on the table, leaving

19-2=17

choices.

Thus, the correct answer is B.