2017 AMC 10A Problem 10

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: 1140

Small hint:

Use the quadrilateral inequality for the longest side

Big hint:

After finding the possible integer lengths, remove rods already on the table

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.