2017 AMC 12A 第 6 题

Question

Joy 有

30

根细杆，长度分别为从

1

cm 到

30

cm 的每个整数，且每种长度各一根。她把长度为

3

cm、

7

cm 和

15

cm 的杆放在桌上。她接着想从剩下的杆中选一根，和这三根一起组成一个面积为正的四边形。她可以选择多少根剩余的杆作为第四根？

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

答案：B

知识点：三角不等式区间内整数计数

难度评级：1350

小提示：

四条长度能组成四边形，当且仅当最长边小于另外三边之和

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

大提示：

这给出

5\lt n\lt 25

；然后去掉已经在桌上的杆

This gives

5\lt n\lt 25

; then remove the rods already on the table

解答：

四条长度能组成面积为正的四边形，当且仅当最长边严格小于另外三边之和。若第四根杆长为

n

，则需要

15\lt 3+7+n

且

n\lt 3+7+15

，所以

5\lt n\lt 25\text{。}

从

6

到

24

的整数共有

19

个，但长度为

7

和

15

的杆已经在桌上，剩下

19-2=17

种选择。

所以正确答案是 B。

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.

2017 AMC 12A 第 6 题

答案：B

难度评级：1350

小提示：

大提示：

解答：

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