2017 AMC 10A 第 10 题

Question

Joy 有

30

根细杆，长度分别为从

1

厘米到

30

厘米的每个整数。她把长度为

3

厘米、

7

厘米和

15

厘米的细杆放在桌上。她想再选一根细杆，与这三根一起组成一个面积为正的四边形。剩下的细杆中有多少根可以作为第四根？

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：1140

小提示：

对最长边使用四边形不等式。

Use the quadrilateral inequality for the longest side

大提示：

找出可能的整数长度后，删去已经在桌上的细杆长度。

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

解答：

四条边能组成面积为正的四边形，当且仅当最长边小于其余三边之和。

设第四根长度为

x

。必须有

x \lt 3 + 7 + 15\text{；}

当原来最长的细杆仍是最长边时，还必须有

x + 3 + 7 \gt 15\text{。}

化简得

5\lt x\lt 25\text{。}

这个范围内有

25 - 5 - 1 = 19

个

x

的整数值。

不过长度为

7

和

15

的细杆已经被使用，所以

x

不能等于这些值。

这留下

19 - 2 = 17

个可行的

x

值。

所以正确答案是 B。

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.

2017 AMC 10A 第 10 题

答案：B

难度评级：1140

小提示：

大提示：

解答：

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