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

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 $x$ 。必须有且若十五是最长边，则因此这个范围内整数个数为 $25 - 5 - 1 = 19$ ，这些都是 $x$ 的可能值。 $x \lt 3 + 7 + 15$ $x + 3 + 7 \gt 15$ $5\lt x\lt 25.$

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

解答：

其他年份的第 10 题