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：

其他年份的第 6 题