AMC12 2017 A

AMC12 2017 A · Q6

AMC12 2017 A · Q6. It mainly tests Area & perimeter, Geometry misc.

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?

Joy 有 30 根细棒，每根长度为 1 cm 到 30 cm 的每个整数长度各一根。她将长度为 3 cm、7 cm 和 15 cm 的棒放在桌子上。然后她想选择一根第四根棒，与这三根一起形成一个有正面积的四边形。她有多少根剩余的棒可以选择作为第四根棒？

(A) 16 16

(B) 17 17

(D) 19 19

(E) 20 20

Answer

Correct choice: (B)

正确答案：（B）

Solution

Four rods can form a quadrilateral with positive area if and only if the length of the longest rod is less than the sum of the lengths of the other three. Therefore if the fourth rod has length $n$ cm, then $n$ must satisfy $15 < 3+7+n$ and $n < 3+7+15$, that is, $5 < n < 25$. Because $n$ is an integer, it must be one of the 19 integers from 6 to 24, inclusive. However, the rods of lengths 7 cm and 15 cm have already been chosen, so the number of rods that Joy can choose is $19 - 2 = 17$.

四根棒可以形成有正面积的四边形当且仅当最长棒的长度小于其他三根棒长度之和。因此如果第四根棒长度为 $n$ cm，则 $n$ 必须满足 $15 < 3+7+n$ 和 $n < 3+7+15$，即 $5 < n < 25$。因为 $n$ 是整数，它必须是 6 到 24 的 19 个整数之一。然而，长度 7 cm 和 15 cm 的棒已被选择，所以 Joy 可以选择的数量是 $19 - 2 = 17$。

Topics