AMC10 2017 A

AMC10 2017 A · Q10

AMC10 2017 A · Q10. It mainly tests Polygons, 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?

乔伊有30根细杆，每根长度从1 cm到30 cm各一根。她把长度3 cm、7 cm和15 cm的杆放在桌上。然后她想选择第四根杆，与这三根一起形成具有正面积的四边形。她有多少根剩余的杆可以选择作为第四根？

(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 the inequalities \(15 < 3+7+n\) 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 \(19 - 2 = 17\).

四根杆能形成具有正面积的四边形当且仅当最长杆的长度小于其他三根长度之和。因此如果第四根杆长度为\(n\) cm，则\(n\)必须满足不等式\(15 < 3+7+n\)和\(n < 3+7+15\)，即\(5 < n < 25\)。因为\(n\)是整数，必须是6到24的19个整数之一。但是7 cm和15 cm的杆已被选择，所以乔伊能选择的杆数为\(19 - 2 = 17\)。

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