AMC10 2006 A

AMC10 2006 A · Q14

AMC10 2006 A · Q14. It mainly tests Arithmetic sequences basics, Arithmetic misc.

A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the other rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

许多相连的环，每个厚 1 厘米，挂在一个钉子上。最上面的环外径 20 厘米。其他每个环的外径比它上面的环少 1 厘米。最下面的环外径 3 厘米。从最上面环的顶部到底面最下面环的底部距离是多少厘米？

(A) 171 171

(B) 173 173

(D) 188 188

(E) 210 210

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) The top of the largest ring is 20 cm above its bottom. That point is 2 cm below the top of the next ring, so it is 17 cm above the bottom of the next ring. The additional distances to the bottoms of the remaining rings are 16 cm, 15 cm, ..., 1 cm. Thus the total distance is $20+(17+16+\cdots+2+1)=20+\dfrac{17\cdot18}{2}=20+17\cdot9=173\text{ cm}.$ OR The required distance is the sum of the outside diameters of the 18 rings minus a 2-cm overlap for each of the 17 pairs of consecutive rings. This equals $(3+4+5+\cdots+20)-2\cdot17=(1+2+3+4+5+\cdots+20)-3-34=\dfrac{20\cdot21}{2}-37=173\text{ cm}.$

(B) 最大圆环的顶部比其底部高 20 cm。该点比下一个圆环的顶部低 2 cm，因此它比下一个圆环的底部高 17 cm。其余圆环到底部的额外距离分别为 16 cm、15 cm、……、1 cm。因此总距离为 $20+(17+16+\cdots+2+1)=20+\dfrac{17\cdot18}{2}=20+17\cdot9=173\text{ cm}.$ 或者所求距离等于 18 个圆环外直径之和，减去 17 对相邻圆环每对重叠的 2 cm。即 $(3+4+5+\cdots+20)-2\cdot17=(1+2+3+4+5+\cdots+20)-3-34=\dfrac{20\cdot21}{2}-37=173\text{ cm}.$

Topics