2014 AIME I 第 1 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

1.

一只运动鞋鞋带穿过的 88 个鞋眼都在一个长方形上,较长的两边各有四个等距的鞋眼。 这个长方形宽 5050 毫米,长 8080 毫米,并且每个顶点都有一个鞋眼。鞋带必须先在长方形一条宽边的两个顶点鞋眼之间穿过, 然后在相邻鞋眼之间交叉穿插,直到到达另一条宽边上的两个鞋眼,如图所示。穿过最后这两个鞋眼之后, 鞋带两端还必须各至少再伸出 200200 毫米,以便打结。求这根鞋带的最小长度,单位为毫米。

The 88 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 5050 mm and a length of 8080 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.

答案:790
知识点:勾股定理周长
难度评级:1890
解答:

每条 8080 毫米边上的四个鞋眼等距排列,且两端在顶点,所以同一边上相邻鞋眼相距 803\frac{80}{3} 毫米。鞋带由一段横跨 5050 毫米宽度的线段、六段交叉线段(经过宽边连接后, 两股鞋带各再交叉三次到达上端)以及两段至少 200200 毫米的自由末端组成。要使总长最短,每一段都应是直线段。

每段交叉线段横跨整个宽度,并沿长边方向上升一个间隔,所以长度为 502+(803)2=289009=1703. \begin{aligned} &\sqrt{50^2 + \left(\tfrac{80}{3}\right)^2} \\ &= \sqrt{\tfrac{28900}{9}} = \frac{170}{3}. \end{aligned}

最小长度为 50+6170350 + 6 \cdot \frac{170}{3} +2200+ 2 \cdot 200 =50+340+400= 50 + 340 + 400 =790= 790

The four eyelets on each 8080 mm side are equally spaced with one at each vertex, so consecutive eyelets on a side are 803\frac{80}{3} mm apart. The lace consists of one segment across the 5050 mm width, six crisscross pieces (after the width crossing, each of the two strands makes three crossings to reach the top), and two free ends of at least 200200 mm each. The lace is shortest when every piece is a straight segment.

Each crisscross piece spans the full width and rises one gap, so its length is 502+(803)2=289009=1703. \begin{aligned} &\sqrt{50^2 + \left(\tfrac{80}{3}\right)^2} \\ &= \sqrt{\tfrac{28900}{9}} = \frac{170}{3}. \end{aligned}

The minimum length is 50+6170350 + 6 \cdot \frac{170}{3} +2200+ 2 \cdot 200 =50+340+400= 50 + 340 + 400 =790.= 790.

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