一道英数题Two ants start at point A and walk at the same pace. One ant walks around a 3 cm by 3 cm square whilst the other walks around a 6 cm by 3 cmrectangle. What is the minimum distance, in centimetres, any one mustcover before they meet aga
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/24 21:33:47
一道英数题Two ants start at point A and walk at the same pace. One ant walks around a 3 cm by 3 cm square whilst the other walks around a 6 cm by 3 cmrectangle. What is the minimum distance, in centimetres, any one mustcover before they meet aga
一道英数题
Two ants start at point A and walk at the same pace. One ant walks around a 3 cm by 3 cm square whilst the other walks around a 6 cm by 3 cmrectangle. What is the minimum distance, in centimetres, any one mustcover before they meet again?
我忘记上传图了
一道英数题Two ants start at point A and walk at the same pace. One ant walks around a 3 cm by 3 cm square whilst the other walks around a 6 cm by 3 cmrectangle. What is the minimum distance, in centimetres, any one mustcover before they meet aga
翻译:两只蚂蚁从A点开始用相同速度移动,一只蚂蚁走3cm*3cm正方形路线,另一只走6cm*3cm矩形(矩形的单词打错咯~).如果两只蚂蚁要相遇,它们至少走过多少路程?
解法:
首先想象这样一个地形:日
两只蚂蚁同时从“日”右上角出发,顺时针旋转,但一只蚂蚁走小口,一只蚂蚁走大口.两个蚂蚁经过每一小直线的速度一样.用两个手指当作两个蚂蚁,慢慢画圈,看什么时候两个手指会到同一点就行了.画画就能看出来,走过30cm(也就是走小口的蚂蚁绕了两圈半),两只蚂蚁相遇.
英语答案的话这样写:The answer is 30cm.
---------
那么上面的解法就完全错了.
稍微改动下:把两个手指放在A点,按两个蚂蚁的方式转动,第二个蚂蚁走长边时第一个蚂蚁走两条边,转一转就能转出来了,第一个蚂蚁走三圈时第二个蚂蚁正好走两圈,都在A点.答案是36cm
The answer is 36cm.
PS:专业一点的解法:两个蚂蚁相遇,也就是他们都经过了整数圈(因为整数圈后在A点).小正方形周长12cm,大矩形周长18cm,找12和18的最小公倍数,也就是36
楼上的没看楼主的图吧,两个移动轨迹只有一个交点,怎么可能出现半圈的情况。就算是日字型轨迹,也不可能是30,同向出发就有很长一段时间同行,不符合meet again,反向出发在第15cm处就相遇了,自己验证一下。
翻译:
两个蚂蚁同时在A点以同样速度爬行,一个蚂蚁围绕一个长宽的3厘米的正方形,另一个蚂蚁围绕一个长6厘米宽3厘米的长方形。他们再次相遇的时候,任何一只蚂蚁爬过的距离至少是...
全部展开
楼上的没看楼主的图吧,两个移动轨迹只有一个交点,怎么可能出现半圈的情况。就算是日字型轨迹,也不可能是30,同向出发就有很长一段时间同行,不符合meet again,反向出发在第15cm处就相遇了,自己验证一下。
翻译:
两个蚂蚁同时在A点以同样速度爬行,一个蚂蚁围绕一个长宽的3厘米的正方形,另一个蚂蚁围绕一个长6厘米宽3厘米的长方形。他们再次相遇的时候,任何一只蚂蚁爬过的距离至少是多少厘米?
一个蚂蚁一圈爬3*4=12厘米,另一个爬(3+6)*2=18厘米,所以12和18的最小公倍数就是他们要移动的距离,答案就是36。第一个蚂蚁爬了3圈,第二个爬了两圈。
Answer:
One ant covers 3*4=12 cm and the other covers (3*6)*2=18 cm in each of their rounds around the rectangles.
When they meet again, they both should have covered an integer multiple of 12 or 18. Therefore, the minimum distance any one must cover is the least common multiple of 12 and 18, which is 36 in centimeters.
收起