(For me, it is pretty hard. Hope you high hands find it interesting.)
Four roads on a plane, each a straight line, are in general position so that no two are parallel and no three pass through the same point. Along each road walks a traveler at a constant speed. Their speeds, however, may not be the same. It's known that traveler #1 met with Travelers #2, #3, and #4. #2, in turn, met #3 and #4 and, of course, #1. Please show that #3 and #4 have also met.
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Sean 你真是太高了，情趣生活的那个Sean 是不是你？几何很厉害。 欢迎来WXC， 最近有一些悬而未决的几何题。 这个题目也在WXC贴出了，至今无解。有个梦回汉朝显然会做，他曾经贴了一句话提示：“ 物理方法”，被我错误的删掉了。后来（当然是过了很久）我理解了他的方法，也很巧妙（用一些物理术语描述起来或者更简单）。所以我觉得这题目还不错， 大概表面上看来不同的方法很多。它的那个解法我就不贴出了，一来不是我的，二来我怕理解错了。 |
Rotate the the coordinate axes so that the projections of the speed V1 and v2 on the x axis are equal. Since #1 and #2 meet, their x coordinates are always equal. Also since #3 and #4 meet both #1 and #2 at different points, their x coordinates must be equal (to that of #1 and #2), too. Since their paths meet (this condition is redundent and can be deducted from the order of the events), they will arrive at the intersection at the same time. |
In order for other guys to meet him, in #1's point of view, other guys must come towords him. Still in #1's point of view, in order for #3 and #4 to meet #2, #2, #3 and #4 should be on the same line (note that the roads are in general position). Then the problem is simplified to be 1-dimensional. |
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