这是上世纪末IBM的一个puzzle (不是用来interview 的， 是IBM给全世界数学爱好者和他们自己的数学家闲暇时间做的）。 他们没有找到简单办法（IBM Comment: Although we have a solution for this problem, it is not a simple one）。没有其他人在一个月时间内给出解答。问题在WXC贴出后，当天就出现了两个简单解法。(Yes, one is by me. - fzy )
We'll prove angle AFD = angle BDE = angle CEF. If two of the angles are the same, it's easy to prove all three are equal. WLOG, suppose angle AFD < angle BDE < angle CEF.
Consider the distance from D to AC, d(D, AC) and the distance from E to AB, d(E, AB), we have d(D, AC) < d(E, AB), therefore angle A < angle B. Similarly, angle B < angle C.
Now we have angle ADF > angle BED > angle CFE (sum of angles of a triangle is 180 degrees).
But angle ADF + angle BDE = 120 degrees = angle AFD + angle CFE, which is a contradiction.
What if there is an obtuse angle in A, B, C? Maybe you need to discuss a few cases.
www.ddhw.com
When this problem appeared in WXC on March 1, two correct solutions were obtained that day (one by fzy, one by mathfun). The person who posted the problem said that his friend had already found a simple solution, and later he revealed that the method was almost the same as fzy's (I personally think this is the simplest; mathfun's is simple too compared to IBM's). A few days later, another guy, coro, came up with another proof. It is a little more complicated and only handles one case, but I think it is correct for that case.
IBM's method discusses five cases. A high school student can not understand their method.
哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
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发表于 2005-3-11 17:49:48
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