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Capture the Flag (Posted on 2008-05-04) Difficulty: 4 of 5

Let O designate the centre of an equilateral triangle. Points U-Z are chosen at random within the triangle. We have learnt that points U,V,W are each nearer to a (possibly different) vertex than to O; while X,Y are each closer to O than to any of the vertices.

Show that triangle XYZ is more likely than triangle UVW to contain the point O within its interior.

See The Solution Submitted by FrankM    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Extra Credit (Triangle XUV) | Comment 8 of 11 |
Frank:

In case you are interested, I worked out and posted my earlier solution before seeing your solution, although it was your comment about Cauchy inversions that jogged my thinking process.

As for the case where exactly one point is closer to the center than to the vertex (such as triangle XUV), I have this as probability 2/9, same as triangle UVW.  I'll try to post my solution tomorrow (no time tonight), but the logic was not as difficult for me as triangle XYZ was.  Feel free to post before I do, if you confirm my probability of 2/9.





  Posted by Steve Herman on 2008-05-07 23:35:23
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