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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)

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Some Thoughts re(2): Extra Credit (Triangle XUV) | Comment 10 of 11 |
(In reply to re: Extra Credit (Triangle XUV) by Charlie)

Ah yes, thanks Charlie.  I see my mistake.

2/3 of the time U and V are near different vertexes.  And 1/3 of those times X is located such that XUV contains O.  (1/3)*(2/3) = 2/9, which is the number I gave.  (Incidentally, this is true even if X is not in the triangle, or is closer to a vertex than the center.  And yes, I know I haven't given this proof yet).

What I overlooked was that XUV can contain O even if U and V are closest to the same vertex (which happens with probability 1/3).  If Charlie is right (and he usually is), then in 1/30 of these cases XUV contains 0. 

2/9 + (1/3)*(1/30) = 21/90 = 7/30.  I have yet to calculate the 1/30, and don't see an obvious way to do it.

  Posted by Steve Herman on 2008-05-08 22:28:49

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