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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.

  Submitted by FrankM    
Rating: 3.5000 (2 votes)
Solution: (Hide)

Let DGJ be an equilateral triangle. Mark the points E and F dividing the length DG into three equal segments. Mark the analogous points HI along GJ and KL along JD. Connect FH, KI and EL to construct the regular hexagon DFHIKL (henceforth called M) and the three small equilateral triangles DEL, FGH, and IJK. (Henceforth called TD, TG and TJ). O, the centroid of DGJ, is simultaneously the centroid of M.

Case UVW:

Since U, V and W each lie closer to a vertex than to O, it follows that each point lies somewhere within one of the three small triangles: TD, TG or TJ. From examination of diagram 1, it is immediately apparent that O will lie in UVW iff the three points are distributed, one each, in these small triangles. Since these smaller triangles have equal area, the probability of O occuring within UVW is 2/9, i.e., it is the probability that three items distributed randomly amongst three containers gives one item in each container.

Case XYZ:

Since X,Y are closer to O than to any of the vertices, X,Y lie within the hexagon M. Draw the segments XO and ZO, extending them past O until they intersect hexagon M at the points X� and Z�, respectively. Then we can see that the triangle XYZ will contain O iff Y lies (within M) inside the sector X�OZ�.

Define X* = -X as the point on OX� such that OX* = OX. Now repeat the procedure of the preceeding paragraph to construct the sector X*�OZ', marking the locus of points Y such that X*YZ contains O. Since XYZ and X*YZ have no area in common, O can appear in at most one of these triangles (see diagram 2). Further, since the angles X�OZ� and X*�OZ� sum to 180 degrees, the probability of O occuring within one of the two triangles is equal to the area of sector X*�FZ�ELX�X*� divided by the area of M. That is: one half. Finally, since X* is uniquely and reflexively determined from X, it follows that the probability of O occuring within XYZ is 1/4.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(3): Extra Credit (Triangle XUV)Charlie2008-05-09 12:12:35
Some Thoughtsre(2): Extra Credit (Triangle XUV)Steve Herman2008-05-08 22:28:49
re: Extra Credit (Triangle XUV)Charlie2008-05-08 10:29:18
Extra Credit (Triangle XUV)Steve Herman2008-05-07 23:35:23
A remarkable insightFrankM2008-05-07 06:13:20
SolutionThe missing insightSteve Herman2008-05-06 19:54:39
Not foolish at allFrankM2008-05-05 22:42:50
Some ThoughtsFools rush in where wise men fear to go (spoiler)Steve Herman2008-05-05 22:30:23
Very pretty (so far)FrankM2008-05-05 06:01:22
Solutionsemi-analytic; semi-computerCharlie2008-05-04 18:26:16
Hints/TipsSolving UVW (partial spoiler)Steve Herman2008-05-04 15:26:12
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