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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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re(3): Extra Credit (Triangle XUV) Comment 11 of 11 |
(In reply to re(2): Extra Credit (Triangle XUV) by Steve Herman)

In order for the triangle to include the point O when one point to is nearer to the center than to a vertex and the other two points are in the same triangle nearest one vertex, the conditional probability would be based on the fact that the point that's nearest point O be in a small triangular subset of the inner hexagon that has an edge along the inner 1/3 of a side of the larger triangle (probability 1/2), and that the common vertex that the other two points are near would be in the triangle farthest from (that is, opposite) the one with the first point (probability 1/3). These together make a probability of 1/6 that this necessary condition be met. But since it's not a sufficient condition, it also depends on where within these smaller triangles the points fall.

If this further conditional probability is 1/5, then the overall conditional probability would be the 1/30 previously referred to, making the overall probability the 7/30 also previously mentioned.

The following program tests the conditional probability, given the presence in the smaller triangles mentioned, that point O would be included in the triangle:

DEFDBL A-Z
x0 = .5: y0 = SQR(3) / 6

'SCREEN 12

FOR i = 1 TO 1000000
  DO
   x1 = RND(1): y1 = RND(1)
  LOOP UNTIL y1 / x1 < SQR(3) AND y1 < SQR(3) * (1 - x1)
  y1 = y1 / 3: x1 = (x1 - x0) / 3 + x0
  ' PSET (x1 * 400, (1 - y1) * 400), 12
  DO
   x2 = RND(1): y2 = RND(1)
  LOOP UNTIL y2 / x2 < SQR(3) AND y2 < SQR(3) * (1 - x2)
  y2 = y2 / 3 + 2 * y0: x2 = (x2 - x0) / 3 + x0
  ' PSET (x2 * 400, (1 - y2) * 400), 14
  DO
   x3 = RND(1): y3 = RND(1)
  LOOP UNTIL y3 / x3 < SQR(3) AND y3 < SQR(3) * (1 - x3)
  y3 = y3 / 3 + 2 * y0: x3 = (x3 - x0) / 3 + x0
  'PSET (x3 * 400, (1 - y3) * 400), 9
 
   m = (y2 - y1) / (x2 - x1)
   a = y1 - m * x1
   test1 = y3 - (m * x3 + a)
   test2 = y0 - (m * x0 + a)
  
   m = (y3 - y1) / (x3 - x1)
   a = y1 - m * x1
   test3 = y2 - (m * x2 + a)
   test4 = y0 - (m * x0 + a)
 
   m = (y3 - y2) / (x3 - x2)
   a = y2 - m * x2
   test5 = y1 - (m * x1 + a)
   test6 = y0 - (m * x0 + a)
 
   IF test1 * test2 > 0 AND test3 * test4 > 0 AND test5 * test6 > 0 THEN
     hit = hit + 1
   END IF
   ct = ct + 1
   PRINT hit, ct, hit / ct
NEXT

However, it finds 196,077 hits out of 1000000 trials, 3923 short of the 200,000 that would be expected if the probability were 1/5.

The 7/30 guess was based on a heuristic probability of 0.233, and the idea that 0.23333... (i.e., 7/30) was more likely than 233/1000. Perhaps the probability is not even a rational number.

But 2/9 + (1/3)*(1/6)*.196 comes out to .2331111..., so the heuristic .196 found by the current program is consistent with the .233 found by the overall Monte Carlo simulation together with Steve Herman's new analysis.

 


  Posted by Charlie on 2008-05-09 12:12:35
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