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.

In this program, I've endeavored to simulate the situation of one point closer to the center, and the other two points closer to a vertex or vertices, but the tally at the end is 233,070 out of 1,000,000 tries. That's 10,848 more than would be expected from a 2/9 probability. It's more consistent with 7/30.

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 < SQR(3) / 3 AND y1 / x1 < SQR(3) AND y1 < SQR(3) * (1 - x1) AND y1 > SQR(3) * (1 / 3 - x1) AND y1 > SQR(3) * (x1 - 2 / 3)
  ' PSET (x1 * 400, (1 - y1) * 400), 12
  DO
   x2 = RND(1): y2 = RND(1)
  LOOP UNTIL (y2 > SQR(3) / 3 OR y2 < SQR(3) * (1 / 3 - x2) OR y2 < SQR(3) * (x2 - 2 / 3)) AND y2 / x2 < SQR(3) AND y2 < SQR(3) * (1 - x2)
  ' PSET (x2 * 400, (1 - y2) * 400), 14
  DO
   x3 = RND(1): y3 = RND(1)
  LOOP UNTIL (y3 > SQR(3) / 3 OR y3 < SQR(3) * (1 / 3 - x3) OR y3 < SQR(3) * (x3 - 2 / 3)) AND y3 / x3 < SQR(3) AND y3 < SQR(3) * (1 - x3)
  ' 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