Let O designate the centre of an equilateral triangle. Points UZ 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.
To demarcate the areas involved, divide each side of the equilateral triangle into 3 equal parts. The points of division form a central regular hexagon inscribed within the triangle, leaving three small equilateral triangles, each with one of the original triangle's vertices as a vertex of its own.
The three points U,V and W are each located in one or another of the small vertex triangles.
If all three of the points U,V,W are within the same small triangle, the triangle they form cannot contain O.
If any two of the three are within the same small triangle, the probability is zero that the triangle formed would even include O on its perimeter. (That would be if one of the two points was at the vertex of the hexagon and the point not in the same small triangle were at the opposite vertex of the hexagon.)
That leaves the situation of each of the three points U,V,W being within a different one of the three small equilateral triangles. This has probability (2/3)*(1/3) = 2/9, and whenever this situation holds, O will be included within the formed triangle. There is probability zero that this will be on the perimeter of the formed triangle, occurring only if two of the points are as close together as possible while being in separate small triangles, and the third point is also at one of the two hexagon vertices that are on the remaining small triangle.
So the probability for U,V,W is 2/9.
The following program simulates the situation for X,Y,Z. The graphic statements, commented out via apostrophe, were for verifying where the points were being randomly placed. The iteration is done one million times, out of which, on 250,477 occasions, the formed triangle enclosed point O, strongly indicating the probability for X,Y,Z is 1/4, which is larger than that for U,V,W.
This is a modification from my program for Three Points in a Square.
DEFDBL AZ
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 AND y2 / x2 < SQR(3) AND y2 < SQR(3) * (1  x2) AND y2 > SQR(3) * (1 / 3  x2) AND y2 > SQR(3) * (x2  2 / 3)
' 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)
' 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
A variation of this program verifies the 2/9 probability found analytically, above, for U,V,W:
DEFDBL AZ
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 OR y1 < SQR(3) * (1 / 3  x1) OR y1 < SQR(3) * (x1  2 / 3)) AND y1 / x1 < SQR(3) AND y1 < SQR(3) * (1  x1)
' 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
where the result was 222,234 occasions of U,V,W enclosing point O, out of a million sets of U,V,W; definitely close enough to the expected 222,222.22....
By the way, if the commentedout graphic statements are restored to activity, the program will not run under Windows Vista. As of Vista, Microsoft has removed the capability of running DOS graphics programs.
Edited on May 4, 2008, 6:29 pm

Posted by Charlie
on 20080504 18:26:16 