Three points have been chosen randomly from the vertices of a
tesseract (4-cube).
What is the probability that they form (a) an acute triangle; (b) a right triangle?
There are 16 vertices on a tesseract. Taken 3 at a time, there are 560 combinations that can be chosen.
The program numbers the vertices 0 through 15 and assigns w, x, y and z positions at all combinations of 0 and 1:
w x y z
0 0 0 0
1 0 0 0
0 1 0 0
1 1 0 0
0 0 1 0
1 0 1 0
0 1 1 0
1 1 1 0
0 0 0 1
1 0 0 1
0 1 0 1
1 1 0 1
0 0 1 1
1 0 1 1
0 1 1 1
1 1 1 1
It checks all 560 possible triplets of these vertices, using the law of cosines to compute each angle. To guard against rounding error leading to non-right angles, if a cosine falls within a small allowable range it's counted as zero, for a right angle. A check is made on all sums of the angles of the triangle and they each indeed come out to 180 degrees total.
The final count is 400 right triangles and 160 acute triangles, giving probabilities of 400/560 = 5/7 of being right and 160/560 = 2/7 of being acute.
DEFDBL A-Z
pi = ATN(1) * 4
DIM w(15), x(15), y(15), z(15)
CLS
FOR i = 0 TO 15
w(i) = i MOD 2: PRINT w(i);
x(i) = (i \ 2) MOD 2: PRINT x(i);
y(i) = (i \ 4) MOD 2: PRINT y(i);
z(i) = (i \ 8) MOD 2: PRINT z(i)
NEXT
FOR a = 0 TO 13
FOR b = a + 1 TO 14
FOR c = b + 1 TO 15
d1 = SQR((w(a) - w(b)) ^ 2 + (x(a) - x(b)) ^ 2 + (y(a) - y(b)) ^ 2 + (z(a) - z(b)) ^ 2)
d2 = SQR((w(c) - w(b)) ^ 2 + (x(c) - x(b)) ^ 2 + (y(c) - y(b)) ^ 2 + (z(c) - z(b)) ^ 2)
d3 = SQR((w(a) - w(c)) ^ 2 + (x(a) - x(c)) ^ 2 + (y(a) - y(c)) ^ 2 + (z(a) - z(c)) ^ 2)
ca1 = (d2 ^ 2 + d3 ^ 2 - d1 ^ 2) / (2 * d2 * d3)
ca2 = (d1 ^ 2 + d3 ^ 2 - d2 ^ 2) / (2 * d1 * d3)
ca3 = (d2 ^ 2 + d1 ^ 2 - d3 ^ 2) / (2 * d2 * d1)
IF ABS(ca1) < .000000001# THEN ca1 = 0
IF ABS(ca2) < .000000001# THEN ca2 = 0
IF ABS(ca3) < .000000001# THEN ca3 = 0
sa1 = SQR(1 - ca1 * ca1)
sa2 = SQR(1 - ca2 * ca2)
sa3 = SQR(1 - ca3 * ca3)
IF ca1 = 0 THEN a1 = 90: ELSE a1 = 180 * ATN(sa1 / ca1) / pi
IF ca2 = 0 THEN a2 = 90: ELSE a2 = 180 * ATN(sa2 / ca2) / pi
IF ca3 = 0 THEN a3 = 90: ELSE a3 = 180 * ATN(sa3 / ca3) / pi
PRINT a1 + a2 + a3;
IF ca1 = 0 OR ca2 = 0 OR ca3 = 0 THEN
rtCt = rtCt + 1
ELSEIF ca1 > 0 AND ca2 > 0 AND ca3 > 0 THEN
acCt = acCt + 1
ELSE
obCt = obCt + 1
END IF
NEXT
NEXT
NEXT
PRINT rtCt, acCt, obCt
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Posted by Charlie
on 2010-12-21 13:48:39 |
computer solution
