Three points have been chosen randomly from the vertices of a n-sided regular polygon.
Determine the probability (in terms of n) that they form (a) an acute triangle; (b) a right triangle?
n p(acute) p(right)
3 1.000000000 0.00000000
4 0.000000000 1.00000000
5 0.666666667 0.00000000
6 0.200000000 0.60000000
7 0.600000000 0.00000000
8 0.285714286 0.42857143
9 0.571428571 0.00000000
10 0.333333333 0.33333333
11 0.555555556 0.00000000
12 0.363636364 0.27272727
13 0.545454545 0.00000000
14 0.384615385 0.23076923
15 0.538461538 0.00000000
16 0.400000000 0.20000000
17 0.533333333 0.00000000
18 0.411764706 0.17647059
19 0.529411765 0.00000000
20 0.421052632 0.15789474
21 0.526315789 0.00000000
22 0.428571429 0.14285714
23 0.523809524 0.00000000
24 0.434782609 0.13043478
25 0.521739130 0.00000000
26 0.440000000 0.12000000
27 0.520000000 0.00000000
28 0.444444444 0.11111111
29 0.518518519 0.00000000
30 0.448275862 0.10344828
31 0.517241379 0.00000000
32 0.451612903 0.09677419
33 0.516129032 0.00000000
34 0.454545455 0.09090909
35 0.515151515 0.00000000
36 0.457142857 0.08571429
37 0.514285714 0.00000000
38 0.459459459 0.08108108
39 0.513513514 0.00000000
40 0.461538462 0.07692308
41 0.512820513 0.00000000
42 0.463414634 0.07317073
43 0.512195122 0.00000000
44 0.465116279 0.06976744
45 0.511627907 0.00000000
46 0.466666667 0.06666667
47 0.511111111 0.00000000
48 0.468085106 0.06382979
49 0.510638298 0.00000000
50 0.469387755 0.06122449
DEFDBL A-Z
CLS
FOR n = 3 TO 50
trict = 0: rtct = 0: obtct = 0
FOR s1 = 1 TO n - 1
FOR s2 = 1 TO n - s1 - 1
c1 = s1 * 360 / n
c2 = s2 * 360 / n
c3 = (n - s1 - s2) * 360 / n
a1 = 180 - (c1 + c2) / 2
a2 = 180 - (c2 + c3) / 2
a3 = 180 - (c3 + c1) / 2
trict = trict + 1
IF ABS(a1 - 90) < .00000001# OR ABS(a2 - 90) < .00000001# OR ABS(a3 - 90) < .00000001# THEN
rtct = rtct + 1
END IF
IF a1 > 90.00000000999999# OR a2 > 90.00000000999999# OR a4 > 90.00000000999999# THEN
obtct = obtct + 1
END IF
NEXT s2
NEXT
PRINT USING "### #.######### #.########"; n; (trict - obtct - rtct) / trict; rtct / trict
NEXT n
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Posted by Charlie
on 2011-08-14 16:58:33 |