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 Triangular Cubes III (Posted on 2011-06-12)
Three points have been chosen randomly from the vertices of a n-dimensional hypercube.

Determine the probability (in terms of n) that they form (a) an acute triangle; (b) a right triangle?

 No Solution Yet Submitted by K Sengupta No Rating

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Using n coordinates to define a point, define the vertices as the 2^n points having different permutations containing only '0's and '1's, for their coordinates.
First consider only those triangles AOB where O is the origin and vertices A, B have position vectors a, b. Angle AOB will be a right angle iff a.b = 0.

i.e.        a1b1 + a2b2 + ..... + anbn = 0                                (1)

A point A whose set of coordinates contains exactly r '1's can be chosen in nC r different ways.

For (1) to be true, the coordinates of B corresponding to these '1's must be zeros while all others can be '0's or '1's, with the exception that they can't all be 0, since B can't be at O. So B can be chosen in 2^(n-r) - 1 ways.

The total number of different triangles with right angles at O is therefore

[Sumr = 1 to n(nCr (2n-r - 1))]/2

The division by 2 is to eliminate duplicates with A and B interchanged, and the summation starts at r = 1 since r = 0 would allow A to be at O.

The same number of different triangles will have right angles at each other vertex, so multiplying by 2^n gives the total number of triangles with right angles. Then modifying the limits gives:

No. of right angled triangles:    R  = 2n-1 [Sumr =0 to n( nCr (2n-r - 1))  -  (2n - 1)]

R  =  2n-1 [Sumr=0 to n(nCr 2n-r)  -  Sumr=0 to n(nCr)  -  (2n - 1)]

Using binomial considerations:     R  =  2n-1 [(1 + 2)n  -  (1 + 1)n  -  2n + 1]

R  =  2n-1 (3n  -  2n+1  +  1)

The total no. of triangles is          T  =  Bin(2n,3)  = 2n(2n - 1)(2n - 2)/6

Thus probability of a right angled triangle = R/T

= 3(3n - 2n+1 + 1)/[(2n - 1)(2n - 2)]

giving:              n                      2          3          4          5          6          ..
P(right angled)               1          6/7       5/7       18/31    43/93
P(acute angled)              0          1/7       2/7       13/31    50/93

Edited on June 21, 2011, 6:18 pm

Edited on June 21, 2011, 6:20 pm

Edited on June 21, 2011, 6:21 pm

Edited on June 21, 2011, 6:24 pm
 Posted by Harry on 2011-06-21 18:15:38

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