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?
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