Four 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) a cyclic quadrilateral; (b) a rhombus.

Any subset of 3 or more points chosen from the regular polygon will form a cyclic polygon.  This is simply because every regular polygon has a circumscribed circle.  So the probability for part (a) is 1.

The only rhombus that has a circumscribed circle is a square.  The regular polygon only has a square among its vertices if n is a multiple of 4.  If n is not a multiple of 4 the probability is 0.

So assuming n is a multiple of 4 the probability is 1*3/(n-1)*2/(n-2)*1/(n-3) or (n/4)/C(n,4).

The first way comes from thinking of the points as picked one at a time.   The first point can be any but the next is constrained to be one of the three others that would make a square etc...
The second way comes from there being n/4 possible squares to be made and C(n,4) is the total number of ways of choosing 4 points.

Either way simplifies to 6 / (n^3 - 6n^2 + 11n - 6)

Posted by Jer on 2011-12-06 11:10:30