[bigpooch]

I didn't know this was a Putnam problem and this had me
stay up at night when you stated there was an elegant
solution. After two hours (usually I try not to spend
more than an hour to find the right idea), here is my
solution. I didn't spend time on finding anything more
efficient or more "elegant".

[BTW, if this really is a Putnam question, there must be
solutions posted somewhere, so maybe somebody can post a
different "elegant" solution if there is one to be found]

Solution
========

Let P(n) be the probability that the convex polygon formed
by n points uniformly randomly distributed on a circle have
the property of posessing an interior acute angle. As pzhon
noted, this is equivalent to there existing a semicircle for
which there exists only one point. Suppose n>=5. The case
for n=4 is trivial (every quadrilateral that can be
circumscribed by a circle has an acute angle except for the
case of the square, but four random points produce a square
with probability zero; thus P(4)=1).

Let the points on the circle be labelled A[j] for j=1 to n
counterclockwise starting with A[1]. Also denote that A[k]
is the same as A[j] if k and j are congruent modulo n, just
for ease of notation (e.g., A[0] is the same point as A[n]).

The angle formed by A[j],A[j+1] and A[j+2] is acute if
A[j+2] is at least a semicircle away. Now, A[j+1] could be
that far away in the counterclockwise direction as well
(this occurs with probability 1/[2^(n-1)]) or A[j+1] is just
within a semicircle and the other points are not (this
occurs with probability (n-1)/[2^(n-1)]). Altogether, then
A[j] "starts" an acute angle with probability n/[2^(n-1)].

This is true for every j and there are n points.

By inclusion-exclusion, the probability that SOME point
"starts" an acute angle is then [let sigma denote summation
over indices k,k1,k2,... where k1<k2<...]:

P(n)= sigma(P(A[k] "starts" an acute angle)
-sigma(P(A[k1] and P(A[k2] both "start" an acute angle)

since the higher order terms aren't possible (you can't have
more than two acute angles for an inscribed n-gon for n>=5).

For two points to "start" an acute angle, they must be
adjacent (otherwise, there would be two subtended arcs that
sum to over 2*pi, so they would have to overlap), so the
last summation has essentially only n terms.

Case a)
------

If A[j] "starts" an acute angle, then what is the
probability that A[j-1] does also? If A[j+1] is more than
a semicircle away, it always does and from above this
happens with probability 1/[2^(n-1)].

Case b)
------

If not, consider a circle with circumference 2 (so that
A[j+1] may be on an interval of unit length) and denote the
position of A[j+1] by u which is the distance along the
circle away from A[j] so that u is in (0,1) and u is a
uniform random variable. For A[j-1] to "start" an acute
angle, then this point, among the (n-2) other points must
be at least (1-u) away from A[j] or the probability of
all these points being at least far enough so that A[j-1] is
at least a distance of 1 or more from A[j+1] is then
u^(n-2) since all these points are uniform random variables
restricted on the "other half" of the circle.

Then, the probability that A[j-1] "starts" an acute angle is
the integral (from u=0 to 1) {[u^(n-2)] du} which has value
1/(n-1). This is GIVEN that A[j+1] is within a semicircle
away, so that the probability of this occuring is then
[1/(n-1)]*[(n-1)/[2^(n-1)]]=1/[2^(n-1)].

Thus, P(A[j-1) and A(j) both "start"...) is the sum of the
probabilities in a) and b) or 2/[2^(n-1)] and

P(n) = n*(n/[2^(n-1)]) - n*(2/[2^(n-1)]) or

P(n) = n(n-2)/[2^(n-1)].

Anyone care to check this?