Find the probability that angle PQR is obtuse.?

Question

A square has corners ABCD labeled CW. Point P is randomly chosen on side AB, point Q is randomly chosen on side BC, and point R is randomly chosen on side CD.

Calculate the probability that angle PQR is obtuse.

The answer is actually 4/9

Answer 1

Here's my *corrected* derivation, the first was incorrect:

Assume without loss of generality that the square has sides of length 1.

P(obtuse|BQ=z)
=P(z*(1-z)>BP)+P(CR =z*(1-z)+P(CR =z*(1-z)
+int(int(1,CR=0..z*(1-z)/BP),
BP=z*(1-z)..1)
=z*(1-z)-z*ln(-z*(z-1))
+z^2*ln(-z*(z-1))

Thus P(obtuse)
=int(P(obtuse|BQ=z),z=0..1)
=int(z*(1-z)-z*ln(z*(1-z))
+z^2*ln(z*(1-z)),z=0..1)
=4/9

Answer 2

The probability is 0.44444...

Unfortunately the proper mathematical solution eludes me, and I arrived at this figure by writing a number-crunching program. As I increased the number of points along each side of the square at which to place P,Q and R, the answer seemed to tend to 0.44 recurring