
Re: Random Triangle Problem
Posted:
Jul 28, 1997 4:18 AM


eclrh@sun.leeds.ac.uk (Robert Hill) wrote: (snip) Tony Richards wrote: >> >> So my final answer is 0.75 + pi/48, and not 0.75.<< (typo corrected) > >Considering separately the three terms for an obtuse angle at the first, second >and third points, I get that (conditionally on x) the probabilities are > >obtuse at first point: 1/2 >obtuse at second point: 1/2  x/(2L) >obtuse at third point: pi x^2 / (16L^2) >all angles acute: x/(2L)  pi x^2 > >which after integration over x become respectively > >obtuse at first point: 1/2 >obtuse at second point: 1/4 >obtuse at third point: pi/48 >all angles acute: 1/4  pi/48 > >Surely a correct answer would give equal probabilities for obtuseness at >the three vertices.
First, thanks for taking the trouble to read my post and point out the typos. I stick by my answer, 0.75+pi/48.
The reason that the probabilities for obtuse at each vertex are different must be because of the bias introduced as a result of placing the origin at the first point, making its location nonrandom in the plane (which, by making L> infinity, becomes infinite eventually). The other two points are then randomly located with respect to this first fixed point. So it is not surprising that the probabilities for each vertex are different.
 Tony Richards 'I think, therefore I am confused' Rutherford Appleton Lab ' UK '

