
Re: Random Triangle Problem
Posted:
Jul 28, 1997 4:59 AM


Charles H. Giffen (chg4k@virginia.edu) wrote: : : An argument involving the incircle can be given, but one must be : more careful (since one is dealing with a distribution conditional : on the requirement that the three points yield a triangle with the : circle as its incircle)  the calculations are rather more tedious, : but do produce the same result: P(obtuse) = 3/4. : Noted the same thing. BTW, as a chemist a "mixing triangle" (to draw three quantities for which x+y+z=1) is well known to me. If you use this on random triangle angles, then outcircle and incircle immediately give P=3/4. (Shall I elaborate?) But let's try to complicate things as much as possible :) Assume random quantities of y=s/R and x=r/R, where s,r,R are half circumference,incircle radius and outcircle radius respectively. 0<x<1/2. y is bound by the equation (y^2+x^210x2)^2<(12x)^3*4. (A triangle cannot be "more than isosceles") The dividing line of the bounded yxarea is y=2+x. (Below are the obtuse, above the acute) Can anybody compute the two elliptic integrals arising?  Hauke Reddmann <:EX8 fc3a501@math.unihamburg.de PRIVATE EMAIL fc3a501@rzaixsrv1.rrz.unihamburg.de BACKUP reddmann@chemie.unihamburg.de SCIENCE ONLY

