
Re: Random Triangle Problem
Posted:
Jul 30, 1997 1:53 AM


eclrh@sun.leeds.ac.uk (Robert Hill) writes of some fascinating simulation data!
> I generated 1 million pseudorandom triangles
> (1) vertices independent uniform inside a square: > obtuse at first vertex at second at third all acute > 241781 241998 241802 274419 > > (2) vertices independent uniform inside a circle: > 248534 248198 248619 254649 > > (3) vertices independent uniform inside an equilateral triangle: > 249377 249031 249279 252313
These are all *hugely* away from 3/4 to 1/4. The best has a chisquare of about 20 for 1 df. MADLY significant.
> I can't understand why the answer for a circle comes between > that for a square and that for a triangle! Can anyone?
I certainly can't. It seems most counterintuitive. Did you write them down in the wrong order maybe?
> Or is it just a bug in my program?
I'd hesitate to say so. For one thing there's no significant differences between the three corners; (in spite of that other twit who thinks there *should* be! :) ) If it's not too difficult, could you repeat the simulation with a regular pentagon, hexagon, octagon... see if the results tend toward the circular case, as they ought.
> (4) x and y coordinates of each vertex independent normal N(0,1) > (if my memory is correct this distribution is isotropic, > i.e. for any fixed a it makes x cos a + y sin a also N(0,1)): > 249791 249989 249974 250246
> answer for this case may be exactly 3/4. So I reran it with > 20 million triangles and got > > 4999137 5002204 4999415 4999244
Now this is absolutely amazing! These figures are spot on the 1:3 ratio. The chisq(1) value is 0.15 . Almost too *good*! (Hello Gregor Mendel!)
The normal distribution has a good record of being the most "truly random" in various contexts; but even so, I'm amazed to see it do the job here! What I'm wondering is how much is due to mere circular symmetry and infinite support. So a good one to compare it with would be Cauchylike.
Robert: can you run off a few more million simulations using a 2D Cauchy; ====== and report back to us please? You can do it with...
@ is uniform on [0,pi]; U is uniform on [0,1] (and indedendent of course).
Then R = tan(pi(U0.5)) and x = R cos@ y = R sin@ .
Hope you can. Thanks in advance.
> Namely, if the vertices of a triangle > are chosen independently and unifomly ON a given circle, then the 3/4 > answer follows rigorously.
> So the 3/4 answer for the whole plane is essentially equivalent > to the statement that the vertices of a random triangle are > uniformly and independently distributed on its circumcircle.
> Or that the points of contact of the sides with the incircle are > independently and uniformly distributed on the incircle.
Yes, an amusing coda. :) A disguised form of my three slopes answer. (Similar identifications hold in the Bertrand problem.)
 Bill Taylor W.Taylor@math.canterbury.ac.nz  I do not claim my argument is logical, but simply that I'm right. 

