Re: points in a square

Simon Tatham wrote:
Mark P <usenet@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

Still no takers on this one? FWIW, it can be solved without resorting to messy integrations, piecewise defined functions, etc. There is elegance to be found.


Hmm. I think I can generalise my result for the circle.

I've already proved that only the _bearing_ of each point from the
centre is relevant. So we can stop seeing solid circles and squares,
and start just seeing a clock face with three randomly chosen points
_on the circumference_. The only difference between the circle and
square cases is that the probability distribution is different.

However. Suppose we _transform_ the interval [0,2pi) into itself in
a non-uniform way by a function f, so that some pieces of the
interval are stretched out and others are squashed up. We can do
this in such a way that selecting three points with uniform
probability from [0,2pi) and then feeding them through f corresponds
exactly to selecting three points from [0,2pi) with the probability
distribution corresponding to the square.

Now the cunning bit is: because the square is symmetric in
180-degree rotation, pairs of points on the interval which are
diametrically opposed remain diametrically opposed under the
transformation f. Therefore, any three points a,b,c are on the same
side of some diameter (<=> the centre point is not contained within
their convex hull) iff f(a),f(b),f(c) are also on the same side of
some diameter. Hence, the probability for a square, or for any
2n-sided polygon, or indeed for any arbitrarily weird shape at all
which is symmetric in 180-degree rotation and star-convex about its
centre, is identical to that for the circle - i.e. 1/4.

Well done (except that I'm not sure about the star-convex requirement-- care to explain this?).

My own approach involves no transformations but simply notes the symmetry in choosing the second point to be either Y or Y reflected about C. Regardless of X, between these _two_ choices of Y, the _total_ area in which Z can be placed to enclose C is exactly 1/2 (a picture helps here). Thus the average over all values of Y gives 1/4, regardless of X, so this is indeed the average over all choices. It's easy to see that this works for any regular 2n-gon too.


However, this proof doesn't work for a regular polygon with an _odd_
number of sides, whereas you said you had one that would work for
all n.

Actually I never said such a thing, I only suggested it as extra credit. :) I too have no clean solution for odd n.
.

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