You choose a random point, uniformly, within a regular hexagon, with unit side length.
What's the average distance to the six sides?
Label the points:
A .5, sqrt(3)/2
B 1, 0
C .5, -sqrt(3)/2
D -.5, -sqrt(3)/2
E -1,0
F -.5, sqrt(3)/2
O 0,0
Concentrate on triangle ABO as in Larry's posts.
All points within the triangle will have distances to AB and DE add up to sqrt(3).
When x < .5, distances to FA and DC will add up to sqrt(3), but
x > .5, distance to FA is distance to A and distance to DC is distance to C.
When y is below line FB the distances to AB and DE add up to sqrt(3), otherwise
distance to EF is distance to F and distance to CB is distance to B.
The slope of FB is -sqrt(3)/2 / (3/2) = -1/sqrt(3) and
FB goes through -.5, sqrt(3)/2, so the dividing line is
y = sqrt(3)/2 - (x+.5)/sqrt(3)
so
A = [.5, sqrt(3)/2];
B = [1, 0];
C = [.5, -sqrt(3)/2];
D = [-.5, -sqrt(3)/2];
E = [-1,0];
F = [-.5, sqrt(3)/2];
O = [0,0];
slopeFB=-1/sqrt(3);
granu=1/1000;
sumdist=0; numdist=0;
for x=granu:granu:1
for y=granu:granu:1
if y<sqrt(3)*x && y<sqrt(3)*(1-x)
sumdist=sumdist+sqrt(3);
numdist=numdist+2;
if x<.5
sumdist=sumdist+sqrt(3);
numdist=numdist+2;
else
sumdist=sumdist+norm(A-[x,y])+norm(C-[x,y]);
numdist=numdist+2;
end
if y<sqrt(3)/2+slopeFB*(x+.5) % slopeFB is negative
sumdist=sumdist+sqrt(3);
numdist=numdist+2;
else
sumdist=sumdist+norm(F-[x,y])+norm(B-[x,y]);
numdist=numdist+2;
end
end
end
end
sumdist/numdist
finds an average of:
0.874441697155773
when the granularity (granu) is changed to 1/2000:
0.874441259261843
The norm function in Matlab is the square root of the sum of the squares.
With granularity 1/1000, sumdist is 2269239.16392143 and numdist is 2595072.
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Posted by Charlie
on 2023-10-26 07:58:07 |
my numerical integration
