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Topic: Bisecting A Tetrahedron.
Replies: 6   Last Post: Oct 8, 2010 4:21 PM

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Patrick D. Rockwell

Posts: 57
Registered: 12/12/04
Re: Bisecting A Tetrahedron.
Posted: Oct 8, 2010 4:21 PM
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I finally figured out the answer that I wanted for
any value of N.


First of all, if you break the stick into 3 pieces,
the answer is 0.04 for d=0.4. The original intent of
my post was to solve this problem for any value of N
where N is the number of pieces that the stick in being
broken into.

If you have a triangle whose altitude H is 1 and
you draw 3 lines within it, each a distance of 0.4
from the side of the triangle to which they are parallel,
then they form a smaller triangle whose altitude is 0.2 H,
therefore it's area is 0.2*0.2=.04 of the original triangle.
The reason why is because the original triangle has its
center at H/3 or 1/3. It shares this center with the smaller
triangle and since the other triangle has its lines at a
distance of 0.4 from the lines of the larger one, so the
center is a distance of (0.4-1/3)/(1/3)=0.2 from each line
of the smaller one.

In trying to solve the broken stick problem for higher
dimensions, and larger values of N, I discovered this.

N as stated is the number of pieces that the stick is
being broken into. In this example, N=4. K is the
dimensionality of the simplex we are working with. For
this example, the simplex is a tetrahedron so, K=3
(K always equals N-1). For information on what a simplex
is, go to http://en.wikipedia.org/wiki/Simplex

So, if you break the stick into 4 pieces and want to know
the probability that the largest piece will be no more than
d of the length of the original stick, where 1/n<=d<1 then
there are 3 regions of interest.



(2/N)<=d<1
(1/k)<=d<(2/N)
(1/N)<=d<(1/k)



This problem can be modeled with two tetrahedrons, one inside
the other with its vertices pointing toward the faces of the
other. If my terminology is correct, one of these represents
all of the ways that the stick can be broken into 4 pieces
while the other represents all of the ways in which the stick
can be broken into 4 pieces so that the largest piece is say,
0.4 the size of the original stick. The first tetrahedron I call
the sample tetrahedron, and the second one I call the event
tetrahedron. For brevity's sake throughout this post I should
refer to the sample tetrahedron as tetrahedron A and the
event tetrahedron as tetrahedron B.

If you ask the question "What is the probability that the largest
piece is between 2/n and 1 times the length of the original stick"
then tetrahedron B is larger then tetrahedron A, but tetrahedron
A still has its 4 vertices poking outside the faces of tetrahedron
B. So you can solve that probability with



1-N*(1-d)^(N-1) where N=4. If d=0.51 then the answer is 1-4*0.49^3
=0.529404.



If d is between 1/N and 1/k then tetrahedron B has NONE of its
vertices poking out of the faces of tetrahedron A and so, you
just take its volume and that will be the answer. For example,

if you break a stick into 4 pieces, what is the probability that
the largest piece is no more than 0.3 of the original stick?

Both of these tetrahedra share the same center. The distance
between the center and one face of tetrahedron A is 1/N or 0.25.
The distance between the center and one face of tetrahedron B is
0.3-0.25 so the tetrahedron B has an altitude equal to (0.3-0.25)/0.25
= 0.2 of tetrahedron A, therefore 0.2^3 of its volume and the answer
is 0.008.

Now, if d falls into the range of (1/k)<=d<(2/N) then it gets just
a little bit more tricky. If I break a stick into 4 pieces, I want
to know the probability that the largest piece is no more than 0.4
of the original stick.

0.4 is between 1/3 and 2/4. Tetrahedron B is smaller than the
tetrahedron A, but its 4 vertices are sticking through its faces. We
want to know what the volume shared by both is in relation to
tetrahedron A.

First, lets determine their respective altitudes. (0.4-.25)/0.25=0.6
so tetrahedron B has 0.6 the altitude of the other, and 0.216 of
its volume. If you draw a line from each face to the vertex of
tetrahedron B, 2/3 of it's length is inside of tetrahedron A and 1/3
is outside so we can determine the proportion of tetrahedron B is
shared with tetrahedron A.



1-4*(1/3)^3=1-4/27=23/27



So 23/27 of tetrahedron B is shared with tetrahedron A but we want to
know want proportion of tetrahedron A this is. To find out we multiply
the above answer times 0.216 (the volume of tetrahedron B) and we get
0.184.

As I typed this I realized that you could just get the answer with



.216-4*(0.2^3)=0.216-4*0.008=0.184




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