Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: Bisecting A Tetrahedron.
Posted:
Oct 8, 2010 4:21 PM


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 (0.41/3)/(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 N1). 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
1N*(1d)^(N1) where N=4. If d=0.51 then the answer is 14*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 0.30.25 so the tetrahedron B has an altitude equal to (0.30.25)/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.
14*(1/3)^3=14/27=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
.2164*(0.2^3)=0.2164*0.008=0.184



