# How to divide a n dimensional simplex in n+1 equal parts [closed]

I want to divide a n dimensional simplex (convex hull of n+1 points) in n+1 parts. E.g. consider the simplex in 2-D which is the convex hull of points (0,0,1), (0,1,0) amd (1,0,0). It can be divided in three parts as $P_1 = \{(0,1,0),(1/2,1/2,0),(0,1/2,1/2),(1/3,1/3,1/3)\}$

$P_2 = \{(1,0,0),(1/2,1/2,,0),(1/2,0,1/2),(1/3,1/3,1/3)\}$

$P_3 = \{(0,0,1),(1/2,0,1/2),(0,1/2,1/2),(1/3,1/3,1/3)\}$

I can somehow arrive at these partitions by considering intersection of some halfspaces (line joining the vertex to the midpoint of the other side). However, I am unable to extend this to higher dimensions.

Is there an algorithm in literature which does the same partitions. Can someone point me to some reference.

Thanks

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## closed as off topic by Andrej Bauer, Ryan Budney, Igor Pak, Andy Putman, Dan PetersenNov 23 '12 at 8:34

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You're looking for "Voronoi cells" :-) –  Gjergji Zaimi Nov 22 '12 at 23:59
This is not a research level question. –  Andrej Bauer Nov 23 '12 at 0:07

## 2 Answers

In fact, there are much more such divisions. The group of simplex is $S_{n+1}$, so it contains a subgroup $H=\langle (123\dots n+1)\rangle$ of order $n+1$ acting transitively on the hyperfaces. Now take a ray $r$ from the center $O$ of the simplex, and let $\{r_1,\dots,r_{n+1}\}$ be its orbit under the action of $H$. Every $n$ of these rays generate a polyhedral cone, the union of these cones is the whole space (unless something degenerates), and $H$ permutes these cones transitively. Hence $H$ also permutes the intersections of these cones with the simplex, and they are congruent. So, we obtain a desired division.

The pictures below show two types of such divsion of a regular tetrahedron.

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The magic word is "barycenter". The convex hull of the barycenter and any one face of the simplex has the right volume.

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just look up barycentric subdivision please before you start second guessing the answer. It is what you want. –  Andrej Bauer Nov 23 '12 at 0:07