# Barycentric coordinates

### Geometric Probability

Two points are selected at random on a straight line segment of length 1. What is the probability that a triangle can be constructed out of thus obtained three segments?

Let u, v, and w stand for the lengths of the three segments (one or two of them may be 0.) The three segments serve as the sides of a triangle iff

 (1) x1 < x2 + x3 x2 < x1 + x3 x3 < x1 + x2

From the conditions of the problem, x1 + x2 + x2 = 1. To each of the inequalities in (1) add its left-hand side to obtain

 (2) x1 < 1/2 x2 < 1/2 x3 < 1/2

The argument is obviously reversible. Therefore also, (2) implies (1). Let us consider the triple (u,v,w) as the barycentric coordinates of a point relative to some fixed triangle ABC. As we already know, points that satisfy (2) lie inside the triangle MaMbMc, where Ma, Mb, Mc are the midpoints of the sides BC, AC, and AB, respectively. The area of the triangle MaMbMc is one fourth that of ABC. Which shows that 1/4 is the sought probability. The barycentric coordinates can be set up in a more general space Rn, n>0. For n=1, it takes 2 distinct points A and B and two coordinates u and v. Every point K on the line R1 is then uniquely represented as K = uA + vB with u + v = 1. In R3, it takes 4 points. More generally, to define the barycentric coordinates in Rn one needs (n+1) points that do not lie in a space of a lesser dimension. For example, in R3, four vertices of a (nondegenerate) tetrahedron define a set of barycentric coordinates. However, no four points in R3 that belong to the same plane (R2) do; for 3>2!

For (n+1) points Ai, i = 0,...,n, the set of all points x0A0 + ... + xnAn with x0 + ... + xn = 1, is known as an n-dimensional simplex. So that respectively a segment is a 1-dimensional, a triangle is a 2-dimensional, and a tetrahedron is a 3-dimensional simplex. Simplexes play a fundamental role in Algebraic Topology. Let's return to our problem. Choose n points at random on a segment of length 1. What is the probability that an (n+1)-gon (a polygon with (n+1) sides) can be constructed from the (n+1) thus obtained segments? Condition (1) generalizes to

 (1') xi < x0 + ... + xn - xi = 1 - xi

for all i = 0, ..., n. Which is obviously equivalent to

 (2') xi < 1/2,

i = 0, ..., n. Generally speaking, areas change as the square of the linear size. Volumes change as the cube of the linear size. By analogy, bodies in Rn are endowed with the n-dimensional volume that changes as the n-th power of the linear size. The body described by (2') is obtained from the basic simplex by removing smaller simplexes, one at every vertex. Each of these (n+1) smaller simplexes has the n-volume equal to (1/2)n of the n-volume of the basis simplex. Therefore, the probability that the barycentric coordinates of a point satisfy (2') is

 (3) p = 1 - (n+1)(1/2)n,

The latter expression tends to 1 as n grows to infinity. It is easier to find segments to construct a many sided polygon than it is to find the sides of a triangle, which is rather natural.

### Reference

1. J. Coffman, What To Solve?, Clarendon Press, Oxford, 1996. Fourth printing ### Barycenter and Barycentric Coordinates ### Geometric Probability 