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

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

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 M_aM_bM_c, where M_a, M_b, M_c are the midpoints of the sides BC, AC, and AB, respectively. The area of the triangle M_aM_bM_c 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 A_i, i = 0,...,n, the set of all points x₀A₀ + ... + x_nA_n with x₀ + ... + x_n = 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

   1. 3D Quadrilateral - a Coffin Problem
   2. Barycentric Coordinates
   3. Barycentric Coordinates: a Tool
   4. Barycentric Coordinates and Geometric Probability
      • Stick Broken Into Three Pieces (Trilinear Coordinates)
      • Stick Broken Into Three Pieces (Cartesian Coordinates)
   5. Ceva's Theorem
   6. Determinants, Area, and Barycentric Coordinates
   7. Maxwell Theorem via the Center of Gravity
   8. Medians in a Quadrilateral
   9. Simultaneous Generalization of the Theorems of Ceva and Menelaus
      • Theorems of Ceva and Menelaus, an Illustrated Generalization
   10. Three glasses puzzle
   11. Van Obel Theorem and Barycentric Coordinates
   

   

   Copyright © 1996-2009 Alexander Bogomolny