At the end of the discussion on Ceva's Theorem, we arrived at the conclusion that, for any point K inside ABC, there exist three masses w_A, w_B, and w_C such that, if placed at the corresponding vertices of the triangle, their center of gravity (barycenter) coincides with the point K. August Ferdinand Moebius (1790-1868) defined (1827) w_A, w_B, and w_C as the barycentric coordinates of K. (It's possible to generalize and consider also negative masses for points outside the triangle. This is not necessary for my purposes.) As defined, the barycentric coordinates are not unique. Masses kw_A, kw_B, and kw_C have exactly same barycenter for any k > 0. Thus barycentric coordinates are a form of general homogeneous coordinates that are used in many branches of mathematics (and even computer graphics). With one additional condition

the barycentric coordinates are defined uniquely for every point inside the triangle. (Barycentric coordinates that satisfy (*) are known as areal coordinates because, assuming the area of ΔABC is 1, the weights w are equal to the areas of triangles KBC, KAC, and KAB.) Since the center of gravity of any two points lies on the connecting segment, w_A = 0 for points on BC, w_B = 0 for points on AC, and w_C = 0 on AB. Vertices A,B,C have coordinates (1,0,0), (0,1,0), and (0,0,1), respectively.

The usual way to define the barycenter of three points with given masses (let's call such points material) is first to place the sum of masses of any two points at their barycenter. Now repeat the same (1-dimensional) operation pairing the new and the third remaining point. (The argument is formalized within affine geometry. I prefer to keep the discussion intuitive.) Assuming (*), if w_A is kept fixed, so will be the sum w_B + w_C. It then follows that for two possible positions D and D' of the barycenter of B and C, we have DK/KA = w_A/(w_B + w_C) = D'K'/K'A. Therefore, KK' is parallel to BC. In other words, the equation w_A = const describes the lines parallel to BC. As we already noted, the equation of BC is w_A = 0. A similar relationship exists between w_B and AC and w_C and AB.

If M_b and M_c are midpoints of AC and AB, respectively, then the equation of M_bM_c is w_A = 1/2. Similarly, M_aM_c = {(w_A, w_B, w_C): w_B = 1/2} and M_aM_b = {(w_A, w_B, w_C): w_C = 1/2}.

Let's have a look at the diagram on the right. In the blue triangle, all three coordinates are less than 1/2. Therefore, the first digit of their binary representation is zero. In the red triangles, two coordinates are less than 1/4 while the third is between 1/2 and 3/4. Therefore, in the red triangles, all three coordinates have their second binary digit zero. This leads to the trema removal procedure for constructing the Sierpinski gasket and provides a clue to its description in the barycentric coordinates.

Barycentric coordinates arise naturally whenever variable quantities have a constant sum. Three glass problem, where we are pouring water from one glass to another under the unrealistic assumption that in the process no drop of water is going to be spilled, is a salient example. The problem illustrates beautifully the concept of barycentric coordinates.

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. Three glasses puzzle
10. Van Obel Theorem and Barycentric Coordinates

Copyright © 1996-2008 Alexander Bogomolny