What is Barycenter? — Mathematics

Definition 17.3 University Mathematics — Year 2 · Chapter 17 — Affine Spaces

Let $(A_i, \lambda_i)_{i \leq k}$ be weighted points with $\sum\lambda_i \neq 0$ . The barycenter $G = \operatorname{bar}\bigl((A_i, \lambda_i)\bigr)$ is the unique point with

\sum_i \lambda_i\, \vect{GA_i} = 0 \qquad\text{equivalently}\qquad \vect{OG} = \frac{1}{\sum\lambda_i}\sum_i \lambda_i\,\vect{OA_i} \quad (\text{any } O).

Barycenters are associative (subgroups of points may be replaced by their partial barycenter with the summed weight) and invariant under rescaling of all weights.

Examples

Example 17.7 (Classical barycenter geometry)

The centroid of a triangle $ABC$ is the barycenter $G = \operatorname{bar}(A,1; B,1; C,1)$ . Associativity with the midpoint $A' = \operatorname{bar}(B, 1; C, 1)$ shows

G = \operatorname{bar}(A, 1;\ A', 2) :

$G$ lies on the median $AA'$ at two-thirds of it — and likewise for the other two medians: the three medians are concurrent, in one line of barycentric calculus.

Example 17.8 (The bimedians of a quadrilateral)

Let $ABCD$ be any quadrilateral (planar or not!) and consider its bimedians: the segments joining the midpoints of opposite sides, $M_{AB}M_{CD}$ and $M_{BC}M_{DA}$ . Introduce the barycenter $G$ of $(A,1; B,1; C,1; D,1)$ and group the weights two ways:

G = \operatorname{bar}\bigl(M_{AB}, 2;\ M_{CD}, 2\bigr) = \operatorname{bar}\bigl(M_{BC}, 2;\ M_{DA}, 2\bigr) :

$G$ is the midpoint of both bimedians — so the two bimedians always bisect each other, and the quadrilateral of the four midpoints is a parallelogram (its diagonals are the bimedians). No case analysis, no coordinates, and the argument survives unchanged for a skew quadrilateral in $\R^3$ , where a picture-based proof would already be delicate: associativity does not care about dimension.

Example 17.12 (Epigraphs are convex sets)

The region $C = \{(x, y) : y \geq x^2\}$ above the parabola is convex: for $(x_1, y_1), (x_2, y_2) \in C$ and $t \in \intcc01$ , the convexity inequality of the square function gives

\bigl((1-t)x_1 + tx_2\bigr)^2 \leq (1-t)x_1^2 + tx_2^2 \leq (1-t)y_1 + ty_2 ,

so the barycenter stays above the parabola. The computation is general: $\{y \geq f(x)\}$ is convex exactly when $f$ is a convex function — convex sets and convex functions (Chapter 8) are two faces of one notion, epigraphs being the dictionary. This is the geometric reason support lines exist for convex functions, the fact that will prove Jensen’s inequality in Chapter 22.

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