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created: 2022-08-06T19:50:01 (UTC -04:00)
tags: []
source: https://math.stackexchange.com/questions/1423664/barycenter-of-a-triangle

author: JohnverJohnver

linear algebra - Barycenter of a triangle

Let $A$$A$, $B$$B$, and $C$$C$ denote vertices of a triangle in the plane. Let $O$$O$ be some point in the plane. Prove that $OM = \frac{1}{3}(OA+OB+OC)$$OM = \frac{1}{3}(OA+OB+OC)$, where $M$$M$ denotes the barycenter of the triangle, and all objects in the equation are geometrical vectors.

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The barycenter, or centroid, of a triangle happens to be the mean of the three vertices, but the definition is the center of mass of the whole triangle.

To see why the barycenter is the mean of the vertices, consider a triangle with two vertices on the $x$$x$-axis:

The mean of the $y$$y$-position would be

$$
\frac{\int_0^1\overbrace{{\ \ \ \ }th{\ \ \ \ }}^y\,\overbrace{(1-t)b\,h\,\mathrm{d}t}^{\mathrm{d}A}}{\int_0^1\underbrace{(1-t)b\,h\,\mathrm{d}t}_{\mathrm{d}A}}
=\frac{bh^2\int_0^1t(1-t)\,\mathrm{d}t}{bh\int_0^1(1-t)\,\mathrm{d}t}
=\frac13h
$$

That is, the distance from the side opposite each vertex to the barycenter is $\frac13$$\frac13$ the distance of the vertex from the side opposite.

Thus, the barycentric coordinates of the center of mass would be

$$
M=\frac13A+\frac13B+\frac13C
$$

Writing this as offsets from a point $O$$O$, we get

$$
OM=\frac13OA+\frac13OB+\frac13OC
$$