Barycenter of a triangle

Question

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

I have been trying to prove this statement for hours now. I typically end up with statements like $AM=AM$ or $0=0$ and am really struggling. I can prove this using the center of mass concept in physics, where each point is assigned the same weight, but I cannot prove it purely geometrically. Also, I can prove geometrically that $OM = \frac{1}{2}(OA+OB)$ for a line $AB$ with midpoint $M$, but the 2D triangle seems to be a bit tougher.

Answer 1

Consider using barycentric coordinates. For simplicity, set a coordinate system so that point $A$ is located at the origin $r_1 = (0,0)$, point $B$ is located at some point $r_2 = (a,0)$ on the x-axis, and $C$ is at some point $C = (b,c)$. The point $O$, whose corresponding vector I denote by $v$, is expressible as the linear combination $$v =\lambda_2 r_2 + \lambda_3 r_3,$$ where $\lambda_i \in \mathbb{R}$, and $\lambda_2 + \lambda_3 = 1.$ Note that the centroid point $M$, denoted by $u$, has coordinates $$ u = \frac{1}{3} r_2 + \frac{1}{3} r_3.$$ So $$OM = ||v - u||$$ Meanwhile, $$OA + OB + OC = || v|| +|| v - r_2|| + ||v - r_3 ||$$ Expanding these out explicitly should lead you to the desired identity.

Answer 2

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 $x$-axis:

The mean of the $y$-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 $\frac13$ 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 $O$, we get $$ OM=\frac13OA+\frac13OB+\frac13OC $$