Well, if you take degenerate triangle $AAA$ where $A$ is any point on the plane then it's obvious.

As I saw the official solution they don't want degenerate triangles. So I leave the proof to you, as there is no sense in re-writing.

just create an equilateral BCD with centroid A and we have f(B)+f(C)+f(D)=f(A) however if we consider the centroid of these three triangles ABC,ABD,ACD we'll discover that A is the centroid of these centroids so that f(A)=3f(A)+2f(B)+2f(C)+2f(D)=5f(A) which leads us to the solution

Double post: https://artofproblemsolving.com/community/c6h1826975p12222324

WolfusA wrote: Well, if you take degenerate triangle $AAA$ where $A$ is any point on the plane then it's obvious. WolfusA wrote: As I saw the official solution they don't want degenerate triangles. So I leave the proof to you, as there is no sense in re-writing. Yeah. Comment from organizer (on ARMO 2019): triangle $ABC$ must be (non-degenerated) triangle. It means that $A$, $B$ and $C$ must be pairwise different non-collinear points.

Wizard_32 wrote: Double post: https://artofproblemsolving.com/community/c6h1826975p12222324 No, it's post about 11.1, and this 'double post' is about 10.1. But it's another point that they were literaly the same)

Let $M_{PQR}$ denotes the centroid of $\triangle PQR.$ For every point $A$ we can construct $X,Y,Z$ such that $M_{XYZ}=A.$ Let $X',Y',Z'$ be midpoints of $YZ,ZX,XY,$ hence $$ f(X)+f(Y)+f(Z)=f(A)=f(M_{XY'Z'})+f(M_{X'YZ'})+f(M_{X'Y'Z})=$$$$=(f(X)+f(Y)+f(Z))+2(f(X')+f(Y')+f(Z'))\implies f(A)=f(X')+f(Y')+f(Z')=0\quad \blacksquare$$

For any point $P$ on the plane, construct a triangle $ABC$ with centroid $P$. Then $f(A)+f(B)+f(C)=f(P)$. Let $M_A, M_B, M_C$ be the midpoints of sides $BC, CA, AB$ respectively. Then $f(M_A)+f(M_B)+f(M_C)=f(P)$ Also, Let $M_1, M_2, M_3$ be the centroids of $\triangle AM_BM_C, \triangle BM_CM_A, \triangle CM_AM_B$ respectively. Then we have, \begin{align} f(M_1)=f(A)+f(M_B)+f(M_C)\\ f(M_2)=f(B)+f(M_C)+f(M_A)\\ f(M_3)=f(C)+f(M_A)+f(M_B) \end{align}Also, $f(M_1)+f(M_2)+f(M_3)=f(P)$ Adding $(1), (2), (3)$, we get, $$f(A)+f(B)+f(C)+2(f(M_A)+f(M_B)+f(M_C))=f(M_1)+f(M_2)+f(M_3)=f(P)$$$$\implies f(P)+2f(P)=f(P)\implies f(P)=0$$ [Also, a similar hexagonal arrangement can also be made instead of triangular arrangement.]