Pretty easy if you've seen this sort of thing before. Let $M$ be arbitrary, and let $\triangle ABC$ be any equilateral triangle with centroid $M$. Denote by $D$, $E$, and $F$ the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ respectively, and set $M_A$, $M_B$, $M_C$ as the centroids of $\triangle AFE$, $\triangle BFD$, and $\triangle CDE$. Observe that by homothety, $M$ is also the centroid of $\triangle DEF$ and $\triangle M_AM_BM_C$. Then \begin{align*} 3f(M) &= f(A) + f(B) + f(C) + 2(f(D) + f(E) + f(F))\\ &= [f(A) + f(E) + f(F)] + [f(B) + f(F) + f(D)] + [f(C) + f(D) + f(E)]\\ &= f(M_A) + f(M_B) + f(M_C) = f(M), \end{align*}whence $f(M) = 0$.

Choose a triangle $ABC$ and let $N$ be the midpoint of $BC.$ Define $M \in AN$ with $AM : MN=2:1.$ Note that $M$ is the centroid of $\triangle ABC.$ Now choose two points $A',M'$ such that $M' \in A'N$ with $A'M':M'N=2:1$ and $A,N,A'$ are not collinear. Then $M'$ is the centroid of $\triangle A'BC.$ Reflect $M'$ about $N$ to get $P.$ Then note that $M$ is the centroid of $\triangle APM'$ and so $$f(M')+f(P)=f(M)-f(A)=f(B)+f(C)=f(M')-f(A')$$Hence $f(A')+f(P)=0.$ Since we can arbitrarily choose $A',N,$ we are done. $\square$

djmathman wrote: Pretty easy if you've seen this sort of thing before. Let $M$ be arbitrary, and let $\triangle ABC$ be any equilateral triangle with centroid $M$. Denote by $D$, $E$, and $F$ the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ respectively, and set $M_A$, $M_B$, $M_C$ as the centroids of $\triangle AFE$, $\triangle BFD$, and $\triangle CDE$. Observe that by homothety, $M$ is also the centroid of $\triangle DEF$ and $\triangle M_AM_BM_C$. Then \begin{align*} 3f(M) &= f(A) + f(B) + f(C) + 2(f(D) + f(E) + f(F))\\ &= [f(A) + f(E) + f(F)] + [f(B) + f(F) + f(D)] + [f(C) + f(D) + f(E)]\\ &= f(M_A) + f(M_B) + f(M_C) = f(M), \end{align*}whence $f(M) = 0$. Yeah, that was also the solution that I found. I believe that this is the official solution as well.

k.vasilev wrote: Each point $A$ in the plane is assigned a real number $f(A).$ It is known that $f(M)=f(A)+f(B)+f(C),$ whenever $M$ is the centroid of $\triangle ABC.$ Prove that $f(A)=0$ for all points $A.$ Nice problem. Let $ABC$ be any triangle with centroid $G$. Reflect $G$ in the midpoint of $BC$ to get $G'$. Further let $G_1,G_2$ be centroids of $\triangle GG'C,GG'B$ respectively. Note that $G$ is the centroid of $\triangle AG_1G_2$. This gives \[f(G) = f(A)+f(G_1)+f(G_2) =f(A)+2f(G)+2f(G')+f(C)+f(B)=3f(G)+2f(G').\]This gives $f(G)=-f(G')$. Therefore for any two points in the plane say $P,Q$, $f(P)=-f(Q)$ but this forces $f(P) = 0$ for any point $P$ and we are done.

Too easy! Here's my solution: Let $\triangle ABC$ have centroid $G$, and $M$ be the midpoint of $BC$. Reflect $A$ and $G$ in the perpendicular bisector of $A'$ and $G'$ respectively, and observe that $G'$ is the centroid of $\triangle A'BC$. This gives $$f(G) -f(A) =f(B) +f(C) =f(G') -f(A') $$As this argument does not depend on the positions of the points, so we get that for any isosceles trapezoid $ABCD$, we have $f(A) +f(C) =f(B) +f(D) $. Now, again fix $\triangle ABC$, and choose a point $D$ in the plane. Let $E$ be the reflection of $D$ in the midpoint $M$ of $BC$. Then $\triangle ADE$ and $\triangle ABC$ have the same centroid $G$ (since the ratio $AG: GM$ is fixed). So we get $$f(B)+f(C)=f(G)-f(A)=f(D) +f(E) $$Again this can be restated as that if $ABCD$ is a parallelogram, then $f(A) +f(C) =f(B) +f(D) $. Choose a point $X$ on line $AB$ such that $AXCD$ is an isosceles trapezoid. Then $$f(B)+f(D)=f(A) +f(C) =f(X) +f(D) \Rightarrow f(X)=f(B)$$This gives that if $B$ and $X$ are reflections in line through $C$ perpendicular to $CD$, then $f(B) =f(X) $. However, this argument can be applied to all points $C$ and $D$, which gives that $f$ is a constant function. Plugging into the given equation, we get $f(A) =0$ for all points $A$. Hence, done. $\blacksquare$

Try this restricted version I just found: We are given a non-degenarate triangle $DEF$. Now, $f(A)+f(B)+f(C)=f(M)$ where $M$ is the centroid of $ABC$ only when $ABC$ is congruent to $DEF$ and may or may not be so in other choices of $A,B,C$. Prove that $f(A)=0$ for all $A$.

k.vasilev wrote: Each point $A$ in the plane is assigned a real number $f(A).$ It is known that $f(M)=f(A)+f(B)+f(C),$ whenever $M$ is the centroid of $\triangle ABC.$ Prove that $f(A)=0$ for all points $A.$ Let the midpoint of the $BC$ side be $X$. Claim:Let the centroid of the $XBC$ be $X$. Proof:

Complex numbers

$ A=X\implies M=X $ $ f(X)+f(B)+f(C)=f(A)+f(B)+f(C)=f(M)=f(X) $ $ Then $ $ f(B)+f(C)=0 $ $f(B)$ does not depend on $f(C)$ $ Then $ $ f(B)=0 $

^$f$ doesn't need to be continuous at any point in the plane.

Let $P$ be any point in the plane, and set it as the origin of the polar coordinate system. Then, using the equalities $$f(P)=f(1,60^\circ)+f(1,180^\circ)+f(1,300^\circ)=f(2,0^\circ)+f(2,120^\circ)+f(2,240^\circ)$$and $$f(1,\theta)=f(2,\theta)+f(1,\theta+60^\circ)+f(1,\theta-60^\circ),\theta=0^\circ,120^circ,240^\circ$$we have: \begin{align*} -2f(P)&=f(1,0^\circ)+f(1,120^\circ)+f(1,240^\circ)-2\cdot(f(1,60^\circ)+f(1,180^\circ)+f(1,300^\circ))-(f(2,0^\circ)+f(2,120^\circ)+f(2,240^\circ))\\ &=f(2,0^\circ)+f(1,60^\circ)+f(1,300^\circ)+f(2,120^\circ)+f(1,180^\circ)+f(1,60^\circ)+f(2,240^\circ)+f(1,300^\circ)+f(1,180^\circ)-2\cdot(f(1,60^\circ)+f(1,180^\circ)+f(1,300^\circ))-(f(2,0^\circ)+f(2,120^\circ)+f(2,240^\circ))\\ &=0,\end{align*}hence the conclusion. (sorry for the terrible proofwriting)

for point $M$ , we draw three equilateral triangle as picture such that : 1- $M$ is centroid of triangles $ABC,A'B'C',A_1B_1C_1$ 2- $A'$ is centroid of triangle $A_1BC$ and $...$ Now $f(M)=\Sigma f(A)=\Sigma f(A')=\Sigma f(A_1)$ but : $\Sigma f(A)+\Sigma f(B)+\Sigma f(C_1)=\Sigma f(C') \Rightarrow 3f(M)=f(M)$ so $f(M)=0$

Attachments:

this is so bad i love it We start with the following lemma: Lemma: Let $ABC$ be a triangle with centroid $G$. Define $A'$ as the centroid of $BGC$ and $B'$, $C'$ are defined similarly. Then, $G$ is the centroid of $A'B'C'$ Proof. Vectors $\blacksquare$ Now, it follows that $f(A')=f(A)+2f(B)+2f(C)$ and cyclic equalities. Thus \[f(A)+f(B)+f(C)=f(G)=5(f(A)+f(B)+f(C))\]so $f(G)=0$. This finishes since every point is the centroid of some triangle.

oops wrote this up a while ago Let $M_0$ be any point in the plane and $ABC$ be a triangle with medians $\overline{AA_0}$, $\overline{BB_0}$, $\overline{CC_0}$, and centroid $M_0$. Additionally, denote by $M_1$, $M_2$, $M_3$ the centroids of $\triangle AB_0C_0$, $\triangle BC_0A_0$, and $\triangle CA_0B_0$, respectively. In complex numbers, we have \begin{align*} \frac{a_0+b_0+c_0}{3}&=\frac{(b+c)/2+(c+a)/2+(a+b)/2}{3} \\ &=\frac{a+b+c}{3} \\ &=m_0 \end{align*}and \begin{align*} \frac{m_1+m_2+m_3}{3}&=\frac{(a+b_0+c_0)/3+(b+c_0+a_0)/3+(c+a_0+b_0)/3}{3} \\ &=\frac{a+b+c}{9}+\frac{2(a_0+b_0+c_0)}{9} \\ &=\frac{m_0}{3} + \frac{2m_0}{3} \\ &=m_0, \end{align*}so $M_0$ is also the centroid of $\triangle A_0B_0C_0$ and $\triangle M_1M_2M_3$. Therefore, we have \begin{align*} f(M_0)&=f(M_1)+f(M_2)+f(M_3) \\ &=(f(A)+f(B)+f(C))+2(f(A_0)+f(B_0)+f(C_0)) \\ &=3f(M_0), \end{align*}ergo $f(M_0) = 0$.

Consider a triangle $\Delta ABC$ with medial triangle $\Delta DEF$ and centroid $G$. Note that $$f(G)=(f(A)+f(E)+f(F))+(f(B)+f(D)+f(F))+(f(C)+f(D)+f(E))=(f(A)+f(B)+f(C))+2(f(D)+f(E)+f(F))=3f(G).$$Since any point in the plane can be the centroid of a triangle, we have $f \equiv 0$.

Consider a regular hexagon $ABCDEF$, and its center, $G$. We have that $f(G)=f(A)+f(C)+f(E)=f(B)+f(D)+f(F)$, and letting $X$ and $Y$ be the centroids of triangles $ABC$ and $DEF$, respectively, we get that $2f(G)=f(X)+f(Y)$. In particular, notice that $G$ is the midpoint of $X$ and $Y$, so we get that for any points $X$ and $Y$ and midpoint $G$, the equation $2f(G)=f(X)+f(Y)$ must be true. Representing points as complex numbers, we get \[f(x)+f(y)=2f\left(\frac{x+y}{2}\right),\]and the original condition gives us \[f\left(\frac{x+y+z}{3}\right)=f(x)+f(y)+f(z)=2f\left(\frac{x+y}{2}\right)+f(z).\]Now, substituting $x$ for $x+y$, we get \[f\left(\frac{x+z}{3}\right)=2f\left(\frac{x}{2}\right)+f(z).\]Setting $y=0$ in $f\left(\frac{x+y+z}{3}\right)=f(x)+f(y)+f(z)$, we get \[f\left(\frac{x+z}{3}\right)=f(x)+f(z).\]Thus, we have $2f\left(\frac{x}{2}\right)=f(x)$, which implies $f(x)+f(y)=2f\left(\frac{x+y}{2}\right)=f(x+y)$. Now, we have \[f\left(\frac{x+y+z}{3}\right)=f(x)+f(y)+f(z)=f(x+y+z)=f\left(\frac{x+y+z}{3}\right)+f\left(\frac{2(x+y+z)}{3}\right),\]giving $f\left(\frac{2(x+y+z)}{3}\right)=0$. Therefore, $f(x)=0$ for all $x$, as desired.