41st Balkan Mathematical Olympiad

Let $G_1\ne G_2$ be arbitrary points and let $d$ be the perpendicular bisector of $G_1{G}_2$. We shall construct two congruent triangles, symmetric with respect to $d$ and with centroids at $G_1$ and $G_2$. Choose an acute triangle $A_{1}BC$ with centroid $G_1$, and rotate it around $G_1$ if necessary to ensure that $BC\parallel G_1{G}_2$. By applying a homotety centered at $G_1$ if necessary, we may ensure that the midpoint $M$ of $BC$ lies on $d$. Let $H_1,O$ be the orthocenter and circumcenter of $△ABC$. By construction, we have $O\in d$ and $A_1{H}_1\parallel OM\perp BC$. If $A_2,H_2$ are the reflections of points $A_1,H_1$ across $d$, then one sees that $△A_{1}BC\equiv △A_{2}CB$, and $H_2,G_2,O$ are the orthocenter, centroid and cirumcenter of $△A_{2}BC$. Therefore, applying the property of $f$ to $△A_{1}BC$ and $△A_{2}BC$, we have

$$\begin{array}{rl}f(A_1)+f(B)+f(C)& =f(O)+f(G_1)+f(H_1)\\ f(A_2)+f(B)+f(C)& =f(O)+f(G_2)+f(H_2)\end{array}$$ Subtracting, we obtain $$f(A_1)-f(A_2)=f(G_1)-f(G_2)+f(H_1)-f(H_2)(1)$$ Now let's reflect the picture across line $l$, the perpendicular bisector of $A_1{H}_1$ ($l$ is a symmetry axis of rectangle $A_1{A}_2{H}_2{H}_1$), denoting by $X^{\prime }$ the image of point $X$. It follows that $A_1^{\prime }=H_1$, $A_2^{\prime }=H_2$ and triangles $H_1{B}^{\prime }{C}^{\prime }$, $H_2{B}^{\prime }{C}^{\prime }$ are symmetric about $d$. Using this symmetry, relation (1) becomes $$f(H_1)-f(H_2)=f(G_1^{\prime })-f(G_2^{\prime })+f(A_1)-f(A_2)(2)$$ From (2) and (3) we obtain $$f(G_1)-f(G_2)+f(G_1^{\prime })-f(G_2^{\prime })=0(3)$$ The argument up to now holds if we scale the picture. It follows that for any rectangle $XYZT$ which is similar to the rectangle $G_1{G}_1^{\prime }{G}_2^{\prime }{G}_2$, we have $$f(X)-f(T)+f(Y)-f(Z)=0(4)$$ Let us double up the rectangle $G_1{G}_1^{\prime }{G}_2^{\prime }{G}_2$ to a rectangle $G_{1}KLM$ (see the picture), with $D$ and $E$ the midpoints of $KL$ and $LM$. Applying (4) to rectangle $G_2{G}_2^{\prime }EM$, we have $f(G_2)-f(M)+f(G_2^{\prime })-f(E)=0$ which added to (3) yields $$f(G_1)-f(M)+f(G_1^{\prime })-f(E)=0$$ $$f(G_1^{\prime })-f(E)+f(K)-f(L)=0$$ Subtracting the last relations we get $$f(G_1)-f(M)-f(K)+f(L)=0$$ On the other hand, applying (4) to the rectangle $G_{1}KLM$ we have $$f(G_1)-f(M)+f(K)-f(L)=0$$ The last 2 relations imply $f(G_1)-f(M)=0$, and scaling back we finally obtain $f(G_1)-f(G_2)=0$. The choice of $G_1\ne G_2$ being arbitrary, it follows that the function $f$ must be constant. $\square$

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Alternative solution.

Alternative Solution. Assume that $f:\pi \to \mathbb{R}$ satisfies the condition of the problem.

1. Step 1. (Rhombus $60^{\circ }$) Consider a rhombus $ABCD$ with $\angle BAC=60^{\circ }$. Let $M$ and $N$ lie on the diagonal $AC$ such that $AM=MN=NC$. Then since $△ABD$ and $△CBD$ are equilateral we have that $M$ and $N$ are their centroids and therefore:

$$\begin{array}{rl}f(A)+f(B)+f(D)& =3f(M)\\ f(C)+f(B)+f(D)& =3f(N).\end{array}$$ This proves that $f(A)-f(C)=3(f(M)-f(N))$. As a consequence we have that whenever $A,M,N,C$ are collinear and $AM=MN=NC$ then: $$f(A)-f(C)=3(f(M)-f(N)).$$

2. Step 2. Now consider five collinear points $A,B,C,D,E$ with $AB=BC=CD=DE$. Then by Step 1 we have: $$\begin{array}{rl}f(A)-f(D)& =3(f(B)-f(C))\\ f(B)-f(E)& =3(f(C)-f(D)).\end{array}$$ Summing up both equalities we obtain: $$f(A)-f(E)=2(f(B)-f(D)).$$ Thus whenever $A,B,D,E$ are collinear (in this order) with $AB=DE=\frac{1}{2}BD$ it holds: $$f(A)-f(E)=2(f(B)-f(D)).$$

3. Step 3. Consider a trapezoid $ABCD$ with bases $AB\parallel CD$ and $AB=2CD$. Let $O$ be the intersection point of the diagonals $AC$ and $BD$ and $M$ and $N$ be the midpoints of $AO$ and $BO$. Finally let $P=DM\cap AB$ and $Q=CN\cap AB$. To start with note that by construction $AM=MO=OC$ and $BN=NO=OD$ therefore by Step 1: $$\begin{array}{rl}f(A)-f(C)& =3(f(M)-f(O))\\ f(B)-f(D)& =3(f(N)-f(O)).\end{array}$$ Subtracting the second equality from the first one we obtain: $$f(A)-f(B)=f(C)-f(D)+3(f(M)-f(N)).$$ Now note that $AP:CD=AM:MC=1:2=BN:ND=BQ:CD$. Therefore $AP=BQ$ and $PQ=AB-2AP=AB-CD=CD=2AP$. Therefore $A,P,Q,B$ are collinear (in this order) with $AP=QB=1/2PQ$ and by Step 2 we have: $$f(A)-f(B)=2(f(P)-f(Q)).$$ Substituting in the above equality we obtain: $$\begin{array}{rl}2(f(P)-f(Q))& =f(C)-f(D)+3(f(M)-f(N))\text{ or after rearrangement}\\ 2f(P)+f(D)-3f(M)& =2f(Q)+f(C)-3f(N).\end{array}$$ Finally note that $PQCD$ is a parallelogram ($PQ\parallel DC$) and $PM=QN=1/3PD$. Clearly we can reconstruct the points $A=CM\cap PQ$ and $B=DN\cap PQ$ and by Thales's Theorem they will satisfy $AP=1/2CD=BQ$. To summarise for every parallelogram $PQCD$ and points $M\in PD$ and $N\in QC$ with $PM=1/3PD=QN$ we have: $$2f(P)+f(D)-3f(M)=2f(Q)+f(C)-3f(N).$$

4. Step 4. Let $PQCD$ be parallelogram and $M,M^{\prime }\in PD$ are $N,N^{\prime }\in QC$ be such that $PM=M{M}^{\prime }=M^{\prime }D$ and $QN=N{N}^{\prime }=N^{\prime }D$. Applying Step 3 to the parallelogram $CDPQ$ and points $N^{\prime }$ and $M^{\prime }$ we have: $$2f(C)+f(Q)-3f(N^{\prime })=2f(D)+f(P)-3f(M^{\prime }).$$ Summing up we get the result from Step 3 for the original parallelogram $PQCD$ and the points $M$ and $N$ we get: $$3(f(P)+f(D)-f(M)-f(M^{\prime }))=3(f(C)+f(Q)-f(N)-f(N^{\prime }))$$ --- or $f(P)+f(D)-f(M)-f(M^{\prime })=f(C)+f(Q)-f(N)-f(N^{\prime })$ whenever the line segments $PD\parallel QC$ and the points $M,M^{\prime }$ and $N,N^{\prime }$ split $PD$ and $QC$ in three equal parts, respectively. We can express this result as follows. There is a function $g$ such that whenever $A_1,A_2,A_3,A_4$ are collinear and $\vec{v}=\overrightarrow{A_i{A}_{i+1}}$ for $i=1,2,3$, then: $$g(\vec{v})=f(A_1)-f(A_2)-f(A_3)+f(A_4).$$

5. Step 5. Let $\vec{v}\ne 0$ be a vector in the plane and let $\overrightarrow{A_i{A}_{i+1}}=\vec{v}$ for $i=1,2,3,4$. Then by Step 4 we have: $$f(A_1)-f(A_2)-f(A_3)+f(A_4)=g(\vec{v})$$ $$f(A_2)-f(A_3)-f(A_4)+f(A_5)=g(\vec{v})$$ Summing up both equalities we obtain: $$f(A_1)-2f(A_3)+f(A_5)=2g(\vec{v}).$$ Therefore by scaling with $1/2$ we also have $f(A_1)-2f(A_2)+f(A_3)=2g(\vec{v}/2)$. Consequently: $$\begin{array}{rl}g(\vec{v})& =f(A_1)-f(A_2)-f(A_3)+f(A_4)\\ & =(f(A_1)-2f(A_2)+f(A_3))+(f(A_2)-2f(A_3)+f(A_4))=4g(\vec{v}/2).\end{array}$$ Therefore $g(\vec{v}/2)=g(v)/4$ or, after rescaling, $4g(\vec{v})=g(2\vec{v})$.

6. Step 6. Let $H,G$ and $O$ be the orthocenter, centroid and the centre of the circumcircle, respectively, of a triangle. Then, by Euler's Theorem, they are collinear with $G$ belonging to the line segment $OH$ and $OH=2OG$. Conversely, if $O,G$ and $H$ satisfy the above condition, then there is an acute (isosceles) triangle $ABC$ such that $H,G$ and $O$ are the orthocenter, centroid and the centre of the circumcircle of $△ABC$. To see this one can take an arbitrary acute triangle $△A^{\prime }{B}^{\prime }{C}^{\prime }$ with orthocenter, centroid and the centre of the circumcircle, $H^{\prime },G^{\prime }$ and $O^{\prime }$, respectively. Then scale $△A^{\prime }{B}^{\prime }{C}^{\prime }$ so that $O^{\prime }{H}^{\prime }=OH$ and finally translate and rotate the diagram so that $O^{\prime }{H}^{\prime }$ match $OH$. During this transformation, clearly, the triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$ is transformed to a similar triangle $ABC$.

7. Step 7. Let $O,G,H$ be collinear with $\overrightarrow{HG}=2\overrightarrow{GO}$. Let $△ABC$ be arbitrary acute triangle such that $O,G,H$ are its centre of the circumcircle, centroid and orthocenter, respectively. Let $A_1,B_1,C_1$ be the midpoints of $BC,AC$ and $AB$, respectively. Finally, let $A_2,B_2$ and $C_2$ be the midpoints of $B_1{C}_1,C_1{A}_1$ and $B_1{A}_1$, respectively. We denote with $O_i,G_i,H_i$ the centre of the circumcircle, centroid and orthocenter, respectively, of triangle $A_i{B}_i{C}_i$ for $i=1,2$. Note that by Step 5: $$f(A)-2f(C_1)+f(B)=2g(\overrightarrow{A{C}_1}/2)=8g(\overrightarrow{A_1{C}_2}/2)=4(f(A_1)-2f(C_2)+f(B_1)).$$ Similarly we have: $$\begin{array}{rl}f(B)-2f(A_1)+f(C)& =4(f(B_1)-f(A_2)+f(C_1))\\ f(C)-2f(B_1)+f(A)& =4(f(C_1)-f(B_2)+f(A_1)).\end{array}$$ Summing up all three equalities and taking into account that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ and $f(A_i)+f(B_i)+f(C_i)=f(O_i)+f(G_i)+f(H_i)$ we conclude that:

$$f(O)+f(G)+f(H)-f(O_1)-f(G_1)-f(H_1)=4(f(O_1)+f(G_1)+f(H_1)-f(O_2)-f(G_2)-f(H_2)).$$ However, obviously $G=G_1=G_2$ and a simple homothetic argument at $G$ shows that $O=H_1$ and $O_1=H_2$. Therefore: $$f(H)-f(O_1)=4(f(H_1)-f(O_2)).$$ Now note that $O_1$ is the midpoint of $OH$ and similarly $O_2$ is the midpoint of $O_1{H}_1=O_{1}O$. Since $O,G,H$ were arbitrary with $\overrightarrow{HG}=2\overrightarrow{GO}$, we conclude that whenever $\overrightarrow{H{O}_1}=2\overrightarrow{O_1{O}_2}=2\overrightarrow{O_2{H}_1}$ we have: $$f(H)-f(O_1)=4(f(H_1)-f(O_2)).$$ Introducing $M$ to be the midpoint of $H{O}_1$ and replacing $(H,H_1)$ with $(H_1,H)$ we get: $$f(H_1)-f(O_1)=4(f(H)-f(M)).$$ Subtracting from the first equality the second one, we arrive at: $$f(H)-f(H_1)=4(f(H_1)-f(H)-f(O_1)+f(M))\text{ or }5(f(H)-f(H_1))=4(f(M)-f(O_1)).$$ Now, since $\overrightarrow{M{O}_1}=2\overrightarrow{HM}=2\overrightarrow{O_1{H}_1}$ by Step 2 we also have: $$f(H)-f(H_1)=2(f(M)-f(O_1)).$$ This already shows that $10(f(M)-f(O_1))=4(f(M)-f(O_1))$ and therefore $f(M)=f(O_1)$, implying that $f(H)=f(H_1)=f(O)$. Since $O$ and $H$ can be considered arbitrary, as $G$ is uniquely determined by the choice of $O$ and $H$, we conclude that $f$ is constant

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