Let $A', B', C'$ be reflections of $G$ in $M_a, M_b, M_c$ respectively. Then $GA'=2GM_a=GA$, similarily $GB'=GB$ and $GC'=GC$, so $\triangle A'B'C'$ is reflection of $\triangle ABC$ in $G$. We also have that $AGBC'$ and $AGCB'$ are parallelograms. Let $X=\odot A'BC' \cap \odot A'B'C$: $\measuredangle BXC=\measuredangle BXA'+\measuredangle A'XC=\measuredangle BC'A'+\measuredangle A'B'C=\measuredangle B'CA+\measuredangle ABC'=\measuredangle GAC+\measuredangle BAG=\measuredangle BAC\implies X\in \odot ABC$ Similarily we get that $\odot AB'C'\cap \odot A'BC'\in \odot ABC$, so $\odot AB'C', \odot A'BC'$ and $\odot A'B'C$ are concurrent at $X$, hence taking homothety $\mathbb{H}(G,\tfrac{1}{2})$, we get that $\Omega_A, \Omega_B$ and $\Omega_C$ are concurrent in midpoint of $GX$.

Let $N_A, N_B, N_C$ be the midpoints of $\overline{AG}, \overline{BG}, \overline{CG}$, respectively, and let $\triangle T_AT_BT_C$ be the medial triangle of $\triangle M_AM_BM_C.$ Since $G$ is the centroid of both $\triangle ABC$ and $\triangle M_AM_BM_C$, we obtain $N_AG = AG / 2 = M_AG = 2T_AG.$ Therefore, $N_A$ is the reflection of $G$ in $T_A.$ Consequently, $\odot(N_AM_BM_C)$ is the reflection of $\odot(GM_BM_C)$ in $M_BM_C.$ Hence, $\odot(N_AM_BM_C)$ passes through the antigonal conjugate $G^*$ of $G$ w.r.t. $\triangle M_AM_BM_C.$ Similar arguments show that all three circles pass through $G^*.$

My solution: Let $X,Y,Z$ be the midpoint of $AM,BM,CM$ and $S$ be the intersection of $(XM_cM_b)$ and $(YM_cM_a)$. We have $\angle M_bSM_a=\angle M_cSM_a-\angle M_cSM_b=\angle M_aYM_c-(180^o-\angle M_cXM_b)=\angle XMM_b $, then $S\in (M_bZM_a)$

One can prove that the intersection of the three circles is in the conic passing through the midpoints of the sides and the midpoints of $AM,BM,CM$.

Alternative Formulation Let $ABC$ be a triangle and $P$ an arbitrary point. Consider $A',B',C'$ the reflections of $A,B,C$ through point $P$. Then the circumcircles of triangles $BCA',ACB',BAC'$ are concurrent.

More general : Let $P$ a point. $M_a,M_b,M_c$ te midpoints of the sides of $BC,CA,AB$ resp. Let $P_1,P_2,P_3$ the midpoints of $AP,BP,CP$ . prove that the circumcircles $\cal{C}$$_1$, $\cal{C}$$_2$, $\cal{C}$$_3$ of the triangles $P_1M_bM_c,P_2M_aM_c,P_3M_aM_b$ have a common point . consider $K$ the second intersection of $\cal{C}$$_1$ and $\cal{C}$$_2$ $(^*)$ we have to prove that $K$ is on $\cal{C}$$_3$: $(^*)$ means that $\widehat{(M_cK,KM_a)}=\widehat{(M_cP_2,P_2M_a)}, \widehat{(M_cK,KM_b)}=\widehat{(M_cP_1,P_1M_b)} $ thus $\widehat{(M_bK,KM_a)}=\widehat{(M_bP_1,P_1M_c)} +\widehat{(M_cP_2,P_2M_a)} =\widehat{(BP,PC)}+\widehat{(CP,PA)}=\widehat{(BP,PA)} =\widehat{(M_bP_3,P_3M_a)}$ which ends the proof . R HAS

Let $A'B'C'$ be the antimedial triangle of $ABC$ and let $G_1, G_2, G_3$ be the centroids of A'BC etc. Let the circumcircles of $M_cN_aM_b, M_bN_cM_a$ meet at $G$. Then using directed angles note that $$ \angle M_cGM_A = \angle M_cN_aM_b + \angle M_bN_cM_a = \angle G_cAG_b + \angle G_bCG_a = -\angle G_aBG_c = \angle M_cN_bM_a$$whence we are done.

To moderators: The problem that MRF2017 posted is not the All-Russian MO 2016 G11 P8. The official version is: Medians $AM_A,BM_B,CM_C$ of triangle $ABC$ intersect at $M$ . Let $\Omega_A$ be the circle passing through midpoint of $AM$ and tangent to $BC$ at $M_A$. Define $\Omega_B$ and $\Omega_C$ analogously. Prove that $\Omega_A, \Omega_B$ and $\Omega_C$ intersect at one point. Obviously, the official problem is different from MRF2017's, and I believe it is harder as well. However, what MRF2017 posted might be considered as a lemma to the official problem. BTW, I also found some interesting results about the official problem: 1. The concurrent point of $\Omega_A, \Omega_B$ and $\Omega_C$ is the midpoint of $MS$, where $S$ is the Steiner point of triangle $ABC$. 2. Denote by $K_A$ the intersection of $\Omega_B$ and $\Omega_C$ other than the concurrent point, and $K_B,K_C$ are defined analogously. Then the lines $AK_A,BK_B,CK_C$ are concurrent at the symmedian point of triangle $ABC$. 3. Denote by $O_A, O_B, O_C$ the centers of $\Omega_A, \Omega_B$ and $\Omega_C$ respectively, then $M, O_A, O_B, O_C$ are concyclic. I cannot find a proof for neither of them, but I believe they must be true. I hope someone will present solutions for them.

Here is an extension and solution. General problem. Let $ABC$ be a triangle with $D,E,F$ are midpoint of $BC,CA,AB$. $P$ is any point. $XYZ$ is cevian triangle of $P$. $U,V,W$ are midpoints of $PA,PB,PC$. Prove that circle $(UDX),(VEY),(WFZ)$ have a common point. Solution. Let circles $(VDF),(WDE)$ intersect again at $Q$. We get directed angle chasing $$(QE,QF)=(QE,QD)+(QD,QF)=(WE,WD)+(VD,VF)=(PA,PB)+(PC,PA)=(PC,PB)=(UE,UF)\pmod\pi.$$Thus $Q$ lies on $(UEF)$. Now we have $$(QU,QD)=(QU,QF)+(QF,QD)=(EU,EF)+(VF,VD)=(PC,BC)+(XU,PC)=(XU,XD)\pmod\pi.$$From this $Q$ lies on $(UDX)$. Similarly, $Q$ lies on $(VEY),(WFZ)$. We are done.

We proceed with barycentric coordinates. Set $ABC$ as reference triangle, and let $$\Omega_A : -a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0.$$Replacing $M_A=(0:1:1)$ and $\dfrac{A+M}{2}=(4:1:1)$ we get $$v+w=\frac{a^2}{2}$$and $$u=\frac{2b^2+2c^2-a^2}{12}.$$Because $\Omega_A$ is tangent to $BC$ at $M_A=(0, 1/2, 1/2)$, the only solution to $x=0$ and $-a^2yz-b^2zx-c^2xy+ux+vy+wz=0$ is when $y=z=1/2$. From this the quadratic equation $$a^2X^2+(v-w-a^2)X+w=0$$has determinant zero, thus with the results above we get $$v=w=\frac{a^2}{4}$$so $$\Omega_A : -a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{2b^2+2c^2-a^2}{12}x+\frac{a^2}{4}y+\frac{a^2}{4}z\right)=0.$$Consider the point $$P=((2a^2-b^2-c^2)^2:(2b^2-c^2-a^2)^2:(2c^2-a^2-b^2)^2).$$We can see that this point is on $\Omega_A$ and analogously on $\Omega_B$ and $\Omega_C$, hence the result follows. jred wrote: 1. The concurrent point of $\Omega_A, \Omega_B$ and $\Omega_C$ is the midpoint of $MS$, where $S$ is the Steiner point of triangle $ABC$. We can see that $P$ is the midpoint of $MS$ because $$2P-M=((2a^2-b^2-c^2)^2-T:(2b^2-c^2-a^2)^2-T:(2c^2-a^2-b^2)^2-T)=\left(\frac{1}{b^2-c^2}:\frac{1}{c^2-a^2}:\frac{1}{a^2-b^2}\right)$$is the Steiner point of $\triangle ABC$, where $T=a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2$. jred wrote: 2. Denote by $K_A$ the intersection of $\Omega_B$ and $\Omega_C$ other than the concurrent point, and $K_B,K_C$ are defined analogously. Then the lines $AK_A,BK_B,CK_C$ are concurrent at the symmedian point of triangle $ABC$. We see that $$K_A=(2b^2+2c^2-a^2:3b^2:3c^2)$$is on $\Omega_B$ and $\Omega_C$, so $K_A$ is the second intersection of $\Omega_B$ and $\Omega_C$, analogously for $K_B$ and $K_C$. Then, the intersection of $AK_A$, $BK_B$ and $CK_C$ is $(a^2:b^2:c^2)$, which is the symmedian point of $\triangle ABC$.

Another thing to prove could be: Let $O_A, O_B,O_C$ be the centers of the circles $\Omega_A , \Omega_B , \Omega_C$. Then prove that $\Omega_A , \Omega_B , \Omega_C , M$ are concyclic.

Can anyone provide a proof for this?

mjuk wrote: Let $A', B', C'$ be reflections of $G$ in $M_a, M_b, M_c$ respectively. Then $GA'=2GM_a=GA$, similarily $GB'=GB$ and $GC'=GC$, so $\triangle A'B'C'$ is reflection of $\triangle ABC$ in $G$. We also have that $AGBC'$ and $AGCB'$ are parallelograms. Let $X=\odot A'BC' \cap \odot A'B'C$: $\measuredangle BXC=\measuredangle BXA'+\measuredangle A'XC=\measuredangle BC'A'+\measuredangle A'B'C=\measuredangle B'CA+\measuredangle ABC'=\measuredangle GAC+\measuredangle BAG=\measuredangle BAC\implies X\in \odot ABC$ Similarily we get that $\odot AB'C'\cap \odot A'BC'\in \odot ABC$, so $\odot AB'C', \odot A'BC'$ and $\odot A'B'C$ are concurrent at $X$, hence taking homothety $\mathbb{H}(G,\tfrac{1}{2})$, we get that $\Omega_A, \Omega_B$ and $\Omega_C$ are concurrent in midpoint of $GX$. This proof is not actually complete. You have proven, with your last homothety, that the circles $\odot{(DEF)} , \odot{(DM_BM_c)}, \odot{(EM_AM_C)} , \odot{(FM_AM_B)}$ intersect at the midpoint of $GX$, but not the ones the problem asked to. PS: $D,E,F$ are the midpoints of $GA,GB,GC$.

myh2910 wrote: We proceed with barycentric coordinates. Set $ABC$ as reference triangle, and let $$\Omega_A : -a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0.$$Replacing $M_A=(0:1:1)$ and $\dfrac{A+M}{2}=(4:1:1)$ we get $$v+w=\frac{a^2}{2}$$and $$u=\frac{2b^2+2c^2-a^2}{12}.$$Because $\Omega_A$ is tangent to $BC$ at $M_A=(0, 1/2, 1/2)$, the only solution to $x=0$ and $-a^2yz-b^2zx-c^2xy+ux+vy+wz=0$ is when $y=z=1/2$. From this the quadratic equation $$a^2X^2+(v-w-a^2)X+w=0$$has determinant zero, thus with the results above we get $$v=w=\frac{a^2}{4}$$so $$\Omega_A : -a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{2b^2+2c^2-a^2}{12}x+\frac{a^2}{4}y+\frac{a^2}{4}z\right)=0.$$Consider the point $$P=((2a^2-b^2-c^2)^2:(2b^2-c^2-a^2)^2:(2c^2-a^2-b^2)^2).$$We can see that this point is on $\Omega_A$ and analogously on $\Omega_B$ and $\Omega_C$, hence the result follows. jred wrote: 1. The concurrent point of $\Omega_A, \Omega_B$ and $\Omega_C$ is the midpoint of $MS$, where $S$ is the Steiner point of triangle $ABC$. We can see that $P$ is the midpoint of $MS$ because $$2P-M=((2a^2-b^2-c^2)^2-T:(2b^2-c^2-a^2)^2-T:(2c^2-a^2-b^2)^2-T)=\left(\frac{1}{b^2-c^2}:\frac{1}{c^2-a^2}:\frac{1}{a^2-b^2}\right)$$is the Steiner point of $\triangle ABC$, where $T=a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2$. jred wrote: 2. Denote by $K_A$ the intersection of $\Omega_B$ and $\Omega_C$ other than the concurrent point, and $K_B,K_C$ are defined analogously. Then the lines $AK_A,BK_B,CK_C$ are concurrent at the symmedian point of triangle $ABC$. We see that $$K_A=(2b^2+2c^2-a^2:3b^2:3c^2)$$is on $\Omega_B$ and $\Omega_C$, so $K_A$ is the second intersection of $\Omega_B$ and $\Omega_C$, analogously for $K_B$ and $K_C$. Then, the intersection of $AK_A$, $BK_B$ and $CK_C$ is $(a^2:b^2:c^2)$, which is the symmedian point of $\triangle ABC$. How did you guess the coordinates of $P$?

Denote by $D_{A},D_{B},D_{C}$ midpoints of $AM,BM,CM$ respectively. Let $T=\Omega_{A}\cap \odot (D_{A}D_{B}D_{C})\backslash D_{A}$. $$\measuredangle M_{A}TD_{B}=\measuredangle M_{A}TD_{A}+\measuredangle D_{A}TD_{B}=\measuredangle CM_{A}A+\measuredangle ACB=\measuredangle CAM$$$$\measuredangle M_{A}TD_{C}=\measuredangle M_{A}TD_{A}+\measuredangle D_{A}TD_{C}=\measuredangle CM_{A}A+\measuredangle D_{A}D_{B}D_{C}=\measuredangle CM_{A}A+\measuredangle ABC=\measuredangle BAM=\measuredangle M_{A}M_{B}D_{C}$$Thus $T\in \odot (M_{A}D_{C}M_{B})\implies \measuredangle M_{B}TD_{B}=\measuredangle M_{B}TM_{A}+\measuredangle M_{A}TD_{B}=\measuredangle M_{A}MM_{B}+\measuredangle CAM=\measuredangle AM_{B}B,$ so $T\in \Omega_{B}$. Analogously $T\in \Omega_{C},$ and we are done.

Doru2718 wrote: How did you guess the coordinates of $P$? From the equations of $\Omega_A$ and $\Omega_B$, the point $P=(x:y:z)$ satisfies \[\frac{2b^2+2c^2-a^2}{12}x+\frac{a^2}{4}y+\frac{a^2}{4}z=\frac{a^2yz+b^2zx+c^2xy}{x+y+z}=\frac{2c^2+2a^2-b^2}{12}y+\frac{b^2}{4}z+\frac{b^2}{4}x\]which implies $\left(a^2+b^2-2c^2\right)(x-y)=3\left(a^2-b^2\right)z$. Letting $t_A=2a^2-b^2-c^2$ and so on, we can verify $t_A-t_B=3(b^2-a^2)$ and $t_A+t_B=-t_C$, therefore \[\frac{x-y}{z}=\frac{t_A-t_B}{-t_C}=\frac{t^2_A-t^2_B}{t^2_C}\]and the rest is simple.

Nice application of barycentric coordinates and linearity of a power of a point. Define $f(\bullet,\Omega_X,\Omega_Y)$ as the difference of power of $\bullet$ wrt $\Omega_X,\Omega_Y$. Note that \begin{align*} f(A,\Omega_A,\Omega_B)&=\frac{AM_A^2}{3}-\frac{AC^2}{4}=\frac{2c^2-b^2-a^2}{12}\\ f(B,\Omega_A,\Omega_B)&=\frac{BC^2}{3}-\frac{BM_B^2}{3}=\frac{b^2+a^2-2c^2}{12}\\ f(C,\Omega_A,\Omega_B)&=\frac{a^2-b^2}{4}, \end{align*}similarly, \begin{align*} f(A,\Omega_B,\Omega_C)&=\frac{2a^2-c^2-b^2}{12}\\ f(B,\Omega_B,\Omega_C)&=\frac{c^2+b^2-2a^2}{12}\\ f(C,\Omega_B,\Omega_C)&=\frac{b^2-c^2}{4}. \end{align*}Therefore we would like to show that the following is true, $$\begin{vmatrix} 2c^2-b^2-a^2 & b^2+a^2-2c^2 & 3a^2-3b^2\\ 1 & 1 & 1\\ 2a^2-c^2-b^2 & c^2+b^2-2a^2 & 3b^2-3c^2 \end{vmatrix}\neq 0.$$After doing some standard matrix operations, we would like to show that $$\begin{vmatrix} 3c^2-3b^2 & b^2+a^2-2c^2 & 0\\ 1 & 1 & 1\\ 0 & 2b^2+2c^2-4a^2 & 3b^2-3a^2 \end{vmatrix}\neq 0,$$which is equivalent to showing that $$-6 a^4 + 6 a ^2b^2 + 6 a ^2c ^2- 6 b^4 + 6 b^2 c^2 - 6 c^4\neq 0.$$Indeed note that \begin{align*} a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2,\end{align*}where the equality holds iff $a=b=c$, which is not the case.

Let $\measuredangle$ denote directed angles modulo $180^\circ$. Claim: $\Omega_B$, $\Omega_C$, and $(N_AN_BN_C)$ concur. Proof: Let $P=(N_AN_BN_C) \cap (N_AM_BM_C)$. We have \[\measuredangle M_BPN_B=\measuredangle M_BPN_A+\measuredangle N_APN_B=\measuredangle M_BM_CN_A+\measuredangle N_AN_CN_B=\measuredangle CBM+\measuredangle ACB=\measuredangle CM_BB,\]so $\Omega_B$ passes through $P$. Similarly, $\Omega_C$ passes through $P$. $\square$ Using symmetric versions of this claim, $\Omega_A$, $\Omega_B$, and $\Omega_C$ intersect $(N_AN_BN_C)$ again at the same point, as desired. $\square$

Claim: $(M_CM_BE),(DM_AM_B),(M_CM_AF),(DEF)$ concurr at a point $K$

Now we show $(EKM_A)$ is tangent to $BC$ at $M_A$. $\measuredangle M_AKE = \measuredangle M_AKM_B - \measuredangle EKM_B = \measuredangle M_ADM_B - \measuredangle EM_CM_B = \measuredangle AM_AB$ $\blacksquare$