Nice Problem! We know that the Area of one quadrilateral ABCD is $\frac{AC.BD.sen(k)}{2}$, where k is the angle between the diagonals! So, we have to prove that $\frac{AD}{sen(BMC)}= \frac{BE}{sen(CMD)}= \frac{CF}{sen(AMB)}$. Let $A_{1}, B_{1}, ... , F_{1}$ the circuncenters of $MAB, MBC, ..., MFA$. We have that $A_{1}B_{1}\parallel D_{1}E_{1}$, $B_{1}C_{1}\parallel E_{1}F_{1}$ and $F_{1}A_{1}\parallel C_{1}D_{1}$, and $A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}$ is cyclic! So, $A_{1}B_{1}C_{1}$ is congruent to $D_{1}E_{1}F_{1}$, $A_{1}C_{1}= D_{1}F_{1}$ and $A_{1}F_{1}= D_{1}C_{1}$. This implies that $A_{1}C_{1}D_{1}F_{1}$ is one rectangle. Let A' and D' the midpoints of MA and MD. We have $AD = 2A'D'$ So, $\frac{AD}{2sen(BMC)}= \frac{A'D'}{sen(BMC)}= \frac{A_{1}D_{1}}{sen(180-BMC)}= \frac{A_{1}D_{1}}{sen(A_{1}B_{1}C_{1}}$ $= 2R$, where $R$ is the ray of $A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}$ circuncircle! Do the same for the other fractions, and we will get the same result (2R)

We denote $ B = A_2,C = A_3,D = A_4,E = A_5,F = A_6$ for brevity. Let $ P_i$ denote the midpoint of $ MA_i$ and $ O_i$ the circumcenter of $ \triangle MA_iA_{i + 1}$, where the index is taken modulo $ 6$. $ O_iO_{i - 1}$ is the perpendicular bisector of $ MA_i$. Therefore $ O_iO_{i + 1}\parallel O_{i + 3}O_{i + 4}$. Hence $ O_iO_{i + 1}O_{i + 3}O_{i + 4}$ is an isosceles trapezoid, implying $ O_iO_{i + 3} = k$ (constant). Now consider the forces $ \overrightarrow{P_1P_4},\overrightarrow{P_5P_2},\overrightarrow{P_3P_6}$ acting at point $ M$. Let $ R$ denote the circumradius of circle $ O_1O_2O_3O_4O_5O_6$. We have \begin{align*} & \frac {P_{i + 2}P_{i + 5}}{\sin\angle P_iMP_{i + 1}} = \frac {P_{i + 2}P_{i + 5}}{\sin\angle P_iO_iP_{i + 1}} = \frac {P_{i + 2}P_{i + 5}\cdot 2R}{O_{i + 1}O_{i + 5}} = 2R\sin\angle O_{i + 1}O_{i+5}O_{i + 4} = O_{i + 1}O_{i + 4} = k.\end{align*} Hence the forces are in equilibrium, and so \begin{align*}\overrightarrow{P_1P_4} + \overrightarrow{P_5P_2} + \overrightarrow{P_3P_6} = 0 \\ \Rightarrow\overrightarrow{P_2P_5}\times (\overrightarrow{P_1P_4} + \overrightarrow{P_5P_2} + \overrightarrow{P_3P_6}) = 0 \\ \Rightarrow\overrightarrow{P_2P_5}\times (\overrightarrow{P_1P_4} + \overrightarrow{P_3P_6}) = 0.\end{align*} Therefore $ |\overrightarrow{P_2P_5}\times\overrightarrow{P_1P_4}| = |\overrightarrow{P_2P_5}\times\overrightarrow{P_6P_3}|$ implying $ [P_1P_2P_4P_5] = [P_2P_3P_5P_6]$. Similarly we can show that $ [P_1P_2P_4P_5] = [P_3P_4P_6P_1]$. Now the homothety with center $ M$ and ratio $ 2$ maps $ P_i$ to $ A_i$. Hence the conclusion.

Attachments:

e.lopes wrote: This implies that $ A_{1}C_{1}D_{1}F_{1}$ is one rectangle... I think this is not true...

it's similar to an OC's problem. let $O_1,O_2,...,O_6$ be six centers,$O_1O_2$ intersects $O_5O_6$ at $O_7$ by OC's theorem,$\frac{O_7O_2}{FC}=\frac{1}{2sinBMC}=\frac{O_7O_5}{BE}$ since $O_1,...,O_6$ are cyclic,hence $\frac{O_7O_6}{O_7O_1}=\frac{sinAMB}{sinAMF}=\frac{FC}{BE}=\frac{O_7O_2}{O_7O_5}$ so $BEsinAMB=FCsinAMF$ yielding $(ABDE)=(CDFA)$.

by the way,I'd like to know what kind of contest Zhautykov is? why is its 2006 identical to Kazakhstan 2006?

littletush wrote: by the way,I'd like to know what kind of contest Zhautykov is? why is its 2006 identical to Kazakhstan 2006? it's international MO,held in Kazakhstan

sorry what is OC problem

can someone please post another solution

Solved with Elliott Liu, Isaac Zhu, Jeffrey Chen, Kevin Wu, Luke Robitaille, and Raymond Feng. Let \(U\), \(V\), \(W\), \(X\), \(Y\), \(Z\) be the circumcenters of \(\triangle MFA\), \(\triangle MAB\), $\ldots$, \(\triangle MEF\). Then \(\overline{UV}\) and \(\overline{XY}\) are both perpendicular to \(\overline{AD}\), so cyclic hexagon \(UVWXYZ\) has opposite sides parallel. Since \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) are the reflections of \(M\) over the sides of \(UVWXYZ\), take a homothety at \(M\) with scale factor \(1/2\) sending \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) to the projections of \(M\) onto sides \(UV\), \(VW\), \(WX\), \(XY\), \(YZ\). [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair U,V,Y,X,Z,WW,M,A,B,C,D,EE,F; U=dir(120); V=reflect( (0,0),(0,1))*U; Y=dir(250); X=reflect( (U+V)/2,origin)*Y; Z=dir(190); WW=reflect( (U+Z)/2,origin)*X; M=0.1*dir(10); A=foot(M,U,V); B=foot(M,V,WW); C=foot(M,WW,X); D=foot(M,X,Y); EE=foot(M,Y,Z); F=foot(M,Z,U); draw(unitcircle); draw(A--D,linewidth(.4)+dashed); draw(B--EE,linewidth(.4)+dashed); draw(C--F,linewidth(.4)+dashed); draw(U--V--WW--X--Y--Z--cycle); dot("\(U\)",U,U); dot("\(V\)",V,V); dot("\(W\)",WW,WW); dot("\(X\)",X,X); dot("\(Y\)",Y,Y); dot("\(Z\)",Z,Z); dot("\(A\)",A,NW); dot("\(B\)",B,NE); dot("\(C\)",C,dir(C-F)); dot("\(D\)",D,S); dot("\(E\)",EE,W); dot("\(F\)",F,unit(F-C)); dot("\(M\)",M,dir(120)); [/asy][/asy] Observe that \begin{align*} [ABDE]&=\tfrac12AD\cdot BE\cdot\sin\angle(\overline{AD},\overline{BE})\\ &=\tfrac12AD\cdot BE\cdot\sin\angle UZW\\ &=\frac{AD\cdot BE\cdot CF}{2\cdot WZ}, \end{align*}and we are done upon noting that \(UX=VY=WZ\) by isosceles trapezoids \(UVXY\), \(VWYZ\), \(WXZU\).

pohoatza wrote: Let $ABCDEF$ be a convex hexagon and it`s diagonals have one common point $M$. It is known that the circumcenters of triangles $MAB,MBC,MCD,MDE,MEF,MFA$ lie on a circle. Show that the quadrilaterals $ABDE,BCEF,CDFA$ have equal areas. I know this problem, the author is Nairi Sedrakyan. His geometry problems are usually very hard, but most of them are extremely interesting. He is also the author of 1996 IMO geometry problem 5, none of the strongest IMO teams (China, USA, Russia) was able to solve that problem correctly, moreover Chinese IMO team received 0 points on that problem (the funny thing is: that problem is considered as an easy problem for Nairi Sedrakyan’s geometry standards).