Reflect $A$ across the midpoint of $FB$ to $A'$, and similarly construct $C', E'$. We first take care of the case when $A' = C' = E'$. Then construct $XYZ$ such that $B$, $D$, and $F$ lie on $YZ$, $ZX$, and $XY$, with $A'B \perp YZ$, $A'D \perp XZ$, and $A'F \perp XY$. Then the result follows from Erdos-Mordell. Now suppose $A'$, $C'$, $E'$ are not all equal. Then we will show that there is a point $P$ inside $A'C'E'$ such that $FA'PB$, $DE'PF$, and $BC'PD$ are all cyclic. Indeed, note that $B$, $A'$, $C'$ are collinear and analogous for $D$ and $F$. It's easy to see that the circumcircles of $FA'B$, $BC'D$, $DE'F$ will in fact concur. Calling this point of concurrence $P$ and letting $P'$ be the isogonal conjugate of $P$, let $QRS$ be the pedal triangle of $P'$ wrt $ABC$. We have that the figures $A'C'E'P$ and $QRSP'$ are similar. Since $P'$ lies within $QRS$, $P'$ must lie within $A'C'E'$. Now WLOG $A'$ lies between $F$ and $E'$. Then $FE'+DC'+BA' \ge FA'+BE'+BC'+A'P+B'P+C'P$ $\ge BP+DP+FP$ so the desired inequality follows from the case where $A' = C' = E'$ that we proved above.

Here is my solution to this nice problem: See the attached figure. We have $\angle C = \angle F , \angle A = \angle D , \angle B = \angle E$. Since $PXYQ$ is a rectangle, $XY= PQ$. Also $BD \ge XY$ , or $2BD \ge XY + PQ$. This gives $2BD \geq BC\sin B + CD\sin A + EF \sin B + AF \sin A$. Similarly $2 BF \ge AB \sin B + AF \sin C + ED \sin B + CD \sin C$ and $2DF \ge FE\sin C + ED \sin A + BC\sin C + AB \sin C$. Note that \begin{align*} 2 \left(R_{A}+R_{C}+R_{E}\right) &= \frac{BF}{\sin A}+\frac{DF}{\sin B}+\frac{BD}{\sin C}\\ &\ge \frac{1}{2}\left ( \left(AB+ ED \right)\left(\frac{\sin B}{\sin A}+ \frac{\sin A}{\sin B} \right)+ \left ( AF + CD \right ) \left ( \frac{\sin C}{\sin A}+ \frac{\sin A}{\sin C} \right ) + \left ( BC + EF \right )\left ( \frac{\sin C}{\sin B}+ \frac{\sin B}{\sin C} \right )\right ) \\ &\ge AB + ED + AF + CD + BC + EF \\ &= P \end{align*} So $R_{A}+R_{C}+R_{E}\geq\frac{P}{2}$. qed

Attachments:

probability1.01 wrote: Since $P'$ lies within $QRS$, $P'$ must lie within $A'C'E'$. Now WLOG $A'$ lies between $F$ and $E'$. Then $FE'+DC'+BA' \ge FA'+BE'+BC'+A'P+B'P+C'P$ $\ge BP+DP+FP$ so the desired inequality follows from the case where $A' = C' = E'$ that we proved above. Unfortunately, I can't really figure out what's going on here... there are a few obvious typos (e.g. $P\in\triangle{A'C'E'}$, not $P'$), but don't we need to show that $(BP+DP+FP)+(BP+DP+FP)\ge (FE'+DC'+BA')+(FA'+DE'+BC')$? If so, $FA'+DE'+BC'+A'P+E'P+C'P\ge BP+DP+FP$ follows from the triangle inequality, but is not in the direction we want. Of course, I am probably wrong, so can anyone figure out what this should say? In any case, I do find the rephrased problem with $A',C',E'$ ($AB\parallel DE$, etc. iff $FA'E'$, etc. are lines) more natural than the original and easier to work with (at least, after having seen Erdős-Mordell). Let the perpendiculars to $FA'E'$ at $F$, $BC'A'$ at $B$, and $DE'C'$ at $D$ meet at $B',D',F'$ such that $B,B'$ are on opposite sides of $DF$, etc. Since $D'A'=2R_A$, etc. we want to model the proof of Erdős-Mordell to show that $D'A'+F'C'+B'E'$ is at least $(FE'+DC'+BA')+(FA'+DE'+BC')$. If two of $A',C',E'$ coincide, it's easy to show that they all coincide, so we can assume $\triangle{A'C'E'}$ is not degenerate. Then we note that $\triangle{A'C'E'}$ is directly similar to $\triangle{D'F'B'}$ (which is easy using cyclic quads $B'DE'F$, etc.). (*) We can get a lower bound on $D'A'\cdot B'F'$ (interpreted using areas) either by considering $D'A'$ itself or its isogonal; we use the latter for the AM-GM crucial to Erdős-Mordell, so we also use that here to get \[D'A'\cdot B'F' \ge A'F\cdot D'F' + A'B\cdot D'B'.\]However, if we directly sum the three inequalities of this form up, we only account for each of $FA',FE',BC',BA',DE',DC'$ once, while we need two for each for AM-GM to work. For instance, $E'F$ only appears as $E'F\cdot F'B'/F'D'$ and $C'B$ only appears as $C'B\cdot B'F'/B'D'$. So if there's any reasonable hope of area inequalities of this type working, we need \[D'A'\cdot B'F'\ge E'F\cdot D'F' + C'B\cdot D'B'.\]But this is in fact equivalent to our previous inequality: indeed, we have $A'E'/A'C'=D'B'/D'F'$ from (*), whence $A'F\cdot D'F'+A'B\cdot D'B' = E'F\cdot D'F'+C'B\cdot D'B'$ (either directed segments or a little bit of casework confirms that $A'F'-E'F$ and $C'B-A'B$ have the same sign). But summing up the six inequalities of this form, we get (using AM-GM) \[D'A'+D'A'+F'C'+F'C'+B'E'+B'E'\ge FE'+FA'+DC'+DE'+BA'+BC',\]as desired. In fact, this is equivalent to the official solution, but as I mentioned above, I think everything is easier to see with this modified (but equivalent) diagram.

I discoverd by geogebra that with this condition (and it isn't necessarily convex hexagon) we have $area(ACE)=area(BDF)$. Who can prove this?

China team got $6$ zero in this problem

Does anyone know which country proposed this problem?

See this. Problem 5 was "proposed by Nairi Sedrakyan, Armenia."

Construct parallelograms \(FABX\), \(BCDY\), \(DEFZ\), so that \(B\), \(D\), \(F\) lie on lines \(XY\), \(YZ\), \(ZX\). Claim: \[4R_A\ge(FA+CD)\frac{\sin Y}{\sin X}+(AB+DE)\frac{\sin Z}{\sin X}.\] Proof. Considering the component of \(\overline{BF}\) perpendicular to \(\overline{YZ}\), we have \[BF\ge BX\sin Y+FX\sin Z\quad\text{and}\quad BF\ge BY\sin Y+FZ\sin Z.\]Combining gives the inequality \[4R_A=\frac{2BF}{\sin X}=(BX+BY)\frac{\sin Y}{\sin X}+(FZ+FX)\frac{\sin Z}{\sin X},\]as desired. \(\blacksquare\) [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=red; pen sec2=lightred+dashed; pen tri=purple+pink; pen fil=invisible; pen sfil=invisible; pen sfil2=invisible; pen tfil=invisible; pair O1,Dp,Bp,Fp,MD,MB,MF,A,EE,C,F,D,B,Ap,Ep,Cp,O2,MA,ME,MC,X,Y,Z; O1=origin; Dp=dir(120); Bp=dir(210); Fp=dir(330); MD=(Bp+Fp)/2; MB=(Fp+Dp)/2; MF=(Dp+Bp)/2; A=Fp+unit(Fp-Bp)*1.3; EE=Dp+unit(Dp-Fp)*1.5; C=Bp+unit(Bp-Dp)*1.8; F=A+EE-Fp; D=EE+C-Dp; B=C+A-Bp; Ap=B+F-A; Ep=F+D-EE; Cp=D+B-C; O2=circumcenter(Ap,Ep,Cp); MA=(Ep+Cp)/2; ME=(Cp+Ap)/2; MC=(Ap+Ep)/2; X=(A+D)/2; Y=(B+EE)/2; Z=(C+F)/2; filldraw(B--F--D--cycle,tfil,tri+linewidth(.4)+dashed); draw(Ep--D,sec2); draw(Cp--B,sec2); draw(Ap--F,sec2); filldraw(Ap--Ep--Cp--cycle,sfil,sec); filldraw(A--B--C--D--EE--F--cycle,fil,pri); dot("\(C\)",A,E); dot("\(D\)",B,SE); dot("\(E\)",C,SW); dot("\(F\)",D,W); dot("\(A\)",EE,NW); dot("\(B\)",F,NE); dot("\(Y\)",Ap,dir(160)); dot("\(X\)",Ep,NW); dot("\(Z\)",Cp,dir(20)); [/asy][/asy] Summing, we have \begin{align*} 4(R_A+R_C+R_E)&\ge\sum_\mathrm{cyc}(FA+CD)\left(\frac{\sin Y}{\sin X}+\frac{\sin X}{\sin Y}\right)\\ &\ge2(AB+BC+CD+DE+EF+FA), \end{align*}and we are done.

Beautiful problem . Sadly, I needed a hint to construct $X,Y,$ and $Z.$ Let $A'$ be a point such that $ABA'F$ is a parallelogram, and construct $B'$ and $C'$ similarly. Moreover, let $X,Y,$ and $Z$ be the points such that $\overline{XY}\perp\overline{A'B},\overline{YZ}\perp\overline{C'D},$ and $\overline{ZX}\perp\overline{E'F}.$ Notice that $A'X=2R_A,$ so by Mordell's lemma,$$2R_A=A'X\ge \frac{A'B\cdot XZ}{ZY}+\frac{A'F\cdot XY}{YZ}.\quad(1)$$Also, $A',$ $C',$ and $B$ are collinear and $\triangle XYZ\sim\triangle A'C'E',$ so $(A'B-C'B)\cdot ZX=(E'F-A'F)\cdot XY$ and$$2R_A\ge \frac{C'B\cdot XZ}{ZY}+\frac{E'F\cdot XY}{YZ}.\quad(2)$$Adding $(1)$ and $(2)$ to similar inequalities with $R_C$ and $R_E$ noting that $A'F=AB$ etc,\begin{align*}4R_A+4R_C+4R_E&\ge\sum_{\text{cyc}}{\frac{(AF+CD)XZ}{ZY}}+\frac{(AF+CD)ZY}{XZ}\\&=\sum_{\text{cyc}}{(AF+CD)\left(\frac{XZ}{ZY}+\frac{ZY}{XZ}\right)}\\&\ge\sum_{\text{cyc}}{(AF+CD)(2)}\\&=2P.\end{align*}$\square$

One of the problems of all time.