I'm a competitor on this year's Olympiad of Metropolises (the second day of the olympiad was held yeasterday) and I haven't solved this problem during the competition but I think I have made some progress. I'll post here what I submitted (in about 30 minutes after I translate it to English) and I will be very grateful if someone can try to finish it or can tell me are there some solutions which use some of claims I've proven on the competition. Thank you all very much in advance.

I didn't solve this during the olympiad. First I thought of something like this but I did not see the last part of the angles and I tried to do inversion in C. In the end everything was in my draft Suppose that $AB_{1}A_{1}B$, $BC_{1}B_{1}C$, $CD_{1}C_{1}D$, $DE_{1}D_{1}E$ are cyclic. Let $X$ the point of intersection between $AB$ and $ED$, then for angle chasing $\measuredangle BCA = \measuredangle ECD$ and $\measuredangle ABE = \measuredangle DBC$ and $\measuredangle CDB = \measuredangle EDA$. And let $O$ the point of intersection between $BB_{1}$ and $DD_{1}$ then for DDIT: $CO$ and $CC_{1}$ are conjugate isogonals respect $\measuredangle CDB$ and $CX$ and $CC_{1}$ are conjugate isogonals respect $\measuredangle CDB$. Then $C$,$ X$ and $O$ are collinear. Later view that $C_{1}$ and $P$ are isogonal conjugates of the cuadrilateral $XBCD$, where $P$ is the point of intersection between $BD$ and $CX$, then $\measuredangle BPX + \measuredangle CPD = 180$, later $CX\perp BD$. Finally for angle chasing $BCOD_{1}$ is cyclic and $\measuredangle CBP + \measuredangle BCP = 90$, but $\measuredangle ABE = \measuredangle DBC$, then $OD_{1}\perp BA$, after $O$ is the orthocenter of the triangle $BDX$. Then $\measuredangle AA_{1}E= \measuredangle B_{1}BA = \measuredangle OBX =\measuredangle ODE = \measuredangle AE_{1}E$ then $EA_{1}E_{1}A$ is cyclic. Sorry for the typos. Remark: I used this is post #10: https://artofproblemsolving.com/community/c6h287868p8472213

Proposed by Nairi Sedrakyan and Yuliy Tikhonov

Here is a solution using complex numbers. Let $\lambda_1$ denote the complex number with magnitude $1$, in the direction of $\overset {\rightarrow} {AC}$. Define $\lambda_i$ ($i=2,3,4,5$) similarly for $\overset {\rightarrow} {BD}$, $\overset {\rightarrow} {CE}$, $\overset {\rightarrow} {DA}$, $\overset {\rightarrow} {EA}$. Note that the condition of $ABA_1B_1$ cyclic becomes : $$ \frac { \overset {\rightarrow}{AB} \cdot \overset {\rightarrow}{A_1B_1}}{ \overset {\rightarrow}{BA_1} \cdot \overset {\rightarrow}{B_1A}} \in \mathbb R \iff \frac { \overset {\rightarrow}{AB} \cdot \overset {\rightarrow}{CE}}{ \overset {\rightarrow}{CD} \cdot \overset {\rightarrow}{DA}} \in \mathbb R$$$$\iff \frac {(\lambda_2+\lambda_4)\lambda_3}{\lambda_2 \cdot \lambda_4} \in \mathbb R \iff (\overline {\lambda_2}+ \overline {\lambda_4})\cdot \lambda_3 \in \mathbb R$$ However consider the following sum with indices mod $5$. $$ \sum_{i=1}^{5} (\overline {\lambda_{i-1}} + \overline {\lambda_{i+1}}) \cdot \lambda_i = \sum_{i=1}^5 ( \lambda_i \overline {\lambda_{i+1}} + \lambda_{i+1} \overline {\lambda_{i}}) \in \mathbb R$$ We have used the fact that $x\overline y + y \overline x = \vert x+y \vert ^2 - \vert x \vert ^2 - \vert y \vert ^2 \in \mathbb R$ Hence if four of the summands in the original sum is real, the fifth must be real as well, so we are done $\blacksquare$

$Lemma:AD_1\cdot BE_1\cdot CA_1\cdot DB_1\cdot EC_1=AC_1\cdot BD_1\cdot CE_1\cdot DA_1\cdot EB_1$ Proof: It can be easily proved using the law of sines. Now lets multiply both sides by $AC\cdot BD\cdot CE\cdot DA\cdot BE$ and lets call this equation $(1)$. From power of a point we get that, for example, if $ABA_1B_1$ is cyclic, then $DB_1\cdot AD=DA_1\cdot BD$ and analogous results for the other quadrilaterals. We get that if 4 of those quadrilaterals are cyclic, then by $(1)$ the 5-th one should also be cyclic(from power of a point or just similar triangles), which conducts the proof.

MathsLion wrote: In convex pentagon $ABCDE$ points $A_1$, $B_1$, $C_1$, $D_1$, $E_1$ are intersections of pairs of diagonals $(BD, CE)$, $(CE, DA)$, $(DA, EB)$, $(EB, AC)$ and $(AC, BD)$ respectively. Prove that if four of quadrilaterals $AB_{1}A_{1}B$, $BC_{1}B_{1}C$, $CD_{1}C_{1}D$, $DE_{1}D_{1}E$ and $EA_{1}E_{1}A$ are cyclic then the fifth one is also cyclic. A bit too easy for a P6 Notations are $\measuredangle{E_1A_1B_1}=X_1,\measuredangle{A_1B_1C_1}=X_2,\measuredangle{B_1C_1D_1}=X_3,\measuredangle{C_1D_1E_1}=X_4,\measuredangle{D_1E_1A_1}=X_5$.By laws of sine in $A_1B_1C_1D_1E_1$ its easy to see $$\dfrac{AC_1}{AD_1}\cdot\dfrac{BD_1}{BE_1}\cdot\dfrac{CE_1}{CA_1}\cdot\dfrac{DA_1}{DB_1}\cdot\dfrac{EB_1}{EC_1}=\dfrac{\sin{X_4}}{\sin{X_3}}\cdot\dfrac{\sin{X_5}}{\sin{X_4}}\cdot\dfrac{\sin{X_1}}{\sin{X_5}}\cdot\dfrac{\sin{X_2}}{\sin{X_1}}\cdot\dfrac{\sin{X_3}}{\sin{X_2}}=1$$Now WLOG let $(ABA_1B_1),(BCB_1C_1),(CDC_1D_1),(DED_1E_1)$ be cyclic.It suffices to show $(EAE_1A_1)$ cyclic.But by POP we have $$1=\dfrac{DA_1\cdot DB}{DB_1\cdot DA}\cdot\dfrac{EB_1\cdot EC}{EC_1\cdot EB}\cdot\dfrac{AC_1\cdot AD}{AD_1\cdot AC}\cdot\dfrac{BD_1\cdot BE}{BE_1\cdot BD}=\dfrac{EC\cdot CE_1}{EA\cdot CA_1}\Longleftrightarrow CE\cdot CA_1=CA\cdot CE_1 $$so we are done.$\square$ REMARKS-Does changing the condition to only 3 being cyclic also work with some extra conditions?Just wondering

mijail wrote: I didn't solve this during the olympiad. First I thought of something like this but I did not see the last part of the angles and I tried to do inversion in C. In the end everything was in my draft Suppose that $AB_{1}A_{1}B$, $BC_{1}B_{1}C$, $CD_{1}C_{1}D$, $DE_{1}D_{1}E$ are cyclic. Let $X$ the point of intersection between $AB$ and $ED$, then for angle chasing $\measuredangle BCA = \measuredangle ECD$ and $\measuredangle ABE = \measuredangle DBC$ and $\measuredangle CDB = \measuredangle EDA$. And let $O$ the point of intersection between $BB_{1}$ and $DD_{1}$ then for DDIT: $CO$ and $CC_{1}$ are conjugate isogonals respect $\measuredangle CDB$ and $CX$ and $CC_{1}$ are conjugate isogonals respect $\measuredangle CDB$. Then $C$,$ X$ and $O$ are collinear. Later view that $C_{1}$ and $P$ are isogonal conjugates of the cuadrilateral $XBCD$, where $P$ is the point of intersection between $BD$ and $CX$, then $\measuredangle BPX + \measuredangle CPD = 180$, later $CX\perp BD$. Finally for angle chasing $BCOD_{1}$ is cyclic and $\measuredangle CBP + \measuredangle BCP = 90$, but $\measuredangle ABE = \measuredangle DBC$, then $OD_{1}\perp BA$, after $O$ is the orthocenter of the triangle $BDX$. Then $\measuredangle AA_{1}E= \measuredangle B_{1}BA = \measuredangle OBX =\measuredangle ODE = \measuredangle AE_{1}E$ then $EA_{1}E_{1}A$ is cyclic. Sorry for the typos. Remark: I used this is post #10: https://artofproblemsolving.com/community/c6h287868p8472213 I was ready to post my attempt but now I see that you have finished it already. Thank you very very much. I have proven that $C,X,O$ are collinear that $BCOD_1$ is cyclic. Also I have angle chasing some that implies $C_1$ and $P$ being isogonal conjugates with respect to $XBCD$ so I hope for some points now.

It's easy if you know some isogonal properties. Assume that $BC_1B_1C, CD_1C_1D, DE_1,D_1E, EA_1E_1A$ are cyclic. We will prove that $AB_1A_1B$ is also cyclic. Define $P=CC_1\cap EE_1$, $Q=BC\cap AE$, $R=DP\cap CE$. $\textbf{Claim 1:}$ $\angle BCA=\angle ECD$, $\angle CDB=\angle EDA$, and $\angle DEC=\angle AEB$. $\textit{Proof:}$ We only need to prove that $\angle CDB=\angle EDA$ as we can prove the other 2 similarly. Notice that because $CD_1C_1D$ and $DE_1D_1E$ cyclic, we have \[\angle CDC_1=\angle AD_1C=\angle BD_1E=\angle EDE_1\]which implies $\angle CDB=\angle EDA$. (Claim 1 is proven) $\textbf{Claim 2:}$ $D,P,Q$ are collinear and the common line is isogonal to $DD_1$ wrt $\angle CDE$. $\textit{Proof:}$ Note that $DE_1$ is isogonal to $DC_1$ wrt to $\angle CDE$, hence by isogonal line lemma, we have $DP$ is isogonal to $DD_1$ wrt $\angle CDE$. Notice also that $DB$ is isogonal to $DA$ wrt to $\angle CDE$, by isogonal line lemma again, we have $DQ$ is isogonal to $DD_1$ wrt $\angle CDE$. The two above facts imply the claim. (Claim 2 is proven) $\textbf{Claim 3:}$ $DCE_1P$ and $DEC_1P$ are cyclic. $\textit{Proof:}$ Notice that \[\angle E_1DP=\angle C_1DD_1=\angle C_1CE_1=\angle PCE_1\implies DCE_1P \text{ is cyclic}\]Similarly, \[\angle C_1DP=\angle E_1DD_1=\angle E_1ED_1=\angle PEC_1\implies DEC_1P \text{ is cyclic}\](Claim 3 is proven) $\textbf{Claim 4:}$ $DP$ is perpendicular to $CE$. $\textit{Proof:}$ Notice that $\angle CDR=\angle CDP=\angle EDD_1$, $\angle DCR=\angle DCE=\angle BCD_1$, and $\angle DER=\angle DEC=\angle ACD_1$ implies that $R$ and $D_1$ is the isogonal conjugate of $QCDE$. Hence, we have $\angle CRD+\angle ERQ=180^{\circ}$ which equivalent to $\angle DRC=90^{\circ}$ as $\angle DRC=\angle ERQ$. (Claim 4 is proven) Now we only need to do some angle chasing. Define $\angle DCE=x, \angle CDB=y, \angle DEC=z, \angle ECC_1=w$. As $DR\perp CE$, we have $\angle PDB=90^{\circ}-(x+y)$. We have $\angle CDC_1=180^{\circ}-\angle DCE-\angle EDC_1-\angle DEC=180^{\circ}-(x+y+z)$. So, $\angle CC_1D=180^{\circ}-\angle CDC_1-\angle C_1CD=y+z-w$. We have $\angle PCE_1=\angle PDE_1=90^{\circ}-(x+y)$ So, $\angle CBD=180^{\circ}-\angle BCE_1-\angle DCP-\angle PCE_1-\angle BDC=90^{\circ}-x-w$ Therefore, \[\angle A_1BB_1=\angle CBB_1-\angle CBD=\angle CC_1D-\angle CBD=(y+z-w)-(90^{\circ}-x-w)=x+y+z-90^{\circ}\]Similarly, we can also show that $\angle A_1AB_1=x+y+z-90^{\circ}$. Hence, $\angle A_1BB_1=\angle A_1AB_1$. So, $AB_1A_1B$ is also cyclic.

Here is the solution I gave during the contest: Assume all but \(EA_1E_1A\) are given to be cyclic, and write \(\angle(BD,CE)=\alpha\) (and its cyclic equivalents). The main idea is to find some condition \(P(A,B)\) equivalent to \(A, B_1, A_1, B\) concyclic, so that $\sum_{cyc} P(A,B)$ or $\prod_{cyc} P(A,B)$ is an identity. So here it is: Lemma: \(A, B_1, A_1, B\) concyclic \(\Leftrightarrow \frac {\sin\alpha}{\sin \beta}=\frac{DB}{DA} \) (and its cyclic equivalents). Proof: By power of a point, we have \(A, B_1, A_1, B\) concyclic \(\Leftrightarrow DA_1 \cdot DB = DB_1 \cdot DA\). Now by sine law on \(\bigtriangleup DA_1B_1\), we have \(\frac {\sin\alpha}{\sin \beta}=\frac{DB_1}{DA_1} \). Combining yields the desired result. \(\square\) Multiplying the equation in the lemma for all 4 given cyclic quadrilaterals, we obtain \(\frac {\sin\alpha}{\sin \epsilon}=\frac{CE}{CA}\). Applying the lemma again on \(EA_1E_1A\) with this equality yields \(EA_1E_1A\) is cyclic. \(\blacksquare\) Remark: When I attempted this problem, I was one hour in with P4,5 in my pocket, so I was hoping if I could bash this out using coordinates or complex numbers. The idea of finding such a condition \(P(A,B)\) came to me naturally, and turned out I didn't even need any bashing! Weird to see something like this in P6, though.

Metropolises 2020 P6 wrote: In convex pentagon $ABCDE$ points $A_1$, $B_1$, $C_1$, $D_1$, $E_1$ are intersections of pairs of diagonals $(BD, CE)$, $(CE, DA)$, $(DA, EB)$, $(EB, AC)$ and $(AC, BD)$ respectively. Prove that if four of quadrilaterals $AB_{1}A_{1}B$, $BC_{1}B_{1}C$, $CD_{1}C_{1}D$, $DE_{1}D_{1}E$ and $EA_{1}E_{1}A$ are cyclic then the fifth one is also cyclic. Alright, it seems that they wanted to follow the recent example of USA TST 2020 P6, that is posing a trivial Problem 6 Geometry. Let $\angle E_1A_1B_1=\alpha, \angle A_1B_1C_1=\beta, \ldots$ Assume that all quadrilaterals are cyclic, except for $AB_1A_1B$. Then, $$\frac{\sin \epsilon}{\sin \alpha}=\frac{CA_1}{CE_1}=\frac{CA}{CE}$$and cyclically we obtain four similar relations. Now note the deep fact $$\frac{\sin \epsilon}{\sin \alpha} \cdot \frac{\sin \alpha}{\sin \beta} \cdot \frac{\sin \beta}{\sin \gamma} \cdot \frac{\sin \gamma}{\sin \delta} \cdot \frac{\sin \delta}{\sin \epsilon}=1=\frac{CA}{CE} \cdot \frac{DB}{DA} \cdot \frac{EB}{EC} \cdot \frac{AD}{AC} \cdot \frac{BE}{BD}$$From each side of this equality we may cancel out all but one fractions, resulting in $$\frac{\sin \angle a}{\sin \angle \beta}=\frac{AD}{BD} \Rightarrow \frac{AD}{BD}=\frac{B_1D}{A_1D} \Rightarrow AD \cdot A_1D=BD \cdot B_1D$$hence we infer that $ABA_1B_1$ is cyclic, as desired. Boom.

A horrible solution. Suppose $ABB_1A_1, BCB_1C_1, CDC_1D_1, DED_1E_1$ are cyclic, we now show that $AA_1EE_1$ is cyclic as well. Claim. $EB,ED$ are isogonals w.r.t. $\angle ABC$. Proof. By sprial sim. lemma, $\triangle BAA_1\sim\triangle BC_1C$, hence $\angle ABD=\angle EBC$ as desired. $\blacksquare$ Similarly, $EC,AC$ are isogonals w.r.t. $\angle DCB$ and $DA,DB$ are isogonals w.r.t. $\angle EDC$. Now, applying Ceva's theorem to $\triangle DBC$ and point $E,A$ we have \begin{align*} \frac{\sin\angle ECD}{\sin\angle ECB}\cdot\frac{\sin\angle EBC}{\sin\angle EBD}\cdot\frac{\sin\angle EDB}{\sin\angle EDC}&=1\\ \frac{\sin\angle ACB}{\sin\angle ACD}\cdot\frac{\sin\angle CDA}{\sin\angle ADB}\cdot\frac{\sin\angle ABD}{\sin\angle ABC}&=1 \end{align*}Notice that $\angle ECD=\angle ACB, \angle CDA=\angle EDB, \angle ECB=\angle ACD,\angle EBC=\angle ABD$, and hence $$\sin\angle EBD\sin\angle EDC=\sin\angle ADB\sin\angle ABC$$Therefore, \begin{align*} \frac{CA_1}{CE_1}&=\frac{\sin\angle CE_1D}{\sin\angle CA_1B}\\ &=\frac{\sin\angle DEB}{\sin\angle DAB}\\ &=\frac{\sin\angle DEB}{DB}\cdot\frac{DB}{\sin\angle DAB}\\ &=\frac{\sin\angle DBE}{DE}\cdot\frac{AB}{\sin\angle ADB}\\ &=\frac{AB\sin\angle ABC}{DE\sin\angle EDC}\\ &=\frac{AC\sin\angle ACB}{EC\sin\angle DCE}\\ &=\frac{AC}{EC} \end{align*}Hence $CA\times CE_1=CA_1\times CE$, and $A,E_1,A_1,E$ are concyclic as desired.