What is $F$?

$F\equiv P$!

Nguyenhuyhoang wrote: What is $F$? You are right, $F$ must be $P$, I have edited.

since $EPA$ and $EPR$ are similar we have $EC^2=EP^2=ER*EA$ assume that $R$ is not the orthocentre of $ABD$ than let $R'$ be the orthocentre. now let $C'$ be the point on $AE$ such that $DC'B$ is right. let $Q'$ be the intersection of $BR',AD$ if $A-R'-R$ than $A-Q'-Q$ so $\angle BQA$ is acute and than $\angle ECB=\angle BQA$ is acute so $E-C'-C$ now we have that $EC'^2=ED*EB$ since $EC'D$ and $EC'B$ are similar. but $ED*EB=ER'*EA$ since $EDR$ and $EAB$ are similar. so $EC'^2=ER'*EA$ one more important fact is that $A-R-E$ otherwise $E$ would be inside $\odot APR$ but that's impossible since $EP$ is tangent to that circle so $R'-R-E$ so $EC^2>EC'^2=ER'*EA>EA*ER$ but that's imposssible since $EC^2=ER*EA$ . the case $A-R-R'$ is done similarly so $R=R'$ now $\angle BCD=\angle AQ'B=90$

Let $\Pi$ be the circumcircle of $\triangle BCD$. Let $\Pi$ meet $[AE]$ at $X$. Let the perpendicular to $BD$ from $Q$ meet $BD$ at $Y$ and $\Pi$ at $Z$. Let $DZ$ and $AC$ meet at $W$. Since $QZ \parallel XC$, $XQZC$ is an isosceles trapezoid. So if we want to prove that $\angle BCD = 90^\circ$. $ \angle BCD = 90^\circ \Leftrightarrow BD $ is a diameter of $\Pi$ $ \Leftrightarrow BD $ bisects $ XC$. Since $\angle QBD = \angle QZD = \angle AWD$, $B,R,D,W$ are concyclic. The power of $E$ with respect to $(BRDW)$ is $BE\cdot ED = ER \cdot EW$. The power of $E$ with respect to $\Pi$ is $BE \cdot ED = EC \cdot EX$. Merging with above, we get $ER \cdot EW = EC \cdot EX$. The power of $E$ with respect to $(APR)$ is $ER \cdot AE = EP^2 = EC^2$. By dividing side by side, we get $\frac {EW}{AE} = \frac {EX}{EC}$. Since $QZ \parallel AW$, we have $\dfrac{EW}{AE} = \frac{YZ}{QY}$. If we merge these two, we get $\frac {XE}{YZ} = \frac {EC}{QY} = k$. So the diagonal $QC$ meets $EY$ with a ratio $k$. Similarly, the diagonal $XZ$ meets $EY$ with a ratio $k$. So $QC$ and $XZ$ meet at $EY$. Since $XCZQ$ is an isosceles trapezoid, the diagonals are equal and meet each other at the symmetry axis of the trapezoid. $EY$ is symmetry axis of the trapezoid, and then, $BD$ bisects $XC$, and then, the center of $\Pi$ is on $BD$, and then $BD$ is the diameter of $\Pi$, and then $\angle BCD = 90^\circ$. Comment: I think the problem is above junior level. Or there is a simpler solution. None of the contestants solved this problem. Moreover, they got at most $2$ out of $7$.

Let $F$ is a point on $AC$ so that $EF=EC$, because $AC \perp BD$ thus $F,D,B,C$ are concyclic$(1)$. Furthermore, we also have $EA.ER=EP^2=EF.EC=EB.ED$(from $(1)$) From here we have $A$ is orthocenter of triangle $BRD$, $\Rightarrow \widehat{BQD}=90 \Rightarrow \widehat{BCD}=90$ I dont know if I have made any mistake but I dont think this is a tough problem Extend this problem a bit:We have $(RAFC)=-1$ and $\widehat{FPC}=90$ so we have $PF$ bisects $\widehat{APR}$

Nguyenhuyhoang wrote: Let $F$ is a point on $AC$ so that $EF=EC$, because $AC \perp BD$ thus $F,D,B,C$ are concyclic$(1)$. Furthermore, we also have $EA.ER=EP^2=EF.EC=EB.ED$(from $(1)$) From here we have $A$ is orthocenter of triangle $BRD$, $\Rightarrow \widehat{BQD}=90 \Rightarrow \widehat{BCD}=90$ I dont know if I have made any mistake but I dont think this is a tough problem Extend this problem a bit:We have $(RAFC)=-1$ and $\widehat{FPC}=90$ so we have $PF$ bisects $\widehat{APR}$ i think it's incorrect as you can't get that $FDBC$ are con-cyclic. for this you must have $\angle BCD=90$ which is what needs to be proven. as the problem it self it definitly isn't meant for P2 maybe P3,P6 since this is the junior NMO.

Let $(ARD)$ meet $BD$ at $F$. The power of $E$ with respect to $(ARFD)$ is $ER \cdot AE = EF \cdot ED$. The power of $E$ with respect to $(ARP)$ is $ER\cdot AE = EP ^2 = EC^2$. So $EF \cdot ED = EC^2$ yields that $(FCD)$ is tangent to $CE$ or in other words $\angle ECF = \angle EDC$. Also we have $\angle ADE = \angle ERF$. Since $\angle QDB + \angle BDC = \angle FRC + \angle RCF$, we have $\angle RBC = \angle RFC$. This yields $BCFR$ is deltoid. (If you cannot see this easily, take reflection of $B$ with respect to $RC$. Call it $B'$ . Since $\angle RBC = \angle RB'C = \angle RFC$, $B'$ is on $BD$, $F=B'$.) So $\angle BCR = \angle RCF = \angle BDC$. Since $\angle BEC = 90^\circ$, $\angle BCD = 90^\circ$.

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Let $QB\cap DC=T,DQ\cap CB=S$ and the perpendicular from $T$ to $DB$ meets $DB,DS,CB$ at $L,N,M$.$QC\cap DB=F$ We have $EC^2=EA.ER$ \[\frac{TL}{LN}=\frac{EC}{EA}=\frac{ER}{EC}=\frac{TL}{LM}\]by homothethy. So $LN=LM\implies M=N\implies ST\perp DB$ Let $O$ be the circumcenter of $AQBC$. We know that $E,O,S,T$ are orthogonal because of Brocard Theorem. $EO\perp ST$ and $EB\perp ST$ gives us that $F,O,B$ are collinear. So $DB$ is the diameter which means that $\angle DCB=90 \blacksquare$