IMO 2019 Shortlisted Problems

Step 1. The external bisector of $\angle BAC$ is the line through $A$ perpendicular to $IA$. Let $DI$ meet this line at $L$ and let $DI$ meet $\omega$ at $K$. Let $N$ be the midpoint of $EF$, which lies on $IA$ and is the pole of line $AL$ with respect to $\omega$. Since $AN\cdot AI=A{E}^2=AR\cdot AP$, the points $R$, $N,I$, and $P$ are concyclic. As $IR=IP$, the line $NI$ is the external bisector of $\angle PNR$, so $PN$ meets $\omega$ again at the point symmetric to $R$ with respect to $AN$ - i.e. at $K$. Let $DN$ cross $\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\omega$ meet on the polar line of the intersection of the diagonals with respect to $\omega$. Since $L$ lies on the polar line $AL$ of $N$ with respect to $\omega$, the line $PS$ must pass through $L$. Thus it suffices to prove that the points $S,Q$, and $P$ are collinear.



Step 2. Let $\Gamma$ be the circumcircle of $△BIC$. Notice that $$\begin{array}{rl}\angle (BQ,QC)=\angle & (BQ,QP)+\angle (PQ,QC)=\angle (BF,FP)+\angle (PE,EC)\\ & =\angle (EF,EP)+\angle (FP,FE)=\angle (FP,EP)=\angle (DF,DE)=\angle (BI,IC)\end{array}$$ so $Q$ lies on $\Gamma$. Let $QP$ meet $\Gamma$ again at $T$. It will now suffice to prove that $S,P$, and $T$ are collinear. Notice that $\angle (BI,IT)=\angle (BQ,QT)=\angle (BF,FP)=\angle (FK,KP)$. Note $FD\perp FK$ and $FD\perp BI$ so $FK\parallel BI$ and hence $IT$ is parallel to the line $KNP$. Since $DI=IK$, the line $IT$ crosses $DN$ at its midpoint $M$.

Step 3. Let $F^{\prime }$ and $E^{\prime }$ be the midpoints of $DE$ and $DF$, respectively. Since $D{E}^{\prime }\cdot E^{\prime }F=D{E}^{\prime 2}=B{E}^{\prime }\cdot E^{\prime }I$, the point $E^{\prime }$ lies on the radical axis of $\omega$ and $\Gamma$; the same holds for $F^{\prime }$. Therefore, this radical axis is $E^{\prime }{F}^{\prime }$, and it passes through $M$. Thus $IM\cdot MT=DM\cdot MS$, so $S,I,D$, and $T$ are concyclic. This shows $\angle (DS,ST)=\angle (DI,IT)=\angle (DK,KP)=\angle (DS,SP)$, whence the points $S,P$, and $T$ are collinear, as desired.



---

Alternative solution.

We start as in Solution 1. Namely, we introduce the same points $K,L,N$, and $S$, and show that the triples $(P,N,K)$ and $(P,S,L)$ are collinear. We conclude that $K$ and $R$ are symmetric in $AI$, and reduce the problem statement to showing that $P,Q$, and $S$ are collinear.

Step 1. Let $AR$ meet the circumcircle $\Omega$ of $ABC$ again at $X$. The lines $AR$ and $AK$ are isogonal in the angle $BAC$; it is well known that in this case $X$ is the tangency point of $\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $XI$ crosses $\Omega$ again at the midpoint $M^{\prime }$ of $\mathrm{arc}BAC$.

Step 2. Denote the circles $BFP$ and $CEP$ by ${\Omega }_B$ and ${\Omega }_C$, respectively. Let ${\Omega }_B$ cross $AR$ and $EF$ again at $U$ and $Y$, respectively. We have $$\angle (UB,BF)=\angle (UP,PF)=\angle (RP,PF)=\angle (RF,FA),$$ so $UB\parallel RF$.



Next, we show that the points $B,I,U$, and $X$ are concyclic. Since $$\angle (UB,UX)=\angle (RF,RX)=\angle (AF,AR)+\angle (FR,FA)=\angle \left(M^{\prime }B,M^{\prime }X\right)+\angle (DR,DF),$$ it suffices to prove $\angle (IB,IX)=\angle \left(M^{\prime }B,M^{\prime }X\right)+\angle (DR,DF)$, or $\angle \left(IB,M^{\prime }B\right)=\angle (DR,DF)$. But both angles equal $\angle (CI,CB)$, as desired. (This is where we used the fact that $M^{\prime }$ is the midpoint of arc $BAC$ of $\Omega$.)

It follows now from circles BUIX and BPUFY that $$\begin{array}{rl}\angle (IU,UB)=\angle (IX,BX)=\angle \left(M^{\prime }X,BX\right)=& \frac{\pi -\angle A}{2}\\ & =\angle (EF,AF)=\angle (YF,BF)=\angle (YU,BU),\end{array}$$ so the points $Y,U$, and $I$ are collinear.

Let $EF$ meet $BC$ at $W$. We have $$\angle (IY,YW)=\angle (UY,FY)=\angle (UB,FB)=\angle (RF,AF)=\angle (CI,CW),$$ so the points $W,Y,I$, and $C$ are concyclic.

Similarly, if $V$ and $Z$ are the second meeting points of ${\Omega }_C$ with $AR$ and $EF$, we get that the 4-tuples ( $C,V,I,X$ ) and ( $B,I,Z,W$ ) are both concyclic.

Step 3. Let $Q^{\prime }=CY\cap BZ$. We will show that $Q^{\prime }=Q$.

First of all, we have $$\begin{array}{rl}\angle \left(Q^{\prime }Y,Q^{\prime }B\right)=\angle (CY,ZB)& =\angle (CY,ZY)+\angle (ZY,BZ)\\ & =\angle (CI,IW)+\angle (IW,IB)=\angle (CI,IB)=\frac{\pi -\angle A}{2}=\angle (FY,FB)\end{array}$$ so $Q^{\prime }\in {\Omega }_B$. Similarly, $Q^{\prime }\in {\Omega }_C$. Thus $Q^{\prime }\in {\Omega }_B\cap {\Omega }_C=\{P,Q\}$ and it remains to prove that $Q^{\prime }\ne P$. If we had $Q^{\prime }=P$, we would have $\angle (PY,PZ)=\angle \left(Q^{\prime }Y,Q^{\prime }Z\right)=\angle (IC,IB)$. This would imply $$\angle (PY,YF)+\angle (EZ,ZP)=\angle (PY,PZ)=\angle (IC,IB)=\angle (PE,PF),$$ so circles ${\Omega }_B$ and ${\Omega }_C$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\prime }=Q$.



Step 4. Now we are ready to show that $P,Q$, and $S$ are collinear.

Notice that $A$ and $D$ are the poles of $EW$ and $DW$ with respect to $\omega$, so $W$ is the pole of $AD$. Hence, $WI\perp AD$. Since $CI\perp DE$, this yields $\angle (IC,WI)=\angle (DE,DA)$. On the other hand, $DA$ is a symmedian in $△DEF$, so $\angle (DE,DA)=\angle (DN,DF)=\angle (DS,DF)$. Therefore, $$\begin{array}{rl}\angle (PS,PF)=\angle (DS,DF)=\angle (DE,DA)=& \angle (IC,IW)\\ & =\angle (YC,YW)=\angle (YQ,YF)=\angle (PQ,PF)\end{array}$$ which yields the desired collinearity.