Lukaluce wrote: Let $ABC$ be a triangle such that $AB < AC$. Let the excircle opposite to A be tangent to the lines $AB, AC$, and $BC$ at points $D, E$, and $F$, respectively, and let $J$ be its centre. Let $P$ be a point on the side $BC$. The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$. Let $R$ be the foot of the perpendicular from $A$ to the line $FJ$. Prove that the points $P, Q$, and $R$ are collinear. (The excircle of a triangle $ABC$ opposite to $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.) $angle DQE=\angle DQP+\angle PQE=\angle ABC+\angle ACB=180-\angle DAE$ so $D,Q,E,A$ are concyclic. More over $\angle ADI_A=\angle ARI_A=\angle AEI_A=90$ so $D,A,R,E,I_A,Q$ are concyclic. No w we have that: $\angle DQP=\angle ABC=\angle DI_AR=\angle DQR$ So $Q,P,R$ are collinear

Following last year, we have another high class JBMO Geometry problem by Bozhidar Dimitrov (Strudan_Borisov) from Bulgaria; or as several leaders said: bomba. The original submission also has a harder version, which hides the point $R$ and requires to show that $PQ$ passes through a fixed point as $P$ varies - very beautiful, but of course not suitable for juniors. I only think it should have been P1, but oh well.

From the cyclic quadrilaterals, we get that: $\angle DQE = \angle DQP + \angle EQP = \angle ABC + \angle ACB = 180 - \angle BAC = 180 - DAE$, hence $Q$ lies on $(ADE)$. Also, $\angle ARJ = \angle AEJ = \angle ADJ = 90$, so $A,R,E,J,D$ are concyclic in a circle with diameter $AJ$ and hence $A,R,E,J,Q,D$ are all concyclic. Now we have that: $AR \perp FJ => AR \parallel BC$. From the cyclic quadrilaterals and using that $AR \parallel BC$: $\angle RQE = \angle RAE = \angle RAC = \angle ACB = 180 - \angle PCE = \angle PQE$ and $P,Q,R$ are collinear, as needed. Note that the same holds for any point $J$, not just the excenter, where $D,E,F$ are the projections of $J$ in the sides of the triangle.

Just using angle chasing.... $\angle QDB=\angle QPC= 180- \angle CEQ$ so, $DAEQ$ is cyclic. Also,$JDAE$ cyclic $\implies$ $JEADQ$ is cyclic.Also we know $AREJ$ cyclic.So,by written some angles we deduce $RQ \cap BC =P$.

Kudos to Bozhidar Dimitrov for the cute problem! Here's a quick solution: Introduce $Q'$ as the second intersection of $\overline{RP}$ and the cyclic $ARDJE$. Then \[\angle DQP=\angle RAD=\beta=\angle ABP\]hence $BPQD$ is cyclic. Similarly, $CPQE$ is cyclic as well, so $Q=Q'$, as desired.

Notice that $ARDJE$ is cyclic. Define $Q'$ as the second intersection of $PR$ and $(ARDJE)$. Then $$\angle PQ'D=180^{\circ}-\angle BAR=\angle ABC$$$$\angle PQ'E=\angle RAC=\angle ACB$$Thus $BPQ'D$ and $CPQ'E$ are concyclic, so $Q=Q'$, finishing the problem.

Since, $\angle ARJ = \angle JDA= \angle AEJ= 90 A$, $R$, $E$, $J$ and $D$ lie on a circle. Now, $JF \perp$ both $BC$ and $AR \implies BC\parallel AR$ $\implies \angle RAB + \angle ABC = 180= \angle RAB + \angle PQD$ $\implies Q \in (AERJD)$ $\implies \angle FPR = \angle PRA =180-\angle ADQ=\angle BPQ$ $\implies P$, $Q$ and $R$ lie on a single line.

Surprisingly very easy. I think the original submission (showing $PQ$ passes through a fixed point) wouldn't have been too hard for the Problem 2 spot. We first prove the following key claim. Claim : Points $A$ , $D$ , $E$ , $J$ , $R$ and $Q$ lie on the same circle. Proof : It is immediate that $ADJE$ is cyclic, and that $R$ lies on this circle as well, due to the right angles. Further, \[\measuredangle DQE = \measuredangle DQP + \measuredangle PQE = \measuredangle ABC + \measuredangle BCA = \measuredangle BAC = \measuredangle DAE \]so $Q$ also lies on this circle, finishing the proof of the claim. Now, let $P' = \overline{QR} \cap \overline{BC}$. We simply note that since $AR \parallel BC$ (since $JF \perp BC$ quite clearly), we have \[\measuredangle DQP' = \measuredangle DQR = \measuredangle DAR = \measuredangle CBA = \measuredangle P'BA\]so $P'$ lies on $(DBQ)$. Thus, $P'=P$ and it immediately follows that points $P, Q$, and $R$ are collinear, as desired.

Something more general holds: if $E$ and $F$ are fixed points on $AB$ and $AC$ respectively and $D$ varies on $BC$, then the radical axis of $(BDE)$ and $(CDF)$ passes through a fixed point. For simplicity, assume $D, E$ and $F$ lie on the sides of $\triangle ABC$ (the same argument works in general). Let $X$ be the point on $(AEF)$ such that $AX\parallel BC$, which is clearly fixed. We claim that $X$ is the desired point. If $(BDE)$ and $(CDF)$ meet a second time at $K,$ then $K$ lies on $(AEF)$ by Miquel's theorem. The collinearity follows because $$\angle XKF=\angle XAF=\angle ACB=180^{\circ}-\angle FKD.$$ In fact, an even further generalization can be found in post #9 here.

Claim : $A,R,E,J,Q,D$ are concyclic Proof : Clearly, the points $R,E,D$ lie on the circle with $AJ$ as diameter since they subtend $90^{\circ}$ angle at $AJ$. Now, since $BPQD$ and $CPQE$ are cyclic, $\angle DQP = \angle B$ and $\angle EQP = \angle C$ $\implies \angle DQE = 180^{\circ} - \angle A$ Let $P'$ be the intersection of $QR$ with $BC$ Claim : $P = P'$ Proof : $\angle BPQ = 180^{\circ} - \angle BDQ = \angle ARQ = \angle ARP' = \angle CP'R = \angle BP'Q$ This means that $P$ and $P'$ lie on the same line and subtend the same angle at $BQ \implies \angle PQP' = 0 \implies \angle P = P'$. And we are done.

I solved it in 2 minutes it's very easy for jbmo

By Miquel Theorem $Q$ lies on $(ADJE)$. Because it's diameter is $AJ$, $R$ also lies on it. Now $\angle DQP=\angle ABC$ and $\angle DQR = \angle DJR=\angle DJF=\angle BAC$, so $P$, $Q$ and $R$ are collinear

trigadd123 wrote: Something more general holds: if $E$ and $F$ are fixed points on $AB$ and $AC$ respectively and $D$ varies on $BC$, then the radical axis of $(BDE)$ and $(CDF)$ passes through a fixed point. Which, we just now realized, appears at the generalization of ISL 2018 G2 by Daniel Zhu (in post 8).

$JDAER$ is clearly cyclic and moreover $AR\parallel BC$ as they are both perpendicular to $RJ$ Claim: $Q$ lies on $(JDAER)$ Proof: \[\angle ADQ=\angle BDQ=\angle QPC=180^\circ- \angle CEQ=180^\circ- \angle AEQ \]$\implies DAEQ$ cyclic $\implies JDAERQ$ cyclic. $\blacksquare$ From here we can finish with 2 ways $1$. \[ \angle DQP = \angle ABP \overset{AR \parallel BC}{=} 180^\circ -\angle BAR = 180^\circ - \angle DAR = \angle DQR \].$\blacksquare$ $2$. \[\angle EQP = \angle ACB = \angle CAR = \angle EAR =\angle EQR\]$\blacksquare$