In triangle $\triangle ABC$ points $M,N$ lie on $BC$ st : $\angle BAM= \angle MAN= \angle NAC$ . Points $P,Q$ are on the angle bisector of $BAC$, on the same side of $BC$ as A , st : $$\frac{1}{3} \angle BAC = \frac{1}{2} \angle BPC = \angle BQC$$Let $E = AM \cap CQ$ and $F = AN \cap BQ$ . Prove that the common tangents to $(EPF), (EQF)$ and the circumcircle of $\triangle ABC$ , are concurrent.
Problem
Source: Iran MO 2023 3rd round , geometry exam P3
Tags: geometry, angle bisector, circumcircle
DottedCaculator
23.08.2023 22:45
$\angle NAC=\frac13A=\angle FQC$ implies $QFAC$ cyclic. Similarly, $QEAB$ cyclic, so $\angle BEC=180^{\circ}-\angle BEQ=\frac12A$. Therefore, $\angle BEC=\angle BFC=\frac A2$, so the circumcenter of $BCEF$ is the arc midpoint of $BAC$, $X$. By radical axis, $AQ$, $BE$, $CF$ concur. Now, let $P'=BE\cap CF$. Then, $ABP'F$ cyclic since $\angle P'BA=\angle CQA=\angle P'FA$. Therefore, $\angle BP'A+\angle AP'C=\angle BFA+\angle CEA=\angle BQC+\angle EAF=\frac23A=\angle BPC$, so $P=P'$. Now, the problem follows from inverting about $(BCEF)$.
v4913
23.08.2023 22:51
Claim 1: $\triangle{ABP} \sim \triangle{AQC}, \triangle{APC} \sim \triangle{ABQ}$. Proof: Let $Q^*$ be on line $AP$ such that $AP \cdot AQ^* = AB \cdot AC$; then we will show that $Q^* = Q$. Note that when inverting about the circle centered at $A$ with radius $\sqrt{AB \cdot AC}$, $B$ and $C$ swap, as do $P$ and $Q^*$, so $\angle{AQ^*B} + \angle{AQ^*C} = \angle{ACP} + \angle{ABP} = \theta$, if $\theta = \frac{1}{3} \cdot \angle{BAC}$. Thus, $Q^* = Q$. Claim 2: $FPAB, EPAC$ are cyclic. Proof: Let $C' = CP \cap QB$; then it we claim that $\angle{CC'Q} = \angle{PAB}$, which is true if and only if $\angle{BQC} + \angle{PCQ} = \frac{3}{2} \theta$, so it suffices to show that $\angle{PBQ} = \angle{PCQ}$, which by Law of Sines on $\triangle{QBP}, \triangle{QCP}$ is equivalent to $\frac{PC}{PB} = \frac{\sin \angle{AQC}}{\sin \angle{AQB}}$, but this is obvious since by Law of Sines on $\triangle{APB}$ and $\triangle{APC}$, $\frac{PC}{PB} = \frac{\sin \angle{ABP}}{\sin \angle{ACP}}$. So, $\angle{PBQ} = \angle{PCQ} \implies \angle{CC'Q} = \angle{BAP} \implies QC'AC$ is cyclic. Now note that since $\angle{C'PB} = 180^{\circ} - 2\theta = \angle{C'AB}$, $C' = F$ so $FPAB, EPAC$ are cyclic. Since $\angle{FQE} = \angle{FAE} = \theta$ and $\angle{FPE} = 2\theta$, if $O_P, O_Q$ are the centers of $(EFP), (EFQ)$ respectively then the exsimilicenter $X$ is the point such that $XE = XF$ and $XE$ is an external bisector of $\angle{O_PEO_Q}$, so by angle chasing $\angle{EXF} = \theta \implies X$ is the midpoint of $\widehat{EAF}$ of $(AEF)$. It suffices to show that there is a spiral similarity at $X$ taking $EU$ to $FV$ if $U, V = AM, AN \cap (ABC)$, which is equivalent to $EU = FV$. By Law of Sines on $\triangle{FCV}, \triangle{EBU}$, $\frac{FV}{CV} = \frac{\sin \angle{PCV}}{\sin \angle{ABP}}$, and similarly $\frac{EU}{BU} = \frac{\sin \angle{PBU}}{\sin \angle{ACP}}$. Since $\frac{\sin \angle{ABP}}{\sin \angle{ACP}} = \frac{PC}{PB}$, it suffices to show that $\frac{\sin \angle{PCV}}{\sin \angle{PBU}} = \frac{PC}{PB}$, which is obvious since $BU \cap CV$ is the intersection of the tangents to $(PBC)$ at $B$ and $C$. $\square$
VicKmath7
24.08.2023 00:03
Redefine $P=BE \cap CF$; by the isogonal line lemma for the pairs $(AF, AE)$ and $(AB, AC)$, we obtain that $AP$ and $AQ$ are isogonal in $\angle BAC$, meaning that $P \in AQ$. We have $\angle BAM=\angle CAN=\angle BQC$, so $BAEQ$ and $CAFQ$ are cyclic, which implies that $BFEC$ is cyclic by PoP. We have $\angle FAP=\angle FCE=\angle FBP$, so $BAPF$ is cyclic and thus $\angle BPC =\angle BAN$ and the new definition of $P$ is correct.
If $D$ is the exsimilicenter, then it's known (e.g by 2017 G7, ARO 2023 10.7, etc) that $\angle EDF=\angle EPF-\angle EQF=\angle EAF$, so $D\in (AEF)$. Actually, this means that $AD$ is the external angle bisector of $\angle EAF$. If $PQ \cap EF=L$ and $BC \cap EF=R$, then by Ceva-Menalaus $(EF;LR)=-1$, which implies that $D, A, R$ are collinear. So, $RB \cdot RC=RF \cdot RE=RA \cdot RD$, which yields that $D \in (BAC)$, as desired.
sami1618
29.06.2024 18:34
Nice Geometry : Claim: $CAQF$ and $BAQE$ are concyclic $$\angle CAF=180^{\circ}-\angle CAN=180^{\circ}-\angle CQB=\angle CQF$$$$\angle BAE=180^{\circ}-\angle BAM=180^{\circ}-\angle BQC=\angle BQE$$Claim: $X$, the midpoint of arc $\overarc{BAC}$, is the circumcenter of $BCEF$ $$\angle CFB=\angle CFQ=180^{\circ}-\angle CAQ=\frac{1}{2}\angle CAB=\frac{1}{2}\angle CXB$$$$\angle BEC=\angle BEQ=180^{\circ}-\angle BAQ=\frac{1}{2}\angle BAC=\frac{1}{2}\angle BXC$$Claim: The inversion about $X$ with radius $XE$ maps $(EPF)$ to $(EQF)$ Notice that $2\angle FXE=2\angle EPF=\angle FQE$ so the arc midpoints of both circles maps to each other. This claim finishes the problem.
Attachments:
