Given is an acute triangle $ABC$ with circumcenter $O$. The point $P$ on $BC$ such that $BP<\frac{BC} {2}$ and the point $Q$ is on $BC$, such that $CQ=BP$. The line $AO$ meets $BC$ at $D$ and $N$ is the midpoint of $AP$. The circumcircle of $(ODQ)$ meets $(BOC)$ at $E$. The lines $NO, OE$ meet $BC$ at $K, F$. Show that $AOKF$ is cyclic.
Problem
Source: Bulgarian Autumn Tournament 2023, 10.2
Tags: geometry
Helixglich
19.11.2023 14:02
Almost no one got this lol. Let $AB$ intersect $(ABC)$ again at $G$ By menelaus on $\triangle{APD}$ $\overline{N-O-K}$ we get $\frac{AN}{NP}\frac{PK}{KD}\frac{PO}{OA} = 1$ and thus $KD = \frac{OD\cdot PD}{DG}$ Let $FC=x$ by pop we have $FC\cdot FB = FQ\cdot FB$ or $x(x+BC)=(x+DC)(x+CQ)$ solving we get $x = \frac{DC\cdot CQ}{PD}$ and thus $FD = FC+CD = \frac{DC\cdot DB}{PD}$ Finally by pop on $D$ we get that $AOKF$ is cyclic iff $DO\cdot DA = DK\cdot DF$ or $DO\cdot DA = \frac{DO\cdot PD}{DG}\frac{DC\cdot DB}{PD}$ so its left to verify that $DA\cdot DG = DC \cdot DB$ which is pop.
VicKmath7
19.11.2023 18:31
Here is a completely synthetic solution, which relies on the idea to focus on triangle $BRC$, where $AO \cap (BOC)=R$ (this is also the main idea in IGO 2022 A3, and in fact, when $P$ is the foot of the altitude from $A$, then $E$ is the $R$-mixtilinear touchpoint for $\triangle BRC$).
Let $A'$ be the antipode of $A$ and let $AO \cap (BOC)=R$. Since $NO \parallel PA'$, by Reim's theorem we need $APA'F$ being cyclic, or $DP \cdot DF=DA \cdot DA'=DB \cdot DC=DO \cdot DR$, so we need $OPRF$ being cyclic, or that $\angle OFP=\angle ORP$. By shooting lemma, $RDEF$ is cyclic, so $\angle OFP=\angle ERD$, so we need $AR$ being the angle bisector of $\angle PRE$. Since $AR$ bisects $\angle BRC$, it is sufficient to show that $\angle BRP=\angle CRE$.
To use that $EQDO$ is cyclic, let $EQ \cap (BOC)=S$; Reim's implies $RS \parallel BC$, which together with the length condition gives that $RSQP$ is an isosceles trapezoid. Hence, $\angle CRE=\angle CSQ=\angle BRP$, which finishes the problem.
$\textbf{Remark.}$ One can show that $AE$ and $NO$ still meet on $(BOC)$ even when $P$ is not the foot of the perpendicular from $A$ - just invert about $(ABC)$ and if $NO \cap (BOC)=T$, then $(AOKF)$ goes to the line $A, E, T$! So, the current problem is the main component of the proof for the generalization of the IGO problem.
Assassino9931
20.11.2023 00:11
Along with @Helixglich's approach, here is the other pedagogically best solution to this problem, found by a few contestants. To employ the midpoint $N$ in midsegments, let $AO$ intersect the circumcircle of $ABC$ at $T$, so that $NO$ is a midsegment in $APT$. Then $PK \parallel OT$, so $\angle ATP = \angle AON = \angle DOK$ and to show this equals $\angle AFD$. it suffices to show that $APTF$ is cyclic. By Power of a Point, we equivalently want $PD \cdot DF = AD \cdot DT$, but also from the circumcircle of $ABC$ we have $AD \cdot DT = BD \cdot DC$, so it suffices to show $PD \cdot FD = BD \cdot DC$. On the other hand, $FD = \frac{FE \cdot FO}{FQ}$ by Power of a Point on the cyclic $ODQE$ and $FE \cdot FO = FC \cdot FB$ from the circumcircle of $BOC$, so we obtain the relation $FD \cdot FQ = FB \cdot FC$. Now denote $BP = CQ = x$, $FC = y$, $PD = z$, $DQ = t$. Then we are given $(x+y+t)(x+y) = (2x+y+z+t)y$ (equivalent to $yz = x(x+t)$ after opening the brackets) and wish to show $z(x+y+t) = (x+z)(x+t)$ (also equivalent to $yz = x(x+t)$), done.