AoPS - Problem

Source: Bulgarian Autumn Tournament 2023, 10.2

Tags: geometry

VicKmath7

19.11.2023 13:47

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.

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.