Cyclic quadrilateral $ABCD$ with circumcenter $O$ is given. Point $P$ is the intersection of diagonals $AC$ and $BD$. Let $M$ and $N$ be the midpoint of the sides $AD$ and $BC$, respectively. Suppose that $\omega_1$, $\omega_2$ and $\omega_3$ be the circumcircle of triangles $ADP$, $BCP$ and $OMN$, respectively. The intersection point of $\omega_1$ and $\omega_3$, which is not on the arc $APD$ of $\omega_1$, is $E$ and the intersection point of $\omega_2$ and $\omega_3$, which is not on the arc $BPC$ of $\omega_2$, is $F$. Prove that $OF=OE$. Proposed by Seyed Amirparsa Hosseini Nayeri
Problem
Source: Iranian TST 2022 problem 4
Tags: geometry
hakN
02.04.2022 23:21
Let $G = AD \cap BC, H = AB \cap CD$ and $X = \omega_1 \cap \omega_2 \neq P$. Since $X$ is the center of the spiral similarity taking $AD$ to $CB$, it takes $M$ to $N$, and hence $\angle XMG = \angle XMA = \angle XNC$, giving that $X \in (GMN)$, but since $O \in (GMN)$, we have $G,O,M,N,X$ are concyclic. Also $\angle NEX = \angle NGX = \angle CGX = \angle CAX = \angle PAX = \angle PEX$, we have $E,P,N$ are collinear and similarly $F,P,M$ are collinear. Now perform an inversion with center $P$ and radius $-\sqrt{PX \cdot PG}$. Then by ISL 2009 G4, $HP$ is tangent to $(PMN)$, and since $(PMN) \to EF$ after inversion, we have $EF \parallel HP$. By Brocard, $HP \perp GO$, so $EF \perp GO$. Since $GO$ is a diameter, we have $GEOF$ is a deltoid, and thus $OE = OF$.
MrOreoJuice
07.04.2022 19:58
P4?
Redefine $E'=\overline{PN} \cap (BND)$ and $F'=\overline{PM} \cap (AMC)$. I claim that $E' \equiv E$ and $F' \equiv F$. Let $G = \overline{AD} \cap \overline{BC} \in (MON)$.
ProofClaim: $E' \in (APD)$.
Proof. Firstly note that $E'P \cdot PN = BP \cdot PD = AP \cdot PC$ so $E'ANC$ is cyclic, thus the result follows from $\measuredangle AE'P = \measuredangle PCN = \measuredangle ADP$.
Claim: $E' \in (GMN)$.
Proof. Firstly note that $\triangle MDP \sim \triangle NCP$ so $\measuredangle APE' = \measuredangle NPC = \measuredangle MPD$ so $PE$ is $P$ symmedian in $\triangle APD$, then we angle chase
\begin{align*}
\measuredangle E'MG &= \measuredangle E'MA \\
&= -(\measuredangle MAE' + \measuredangle AE'M) \\
&= -(\measuredangle DPE' + \measuredangle PE'D) = \measuredangle E'DP = \measuredangle E'NB = \measuredangle E'NG
\end{align*}
So $E' \equiv E$ and by symmetry the result follows for $F'$. Using the symmedian property it is easy to show by angle chasing that $\overline{AD}$ bisects angle $EMP$ and $\overline{BC}$ bisects angle $FNP$ and since $EMNF$ is cyclic it means $\measuredangle EMA = \measuredangle BNF$.
$$\measuredangle EGO = \measuredangle EMO = \measuredangle EMA + 90^\circ = \measuredangle BNF + 90^\circ = \measuredangle ONF = \measuredangle OGF$$so $GO$ bisects angle $EGF$ and the result follows.
Found this fairly hard for a P1/4, I think the key of the problem lies in noticing the angle equalities because of symmedians and in order to be able to utilize it we need some collinearities for which we need a redefinition of $E,F$ and because there are so many circles it is ``kind of'" motivated to introduce intersections with circumcircles and use POP.
Mahdi_Mashayekhi
01.06.2022 17:32
Let $AD,BC$ meet $K$ and Let $\omega_1$ and $\omega_2$ meet at $S$. Claim $: OSMN$ is cyclic. Proof $:$ Note that $\angle OMK = \angle ONK = \angle 90$ so $OMKN$ is cyclic also Note that $ASD$ and $BSC$ are similar so $\angle AMS = \angle SNC$ so $SMKN$ is cyclic. Note that by Radical Axis Theorem we have $K,P,S$ are collinear Claim $: N,P,E$ and $M,P,F$ are collinear. Proof $:$ Note that $\angle ASP = \angle ADP = \angle KCP$ so $ASP$ and $KCP$ are similar so $\angle NES = \angle NKS = \angle NKP = \angle SAP = \angle SEP$ so $N,P,E$ are collinear. we prove the other part with same approach. Note that $\angle EMA = \angle EMF - \angle AMP = \angle ENF - \angle BNP = \angle BNF$. $\angle ONE = \angle 180 - (\angle 90 + \angle EMA) = \angle 180 - (\angle 90 + \angle BNF) = \angle OMF \implies OE = OF$. we're Done.
khina
01.06.2022 18:53
By an inversion about $AD \cap BC$ fixing $(ADP)$ and $(BCP)$, it suffices to prove that the second points which the line through $P$ and $AB \cap CD$ meet $(ADP)$ and $(BCP)$ at are equidistant from $O$. In fact I will prove that for any line through $P$ intersecting $(ADP)$ and $(BCP)$ again at $E$ and $F$, we have that $OE = OF$ (I remember posting this on the OTIS discord once.) The solution is slick: consider the line isogonal to $PE$ wrt $(ADP)$, and say it intersects $(ADP)$ and $(BCP)$ again at $E'$ and $F'$. Then clearly $EO = E'O$, and $FO = F'O$, but we have that $EE'P$ is directly similar to $F'FP$ by similar figures, so this gives us that $EE'FF'$ is cyclic. Hence $EO = FO$ as desired. edit: im pretty sure this solution is turbo wrong
PHSH
22.06.2022 21:15
Let $U = {AD} \cap {BC}$ and $T={AB} \cap {CD} ,$ then let $K = TU \cap PO$ and $S = {PU} \cap {TO}.$ Since the triangle $\triangle UPT$ is self-polar, we have that ${KPO} \perp {TKU}$ and ${TSO} \perp {UPS}.$ Then we have $$ \measuredangle OKU = \measuredangle OSU = \measuredangle OMU = \measuredangle ONU = \frac{\pi}{2}$$It follows that points $K,S$ and $U$ lies on the circumcircle of $OMN.$ Lemma: Quadrilaterals $KAOC$ and $KBOD$ are cyclic and $KP \cdot PO = AP \cdot PC = DP \cdot PB.$
It's well known that $K$ is the inverse of $P$ with respect to $(ABCD).$ Then $\measuredangle AKO = \measuredangle OAP = \measuredangle OAC.$ Similarly we have $\measuredangle OKC = \measuredangle PCO = \measuredangle ACO$ and hence $$\measuredangle AKC = \measuredangle AKO + \measuredangle OKC = \measuredangle OAC + \measuredangle ACO = \measuredangle AOC$$Implying $KAOC$ is cyclic. Equivalently, we have $KBOD$ cyclic and the result follows by power of a point.
[asy][asy] import graph; import geometry; size(300); pair A,B,C,D,P,T,U,O,K,M,N,S; O = (0,0); A= dir(130); B = dir(60); C = dir(315); D = dir(180); P = extension(A,C,B,D); T = extension(A,B,C,D); U = extension(A,D,B,C); K = extension(T,U, P,O); S = extension(T,O,P,U); M = 0.5*A + 0.5*D; N = 0.5*C + 0.5*B; fill(A--B--C--D--cycle, 0.15*cyan + 0.85*white); fill(A--B--U--cycle, 0.1*green + 0.9*white); fill(A--D--T--cycle, 0.1*green + 0.9*white); draw(circumcircle(O,M,N), blue); draw(unitcircle, red); draw(T--O, dotted); draw(S--U, dotted); draw(U--D); draw(U--C); draw(T--B); draw(T--C); draw(T--U,dotted); draw(A--C); draw(B--D); draw(K--O,dotted); dot("$A$", A, dir(135)); dot("$B$", B, dir(55)); dot("$C$", C, dir(300)); dot("$D$", D, dir(205)); dot("$P$", P, dir(180)); dot("$T$", T, dir(180)); dot("$U$", U, dir(90)); dot("$O$", O,dir(-95)); dot("$K$", K, dir(120)); dot("$S$", S, dir(-90)); dot("$M$", M, dir(30)); dot("$N$", N, dir(-20)); [/asy][/asy] By the Lemma, we have $$ UP \cdot PS = KP \cdot PO = AP \cdot PC = DP \cdot PB $$ Hence, the composition of an inversion centered at $P$ with ratio $R=\sqrt{AP \cdot PB}$ with a reflection by $P$ swaps the pairs $(A,C), (B,D), (K,O)$ and $(U,S).$ Then, under this map, the circles $(APD)$ and $(BPC)$ are send to lines $BC$ and $AD,$ respectively, while $(OMN)$ is fixed. It folows that points $E$ and $F$ are send to $N$ and $M,$ respectively. Therefore, segments $OE$ and $OF$ are swapped with $KN$ and $KM$, respectively, and it follows that $$ OE = \frac{R^2}{PK \cdot PN} \times KM, OF = \frac{R^2}{PK \cdot PM} \times KN $$Implying that $OE = OF$ if and only if $\frac{KM}{KN} = \frac{PM}{PN}.$ Indeed, consider the spiral similarity sending $A$ to $B$ and $D$ to $C;$ its center is the point $K$ and then $\frac{KM}{KN} = \frac{PM}{PN}$ because $M$ is sent to $N$ and $\triangle APD \sim \triangle BPC .~ \blacksquare$
VicKmath7
09.10.2023 16:14
Let $(ADP) \cap (BCP)=R$ and let $AD \cap BC=Q$. Firstly, $(OMN)$ has a diameter $OQ$ and due to spiral similarity $R \in (OMN)$. Notice that it is sufficient to show that $QE=QF$, which transforms to $\angle QME=\angle QNF$. We will show that $\triangle EAD \sim \triangle FBC$, which will finish the problem.
Actually, we claim that $EAPD$ and $FBPC$ are harmonic. This will imply the conclusion, since then $\triangle APD \cup \{E\} \sim \triangle BPC \cup \{F\}$. As of why $EAPD$ is harmonic, let $RE \cap AD=T$ and projecting through $R$ yields that we need $(DA;TQ)=-1$, or that $TA \cdot TD=TM \cdot TQ$. But this is trivial as both sides equal $TE \cdot TR$ by PoP, done.
HoRI_DA_GRe8
07.01.2024 18:11
Invert at $O$ fixing $(ABCD)$.The inverted problem becomes; Inverted Iran TST 2022 P4 wrote: In cyclic quadrilateral $(ABCD)$ with $AC \cap BD=P,AD \cap BC=E,AB \cap CD=F$.If $FP \cap (FAD)=K$ and $FP \cap (FBC)=L$,then prove that $OK=OL$ Since $OE \cap FP$ it suffices to prove $EK=EL$.We start with a claim, Claim : Let $I$ be the miquel point of $(ABCD)$ and let $FP \cap AD=T,FP \cap BC=U$.Then $IT,EK,(FAD)$ and $IU,EL,(FBC)$ are concurrent. Proof : Let $EK$ meet $(FAD)$ at $J$ and similarly define $H$.Inversion at $E$ gives $E,J,H,O,I$ concyclic.Now we do a harmonic chase $$(F,IT \cap (FAC) ;A,D)\stackrel{I}=(ET;AD)\stackrel{K}=(J,F;A,D)=(F,J;A,D)$$This gives $IT \cap (FAC)=J$, the other case ($IU \cap (FBC)=H$) holds similarly $\square$ Now we have, $$(E,O;J,H)\stackrel{I}=(F,P;T,U)=-1$$Which gives $EJ=EH$ since $EO$ is diameter of $(EOJH)$ as $\angle EIO=90$. Again by power of point, $$EJ.EK=EI.EF=EH.EL \implies EK=EL \implies OK=OL \text{ }\blacksquare$$░░░░░▐▀█▀▌░░░░▀█▄░░░ ░░░░░▐█▄█▌░░░░░░▀█▄░░ ░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░░ ░░░░▄▄▄██▀▀▀▀░░░░░░░ ░░░█▀▄▄▄█░▀▀░░ ░░░▌░▄▄▄▐▌▀▀▀░░ ▄░▐░░░▄▄░█░▀▀ ░░ ▀█▌░░░▄░▀█▀░▀ ░░ ░░░░░░░▄▄▐▌▄▄░░░ ░░░░░░░▀███▀█░▄░░ ░░░░░░▐▌▀▄▀▄▀▐▄░░ ░░░░░░▐▀░░░░░░▐▌░░ ░░░░░░█░░░░░░░░█░░░░░░ ░░░░░░█░░░░░░░░█░░░░░░ ░░░░░░█░░░░░░░░█░░░░░░
BVKRB-
17.01.2024 09:29
Bruh saw symmedians and inversion written in the thread, but idk they didn't help a lot Also 300th post! Let $MP \cap (OMN)=F'$ and $NP \cap (OMN) = E'$ and $AD \cap BC =X$ Obviously $XO$ is diameter of $(OMN)$. Now by Brocard we know $X$ lies on the polar of $P$, hence $P$ lies on the polar of $X$, but this corresponds to the radical axis of $(OMN)$ and $OX$ which means $P$ has equal power wrt to both these circles. So $$MP \cdot PF' = AP \cdot PC \implies (MAPC) \text{ is cyclic}$$This in-turn gives us $$\angle CF'P = \angle CF'M=\angle MAC=\angle CBP \implies F' \in (PBC)$$which implies $F'=F$ and similarly $E'=E$. Now $$\angle OFE=\angle ONP=\angle OMP=\angle OEF \implies OE=OF$$where the last line follows from $\triangle APD \sim \triangle BPC \ \blacksquare$
sami1618
18.09.2024 20:05
Let $w_1$ meet $w_2$ again at $Q$. Then $Q$ is the spiral center $AD\mapsto BC$ and thus also maps $M\mapsto N$. The angle of this spiral symmetry is $$\measuredangle(QM,QN)=\measuredangle(AD,BC)=\measuredangle(OM,ON)$$So $Q$ lies on $w_3$. Then $$\angle QEN=\angle QMN=\angle QAC=\angle QAP=\angle QEP$$Thus $E$, $N$, and $P$ are collinear. Similarly, $F$, $M$, and $P$ are collinear. To finish $$\angle OEF=\angle OMF=\angle OMP=\angle ONP=\angle ONE=\angle OFE$$
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