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
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 $OE = OF$.
07.04.2022 19:58
P4?
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.
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
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 $KP \cdot PO = AP \cdot PC = DP \cdot PB.$
[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 $\triangle APD \sim \triangle BPC .~ \blacksquare$
09.10.2023 16:14
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, $$EJ.EK=EI.EF=EH.EL \implies EK=EL \implies OK=OL \text{ }\blacksquare$$░░░░░▐▀█▀▌░░░░▀█▄░░░ ░░░░░▐█▄█▌░░░░░░▀█▄░░ ░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░░ ░░░░▄▄▄██▀▀▀▀░░░░░░░ ░░░█▀▄▄▄█░▀▀░░ ░░░▌░▄▄▄▐▌▀▀▀░░ ▄░▐░░░▄▄░█░▀▀ ░░ ▀█▌░░░▄░▀█▀░▀ ░░ ░░░░░░░▄▄▐▌▄▄░░░ ░░░░░░░▀███▀█░▄░░ ░░░░░░▐▌▀▄▀▄▀▐▄░░ ░░░░░░▐▀░░░░░░▐▌░░ ░░░░░░█░░░░░░░░█░░░░░░ ░░░░░░█░░░░░░░░█░░░░░░ ░░░░░░█░░░░░░░░█░░░░░░
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 $\triangle APD \sim \triangle BPC \ \blacksquare$
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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