In a triangle $ABC$ with $AB=AC$, $\omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $BA$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $OAD$ intersects the line $AC$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $DAEH$ is a parallelogram. Line $EH$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $GB$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $GJ$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle.
Problem
Source: BMO 2018 Shortlist G6
Tags: geometry, circumcircle, parallelogram, cyclic quadrilateral, moving points
04.05.2019 21:27
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31.07.2019 01:54
This problem was proposed by Theoklitos Paragyiou from Cyprus.
06.08.2019 20:01
Here's my solution: Let $K$ be the point on $BC$ such that $OK \parallel AB$, $F$ be the antipode of $B$, and $M$ be a point such that $AMCB$ is an isosceles trapezoid with $AB \parallel CM$. It's easy to show that $H$ lies on $BC$. Also, $$\angle JOA=\angle JEA=\angle BAC=189^{\circ}-\angle AOB \Rightarrow J \in OA$$We claim that $J,G,M$ are collinear. This is easy angle chase, as shown below $$\angle JGA=\angle JEA=\angle BAC=\angle ACM=180^{\circ}-\angle AGM \Rightarrow J \in GM$$This also means that $L$ is simply the antipode of $M$ in $\omega$. Finally note that $G,F,N$ are collinear, and so $$\angle NOA=\angle NGA=\angle FBA=\angle BAO \Rightarrow NO \parallel AB \Rightarrow N \in OK$$Now animate $D$ linearly on line $AB$. Using the claims we have shown till now, we get that $$D \mapsto E \mapsto H \mapsto J \mapsto G \mapsto N$$is a projective map. This means that the degree of all the points $H,G,N$ is one. Thus it suffices to show the result for four positions of $D$ (since $L$ is fixed). When $D=A$, we have $H=G=C$; when $D=B$, we have $H=G=B$. Similarly, when $D$ is such that $J=O$, we get $G=L$; and the fourth position can be taken so that $G$ and $N$ coincide with the point $KO \cap \omega$. Then the result is trivially true for these four positions. Hence, done. $\blacksquare$
15.08.2019 21:13
Noice ON//DB, ON-AC=L, IL//OB-DE=F so LFOE cyclic so LFO=LEO but OJA=OJG and B,O,J collinear so OF perp GK so F center (KGH), also FOD=BGK=OAE=ODF and DOGN isoceles trapezoid hence F is center (NGHK) so done!
11.06.2020 21:06
How can we get that the degree of statement, that one fixed point and $3$ moving points lie on a circle, is a sum of degrees of these $3$ points $+ 1$?
05.11.2020 22:52
Obviously, $\triangle DAO \cong \triangle ECO$. Hence, $B,C,H$ are collinear. There is a circle with center $E$ and pass through $C, H, G$. Let $E'$ be the midpoint of $CH$ and $X = LC \cap NE$. Then $\angle GNE = \angle GAC = \angle GBC$. Hence, $B,G,E',N$ are concyclic and $BN \perp CH$. Let $\angle ABC = \alpha$. We can easily get that $\angle XCH = \angle LGB = 90^{\circ} - \alpha$. Since $\angle XLG = \angle GLC = \angle GNE$, $L,N,G,X$ are concyclic and $\angle XHC = \angle XCH = 90^{\circ} - \alpha$. Therefore, $\angle GHX = \angle CHX - \angle CHG = 90^{\circ} - \alpha - \frac{1}{2} \angle CEG = \angle ACO - \angle OEC = \angle EOC = \angle GNE$. Hence, $L,G,N,X,H$ are concyclic.
24.02.2022 22:26
Claim1 : $OAD$ and $OCE$ are similar. Proof : $OA=OC$, $\angle DAO = \angle DCE$ and $DO = EO$. so we have $B,C,H$ are collinear. Let $BG$ meet $\omega_{1}$ at $K$. $\angle NGK = \angle 90$ so $NE \perp EK$ and also $NE \perp CH$. Note that $CEH$ is isosceles because $CE = DA = HE$ so $NE$ is perpendicular bisector of $CH$. Let $LC$ meet $NE$ at $P$. Claim2 : $GLNP$ is cyclic. Proof : $\angle LGN = \angle 90 - \angle LGB = \angle 90 - \angle LCB = \angle 90 - \angle HCP = \angle NPC = \angle LPN$. Note that $OD = OE$ and $OA = OG$ so $GE = DA = EC = EH$ so $E$ is center of $GCH$. Claim3 : $GNHP$ is cyclic. Proof : $\angle GHP = \angle CHP - \angle CHG = \angle 90 - \angle AGB - \angle CEG/2 = \angle 90 - \angle ACB - \angle OEG = \angle ACO - \angle OEG = \angle COE = \angle DOA = \angle DEA = \angle GNE = \angle GNP$. Now we have $GLNHP$ is cyclic. we're Done.
01.11.2024 14:45
A Great Angle-Chase (would have got the intuition fast with the use of Geogebra though, a ruler and compass suffice ) : Claim: $B, C, H$ are collinear: Proof: Note that: $AD=HE$, $OA=OC$ and $\angle OAD = \angle OCH$ (as $\angle B = \angle C$) which implies $\triangle OAD \cong \triangle OCE$. Notice that: $HE=EC$ which implies $\triangle ABC \sim EHC$ and the claim follows. Let $\angle ABC = \angle ACB = \alpha$, $X = LC \cap NE$, $K = BG \cap (GEA)$. Note that: $NE \perp CH$ and $OE \perp GC$. Note that: $E$ is the center of $\triangle GCH$ as $GE=EC=EH$. Now, we will prove two claims, which will finish the problem: Claim: $GLNX$ is cyclic. Proof: A straight-angle chase gives us: $$\angle LCB = \angle LGB = \angle NGJ = \angle NEJ = 90^\circ - \alpha \implies LCN = \alpha.$$$$\angle LXN = \angle CXN = 90^\circ - \angle XCH = 90^\circ - \angle LCB = \alpha.$$ Claim: $GNHX$ is cyclic. Proof: Again an angle-chase will give us: $$\angle GHX = \angle CHX - \angle CHG = 90^\circ - \alpha - \tfrac{1}{2} \angle CEG $$$$= \angle OCA - \angle OEA = \angle EOC $$$$= \angle AOD = \angle GNE$$where last part comes from $AD=GE$. Hence we conclude.