Given an isosceles triangle $\triangle ABC$, $AB=AC$. A line passes through $M$, the midpoint of $BC$, and intersects segment $AB$ and ray $CA$ at $D$ and $E$, respectively. Let $F$ be a point of $ME$ such that $EF=DM$, and $K$ be a point on $MD$. Let $\Gamma_1$ be the circle passes through $B,D,K$ and $\Gamma_2$ be the circle passes through $C,E,K$. $\Gamma_1$ and $\Gamma_2$ intersect again at $L \neq K$. Let $\omega_1$ and $\omega_2$ be the circumcircle of $\triangle LDE$ and $\triangle LKM$. Prove that, if $\omega_1$ and $\omega_2$ are symmetric wrt $L$, then $BF$ is perpendicular to $BC$.
Problem
Source: China Additional TST for IMO 2020, P2
Tags: geometry, right angle, equal segments, tangent circles
15.01.2021 06:57
Assuming that "(GIE) and (HDI) are symmetric wrt I" means that (GIE) and (HDI) have the same size, and are tangent at I, this problem is wrong?
27.03.2021 20:55
wording has been corrected, enjoy / start solving
30.05.2021 18:47
bumping it is a pity a China TST geo problem to remain unsolved
05.07.2021 22:07
Let $N$ be the antipode of $A$ wrt. $\omega:=\odot(ABC)$. By angle chasing, $\measuredangle BLC = \measuredangle BLK+\measuredangle KLC = \measuredangle BDK+\measuredangle KEC =\measuredangle(AB,DE)+\measuredangle(DE,AC) = \measuredangle BAC$. So $L \in \omega$. Since $\measuredangle MNL=\measuredangle ANL = \measuredangle ABL = \measuredangle DBL = \measuredangle DKL = \measuredangle MKL$. Hence, $L,K,M,N$ are concyclic. Let $M',N'$ be the reflection of $M,N$ across $L$. We get $D,E,L,M',N'$ are concyclic on $\omega_1$. Let $O,O_1$ be the center of $\omega$ and $\omega_1$. We get $\measuredangle M'O_1L = 2 \measuredangle M'N'L = 2 \measuredangle MNL = \measuredangle AOL$. Hence, $\triangle M'O_1L \sim \triangle LOA$. Let $O_1'$ be a point on $\overrightarrow{MA}$ such that $MO_1'\cdot MA = ML \cdot MM' = ME \cdot MF$. We get $M',L,O_1',A$ are concyclic. So, $\measuredangle LAO = \measuredangle LAO_1' = \measuredangle LM'O_1' = \measuredangle LM'O_1$. So, $O_1',O_1,M'$ are collinear. Moreover, Since $D,E,A,O_1'$ are concyclic and $O_1'A$ is external bisector of $\angle DAE$. So, $O_1',O_1$ lie on the perpendicular bisector of $EF$. Clearly, $M'$ doesn't lie on the perpendicular bisector of $DE$. Therefore, $O_1'=O_1$. We can conclude that $O_1$ lies on $AM$. Lemma: In an isosceles triangle $\triangle ABC$ with $AB=AC$. $D$ lies on $BC$ such that $C$ is midpoint of $BD$. $AD$ intersects $\odot(ABC)$ again at $E$. Then, perpendicular bisector of $CE$ intersects $AD$ again at the reflection of $A$ across $E$. Proof: Let perpendicular bisector of $CE$ intersects $AD$ at $F$. Since $\measuredangle CEF = \measuredangle CBA$. $\triangle CEF \sim \triangle CBA$. Since $-1=(B,D;C,\infty) \overset{A}{=} (B,E;C,A)$. We get $\frac{AE}{EC}=\frac{AB}{BC} = \frac{EF}{EC}$ impiles $AE=EF$. Apply the lemma to $\triangle O_1M'L$. We get $A$ is the midpoint of $O_1O$. Let $F'\ne A$ be a point on $AC$ such $O_1F' = O_1A$. We get $\triangle O_1F'A \equiv \triangle OAB$ and they are isosceles. So $O_1OBF'$ become isosceles trapizoid. $BF' \parallel O_1O \perp BC$. Since $\triangle O_1F'A \sim \triangle O_1ED$ and $\triangle O_1EA \sim \triangle O_1ME$. We get $\triangle O_1EA \cup F' \sim \triangle O_1ME \cap D$. Therefore, $\frac{EF'}{F'A} = \frac{MD}{DE}$ but $\frac{EF}{FM} = \frac{MD}{DE}$. So, $F'F \parallel AM \parallel F'F$. $B,F,F'$ are collinear and perpendicular to $BC$. [asy][asy] import graph; import geometry; size(350); pair A,B,C,D,E,M,N,bruh,bruhh,O,O1,MM,NN,F,FF; A=dir(90); B=dir(200); C=dir(-20); N=dir(-90); O=(0,0); M = B/2+C/2; E = (3+2*sqrt(3))/(5+3*sqrt(3))*A + (2+sqrt(3))/(5+3*sqrt(3))*B; D=extension(A,C,M,E); bruh = D/2+E/2; bruhh = rotate(90, bruh)*E; O1 = extension(bruh,bruhh,A,M); path CC = circle(O1,length(O1-D)); path CCC = circumcircle(A,B,C); pair[] LL = intersectionpoints(CC,CCC); pair L=LL[0]; MM=2*L-M; NN=2*L-N; pair bruhhh =dir(160); FF= extension(bruhhh,B,A,C); F=extension(E,M,bruhhh,B); fill(A--B--C--cycle,0.1*cyan+0.9*white); fill(O1--E--D--cycle,0.8*lightgreen+0.2*white); fill(O1--A--FF--cycle,palegreen); fill(O--A--B--cycle,palegreen); fill(O1--MM--L--cycle,0.8*pink+0.2*white); fill(A--O--L--cycle,0.8*pink+0.2*white); draw(unitcircle); draw(circumcircle(L,D,E),orange); draw(A--B--C--cycle); draw(A--D); draw(M--D); draw(FF--B); draw(N--O1); draw(M--MM); draw(N--NN); draw(MM--O1--L,linewidth(0.2)+red); draw(A--L--O,linewidth(0.2)+red); draw(B--O); draw(FF--O1); draw(D--O1); draw(O1--E); dot("$A$",A,dir(60)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot(E,dir(90)); dot("$E$",D,dir(120)); dot("$M$",M,dir(-45)); dot("$O_1$",O1,dir(0)); dot("$L$",L,dir(-100)); dot("$M'$",MM,dir(190)); dot("$N'$",NN,dir(175)); dot("$F'$",FF,dir(40)); dot("$N$",N,dir(N)); dot("$F$",F,dir(230)); dot("$O$",O,dir(0)); [/asy][/asy]
20.05.2022 16:37
Simple angle chasing yields that points $A,B,L,C$ are concyclic.Let $T$ be the midpoint of arc $BC$.Now we establish our main claim: Suppose that $\odot(LKM)$ and $\odot(LDE)$ are tangent at $L$,we prove that $K\in \odot(T,TB)$ . The given tangency yields $\angle DLM+\angle ELK=180^{\circ}$,therefore $\angle DLM=\angle ABK$,similarly we have $\angle ELM=180^{\circ}-\angle ACK$. We have $$\begin {aligned}\frac{DM}{EM}&=\frac{[DML]}{[EML]}\\&=\frac{DL\sin DLM}{EL\sin ELM}\\&=\frac{\sin LED}{\sin LDE}\frac{\sin ACK}{\sin ABK}\\&=\frac{\sin LCK}{\sin LBK}\frac{\sin ACK}{\sin ABK}\\&=\frac{BK\sin KLC}{CK\sin KLB}\frac{BK\sin CAK}{CK \sin BAK}\\&=\frac{BK^2}{CK^2}\frac{\sin DEA}{\sin EDA}\frac {\sin CAK}{\sin BAK}\\&=\frac{AD}{AE}\frac{BK^2}{CK^2}\frac{\sin CAK}{\sin BAK}\end{aligned}$$ Since $AM$ is the external bisector of $\angle EAD$,$\frac{DM}{EM}=\frac{AD}{AE}$,therefore $\frac{BK^2}{CK^2}=\frac{\sin BAK}{\sin CAK}$,which establishes the claim.(Trivial computations left as an exercise to the reader) The rest is easy.By angle chasing,$\angle DLE=\angle KLE-\angle KLD=180^{\circ}-\angle{ACK}-\angle {ABK}=180^{\circ}-B $. $\angle LTM=\angle LTA=\angle LBA=\angle LBD=\angle LKD$,indicating that $L,K,M,T$ concyclic. Therefore the radius of $\odot (LED),\odot(LKM)$ is equal can be represented as $\frac{ED}{\cos B}=\frac{KT}{\cos KMB}$. Whlie $KT\cos B=TB\cos B=BM,ED=MF$,we have $MF \cos FMB=BM$,and the result follows.
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