The circle $\omega_1$ with diameter $[AB]$ and the circle $\omega_2$ with center $A$ intersects at points $C$ and $D$. Let $E$ be a point on the circle $\omega_2$, which is outside $\omega_1$ and at the same side as $C$ with respect to the line $AB$. Let the second point of intersection of the line $BE$ with $\omega_2$ be $F$. For a point $K$ on the circle $\omega_1$ which is on the same side as $A$ with respect to the diameter of $\omega_1$ passing through $C$ we have $2\cdot CK \cdot AC = CE \cdot AB$. Let the second point of intersection of the line $KF$ with $\omega_1$ be $L$. Show that the symmetric of the point $D$ with respect to the line $BE$ is on the circumcircle of the triangle $LFC$.
Problem
Source: Turkey National Olympiad Second Round 2013 P1
Tags: geometry, circumcircle, geometric transformation, reflection, symmetry, cyclic quadrilateral, similar triangles
61plus
29.11.2013 08:33
LeT $CD$ intersect $w_1$ at $K'$. It is easy to show that $\triangle EK'C$~$\triangle AOC$ Hence $K'$ satisfy $ 2\cdot CK'\cdot AC = CE\cdot AB \Rightarrow K'=K$. Let $BE$ intersect $w_1$ at $X$, $CX$ intersect $w_2$ at $Y$. reflection of $D$ across $BE$ be $D'$ As $\angle D'XB=\angle DXB=\angle DXC$, $D',C,X$ collinear. $\angle AXB=90\Rightarrow XE=XF$. $\angle XYD=\frac{1}{2}\angle CXD=\angle CAO$, $\angle YXD=\angle CBD=\angle COA\Rightarrow \triangle YXD$~$\triangle COA$. Hence $XC\cdot XD'=XC\cdot XD=XC\cdot XY=XF^2$ $\angle CLF=\angle CDE=\frac{1}{2}\angle CAE=\angle CFE\Rightarrow$ circumcircle of $\triangle LCF$ tangent to $BE$ at $F$. Therefore $LFCD'$ concyclic.
mathuz
29.11.2013 09:09
Let $D'$ - that symmetry point, $M,N$ are midpoint of the segments $DD',CD$. Since $ 2\cdot CK\cdot AC=CE\cdot AB $ we get that the triangles $OCK$ and $ACE$ are similar. So \[ \angle CLF=\angle CBK=\angle CDE. \] We have that $DNMB$ is cyclic and \[ \angle CD'F=\angle CD'D+\angle FD'D=\angle NMD+\angle FDM=angle DBA +\angle FDM=90^\circ -\angle DEC+90^\circ -\angle DCE=\angle CDE. \] Hence $ \angle CLF = \angle CD'F $ and $D'\in (CLF) $.
Vo Duc Dien
05.12.2013 18:34
Let $D'$ be that symmetric point, $O$ and $O'$ be the circumcenters of $ABC$ and $LFC$, respectively. Observe that $CFED$ is a harmonic cyclic quadrilateral and we have the similar triangles $AED$, $OBK$ and $O'FD'$ and thus $CFLD'$ is cyclic.
electrovector
20.02.2021 10:40
Here is a solution by me and Burak Buğra Önder. I think this was a bit hard for P1 so here is a solution with motivations. Firstly let's focus on what we need to prove. While $C$ and $F$ look more stable, $D'$ and $L$ is unrelatable. So it seems easier to prove $\angle CLF = \angle CD'F$. We will first focus on the angle $CLF$. It is easy to transform the given equation. Let $O$ be the midpoint of $[AB]$. Thus $CK \cdot AC=CE \cdot OA \Rightarrow \frac{CE}{CK} = \frac{AC}{OC} = \frac{AE}{OK} \Rightarrow \triangle CAE \sim \triangle COK \Rightarrow \angle CAE= \angle COK \Rightarrow \angle CBK = \angle CFE$ But we wonder the measure of $\angle CLF$ and it is equivelant to $180- \angle CBK= 180- \angle CFE= \angle CEF + \angle FCE$. If we look the other angle we want to compute, it is $\angle CD'F= \angle ECD' + \angle CEF + \angle D'FE$. Now by equaling them we will figure out if $\angle FCD' = \angle D'FE$ or equals to $BE$ is tangent to the circumcircle of $CLF$ at $F$. Actually it doesn't seem to possible to continue from now on using angle chasing after a few tries, so it makes sense to focus on lengths and we want to figure out the tangency. It is definitely reasonable to take the point where $BE$ and $CD'$ intersect at. Let this point be $T$. Drawing diagram shows us it is also on $w_1$. So define point $T'$ as the intersection of $BE$ and $w_1$. Hence we have $\angle D'T'B= \angle DT'B= \angle DCB =\angle CDB= \angle CT'B \Rightarrow$ $C,T',B$ are collinear which says $T'=T$. Let's make an observation about $T$. We see $\angle ATB = 90$ and this says $AT \perp EF \Rightarrow TE=TF$. So we wanted to prove that $TD' \cdot TC =TF^2$ and we know $TF^2=TE \cdot TF$ which equals the power of point $T$ to $w_2$. Therefore we should intersect $TC$ and $w_2$ and let this point be $P$. We have $TE \cdot TF = TP \cdot TC$ and if it equals to $TD' \cdot TC$, we will have $TP=TD'$. Finally we have $\angle ETP= \angle CTB= \angle CDB= \angle DCB = \angle DTB$ which gives us $TP=TD=TD'$ and we are done.
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