Let $ABC$ be an acute triangle and $\Gamma$ be its circumcircle. Line $\ell$ is tangent to $\Gamma$ at $A$ and let $D$ and $E$ be distinct points on $\ell$ such that $AD = AE$. Suppose that $B$ and $D$ lie on the same side of line $AC$. The circumcircle $\Omega_1$ of $\vartriangle ABD$ meets $AC$ again at $F$. The circumcircle $\Omega_2$ of $\vartriangle ACE$ meets $AB$ again at $G$. The common chord of $\Omega_1$ and $\Omega_2$ meets $\Gamma$ again at $H$. Let $K$ be the reflection of $H$ across line $BC$ and let $L$ be the intersection of $BF$ and $CG$. Prove that $A, K$ and $L$ are collinear.
Problem
Source: 2019 Thailand October Camp TSTST 2.3
Tags: geometry, circumcircle, reflection, collinear
MarkBcc168
15.10.2020 17:44
The solution is divided into three parts.
Part I: The Ratio Bashing
Claim: $FG\parallel BC$
Proof. Notice that $\triangle BDF$ and $\triangle BAC$ are directly similar. Thus
$$\triangle BDA\sim\triangle BFC\implies \frac{FC}{AD} = \frac{BC}{AB}$$Similarly $\tfrac{BG}{AE} = \tfrac{BC}{AC}$. Dividing these two equations yields $\tfrac{CF}{BG} = \tfrac{AC}{AB}$ or $FG\parallel BC$ as desired. $\blacksquare$
By Ceva, we get that$L$ lies on $A$-median of $\triangle ABC$.
Part II : The symmedian
We take a look on the other part of this configuration.
Claim: $AH$ is $A$-symmedian of $\triangle ABC$.
Proof: Construct isosceles trapezoid $ADD_1B$ and $AEE_1C$. Note that $BD_1$ and $CE_1$ are tangents to $\odot(ABC)$ thus $T=BD_1\cap CE_1$ lie on $A$-symmedian of $\triangle ABC$.
However, since $BD_1=AD=AE=CE_1$, we get $BD_1E_1C$ is an isosceles trapezoid too. Thus the power from $T$ to $\Omega_1$ and $\Omega_2$ are equal hence $A,H,T$ are colinear as desired. $\blacksquare$
Part III : The well known lemma
By a well known lemma, the reflection $K$ of $H$ across $BC$ is $A$-HM point of $\triangle ABC$ which clearly lie on the median. Hence we are done.
To avoid citing lemma, we can do the following. Construct parallelogram $ABA_1C$. Thus $K\in\odot(BA_1C)$ therefore
$$\angle KA_1B = \angle KCB = \angle HCB = \angle HAB = \angle A_1AC = \angle AA_1B.$$Hence $K\in AA_1$ as desired.
Krits
13.01.2021 19:46
Let $\Omega_1\cap \Omega_2=\{A,I\}$. I claim that $K,L$ lie on $A-$median of $\triangle ABC$, called $\ell$. Notice that $\triangle BDA \overset{+}{\sim}\triangle BFC$ and $\triangle CEA\overset{+}{\sim}\triangle CGB$ implying $\frac{BG}{CF}=\frac{AB}{AC}$ thus $GF//BC$ and $L$ lies on $\ell$ (Ceva). It is well-known that the reflection of the intersection of $A-$symmedian of $\triangle ABC$ and $(ABC)$ lies on $\ell$ (Can be shown by Power of point). So i'll show that $H$ is that point. Notice that $I$ is center of spiral similarity mapping $BF\to GC$ and $C$ is center of spiral similarity mapping $GI\to BH$. Thus $\triangle IBF\overset{+}{\sim} \triangle IGC\overset{+}{\sim}\triangle HBC$. We deduce that $(H,A;C,B)=-1$ yielding the desired result.
rafaello
12.09.2021 18:22
Same as others, but meh. Firstly note that $\triangle CBF\sim \triangle ABD$ and $\triangle CAE\sim \triangle CBG$. Thus, $\frac{BG}{CF}=\frac{BA}{CA}$, which means that $FG\parallel BC$. Invert at $A$, we get that $\Omega_1\cap \Omega_2$ maps to the point on line $AL$. By Menelaus, $AL$ is the $A$-median, hence $H$ is the point on $(ABC)$ such that $AH$ is the symmedian of $\triangle ABC$. But it is well-known that $K$ lies on $A$-median (as $K$ is actually the $A$-Humpty point of $\triangle ABC$).
estetasks
22.02.2022 22:32
Hello! Sorry for the silly question, but I was wondering if anyone could tell me where can I find this specific contest.(if it is in the contest section)
parmenides51
22.02.2022 23:39
It has been recently added inside Thailand Contests here (last 2 years) older geometry problems may be found with aops links here