In a plane the circles $\mathcal K_1$ and $\mathcal K_2$ with centers $I_1$ and $I_2$, respectively, intersect in two points $A$ and $B$. Assume that $\angle I_1AI_2$ is obtuse. The tangent to $\mathcal K_1$ in $A$ intersects $\mathcal K_2$ again in $C$ and the tangent to $\mathcal K_2$ in $A$ intersects $\mathcal K_1$ again in $D$. Let $\mathcal K_3$ be the circumcircle of the triangle $BCD$. Let $E$ be the midpoint of that arc $CD$ of $\mathcal K_3$ that contains $B$. The lines $AC$ and $AD$ intersect $\mathcal K_3$ again in $K$ and $L$, respectively. Prove that the line $AE$ is perpendicular to $KL$.
Problem
Source: Middle European Mathematical Olympiad 2011 - Individuals I-3
Tags: geometry, circumcircle, geometric transformation, reflection
Luis González
06.09.2011 21:08
$\angle (BA,BC)=\angle BAC+\angle ACB=\angle BAC+\angle DAB=\angle DAC.$ Similarly, we have $\angle (BD,BA)=\angle DAC$ $\Longrightarrow$ $\angle DBC=2\angle DAC.$ Thus, $ED=EC$ and $\angle DEC=\angle DBC=2\angle DAC$ imply that $E$ is the circumcenter of $\triangle ACD.$ Since $KL$ is antiparallel to $CD$ with respect to the lines $AC,AD,$ then it follows that $KL$ is perpendicular to the A-circumdiameter $AE$ of $\triangle ACD,$ as desired.
RSM
06.09.2011 21:42
Under inversion wrt $ A $ suppose $ B,C,D $ goes to $ B',C',D' $. $ AC'B'D' $ is a parallelogram. Suppose, $ E $ goes to $ E' $. Clearly, $ \frac {CE'}{DE'}=\frac {AC}{AD} $. Since $ E' $ lies on $ \odot B'C'D' $, so it is the reflection of $ A $ on $ C'D' $. So $ AE'\perp C'D' $. So $ AE\perp KL $,since $ KL $ is anti-parallel to $ CD $.
vprasad_nalluri
07.09.2011 20:19
$\angle KAL = \angle KAB + \angle BAL = \angle ACB + \angle ADB = \angle BDL + \angle ADB = \angle ADL$ $\Longrightarrow$ $\angle I_1 D L = \angle I_1 A L = 90^{\circ}.$ So $AD$ is the polar of $L$ wrt circle $(I_1)$ and $I_1L \perp AD$ (or AK), $I_2 K \perp AL.$ Follows $E$ is the orthocentre of $\triangle AKL.$ Hence, $AE \perp KL.$ [Moderator edit: LaTeXed]
littletush
27.11.2011 07:51
it suffices to prove that A is the orthocenter of triangle EKL and by angle substitutions it's easy to obtain $AK\bot EL,AL\bot EK$.
lazizbek42
18.01.2022 09:42
$E$ is circumcenter tringle $ADC$
Kimchiks926
20.03.2022 15:06
Observe that: $$ \angle DAB = \angle ACB = \angle BDK = \alpha \quad \text{and} \quad \angle CAB = \angle ADB = \angle BCL =\beta $$Thus $LC$ is tangent to $\mathcal K_2$ and $LD$ is tangent to $\mathcal K_1$ meaning that $LA=LC$ and $KA = KD$. Claim: Point $E$ is the circumcenter of $\odot(ADC)$ Proof: Just note that $DE=EC$, since $E$ is the midpoint of arc $EC$. On the other hand $\angle DAC = \alpha+\beta$, whereas $\angle LAC = \angle DAC = 180^{\circ} -2 \alpha -2\beta$. Thus $\angle DEC = 2(\alpha+\beta)$ implying that $E$ is a circumcenter as desired. To finish the problem observe that points $K, E, I_1$ all lie on the perpendicular bisector of the segment $AD$ and similarly points $L,E,I_2$ all lie on the perpendicular bisector of the segment $AC$. Therefore $E$ is the orthocenter of $\triangle AKL$ meaning that $AE \perp KL$ as desired.
Assassino9931
03.05.2024 13:57
littletush wrote: it suffices to prove that A is the orthocenter of triangle EKL and by angle substitutions it's easy to obtain $AK\bot EL,AL\bot EK$. Indeed, I am fairly sure that one does not have to go through justifying that $E$ is the circumcenter of $ACD$, only pure angle chasing suffices to obtain the orthocenter.