In a plane the circles K1 and K2 with centers I1 and I2, respectively, intersect in two points A and B. Assume that ∠I1AI2 is obtuse. The tangent to K1 in A intersects K2 again in C and the tangent to K2 in A intersects K1 again in D. Let K3 be the circumcircle of the triangle BCD. Let E be the midpoint of that arc CD of K3 that contains B. The lines AC and AD intersect K3 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
06.09.2011 21:08
∠(BA,BC)=∠BAC+∠ACB=∠BAC+∠DAB=∠DAC. Similarly, we have ∠(BD,BA)=∠DAC ⟹ ∠DBC=2∠DAC. Thus, ED=EC and ∠DEC=∠DBC=2∠DAC imply that E is the circumcenter of △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 △ACD, as desired.
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, CE′DE′=ACAD. Since E′ lies on ⊙B′C′D′, so it is the reflection of A on C′D′. So AE′⊥C′D′. So AE⊥KL,since KL is anti-parallel to CD.
07.09.2011 20:19
∠KAL=∠KAB+∠BAL=∠ACB+∠ADB=∠BDL+∠ADB=∠ADL ⟹ ∠I1DL=∠I1AL=90∘. So AD is the polar of L wrt circle (I1) and I1L⊥AD (or AK), I2K⊥AL. Follows E is the orthocentre of △AKL. Hence, AE⊥KL. [Moderator edit: LaTeXed]
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⊥EL,AL⊥EK.
18.01.2022 09:42
E is circumcenter tringle ADC
20.03.2022 15:06
Observe that: ∠DAB=∠ACB=∠BDK=αand∠CAB=∠ADB=∠BCL=βThus LC is tangent to K2 and LD is tangent to K1 meaning that LA=LC and KA=KD. Claim: Point E is the circumcenter of ⊙(ADC) Proof: Just note that DE=EC, since E is the midpoint of arc EC. On the other hand ∠DAC=α+β, whereas ∠LAC=∠DAC=180∘−2α−2β. Thus ∠DEC=2(α+β) implying that E is a circumcenter as desired. To finish the problem observe that points K,E,I1 all lie on the perpendicular bisector of the segment AD and similarly points L,E,I2 all lie on the perpendicular bisector of the segment AC. Therefore E is the orthocenter of △AKL meaning that AE⊥KL as desired.
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⊥EL,AL⊥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.