Circles ω1 , ω2 are externally tangent at point M and tangent internally with circle ω3 at points K and L respectively. Let A and B be the points that their common tangent at point M of circles ω1 and ω2 intersect with circle ω3. Prove that if ∠KAB=∠LAB then the segment AB is diameter of circle ω3. Theoklitos Paragyiou (Cyprus)
Problem
Source: 2013 JBMO Shortlist G2
Tags: geometry, JBMO
20.04.2020 13:45
Solution?
05.07.2020 11:25
Let all tangents at K,L,M intersect on X. From the construction it is obvious that X,A,B points are collinear and XK=XL. From the given condition KB=BL. Combining this two gives us XKBL is a deltoid and so XB⊥KL. By angle chasing ∠BAL=∠KAB=∠KLB. Let KL∩AB={P}. Then we have △BAL∼△BLP. Since ∠LAB=90, ∠ALB=90 which implies [AB] is diameter.
24.11.2022 09:43
Let T be the intersection of tangents to circles ω1 and ω2 from points K and L respectively. Obviously, the line AB passes through T. Note that ∠BKL=∠BAL, ∠BKT=∠KAB, meaning BK is the angle bisector of ∠TKL. By the same way we get that LB is the angle bisector of ∠KLT. So B is the incenter of triangle TKL, meaning TB is the angle bisector of ∠KTL. However the center O of ω3 lies on the angle bisector of ∠KTL, hence A, O, B are collinear. This is the same as saying that AB is the diameter of ω3
01.07.2024 20:24
By Archimedes lemma: KA/KB=MA/MB=LA/LB, KB=LB => KA=LA, so triangles are equal and <ALB=<AKB=180-<ALB => <ALB=90 (sorry, i don’t know LateX)