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Turkey — Team Selection Test

Geometry Difficulty 8.3 Shortlist Prove it Turkey

The circles ω1\omega_1 and ω2\omega_2 which do not intersect and which have different sizes are tangent to the line ℓ\ell at KK and LL, and are tangent to the circle Γ\Gamma at MM and NN respectively, such that all three circles lie on the same side of ℓ\ell. A circle which passes through KK and LL intersects Γ\Gamma at AA and BB. The reflections of MM and NN over ℓ\ell are RR and SS respectively. Show that the points A,B,R,SA, B, R, S are concyclic.

Solution

Denote the tangent lines to Γ\Gamma at MM and NN by mm and nn, and let
α:=∠LKM=∠(KM,m),β:=∠(m,MN)=∠(MN,n),γ:=∠(n,NL)=∠NLK. \alpha := \angle LKM = \angle (KM, m), \\ \beta := \angle (m, MN) = \angle (MN, n), \quad \gamma := \angle (n, NL) = \angle NLK.
The internal angles of KMNLKMNL add up to 360∘360^\circ, thus α+β+γ=180∘\alpha + \beta + \gamma = 180^\circ, hence KMNLKMNL is cyclic. Now the lines AB,KL,MNAB, KL, MN are the pairwise radical axes of the circles KLMN,ABMN,ABKLKLMN, ABMN, ABKL, thus they concur at a point TT. The line RSRS is the reflection of MNMN in KLKL, therefore T∈RST \in RS and then TR⋅TS=TM⋅TN=TA⋅TBTR \cdot TS = TM \cdot TN = TA \cdot TB, hence ABRSABRS is cyclic.

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