MathLabs

Problem 5

Circles ω\omega and Ω\Omega meet at points AA and BB. Let MM be the midpoint of the arc ABAB of circle ω\omega (MM lies inside Ω\Omega). A chord MPMP of circle ω\omega intersects Ω\Omega at QQ (QQ lies inside ω\omega). Let ℓP\ell_P be the tangent line to ω\omega at PP, and let ℓQ\ell_Q be the tangent line to Ω\Omega at QQ. Prove that the circumcircle of the triangle formed by the lines ℓP\ell_P, ℓQ\ell_Q, and ABAB is tangent to Ω\Omega.
Step 3 of 4: Locate the homothety center
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D=XR∩Ω,XF2=XP2=XA⋅XB=XD⋅XRD=XR\cap\Omega,\quad XF^2=XP^2=XA\cdot XB=XD\cdot XR
Detailed analysis

Let D be the second intersection of XR with Omega. Power of X gives the displayed equality. Thus triangles XDF and XFR are similar, which yields the angle relation showing that D,Y,Q,F are concyclic. The resulting cyclic angles put Y,D,S on one line, so D is the homothety center.