Let ω1 and ω2 be two circles, t and t′ two tangents to the circles. We denote T1 and T2 the points of tangency of t with ω1 and ω2 respectively, and T1′ and T2′ the points of tangency with t′. Let M be the midpoint of [T1T2], P1 and P2 the intersections of (MT1′) and (MT2′) with ω1 and ω2 respectively. Show that the quadrilateral P1P2T1′T2′ is cyclic.
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Official solution
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M is on the radical axis of ω1 and ω2 because Pω1(M)=MT12=MT22=Pω2(M). Therefore, MP1⋅MT1′=Pω1(M)=Pω2(M)=MP2⋅MT2′, which concludes by the power of point M.
Source: NuminaMath-1.5,
licensed Apache-2.0.
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