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Turkey NMO 2012 Problem 2

Source: Turkey NMO 2012

November 24, 2012
trigonometrygeometrycircumcircleprojective geometrygeometry proposed

Problem Statement

Let ABCABC be a isosceles triangle with AB=ACAB=AC an DD be the foot of perpendicular of AA. PP be an interior point of triangle ADCADC such that m(APB)>90m(APB)>90 and m(PBD)+m(PAD)=m(PCB)m(PBD)+m(PAD)=m(PCB). CPCP and ADAD intersects at QQ, BPBP and ADAD intersects at RR. Let TT be a point on [AB][AB] and SS be a point on [AP[AP and not belongs to [AP][AP] satisfying m(TRB)=m(DQC)m(TRB)=m(DQC) and m(PSR)=2m(PAR)m(PSR)=2m(PAR). Show that RS=RTRS=RT
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