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Print74th NMO Selection Tests for JBMO
Romania geometry
Problem
Let be a triangle with and be its circumcircle. The tangent at to the circle intersects line at , and the line through parallel to meets again the circle at . Line intersects at and the circle for the second time at . On the line , consider the point such that , and are con-cyclic. Let and be the intersection points of line with and , respectively. Prove that is a cyclic quadrilateral. JBMO Shortlist 
Solution
so ANBT is also cyclic.
Using the fact that , it follows that , so line AB is the tangent at A to the circumcircle of AST, (1).
Also, from it results that , so . By power of point F to circle , we infer that . Multiplying the last two equalities, we get .
From (1) it follows that , so S, T, D, G belong to the same circle. Hence, STDG is a cyclic quadrilateral.
Using the fact that , it follows that , so line AB is the tangent at A to the circumcircle of AST, (1).
Also, from it results that , so . By power of point F to circle , we infer that . Multiplying the last two equalities, we get .
From (1) it follows that , so S, T, D, G belong to the same circle. Hence, STDG is a cyclic quadrilateral.
Techniques
Cyclic quadrilateralsTangentsAngle chasing