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PrintThe 37th Korean Mathematical Olympiad Final Round
South Korea geometry
Problem
Let be the circumcircle of a triangle . Choose a point on the extension of segment towards . Points and are two distinct points on circle such that lines and are both tangent to circle . Segment intersects side at point . Choose a point on the arc of circle that doesn't contain point . Let the intersection point of line and circle be . Lines and intersect at a point (other than ). Let be the intersection point of line and circle . Prove that the points are collinear.
Solution
Let be the intersection point of line and . For cyclic quadrilateral , is the intersection point of two sides and , and is the polar line of with respect to the circle . Therefore, the intersection point of and should lie on , which means the points are collinear.
Now use Pascal's theorem on hexagon , and we can conclude that three points , , and are collinear, which implies that lies on line . Therefore, is same with , and which implies that are collinear, as desired.
Now use Pascal's theorem on hexagon , and we can conclude that three points , , and are collinear, which implies that lies on line . Therefore, is same with , and which implies that are collinear, as desired.
Techniques
Cyclic quadrilateralsTangentsPolar triangles, harmonic conjugates