2013 APMO Problems/Problem 5
Let be a quadrilateral inscribed in a circle , and let be a point on the extension of such that and are tangent to . The tangent at intersects at and the line at . Let be the second point of intersection between and . Prove that , , are collinear.
Let . Note that the tangents at and concur on at , so is harmonic, hence the tangents at and concur on at , say.
Now apply Pascal's Theorem to hexagon to find that , and are collinear. Now note that and both lie on the tangent at , hence also lies on the tangent at . It follows that . So and are in fact the same point. Since lies on by definition, it follows that , are indeed collinear, and thus the problem is solved.
We use complex numbers. Let be the unit circle, and let the lowercase letter of a point be its complex coordinate.
Since lies on the intersection of the tangents to at and , we have . In addition, lies on chord , so . This implies that , or .
lies on the tangent at , and lies on , so .
lies on chord and on the tangent at . Therefore we have and . Solving for yields are collinear, so we have , or We must prove that are collinear, or that or Cross-multiplying, we have which is true.
Set , , and , where . Note that since is harmonic, we have collinear and with But is harmonic; therefore .
Use barycentrics on , in that order. It is easy to derive that and . Clearly, line has equation , and the tangent from has equation , so we get that . Line has equation , so we also get that . Line has equation , and line has equation , so we quickly derive that has coordinates , and it is easy to verify that this lies on the circumcircle , so we're done.