Difference between revisions of "2024 AMC 8 Problems/Problem 15"
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Revision as of 12:02, 23 January 2024
Problem
Solution
We prove that circles and centered at (the intersection point and have a common chord. Let be the intersection point of the tangent to the circle at the point and the line is inverse to with respect to the circle centered at with radius Then the pairs of points and and are inverse with respect to , so the points and are collinear. Quadrilaterals containing the pairs of inverse points and and and are inscribed, is antiparallel to with respect to angle (see ). Consider the circles centered at centered at and Denote . Then is cyclic), is cyclic, is antiparallel), is the point of the circle Let the point be the radical center of the circles It has the same power with respect to these circles. The common chords of the pairs of circles where intersect at this point. has power with respect to since is the radical axis of has power with respect to since containing is the radical axis of and Hence has power with respect to Let be the point of intersection Since the circles and are inverse with respect to then lies on and lies on the perpendicular bisector of The power of a point with respect to the circles and are the same, the points and coincide. The centers of the circles and ( and ) are located on the perpendicular bisector , the point is located on the perpendicular bisector and, therefore, the points and lie on a line, that is, the lines and are concurrent.
Let be bisector of the triangle , point lies on The point on the segment satisfies . The point on the segment satisfies Let be the intersection point of the tangent to the circle at the point and the line Let the circle be centered at and has the radius Then the pairs of points and and are inverse with respect to and and are antiparallel with respect to the sides of an angle Let the point is symmetric to with respect to bisector Symmetry of points and implies Similarly, we prove that and are antiparallel with respect to angle and the points in triangles and coincide. Hence, and are antiparallel and is cyclic. Note that and so is tangent to the circle that is, the points and and are inverse with respect to the circle vladimir.shelomovskii@gmail.com, vvsss
(This was on the page of the 2021 IMO problem 3) Yes this person actually solved it. I just copy pasted for people that want to know. -Multpi12)