2019 USAMO Problems/Problem 2
Contents
Problem
Let be a cyclic quadrilateral satisfying . The diagonals of intersect at . Let be a point on side satisfying . Show that line bisects .
Solution
Let . Also, let be the midpoint of . Note that only one point satisfies the given angle condition. With this in mind, construct with the following properties:
(1)
(2)
Claim:
Proof: The conditions imply the similarities and whence as desired.
Claim: is a symmedian in
Proof: We have as desired.
Since is the isogonal conjugate of , . However implies that is the midpoint of from similar triangles, so we are done.
Solution 2
Let be the circle centered at with radius
Let be the circle centered at with radius
We denote and inversion with respect to and respectively. Let be the circle
straight line, therefore straight line, therefore is cyclic points and are coincide.
Denote
Suppose, we move point from to Then decreases monotonically, increases monotonically. So, there is only one point where
is cyclic. is trapezoid.
It is known that the intersection of the diagonals, intersection point of the lines containing the lateral sides of the trapezoid and the midpoints of two parallel sides are collinear.
vladimir.shelomovskii@gmail.com, vvsss
See also
2019 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.