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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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 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 |