2019 USAMO Problems/Problem 2
Problem
Let be a cyclic quadrilateral satisfying
. The diagonals of
intersect at
. Let
be a point on side
satisfying
. Show that line
bisects
.
Solution
Realize that there is only one point on
satisfying the conditions, because
decreases and
increases as
moves from
to
. Therefore, if we prove that there is a single point
that lies on
such that
, and that
bisects
, it must coincide with the point from the problem, so we will be done.
Since , there is some
on
such that
Thus,
and
. Thus we have that
and
, meaning that
.
We next must show that bisects
. Define
as the intersection of
and
and
as the intersection of
and
. We know that
and
are cyclic, because
where
represents an angle which is measured
. Furthermore, quadrilateral
is also cyclic, because
and
, and these are equal.
As the quadrilaterals are cyclic, we have that , meaning that
. Thus
is a trapezoid whose legs intersect at
and whose diagonals intersect at
. Therefore, line
bisects the bases
and
, as wished. ~ciceronii
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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 |