2006 USAMO Problems/Problem 6
Problem
Let be a quadrilateral, and let
and
be points on sides
and
, respectively, such that
. Ray
meets rays
and
at
and
respectively. Prove that the circumcircles of triangles
and
pass through a common point.
Solution
Let the intersection of the circumcircles of and
be
, and let the intersection of the circumcircles of
and
be
.
because
tends both arcs
and
.
because
tends both arcs
and
.
Thus,
by AA similarity, and
is the center of spiral similarity for
and
.
because
tends both arcs
and
.
because
tends both arcs
and
.
Thus,
by AA similarity, and
is the center of spiral similarity for
and
.
From the similarity, we have that . But we are given
, so multiplying the 2 equations together gets us
.
are the supplements of
, which are congruent, so
, and so
by SAS similarity, and so
is also the center of spiral similarity for
and
. Thus,
and
are the same point, which all the circumcircles pass through, and so the statement is true.
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.