2012 USAJMO Problems/Problem 1
Problem
Given a triangle
, let
and
be points on segments
and
, respectively, such that
. Let
and
be distinct points on segment
such that
lies between
and
,
, and
. Prove that
,
,
,
are concyclic (in other words, these four points lie on a circle).
Solution
Since
, the circumcircle of triangle
is tangent to
at
. Similarly, since
, the circumcircle of triangle
is tangent to
at
.
For the sake of contradiction, suppose that the circumcircles of triangles
and
are not the same circle. Since
,
lies on the radical axis of both circles. However, both circles pass through
and
, so the radical axis of both circles is
. Hence,
lies on
, which is a contradiction.
Therefore, the two circumcircles are the same circle. In other words,
,
,
, and
all lie on the same circle.
See also
| 2012 USAJMO (Problems • Resources) | ||
| First Problem | Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAJMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.