2015 IMO Problems/Problem 3
Let be an acute triangle with
. Let
be its circumcircle,
its orthocenter, and
the foot of the altitude from
. Let
be the midpoint of
. Let
be the point on
such that
. Assume that the points
,
,
,
, and
are all different, and lie on
in this order.
Prove that the circumcircles of triangles and
are tangent to each other.
Solution
We know that there is a negative inversion which is at and swaps the nine-point circle and
. And this maps:
. Also, let
. Of course
so
. Hence,
. So:
. Let
and
intersect with nine-point circle
and
, respectively. Let's define the point
such that
is rectangle. We have found
and if we do the same thing, we find:
. Now, we can say:
and
. İf we manage to show
and
are tangent, the proof ends.
We can easily say and
because
and
are the midpoints of
and
, respectively.
Because of the rectangle ,
and
.
Hence, and
so
is on the perpendecular bisector of
and that follows
is isoceles. And we know that
, so
is tangent to
. We are done.
~ EgeSaribas
Really Important Note: This solution is in the "IMO 2015 Solution Notes" which is written by Evan Chen.
There is the link: https://web.evanchen.cc/exams/IMO-2015-notes.pdf
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See Also
2015 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |