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 |