Difference between revisions of "2015 IMO Problems/Problem 3"
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Prove that the circumcircles of triangles <math>KQH</math> and <math>FKM</math> are tangent to each other. | Prove that the circumcircles of triangles <math>KQH</math> and <math>FKM</math> are tangent to each other. | ||
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==Solution== | ==Solution== | ||
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[[File:IMO2015 P3.png|600px|up]] | [[File:IMO2015 P3.png|600px|up]] | ||
− | We know that there is a negative inversion which is at <math>H</math> and swaps the nine-point circle and <math>\ | + | We know that there is a negative inversion which is at <math>H</math> and swaps the nine-point circle and <math>\Gamma</math>. And this maps: |
<math>A \longleftrightarrow F</math>. Also, let <math>M \longleftrightarrow Q`</math>. Of course <math>\triangle HFM \sim \triangle HQ'A</math> so <math>\angle HQ'A = 90</math>. Hence, <math>Q' = Q</math>. So: | <math>A \longleftrightarrow F</math>. Also, let <math>M \longleftrightarrow Q`</math>. Of course <math>\triangle HFM \sim \triangle HQ'A</math> so <math>\angle HQ'A = 90</math>. Hence, <math>Q' = Q</math>. So: | ||
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~ EgeSaribas | ~ 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 | There is the link: https://web.evanchen.cc/exams/IMO-2015-notes.pdf | ||
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{{solution}} | {{solution}} | ||
Latest revision as of 14:02, 1 June 2024
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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 |