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. | ||
− | == | + | ==Solution== |
+ | |||
+ | [[File:IMO2015 P3.png|600px|up]] | ||
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+ | 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: | |
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+ | <math>M \longleftrightarrow Q</math>. Let <math>HA</math> and <math>HQ</math> intersect with nine-point circle <math>T</math> and <math>Q</math>, respectively. Let's define the point <math>L</math> such that <math>TNML</math> is rectangle. We have found <math>M \longleftrightarrow Q</math> and if we do the same thing, we find: | ||
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+ | <math>L \longleftrightarrow K</math>. Now, we can say: | ||
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+ | <math>(KQH) \longleftrightarrow ML</math> and <math>(FKM) \longleftrightarrow (ALQ)</math>. İf we manage to show <math>ML</math> and <math>(ALQ)</math> are tangent, the proof ends. | ||
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+ | We can easily say <math>TN || AQ</math> and <math>AQ = 2.TN</math> because <math>T</math> and <math>N</math> are the midpoints of <math>HA</math> and <math>HQ</math>, respectively. | ||
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+ | Because of the rectangle <math>TNML</math>, <math>TN || ML</math> and <math>TN = ML</math>. | ||
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+ | Hence, <math>ML || AQ</math> and <math>AQ = 2.ML</math> so <math>L</math> is on the perpendecular bisector of <math>AQ</math> and that follows <math>\triangle ALQ</math> is isoceles. And we know that <math>ML || AQ</math>, so <math>ML</math> is tangent to <math>(ALQ)</math>. We are done. <math>\blacksquare</math> | ||
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+ | ~ EgeSaribas | ||
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+ | 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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{{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
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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 |