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== | ||
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+ | [[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: | ||
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+ | <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. | ||
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+ | There is the link: https://web.evanchen.cc/exams/IMO-2015-notes.pdf | ||
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+ | {{solution}} | ||
==See Also== | ==See Also== |
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 |