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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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== The Actual Problem ==
 
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Let <math>ABC</math> be an acute triangle with <math>AB > AC</math>. Let <math>\gamma</math> be its circumcircle, <math>H</math> its orthocenter, and <math>F</math> the foot of the altitude from <math>A</math>. Let <math>M</math> be the midpoint of <math>BC</math>. Let <math>Q</math> be the point on <math>\gamma</math> such that <math>\angle HQA = 90◦</math> and let <math>K</math> be the point on <math>\gamma</math> such that <math>\angle HKQ = 90◦</math> . Assume that the points <math>A</math>, <math>B</math>, <math>C</math>, <math>K</math> and <math>Q</math> are all different and lie on <math>\gamma</math> in this order. Prove that the circumcircles of triangles <math>KQH</math> and <math>FKM</math> are tangent to each other.
 
  
 
==Solution==
 
==Solution==
  
 
[[File:IMO2015 P3.png|600px|up]]
 
[[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>\triange HFM \sim \triange HQ`A</math> so <math>\angle HQ`A = 90</math>. Hence, <math>Q` = Q</math>. So:
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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:
  
 
<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:
 
<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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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>
 
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
  
 
{{solution}}
 
{{solution}}

Latest revision as of 14:02, 1 June 2024

Let $ABC$ be an acute triangle with $AB>AC$. Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HKQ=90^\circ$. Assume that the points $A$, $B$, $C$, $K$, and $Q$ are all different, and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Solution

up

We know that there is a negative inversion which is at $H$ and swaps the nine-point circle and $\Gamma$. And this maps:

$A \longleftrightarrow F$. Also, let $M \longleftrightarrow Q`$. Of course $\triangle HFM \sim \triangle HQ'A$ so $\angle HQ'A = 90$. Hence, $Q' = Q$. So:

$M \longleftrightarrow Q$. Let $HA$ and $HQ$ intersect with nine-point circle $T$ and $Q$, respectively. Let's define the point $L$ such that $TNML$ is rectangle. We have found $M \longleftrightarrow Q$ and if we do the same thing, we find:

$L \longleftrightarrow K$. Now, we can say:

$(KQH) \longleftrightarrow ML$ and $(FKM) \longleftrightarrow (ALQ)$. İf we manage to show $ML$ and $(ALQ)$ are tangent, the proof ends.

We can easily say $TN || AQ$ and $AQ = 2.TN$ because $T$ and $N$ are the midpoints of $HA$ and $HQ$, respectively.

Because of the rectangle $TNML$, $TN || ML$ and $TN = ML$.

Hence, $ML || AQ$ and $AQ = 2.ML$ so $L$ is on the perpendecular bisector of $AQ$ and that follows $\triangle ALQ$ is isoceles. And we know that $ML || AQ$, so $ML$ is tangent to $(ALQ)$. We are done. $\blacksquare$

~ 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
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