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Difference between revisions of "2015 IMO Problems/Problem 3"

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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>
  
 +
~ EgeSaribas
 +
 +
<math>Really İmportant Note:</math> 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
 
{{solution}}
 
{{solution}}
  

Revision as of 13:46, 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.

The Actual Problem

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 $\angle HQA = 90◦$ and let $K$ be the point on $\gamma$ such that $\angle HKQ = 90◦$ . 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

$Really İmportant Note:$ (Error compiling LaTeX. Unknown error_msg) 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
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