Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 8: Line 8:
  
 
==Solution==
 
==Solution==
 +
 +
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>\triange HFM \similar \triange 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>L \longleftrightarrow K</math>. Now, we can say:
 +
 +
<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.
 +
 +
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.
 +
 +
Because of the rectangle <math>TNML</math>, <math>TN || ML</math> and <math>TN = ML</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>\black square</math>
 +
 
{{solution}}
 
{{solution}}
  

Revision as of 13:34, 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

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 $\triange HFM \similar \triange HQ`A$ (Error compiling LaTeX. Unknown error_msg) 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. $\black square$ (Error compiling LaTeX. Unknown error_msg)

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_IMO_Problems/Problem_3&oldid=219943"