2015 IMO Problems/Problem 3

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

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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2015 IMO (Problems) • Resources
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2015 IMO Problems/Problem 3

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