IMO 2011-G5

[Tahmid Hasan]

Let $ABC$ be a triangle with incenter $I$ and circumcircle $\omega$.Let $D$ and $E$ be the second
intersection points of $\omega$ with the lines $AI$ and $BI$,respectively.The chord $DE$ meets $AC$ at a
point $F$,and $BC$ at a point $G$.Let $P$ be the intersection point of the line through $F$ parallel to
$AD$ and the line through $G$ parallel to $BE$.Suppose that the tangents to $\omega$ at $A$ and at $B$ meet
at a point $K$.Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.

[Tahmid Hasan]

Oops,double quote?p.Moderators,delete this post please.

[Tahmid Hasan]

My solution:

Observation $\Rightarrow \angle IAF=\angle IEF=\frac {1}{2}\angle A$,so $AIFE$ is cyclic.Let the circle be $\alpha$.
$\angle IBG=\angle IDG=\frac {1}{2}\angle B$,so $BIGD$ is cyclic.Let the circle be $\beta$.
Let $Q=\alpha \cap \beta \neq I$.
Claim1 $\Rightarrow AKBQ$ is cyclic.
Proof:$\angle AKB+\angle AQB=180^{\circ}-2\angle C+\angle AQI+\angle BQI$ $=180^{\circ}-2\angle C+2\angle C=180^{\circ}$
Claim2 $\Rightarrow K,I,Q$ are collinear.
Proof:$\angle IQB=\angle IDB=\angle ADB=\angle KAB=\angle KQB$ implying $K,I,Q$ are collinear.
Claim3 $\Rightarrow QGPF$ is cyclic and $Q,P,I$ are collinear.
Let $P'=IQ \cap \odot QGF \neq Q$.
now $\angle QP'G=\angle QFG=\angle QIE \rightarrow BE \parallel P'G$
similarly $AD \parallel P'F$ implying $P=P'$.
So we get $K,P \in IQ$ i.e. the radical axis of $\alpha,\beta$.
Now the radical axis of $\omega,\alpha$ is $AE$,the radical axis of $\omega,\beta$ is $BD$ and the radical axis of $\alpha,beta$ is $KP$
concur at the radical centre($O'$) of the three circles.
If $\alpha,\beta,\omega$ are co-axial then $O'=\infty$ meaning $AE \parallel BD \parallel KP$.