75th NMO Selection Tests

The argument hinges on the claim below:

Claim. The lines $AE$ and $BI$ are perpendicular; similarly, $AF$ and $CI$ are perpendicular

Proof. Let $\alpha =\angle BAC$, $\beta =\angle CAB$ and $\gamma =\angle ACB$. Let $DX$ cross the circle $ADC$ again at $D^{\prime }$. Note that $\angle ECX=\angle ACX-\angle ACE=90^{\circ }-\beta /2-\gamma /2=\alpha /2=\angle DAC=\angle D{D}^{\prime }C=\angle X{D}^{\prime }C$. As $\angle CXE=\angle CX{D}^{\prime }$, triangles $XC{D}^{\prime }$ and $XEC$ are similar, so $X{D}^{\prime }\cdot XE=X{C}^2$.

As $XA=XC$, it follows that $X{D}^{\prime }\cdot XE=X{A}^2$, so triangles $X{D}^{\prime }A$ and $XAE$ are similar, so $\angle XAE=\angle A{D}^{\prime }X=\angle A{D}^{\prime }D=\angle ACD=\gamma$. Finally, note that $\angle XAD=\angle XAC-\angle DAC=90^{\circ }-\beta /2-\alpha /2=\gamma /2$, so $\angle IAE=\angle DAE=\angle XAE-\angle XAD=\gamma /2$. As $\angle AIB=90^{\circ }+\gamma /2$, the claim follows.

Let $W$ be the mid-arc point of $CAB$ and let $I^{\prime }$ be the reflection of $I$ across $O$. As $W,X,Y$ are the mid-arc points of $CAB,ABC,BCA$, respectively, their reflections across $O$ are the mid-arc points opposite. These latter form a triangle with orthocentre $I$, so $I^{\prime }$ is the orthocentre of triangle $WXY$. Reflection across $O$ maps lines $XV,WY$ and $WX$ to the perpendicular bisectors of $AI,BI$ and $CI$, respectively, so $XY\perp AI,WY\perp BI$ and $WX\perp CI$. By the claim, $AE\perp IF$ and $AF\perp IE$, so $I$ is the orthocentre of triangle $AEF$ and hence $EF\perp AI$ as well. Triangles $AEF$ and $WXY$ have therefore corresponding parallel sides, so they are homothetic from some point $R$. This homothety maps $I$ to $I^{\prime }$, as they are corresponding orthocentres. Hence the lines $AW,EY,FY$ and $I{I}^{\prime }$ are concurrent at $R$. As $I,O$ and $I^{\prime }$ are collinear and $AW$ is the external bisector of $\angle BAC$, the conclusion follows.