23rd Korean Mathematical Olympiad Final Round

Let $E^{\prime }(\ne A)$ be the intersecting point of line $AH$ and the circumscribed circle of $△ABC$, and $D^{\prime }$ be the intersecting point of line $O{E}^{\prime }$ and side $BC$. We have $$\angle OE{A}^{\prime }=\angle D^{\prime }{E}^{\prime }A=\angle D^{\prime }H{E}^{\prime }$$ since points $H$ and $E^{\prime }$ locate symmetrically about side $BC$. And it also holds that $$\angle OE{A}^{\prime }=\angle OA{E}^{\prime }$$ since line segments $AO$ and $E^{\prime }O$ are radii of the circumscribed circle of $△ABC$. Hence we obtain that $\angle OA{E}^{\prime }=\angle D^{\prime }H{E}^{\prime }$, and thus $AO//H{D}^{\prime }$. By condition $AO//HD$ in the problem we have that $D=D^{\prime }$ and $E=E^{\prime }$, since both points $D$ and $D^{\prime }$ lie on line segment $BC$.

Let $P(\ne D)$ be the intersecting point of line $ID$ and the inscribed circle of $△ABC$, and $Q$ be the intersecting point of line $AP$ and side $BC$. Then by homothety, the inscribed circle touches side $BC$ at $Q$ because that point $A$ is the center of similarity for the inscribed circle and the inscribed circle of $△ABC$. Thus the midpoint $M$ of line segment $BC$ also becomes the midpoint of line segment $QD$. Now let $P^{\prime }$ be the intersecting point of lines $OQ$ and $PD$. Since $QM=MD$ and $OM//P^{\prime }D$, we have $$2OM=P^{\prime }D.$$ Using properties of the Euler line, we have $$2OM=AH.$$ Hence we have $$P^{\prime }D=AH.$$ Since $P^{\prime }D//AH$, rectangle $A{P}^{\prime }DH$ is a parallelogram, and thus $A{P}^{\prime }//HD$. Therefore points $O$ and $Q$ lie on the line $A{P}^{\prime }$. That is, $P=P^{\prime }$. Hence $2OM=P^{\prime }D=PD=2ID$ and $OM=ID$. Therefore, we have $$OI//BC.$$ By the relationship between the circumcenter and the orthocenter of a triangle we have that $\angle BAO=\angle CAE$ and $\angle CAE=90^{\circ }-\angle C$. Thus $$\angle OAH=\angle A-(\angle CAE+\angle BAO)=\angle A-2(90^{\circ }-\angle C).$$ By properties of an inscribed angle in the circumscribed circle of $△ABC$, it holds that $\angle AEC=\angle B$. And $\angle DEA=\angle OAH$ because $\overline{AO}=\overline{OE}$. Thus $$\angle DEC=\angle DEA+\angle AEC=\angle A-2(90^{\circ }-\angle C)+\angle B=\angle C.$$ On the other hand, the point $F$ is the center of the circle that passes through three points $I,D,C$, which means that $FD=FC$. Hence we have $$\angle DFC=180^{\circ }-\angle C.$$ Therefore four points $E,C,F,D$ lie on a circle, and thus $\angle EDC=\angle EFC$. From the fact $OI//BC$ we have that $\angle EDC=\angle EOI$, i.e., $\angle EFC=\angle EOI$. Hence we finally conclude that four points $E,F,I,O$ lie on a circle. $\square$