VN IMO Booklet

(a) Denote $(I)$ as the incircle of triangle $ABC$. Let $D^{\prime },F^{\prime }$ be the tangent points of $(I)$ with $BC,BA$. $DW$ is the diameter of $(J)$. We have $$\frac{AI}{AJ}=\frac{I{F}^{\prime }}{JF}=\frac{I{D}^{\prime }}{JW}$$ hence $A,D^{\prime },W$ are collinear. Denote $D_1$ as the intersection of $I{D}^{\prime }$ and $AD$. We have $JD=JW$, so $I{D}^{\prime }=I{D}_1$. Moreover, $B{D}^{\prime }=CD$ then $LD=L{D}^{\prime }$, therefore $IL\parallel D{D}_1$.

Let $X$ be the midpoint of $D^{\prime }W$. It is easy to see that $DLXJ$ is a rectangle. Thus $\angle XHD=90^{\circ }$, and $XH\parallel JB$. Then we have $XH\perp BI$.

Construct a rectangle $CDJU$. We have $JU\parallel CD$ and $JU=CD$, hence $JU\parallel B{D}^{\prime }$ and $JU=B{D}^{\prime }$. Then $B{D}^{\prime }UJ$ is a parallelogram. We conclude that $D^{\prime }U\parallel XH$.

Let $Y,V,Z$ be the intersection points of $XH$ with $DJ,UW,UC$, respectively. Notice that $V$ is the midpoint of $UW$, and combined with $UZ\parallel YW$, then $V$ is the midpoint of $YZ$.

The cyclic quadrilateral $CEUJ$ has $CE=CD=UJ$, so $UE\parallel CJ$ and $UE\perp DE$. Moreover, $EW\perp ED$ then $W,U$ and $E$ are collinear.

Triangles $BIC$ and $YVW$ have $$\angle YWV=\angle CED=\angle ICB$$ and $$\angle WYV=\angle XYJ=\angle BJD=\angle IBC$$ then $△BIC\sim △YVW$ (a-a). But $V,L$ are midpoints of segments $YZ,BC$ respectively, therefore $△BIL\sim △YZW$ (s-a-s). Hence $\angle YWZ=\angle BLI=\angle BDA=90^{\circ }-\angle WD{U}^{\prime }$. Let $U^{\prime }$ be the intersection of $WZ$ and $AD$, then $\angle D{U}^{\prime }W=90^{\circ }$. Moreover, we have $\angle D{U}^{\prime }W=\angle D{U}^{\prime }Z=90^{\circ }$. Thus $U^{\prime }$ lies on $(CDH)$ and $(J)$.

Analogously, $(BDK)$ passes through $U^{\prime }$. This completes the proof of this part.

(b) Let $S^{\prime },R$ be the second intersection points of $AI,SI$ with $(O)$. We have $S^{\prime }{I}^2=S^{\prime }L\cdot S^{\prime }S$ then $\angle S^{\prime }IL=\angle S^{\prime }SI=\angle SAR$. But we have $IL\parallel AD$ so $\angle S^{\prime }IL=\angle S^{\prime }AD$. Then we conclude that $\angle S^{\prime }AR=\angle S^{\prime }AD$ and $\angle BAR=\angle CAD$.

Construct $D{H}^{\prime }\perp JG$ ($H^{\prime }\in JG$). It is easy to see that $G{H}^{\prime }\cdot GJ=G{D}^2=GE\cdot GF$ then $EFJ{H}^{\prime }$ is a cyclic quadrilateral. We also have $AEJF$ is cyclic, so $A,E,F,J$ are $H^{\prime }$ concyclic. Therefore $\angle A{H}^{\prime }J=90^{\circ }$ and $A,D,H^{\prime }$ are collinear.

On the other hand, we have $\angle PAJ=\angle QAJ=90^{\circ }+\frac{1}{2}\angle BAC=\angle BIC$ and $\angle IBC=\angle IJC$, $\angle ICB=\angle IJB$ then $△IBC\sim △APJ\sim △AJQ$ (a-a).

Denote $P^{\prime },Q^{\prime }$ as midpoints of $JP,JQ$ then $△IBL\sim △AP{P}^{\prime }\sim △AJ{Q}^{\prime }$ and $△ICL\sim △AQ{Q}^{\prime }$. Therefore $A,P^{\prime },J,Q^{\prime }$ are concyclic. Let $L^{\prime }$ be the intersection of $LI$ and $PQ$. With $\angle BLI=\angle JA{Q}^{\prime }=\angle J{P}^{\prime }{Q}^{\prime }=\angle BPQ$, we have $P,B,I,L^{\prime }$ are concyclic. We also have $\angle PBI=90^{\circ }$ therefore $IL\perp PQ$. Moreover, $IL\parallel AD$ and $AD\perp MN$ then $MN\parallel PQ$.

Consider $A^{\prime }\in AT$ such that $M{A}^{\prime }\parallel AT$. Notice that $$\frac{T{A}^{\prime }}{TA}=\frac{TM}{TP}=\frac{TN}{TQ},$$ so $N{A}^{\prime }\parallel AQ$, then $\angle AM{A}^{\prime }=\angle AN{A}^{\prime }=90^{\circ }$.

Thus $\angle BAR=\angle BAT=90^{\circ }-\angle A{A}^{\prime }M=90^{\circ }-\angle ANM=\angle CAD$. Then $A,R,T$ are collinear, the proof is complete.