VN IMO Booklet

(a) Let $I$ be the reflection of $O$ with respect to $P$. Since $O^{\prime }$ is the midpoint of $OH$, it follows that $O^{\prime }P\parallel IH$. Moreover, we have $PO=P{O}^{\prime }$, thus $IO=IH$. Let $S,G$ be midpoints of $AI$ and $AH$, respectively. We have $$SP=\frac{1}{2}AO=\frac{1}{2}R=O^{\prime }D$$ and $SP\parallel AO\parallel O^{\prime }D$, thus $O^{\prime }SPD$ is a parallelogram. It follows that $DP\parallel O^{\prime }S$. Furthermore, we have $SP=\frac{1}{2}R=O^{\prime }{D}^{\prime }$, therefore $S{D}^{\prime }P{O}^{\prime }$ is an isosceles trapezoid which leads to $O^{\prime }P=S{D}^{\prime }$ and $$IH=2O^{\prime }P=2S{D}^{\prime }=I{A}^{\prime }.$$ Thus $I{A}^{\prime }=IH=IO$ which implies $A^{\prime }$ lies on the circle $(I,IO)$. Similarly, $B^{\prime }$ and $C^{\prime }$ also lie on $(I,IO)$. This leads to the conclusion of (a).



(b) Let $R$ be the radius of the circle $(O)$. It is obvious that $GD=R$. Consider the homothetic transformation with center $A$ and ratio $\frac{1}{2}$ which sends $B,C,A^{\prime },X,H,M$ and the perpendicular bisector of $BC$ to $F,E,D^{\prime },U,G,M^{\prime }$ and the perpendicular bisector $EF$, respectively. Then $\frac{MB}{MC}=\frac{M^{\prime }F}{M^{\prime }E}$ and $U$ is the reflection of $D^{\prime }$ with respect to $EF$. Thus $$\frac{MB}{MC}=\frac{M^{\prime }F}{M^{\prime }E}=\frac{GF}{GE}\cdot \frac{UF}{UE}=\frac{\sqrt{R^2-D{F}^2}}{\sqrt{R^2-D{E}^2}}\cdot \frac{D^{\prime }E}{D^{\prime }F}.$$ Similarly, we can calculate $\frac{NC}{NA}$ and $\frac{KA}{KB}$.



Since $D{D}^{\prime }$, $E{E}^{\prime }$, $F{F}^{\prime }$ are concurrent, it follows $$\frac{D^{\prime }F}{D^{\prime }E}\cdot \frac{F^{\prime }E}{F^{\prime }D}\cdot \frac{E^{\prime }D}{E^{\prime }F}=1.$$ Thus $\frac{MB}{MC}\cdot \frac{NC}{NA}\cdot \frac{KA}{KB}=1$, which implies $M,N,K$ are collinear. This is the desired result.