Vietnamese Mathematical Olympiad

a) Considering the figure shown above, the remaining cases are proved similarly. Let $T$ be the midpoint of $NP$. We have $(I)$ as the circle $A$-mixtilinear tangent to triangle $A{A}_b{A}_c$, so according to Sawayama's lemma, $T$ is the incenter of triangle $A{A}_b{A}_c$. By angle chasing, we conclude that $$△T{A}_{b}P\sim △A_{c}TN.$$ It follows that $$A_{b}P\cdot A_{c}N=TP\cdot TN=\frac{N{P}^2}{4}.$$ By AM-GM inequality, one can get $$A_{b}P+A_{c}N\ge 2\sqrt{A_{b}P\cdot A_{c}N}=NP.$$ Similarly, $$P{B}_a+M{B}_c\ge MP,N{C}_a+M{C}_b\ge MN$$ From these, we conclude the required inequality.



b) Let $X$ be the intersection of $B_a{B}_c$ with $C_a{C}_b$, $Y$ be the intersection of $C_a{C}_b$ with $A_b{A}_c$ and $Z$ the intersection of $A_b{A}_c$ with $B_a{B}_c$. Let $△$ be the projection triangle of $H$ corresponding to triangle $ABC$. Then $H$ is the incenter of $△$ and the midpoint $OH$ is the circumcenter of $△$. In this solution, we will show that $I$ is the center of the circle inscribed in the triangle $XYZ$ and that the two triangles $XYZ$ and $△$ have corresponding parallel sides. From there, $IK\parallel OH$.

First, to prove that triangles $XYZ$ and $△$ have corresponding sides parallel, we show that $A_b{A}_{c}CB$ is a cyclic quadrilateral, and then $A_b{A}_c$ is anti-parallel to $BC$ in $\angle BAC$ so it is parallel to the line connecting the foot of the vertex $B$ and $C$. Indeed, considering the inversion of the center $A$, power $A{P}^2$ which is denoted by ${\mathcal{I}}_A$. This inversion preserves $(I)$ and maps $A^{\prime }\mapsto D$.



so the image of $(A{A}_b{A}_c{A}^{\prime })$ passes through $D$ and touches $(I)$ so $${\mathcal{I}}_A:(A{A}_b{A}_c)\mapsto BC.$$ Therefore, the image of $A_b$ lies on $AB$ and $BC$, so $B$ is the image of $A_b$ through ${\mathcal{I}}_A$ which infer $$\overline{A{A}_b}\cdot \overline{AB}=A{P}^2.$$ Similarly, $\overline{A{A}_c}\cdot \overline{AC}=A{N}^2=A{P}^2$ so $A_b{A}_{c}CB$ is a cyclic quadrilateral. So two triangles $XYZ$ and $△$ have corresponding sides are parallel.

Finally, we need to prove that $I$ is the incenter of triangle $XYZ$ or equivalently $XI$ is the angle bisector of $\angle B_{c}X{C}_b$. On the other hand, similar to the above proof, then $B_a{B}_{c}CA$ and $C_a{C}_{b}BA$ are cyclic quadrilaterals, so $$\angle X{C}_b{B}_c=\angle BAC=\angle X{B}_c{C}_b$$ leads to $X{C}_b=X{B}_c$. So it suffices to show that $M$ is the midpoint of $B_c{C}_b$ and $X,M,I$ lie on the bisector $\angle YXZ$. By considering the inversions of centers $B$ and $C$ preserving $(I)$ similar to the above, we can show that $$B{B}_c\cdot BC=B{M}^2,C{C}_b\cdot CB=C{M}^2$$ and it leads to $$M{B}_c=MB-B{B}_c=MB-\frac{M{B}^2}{BC}=\frac{MB\cdot MC}{BC}.$$ The length of $M{C}_b$ can be calculated in the same way and then $M{B}_c=M{C}_b$. Thus $XM$ is perpendicular bisector of $B_c{C}_b$ so $XM\perp BC$ and $I$ lies on $XM$. So $XI$ is the angle bisector of $\angle YXZ$. Similar to the vertices $Y,Z$, we can conclude that $I$ is the incenter of the triangle $XYZ$. This finishes the proof. $\square$