Vietnamese MO

a) Let $(A^{\prime }),(B^{\prime }),(C^{\prime })$ respectively represent the circle passing through point $C$ and touching the line $AB$ at point $A$, the circle passing through point $B$ and touching the line $BC$ at point $B$, and the circle passing through point $C$ and touching the line $CA$ at point $C$.

Let $K$ be the second intersection point of two circles $(A^{\prime })$ and $(B^{\prime })$. We have $$(AK,AB)\equiv (BK,BC)\equiv (CK,CA)(\mathrm{mod}\pi ).$$ Therefore, point $K$ also belongs to circle $(C^{\prime })$. Now, denote by $D,E,F$ respectively the foot of the perpendicular drawn from point $K$ to lines $BC,CA$ and $AB$. According to Erdos inequality, we have $$KA+KB+KC\ge 2(KD+KE+KF).$$

Let $\angle KBC=\angle KAB=\angle KCA=\omega$. Because $$\sin \omega =\frac{KD}{KB}=\frac{KE}{KC}=\frac{KF}{KA}=\frac{KD+KE+KF}{KB+KC+KA}\le \frac{1}{2}$$ so $\omega \le 30^{\circ }$.

The triangles $KA{A}^{\prime }$, $KB{B}^{\prime }$, $KC{C}^{\prime }$ are isosceles triangles at $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ with vertex angle equal to $2\omega \le 60^{\circ }$.

Put $$(\overrightarrow{KA},\overrightarrow{K{A}^{\prime }})\equiv (\overrightarrow{KB},\overrightarrow{K{B}^{\prime }})\equiv (\overrightarrow{KC},\overrightarrow{K{C}^{\prime }})\equiv \varphi (\mathrm{mod}2\pi )$$ $$\text{and }k=\frac{K{A}^{\prime }}{KA}=\frac{K{B}^{\prime }}{KB}=\frac{K{C}^{\prime }}{KC}\ge 1.$$ We denote by $f$ the rotational homothety with center $K$, angle $\varphi$ and coefficient $k$.

Since $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ are images of $A$, $B$, $C$ by $f$ respectively, $△A^{\prime }{B}^{\prime }{C}^{\prime }\sim △ABC$. We deduce that $\frac{S(A^{\prime }{B}^{\prime }{C}^{\prime })}{S(ABC)}=k^2\ge 1$, or $S(A^{\prime }{B}^{\prime }{C}^{\prime })\ge S(ABC)$. The equality occurs if and only if $ABC$ is an equilateral triangle.

b) Since $A^{\prime }$, $C^{\prime }$ are images of $A$, $C$ through $f$ and $C^{\prime }{B}^{\prime }\perp BK$, $C^{\prime }O\perp BC$ respectively, we get $$(C^{\prime }K,C^{\prime }{A}^{\prime })\equiv (CK,CA)\equiv (BK,BC)\equiv (C^{\prime }{B}^{\prime },C^{\prime }O)(\mathrm{mod}\pi ).$$ Therefore $C^{\prime }O$ and $C^{\prime }K$ are isogonal in angle $A^{\prime }{C}^{\prime }{B}^{\prime }$. By similar argument, we have $O$ and $K$ are isogonal conjugate points in triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$. So $K{X}^{\prime }\perp A^{\prime }{B}^{\prime }$, $K{Y}^{\prime }\perp B^{\prime }{C}^{\prime }$ and $K{Z}^{\prime }\perp C^{\prime }{A}^{\prime }$. We also have $AK\perp A^{\prime }{B}^{\prime }$, $BK\perp B^{\prime }{C}^{\prime }$ and $CK\perp C^{\prime }{A}^{\prime }$ so we deduce that three lines $A{X}^{\prime }$, $B{Y}^{\prime }$ and $C{Z}^{\prime }$ concur at $K$.