Iranian Mathematical Olympiad

The idea is to remove some neighborhoods of $A$ and $B$, because near these points we cannot bound $\frac{A{A}^{\prime }}{B{B}^{\prime }}$. Let $Z$ be the intersection of the ray $AB$ with the boundary of $S$ and let $Z{B}^{\prime }$ intersect $A{A}^{\prime }$ in $A^{\prime \prime }$. $A^{\prime \prime }$ is between $A$ and $A^{\prime }$ since $S$ is convex. So, if we let $S^{\prime \prime }$ be the set of those points $X$ such that $A{A}^{\prime \prime }>\frac{1}{3}B{B}^{\prime }$, we have $S^{\prime \prime }\subseteq S^{\prime }$. Hence, it is enough to prove that the area of $S^{\prime \prime }$ is at least $6$. By Menelaus's theorem, we have $$\frac{A{A}^{\prime \prime }}{A^{\prime \prime }X}\cdot \frac{X{B}^{\prime }}{B^{\prime }B}\cdot \frac{BZ}{ZA}=1\Rightarrow \frac{A{A}^{\prime \prime }}{A^{\prime \prime }X}\cdot \frac{X{B}^{\prime }}{B^{\prime }B}=2.$$ To omit $A^{\prime \prime }X$, we write $\frac{A{A}^{\prime \prime }}{A^{\prime \prime }X}$ in terms of $A{A}^{\prime \prime }$ and $AX$ and we treat $B^{\prime }X$ similarly: $$\begin{array}{rl}\Rightarrow \frac{A^{\prime \prime }X}{A{A}^{\prime \prime }}& =\frac{1}{2}\cdot \frac{X{B}^{\prime }}{B^{\prime }B}\\ \Rightarrow \frac{AX}{A{A}^{\prime \prime }}& =1-\frac{1}{2}\left(1-\frac{BX}{B{B}^{\prime }}\right)\\ \Rightarrow \frac{AX}{A{A}^{\prime \prime }}& =\frac{1}{2}\cdot \frac{BX}{B{B}^{\prime }}+\frac{1}{2}.\end{array}$$ Let $D_1$ be the set of those points $X$ such that $\frac{1}{2}\cdot \frac{BX}{B{B}^{\prime }}+\frac{1}{2}\ge \alpha \frac{BX}{B{B}^{\prime }}$ for some constant $\alpha$ (for suitable $\alpha$, $D_1$ is a neighborhood of $B$). If $X\notin D_1$, then $$\begin{array}{l}\frac{AX}{A{A}^{\prime \prime }}<\alpha \frac{BX}{B{B}^{\prime }}\\ \Rightarrow \frac{A{A}^{\prime \prime }}{B{B}^{\prime }}>\frac{1}{\alpha }\cdot \frac{AX}{BX}.\end{array}$$ Let $D_2$ be the set of those points $X$ in the plane such that $\frac{AX}{BX}<\frac{\alpha }{3}$ (for suitable $\alpha$, $D_2$ is a neighborhood of $A$). If $X$ is not in either of $D_1$ or $D_2$, then $$\frac{A{A}^{\prime \prime }}{B{B}^{\prime }}>\frac{1}{\alpha }\cdot \frac{\alpha }{3}=\frac{1}{3}$$ as desired. So $$S-D_1-D_2\subseteq S^{\prime \prime }\subseteq S^{\prime }.$$ Now, we calculate the areas of $D_1$ and $D_2$. If $\alpha <3$, then $D_2$ is the interior part of the Apollonius circle of $A$ and $B$. Also, $$X\in D_1\iff \frac{1}{2}\cdot \frac{BX}{B{B}^{\prime }}+\frac{1}{2}\ge \alpha \frac{BX}{B{B}^{\prime }}\iff \frac{BX}{B{B}^{\prime }}<\frac{1}{2\alpha -1}.$$ So, if $\alpha >1$ then $D_1$ is a neighborhood of $B$ similar to $S$. Now, we take $\alpha =\frac{3}{2}$. We conclude that the area of $D_1$ is ${\left(\frac{1}{2}\right)}^2$ times the area of $S$ which is $\frac{10}{4}=2.5$. Also, if $C$ and $D$ are the intersection points of the boundary of $D_2$ with the line $AB$ and $C$ is between $A$ and $B$, we have $$\begin{array}{rl}\frac{AD}{DB}& =\frac{1}{2}\Rightarrow \frac{AD}{AD+1}=\frac{1}{2}\Rightarrow AD=1\\ \frac{AC}{CB}& =\frac{1}{2}\Rightarrow \frac{AC}{1-AC}=\frac{1}{2}\Rightarrow AC=\frac{1}{3}.\end{array}$$ So the diameter of $D_2$ is $1+\frac{1}{3}=\frac{4}{3}$ and hence, the area of $D_2$ is ${\left(\frac{2}{3}\right)}^2\pi <1.5$. So $$\begin{array}{rl}{T}_X& \ge T_{S-D_1-D_2}\\ & \ge 2T_S-T_{D_1}-T_{D_2}\\ & >10-2.5-1.5=6.\end{array}$$ where $T_K$ represents the area of figure $K$ in the plane. Hence, the assertion is proved. $\square$