Iranian Mathematical Olympiad

The idea is to remove some neighborhoods of $A$ and $B$, because near these points we cannot bound $\frac{AA'}{BB'}$. Let $Z$ be the intersection of the ray $AB$ with the boundary of $S$ and let $ZB'$ intersect $AA'$ in $A''$. $A''$ is between $A$ and $A'$ since $S$ is convex. So, if we let $S''$ be the set of those points $X$ such that $AA'' > \frac{1}{3} BB'$, we have $S'' \subseteq S'$. Hence, it is enough to prove that the area of $S''$ is at least $6$.
By Menelaus's theorem, we have
$$\frac{AA''}{A''X} \cdot \frac{XB'}{B'B} \cdot \frac{BZ}{ZA} = 1 \Rightarrow \frac{AA''}{A''X} \cdot \frac{XB'}{B'B} = 2.$$
To omit $A''X$, we write $\frac{AA''}{A''X}$ in terms of $AA''$ and $AX$ and we treat $B'X$ similarly:
$$\begin{align*} \Rightarrow \quad \frac{A''X}{AA''} &= \frac{1}{2} \cdot \frac{XB'}{B'B} \\ \Rightarrow \quad \frac{AX}{AA''} &= 1 - \frac{1}{2}\left(1 - \frac{BX}{BB'}\right) \\ \Rightarrow \quad \frac{AX}{AA''} &= \frac{1}{2} \cdot \frac{BX}{BB'} + \frac{1}{2}. \end{align*}$$
Let $D_1$ be the set of those points $X$ such that $\frac{1}{2} \cdot \frac{BX}{BB'} + \frac{1}{2} \ge \alpha \frac{BX}{BB'}$ for some constant $\alpha$ (for suitable $\alpha$, $D_1$ is a neighborhood of $B$). If $X \notin D_1$, then
$$\begin{array}{l} \frac{AX}{AA''} < \alpha \frac{BX}{BB'} \\ \Rightarrow \quad \frac{AA''}{BB'} > \frac{1}{\alpha} \cdot \frac{AX}{BX}. \end{array}$$
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
$$\frac{AA''}{BB'} > \frac{1}{\alpha} \cdot \frac{\alpha}{3} = \frac{1}{3}$$
as desired. So
$$S - D_1 - D_2 \subseteq S'' \subseteq S'.$$
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,
$$X \in D_1 \Leftrightarrow \frac{1}{2} \cdot \frac{BX}{BB'} + \frac{1}{2} \ge \alpha \frac{BX}{BB'} \Leftrightarrow \frac{BX}{BB'} < \frac{1}{2\alpha - 1}.$$
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
$$\begin{align*} \frac{AD}{DB} &= \frac{1}{2} \Rightarrow \frac{AD}{AD+1} = \frac{1}{2} \Rightarrow AD = 1 \\ \frac{AC}{CB} &= \frac{1}{2} \Rightarrow \frac{AC}{1-AC} = \frac{1}{2} \Rightarrow AC = \frac{1}{3}. \end{align*}$$
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
$$\begin{align*} T_X &\ge T_{S-D_1-D_2} \\ &\ge 2T_S - T_{D_1} - T_{D_2} \\ &> 10 - 2.5 - 1.5 = 6. \end{align*}$$
where $T_K$ represents the area of figure $K$ in the plane. Hence, the assertion is proved. $\square$