Czech and Slovak Mathematical Olympiad

The line segments $AS_2$ and $BS_1$ lie on the diagonals of the given square, so they are perpendicular to each other and intersect at the center $P$ of the square. We have
$$\begin{aligned} |DS_1| &= r_1 \cdot \sqrt{2}, & |BS_1| &= (a-r_1)\sqrt{2}, & |PS_1| &= \left(\frac{a}{2} - r_1\right)\sqrt{2}, \\ |CS_2| &= r_2 \cdot \sqrt{2}, & |AS_2| &= (a-r_2)\sqrt{2}, & |PS_2| &= \left(\frac{a}{2} - r_2\right)\sqrt{2}. \end{aligned}$$
Therefore, the area of the triangle $AS_1S_2$ is
$$S_{AS_1S_2} = \frac{1}{2}|AS_2| \cdot |PS_1| = (a-r_2)\left(\frac{a}{2}-r_1\right),$$
while the area of the triangle $BS_1S_2$ is
$$S_{BS_1S_2} = \frac{1}{2}|BS_1| \cdot |PS_2| = (a-r_1)\left(\frac{a}{2}-r_2\right).$$
The sum of these areas is
$$S = (a - r_2) \left(\frac{a}{2} - r_1\right) + (a - r_1) \left(\frac{a}{2} - r_2\right) = a^2 - \frac{3}{2}a(r_1 + r_2) + 2r_1r_2.$$
Let $K$ denote the point at which the circle $k_1$ touches the side $AD$, $H$ and $L$ denote the points at which $k_2$ touches the sides $CD$ and $BC$, respectively, and $M$ be the intersection point of the lines $KS_1$ and $HS_2$ (Fig. 1).

Fig. 1
By the Pythagoras' theorem for the triangle $S_1MS_2$, we have
$$(a - r_1 - r_2)^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2.$$
Hence we obtain
$$\begin{aligned} (a - r_1 - r_2)^2 &= 4r_1r_2, \\ a - r_1 - r_2 &= 2\sqrt{r_1r_2}, \\ a &= r_1 + r_2 + 2\sqrt{r_1r_2} = (\sqrt{r_1} + \sqrt{r_2})^2 \ge 4\sqrt{r_1r_2}, \end{aligned}$$
i. e.
$$r_1r_2 \le \frac{a^2}{16}.$$
The length of the segment $DC$ cannot be greater than the length of the polygonal chain $KS_1S_2L$, so
$$2r_1 + 2r_2 \ge a.$$
(This follows from the equality $a = r_1 + r_2 + 2\sqrt{r_1r_2}$ as well since $2\sqrt{r_1r_2} \le r_1 + r_2$, by the AM-GM inequality.) Therefore,
$$S = a^2 - \frac{3}{2}a(r_1 + r_2) + 2r_1r_2 \le a^2 - \frac{3}{4}a^2 + \frac{1}{8}a^2 = \frac{3}{8}a^2.$$
This means that at least one of the areas $S_{AS_1S_2}$, $S_{BS_1S_2}$ is at most $\frac{3}{16}a^2$.

We can set $a = 1$. The difference of the areas of the triangles $AS_1S_2$ and $BS_1S_2$ is (according to the expression from the original solution)
$$S_{AS_1S_2} - S_{BS_1S_2} = (1 - r_2)\left(\frac{1}{2} - r_1\right) - (1 - r_1)\left(\frac{1}{2} - r_2\right) = \frac{1}{2}(r_2 - r_1).$$
Without loss of generality, we can assume that $r_1 \ge r_2$. Then $S_{AS_1S_2} \le S_{BS_1S_2}$. Now, let us calculate the area of the triangle $AS_1S_2$. By the Pythagoras' theorem, we have $(1 - r_1 - r_2)^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2$, hence $\sqrt{r_1} + \sqrt{r_2} = 1$, so $r_2 = (1 - \sqrt{r_1})^2$. Let us denote $x = \sqrt{r_1}$. It follows from the inequalities $r_1 + r_2 \ge \frac{1}{2}$ and $r_1 \ge r_2$ that $r_1 \ge \frac{1}{4}$, and, on the other hand, we have $r_1 \le \frac{1}{2}$ since the circle $k_1$ lies in the square $ABCD$. Hence it follows that $\frac{1}{2} \le x \le \sqrt{\frac{1}{2}}$. The area of the triangle $AS_1S_2$ is
$$\begin{aligned} S_{AS_1S_2} &= (1 - r_2)\left(\frac{1}{2} - r_1\right) = \frac{1}{2} - r_1 - \frac{r_2}{2} + r_1r_2 = \\ &= \frac{1}{2} - x^2 - \frac{1}{2}(1-x)^2 + x^2(1-x)^2 = x^4 - 2x^3 - \frac{1}{2}x^2 + x; \\ S_{AS_1S_2} - \frac{3}{16} &= x^4 - 2x^3 - \frac{1}{2}x^2 + x - \frac{3}{16} = (x - \frac{1}{2})(x^3 - \frac{3}{2}x^2 - \frac{5}{4}x + \frac{3}{8}) = \\ &= (x - \frac{1}{2})\left[x^2\left(x - \frac{3}{2}\right) - \frac{5}{4}\left(x - \frac{3}{10}\right)\right] \le 0 \end{aligned}$$
as we have $\frac{3}{10} < \frac{1}{2} \le x \le \frac{1}{2}\sqrt{2} < \frac{3}{2}$. Therefore, $S_{AS_1S_2} \le \frac{3}{16}$.