62nd Czech and Slovak Mathematical Olympiad

The line segments $A{S}_2$ and $B{S}_1$ 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 $$\begin{array}{rlrlrl}|D{S}_1|& =r_1\cdot \sqrt{2},& |B{S}_1|& =(a-r_1)\sqrt{2},& |P{S}_1|& =\left(\frac{a}{2}-r_1\right)\sqrt{2},\\ |C{S}_2|& =r_2\cdot \sqrt{2},& |A{S}_2|& =(a-r_2)\sqrt{2},& |P{S}_2|& =\left(\frac{a}{2}-r_2\right)\sqrt{2}.\end{array}$$ Therefore, the area of the triangle $A{S}_1{S}_2$ is $$S_{A{S}_1{S}_2}=\frac{1}{2}|A{S}_2|\cdot |P{S}_1|=(a-r_2)\left(\frac{a}{2}-r_1\right),$$ while the area of the triangle $B{S}_1{S}_2$ is $$S_{B{S}_1{S}_2}=\frac{1}{2}|B{S}_1|\cdot |P{S}_2|=(a-r_1)\left(\frac{a}{2}-r_2\right).$$ The sum of these areas is $$S=(a-r_2)\left(\frac{a}{2}-r_1\right)+(a-r_1)\left(\frac{a}{2}-r_2\right)=a^2-\frac{3}{2}a(r_1+r_2)+2r_1{r}_2.$$ 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 $K{S}_1$ and $H{S}_2$ (Fig. 1). Fig. 1 By the Pythagoras' theorem for the triangle $S_{1}M{S}_2$, we have $$(a-r_1-r_2{)}^2+(r_1-r_2{)}^2=(r_1+r_2{)}^2.$$ Hence we obtain $$\begin{array}{rl}(a-r_1-r_2{)}^2& =4r_1{r}_2,\\ a-r_1-r_2& =2\sqrt{r_1{r}_2},\\ a& =r_1+r_2+2\sqrt{r_1{r}_2}=(\sqrt{r_1}+\sqrt{r_2}{)}^2\ge 4\sqrt{r_1{r}_2},\end{array}$$ i. e. $$r_1{r}_2\le \frac{a^2}{16}.$$ The length of the segment $DC$ cannot be greater than the length of the polygonal chain $K{S}_1{S}_{2}L$, so $$2r_1+2r_2\ge a.$$ (This follows from the equality $a=r_1+r_2+2\sqrt{r_1{r}_2}$ as well since $2\sqrt{r_1{r}_2}\le r_1+r_2$, by the AM-GM inequality.) Therefore, $$S=a^2-\frac{3}{2}a(r_1+r_2)+2r_1{r}_2\le a^2-\frac{3}{4}{a}^2+\frac{1}{8}{a}^2=\frac{3}{8}{a}^2.$$ This means that at least one of the areas $S_{A{S}_1{S}_2}$, $S_{B{S}_1{S}_2}$ is at most $\frac{3}{16}{a}^2$.

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Alternative solution.

We can set $a=1$. The difference of the areas of the triangles $A{S}_1{S}_2$ and $B{S}_1{S}_2$ is (according to the expression from the original solution) $$S_{A{S}_1{S}_2}-S_{B{S}_1{S}_2}=(1-r_2)\left(\frac{1}{2}-r_1\right)-(1-r_1)\left(\frac{1}{2}-r_2\right)=\frac{1}{2}(r_2-r_1).$$ Without loss of generality, we can assume that $r_1\ge r_2$. Then $S_{A{S}_1{S}_2}\le S_{B{S}_1{S}_2}$. Now, let us calculate the area of the triangle $A{S}_1{S}_2$. 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 $A{S}_1{S}_2$ is $$\begin{array}{rl}{S}_{A{S}_1{S}_2}& =(1-r_2)\left(\frac{1}{2}-r_1\right)=\frac{1}{2}-r_1-\frac{r_2}{2}+r_1{r}_2=\\ & =\frac{1}{2}-x^2-\frac{1}{2}(1-x{)}^2+x^2(1-x{)}^2=x^4-2x^3-\frac{1}{2}{x}^2+x;\\ S_{A{S}_1{S}_2}-\frac{3}{16}& =x^4-2x^3-\frac{1}{2}{x}^2+x-\frac{3}{16}=(x-\frac{1}{2})(x^3-\frac{3}{2}{x}^2-\frac{5}{4}x+\frac{3}{8})=\\ & =(x-\frac{1}{2})\left[x^2\left(x-\frac{3}{2}\right)-\frac{5}{4}\left(x-\frac{3}{10}\right)\right]\le 0\end{array}$$ as we have $\frac{3}{10}<\frac{1}{2}\le x\le \frac{1}{2}\sqrt{2}<\frac{3}{2}$. Therefore, $S_{A{S}_1{S}_2}\le \frac{3}{16}$.