62nd Czech and Slovak Mathematical Olympiad

a) Let $M$ and $N$ be intersections of the line through $S_1$ parallel to $AD$. Analogously, let $K$ and $L$ be intersections of the line through $S_2$ parallel to $AB$. Let $P$ be the intersection of $KL$ and $MN$ (see Fig. 1). The Pythagoras theorem for $S_{1}P{S}_2$ gives $$\begin{array}{rl}(r_1+r_2{)}^2& =(8-r_1-r_2{)}^2+(9-r_1-r_2{)}^2,\\ (r_1+r_2{)}^2-34(r_1+r_2)+145& =0,\\ (r_1+r_2-5)(r_1+r_2-29)& =0.\end{array}$$ Since $2r_1\le 8$, $2r_2\le 8$, we have $r_1+r_2=5$.

b) Let $Q$ be a foot of a perpendicular to $AB$ from $S_2$, let $R$ be a foot of a perpendicular to $AD$ from $S_1$ and let $T$ be the intersection of $Q{S}_2$ and $R{S}_1$ (Fig. 1).

The area $S$ of $A{S}_2{S}_1$ is given by the difference of the area of rectangle $AQTR$ and areas of right triangles $AQ{S}_2$, $A{S}_{1}R$, and $S_1{S}_{2}T$: $$\begin{array}{rl}S& =(9-r_2)(8-r_1)-\frac{1}{2}{r}_2(9-r_2)-\frac{1}{2}{r}_1(8-r_1)-\frac{1}{2}(9-r_1-r_2)(8-r_1-r_2)\\ & =72-9r_1-8r_2+r_1{r}_2-\frac{9}{2}{r}_2+\frac{1}{2}{r}_1^2-4r_1+\frac{1}{2}{r}_1^2-36+\frac{17}{2}(r_1+r_2)\\ & -\frac{1}{2}(r_1+r_2{)}^2\\ & =36-\frac{9}{2}{r}_1-4r_2=36-\frac{9}{2}{r}_1-4(5-r_1)=16-\frac{1}{2}{r}_1,\end{array}$$ where we used $r_1+r_2=5$. Further, we know $2r_1\le 8$ and $2r_2\le 8$ which implies $1\le r_1,r_2\le 4$, thus $$S=16-\frac{1}{2}{r}_1\in ⟨14,\frac{31}{2}⟩;$$ and the least possible value of the area is $14$, for $r_1=4$ and $r_2=1$, and the greatest value possible is $\frac{31}{2}$, for $r_1=1$ and $r_2=4$.

Fig. 1