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 $S_1PS_2$ gives
$$(r_1 + r_2)^2 = (8 - r_1 - r_2)^2 + (9 - r_1 - r_2)^2, \ (1)$$
$$(r_1 + r_2)^2 - 34(r_1 + r_2) + 145 = 0,$$
$$(r_1 + r_2 - 5)(r_1 + r_2 - 29) = 0.$$
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 $QS_2$ and $RS_1$ (Fig. 1).

The area $S$ of $AS_2S_1$ is given by the difference of the area of rectangle $AQTR$ and areas of right triangles $AQS_2$, $AS_1R$, and $S_1S_2T$:
$$\begin{align*} S &= (9 - r_2)(8 - r_1) - \frac{1}{2}r_2(9 - r_2) - \frac{1}{2}r_1(8 - r_1) - \frac{1}{2}(9 - r_1 - r_2)(8 - r_1 - r_2) \\ &= 72 - 9r_1 - 8r_2 + r_1r_2 - \frac{9}{2}r_2 + \frac{1}{2}r_2^2 - 4r_1 + \frac{1}{2}r_1^2 - 36 + \frac{17}{2}(r_1 + r_2) \\ &\quad - \frac{1}{2}(r_1 + r_2)^2 \\ &= 36 - \frac{9}{2}r_1 - 4r_2 = 36 - \frac{9}{2}r_1 - 4(5 - r_1) = 16 - \frac{1}{2}r_1, \end{align*}$$
where we used $r_1 + r_2 = 5$. Further, we know $2r_1 \le 8$ and $2r_2 \le 8$ which implies $r_1, r_2 \le 4$, thus
$$S = 16 - \frac{1}{2}r_1 \in \left\langle 14, \frac{31}{2} \right\rangle;$$
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

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 $S_1PS_2$ gives
$$\begin{align*} (r_1 + r_2)^2 &= (8 - r_1 - r_2)^2 + (9 - r_1 - r_2)^2, \\ (r_1 + r_2)^2 - 34(r_1 + r_2) + 145 &= 0, \\ (r_1 + r_2 - 5)(r_1 + r_2 - 29) &= 0. \end{align*}$$
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 $QS_2$ and $RS_1$ (Fig. 1).

The area $S$ of $AS_2S_1$ is given by the difference of the area of rectangle $AQTR$ and areas of right triangles $AQS_2$, $AS_1R$, and $S_1S_2T$:
$$\begin{align*} S &= (9-r_2)(8-r_1) - \frac{1}{2}r_2(9-r_2) - \frac{1}{2}r_1(8-r_1) - \frac{1}{2}(9-r_1-r_2)(8-r_1-r_2) \\ &= 72 - 9r_1 - 8r_2 + r_1r_2 - \frac{9}{2}r_2 + \frac{1}{2}r_1^2 - 4r_1 + \frac{1}{2}r_1^2 - 36 + \frac{17}{2}(r_1+r_2) \\ &\quad - \frac{1}{2}(r_1+r_2)^2 \\ &= 36 - \frac{9}{2}r_1 - 4r_2 = 36 - \frac{9}{2}r_1 - 4(5-r_1) = 16 - \frac{1}{2}r_1, \end{align*}$$
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
$$S = 16 - \frac{1}{2}r_1 \in \left\langle 14, \frac{31}{2} \right\rangle;$$
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