
1. There are two touching circles, $k_{1}\left(S_{1}, r_{1}\right.$ and $k_{2}\left(S_{2}, r_{2}\right)$ in an rectangle $A B C D$ with $|A B|=9,|B C|=8$. Moreover $k_{1}$ touches $A D$ and $C D$, while $k_{2}$ touches $A B$ and $B C$.

a) Prove $r_{1}+r_{2}=5$.

b) What is the least and what is the greatest possible area of $A S_{1} S_{2}$ ?

(Pavel Novotný)


Solution. a) Let $M$ and $N$ be intersections of the line through $S_{1}$ parallel to $A D$. Analogously let $K$ and $L$ be intersections of the line through $S_{2}$ parallel to $A B$. Let $P$ be the intersection of $K L$ and $M N$ (see Fig. 1). The Pythagoras theorem for $S_{1} P S_{2}$ gives

$$
\begin{gathered}
\left(r_{1}+r_{2}\right)^{2}=\left(8-r_{1}-r_{2}\right)^{2}+\left(9-r_{1}-r_{2}\right)^{2} \\
\left(r_{1}+r_{2}\right)^{2}-34\left(r_{1}+r_{2}\right)+145=0 \\
\left(r_{1}+r_{2}-5\right)\left(r_{1}+r_{2}-29\right)=0 .
\end{gathered}
$$

Since $2 r_{1} \leqslant 8,2 r_{2} \leqslant 8$, we have $r_{1}+r_{2}=5$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5816354d2b4b35f9c465g-13.jpg?height=771&width=851&top_left_y=1625&top_left_x=568)

Fig. 1

b) Let $Q$ be a foot of a perpendicular to $A B$ from $S_{2}$, let $R$ be a foot of a perpendicular to $A D$ 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 $A Q T R$ and areas of right triangles $A Q S_{2}, A S_{1} R$, and $S_{1} S_{2} T$ :

$$
\begin{aligned}
S= & \left(9-r_{2}\right)\left(8-r_{1}\right)-\frac{1}{2} r_{2}\left(9-r_{2}\right)-\frac{1}{2} r_{1}\left(8-r_{1}\right)-\frac{1}{2}\left(9-r_{1}-r_{2}\right)\left(8-r_{1}-r_{2}\right) \\
= & 72-9 r_{1}-8 r_{2}+r_{1} r_{2}-\frac{9}{2} r_{2}+\frac{1}{2} r_{2}^{2}-4 r_{1}+\frac{1}{2} r_{1}^{2}-36+\frac{17}{2}\left(r_{1}+r_{2}\right) \\
& -\frac{1}{2}\left(r_{1}+r_{2}\right)^{2} \\
= & 36-\frac{9}{2} r_{1}-4 r_{2}=36-\frac{9}{2} r_{1}-4\left(5-r_{1}\right)=16-\frac{1}{2} r_{1},
\end{aligned}
$$

where we used $r_{1}+r_{2}=5$. Further we know $2 r_{1} \leqslant 8$ and $2 r_{2} \leqslant 8$ which implies $r_{1}$, $r_{2} 1 \leqslant r_{1}, r_{2} \leqslant 4$, 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 $r_{2}=4$.
