1. In rectangle $A B C D$ with sides $|A B|=9,|B C|=8$, there are mutually touching circles $k_{1}\left(S_{1}, r_{1}\right)$ and $k_{2}\left(S_{2}, r_{2}\right)$ such that $k_{1}$ touches sides $A D$ and $C D$, and $k_{2}$ touches sides $A B$ and $B C$.

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

b) Determine the smallest and largest possible value of the area of triangle $A S_{1} S_{2}$.
1. a) Through point $S_{1}$, draw a line parallel to side $AD$ and denote its intersections with sides $AB$ and $CD$ as $M$ and $N$. Similarly, through point $S_{2}$, draw a line parallel to $AB$ and denote its intersections with sides $AD$ and $BC$ as $K$ and $L$; the intersection of lines $KL$ and $MN$ is denoted as $P$ (Fig. 1). According to the Pythagorean theorem for triangle $S_{1} P S_{2}$, we have

$$
\left(r_{1}+r_{2}\right)^{2}=\left(8-r_{1}-r_{2}\right)^{2}+\left(9-r_{1}-r_{2}\right)^{2}
$$

and thus

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

Since $2 r_{1} \leqq 8, 2 r_{2} \leqq 8$ (the diameter of the circles, which lie entirely within the given rectangle according to the problem, cannot be greater than the length of side $AD$), it must be that $r_{1}+r_{2}=5$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2cdeaa7a91efbb40eb97g-2.jpg?height=760&width=851&top_left_y=1045&top_left_x=608)

Fig. 1

b) Let $Q$ be the foot of the perpendicular from point $S_{2}$ to side $AB$, $R$ the foot of the perpendicular from point $S_{1}$ to side $AD$, and $T$ the intersection of lines $QS_{2}$ and $RS_{1}$ (Fig. 1). The area $S$ of triangle $AS_{2}S_{1}$ can be calculated by subtracting the sum of the areas of the right triangles $AQS_{2}$, $AS_{1}R$, and $S_{1}S_{2}T$ from the area of rectangle $AQT R$:

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

where we substituted $r_{2}$ from the equation $r_{1}+r_{2}=5$ in the last step. From this and the inequalities $2 r_{1} \leqq 8, 2 r_{2} \leqq 8$, it follows that both radii $r_{1}, r_{2}$ lie in the interval $\langle 1,4\rangle$, and thus

$$
S=16-\frac{1}{2} r_{1} \in\left\langle 14, \frac{31}{2}\right\rangle
$$

the area has the minimum value 14 when $r_{1}=4$ and $r_{2}=1$, and the maximum value $\frac{31}{2}$ when $r_{1}=1$ and $r_{2}=4$.

Alternative Solution. a) Let $\alpha$ be the angle $PS_{2}S_{1}$ (Fig. 1). From the equality $|AB|=|KP|+\left|PS_{2}\right|+\left|S_{2}L\right|$, we get $r_{1}+\left(r_{1}+r_{2}\right) \cos \alpha+r_{2}=9$ which simplifies to

$$
(1+\cos \alpha)\left(r_{1}+r_{2}\right)=9
$$

Similarly, from the equality $|AD|=|MP|+\left|PS_{1}\right|+\left|S_{1}N\right|$, we get

$$
(1+\sin \alpha)\left(r_{1}+r_{2}\right)=8
$$

From the last two equations, we obtain

$$
\begin{aligned}
8(1+\cos \alpha) & =9(1+\sin \alpha) \\
8 \cos \alpha & =1+9 \sin \alpha .
\end{aligned}
$$

By squaring this equation and using the identity $\cos ^{2} \alpha=1-\sin ^{2} \alpha$, we get a quadratic equation for $\sin \alpha$,

$$
\begin{gathered}
64\left(1-\sin ^{2} \alpha\right)=1+18 \sin \alpha+81 \sin ^{2} \alpha \\
145 \sin ^{2} \alpha+18 \sin \alpha-63=0
\end{gathered}
$$

This equation has a discriminant $D=18^{2}+4 \cdot 63 \cdot 145=9(36+28 \cdot 145)=3^{2} \cdot 2^{12}$ and two real roots $-\frac{21}{29}$ and $\frac{3}{5}$. Since $\alpha$ is an interior angle of the triangle, we get

$$
\sin \alpha=\frac{3}{5}
$$

and thus

$$
r_{1}+r_{2}=\frac{8}{1+\sin \alpha}=5
$$

b) The area $S$ of triangle $AS_{2}S_{1}$ can be calculated using the vector product of vectors $S_{2}-A$ and $S_{1}-A$. Choose a Cartesian coordinate system such that $A=[0,0,0]$, $B=[9,0,0]$, $D=[0,8,0]$. Then $S_{2}=\left[9-r_{2}, r_{2}, 0\right]$, $S_{1}=\left[r_{1}, 8-r_{1}, 0\right]$, and the area $S$ is

$$
\begin{aligned}
S & =\frac{1}{2}\left|\left(S_{2}-A\right) \times\left(S_{1}-A\right)\right|=\frac{1}{2}\left[\left(9-r_{2}\right)\left(8-r_{1}\right)-r_{1} r_{2}\right]=\frac{1}{2}\left(72-9 r_{1}-8 r_{2}\right)= \\
& =36-\frac{9}{2} r_{1}-4 r_{2} .
\end{aligned}
$$

We obtained the same expression as in the first solution, and thus by the same method, we find that the area has a minimum value of 14 and a maximum value of $\frac{31}{2}$.

For a complete solution, award 6 points. Award one point for equation (1) or the system (2), (3), another point for determining the sum $r_{1}+r_{2}$, i.e., up to two points for part a). Award two points for a suitable expression of the area of triangle $AS_{2}S_{1}$ and the remaining two points for determining the minimum and maximum values of the area using the inequalities $1 \leqq r_{1} \leqq 4$ or $1 \leqq r_{2} \leqq 4$.