6. Answer: $\Sigma_{\max }=\frac{\pi}{4}\left((\sqrt{2 a}-\sqrt{b})^{4}+6(3-2 \sqrt{2}) b^{2}\right)=4 \pi(19-12 \sqrt{2}) \approx 25.5$

Option 0

In rectangle $ABCD$ with sides $AD=a, AB=b(b<a<2b)$, three circles $K, K_{1}$, and $K_{2}$ are placed. Circle $K$ is externally tangent to circles $K_{1}$ and $K_{2}$, and also to the lines $AD$ and $BC$. Circles $K_{1}$ and $K_{2}$ are also tangent to the sides $AD, AB$ and $AD, CD$ respectively. Find the maximum and minimum possible value of the sum of the areas of the three circles.
Answer: $\quad S_{\max }=\pi\left(R^{2}+\rho^{2}+r^{2}\right)_{\max }=\frac{\pi}{4}\left(6(3-2 \sqrt{2}) b^{2}+(\sqrt{2 a}-\sqrt{b})^{4}\right)$

$$
S_{\min }=\pi\left(R^{2}+\rho^{2}+r^{2}\right)=\pi\left(\frac{b^{2}}{4}+\frac{(\sqrt{a+b}-\sqrt{b})^{4}}{2}\right)
$$

Solution.

Let $\rho, r, R$ be the radii of circles $K_{1}, K_{2}, K$. $b=2 R$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3f94d691aa01194470c9g-04.jpg?height=300&width=526&top_left_y=455&top_left_x=685)

$A E=x-$ variable, $x \in[R ; a-R], T E=P N=x-\rho=2 \sqrt{R \rho}$,

$E L=Q M=2 \sqrt{R r}=a-x-r, O P=R+\rho, O Q=R+r, O M=R-r, O N=R-\rho$.

The conditions $\left\{\begin{array}{c}\rho+2 \sqrt{R} \cdot \sqrt{\rho}-x=0 \\ r+2 \sqrt{R} \cdot \sqrt{r}-a+x=0 \\ \rho^{2}+r^{2} \rightarrow \max \end{array}\right.$ lead to the dependence of $r$ and $\rho$ on $x$ in the form $\left\{\begin{array}{c}\rho=(\sqrt{x+R}-\sqrt{R})^{2} \\ \rho=(\sqrt{a-x+R}-\sqrt{R})^{2}\end{array}\right.$. Denoting $\varphi(x)=(\sqrt{x+R}-\sqrt{R})^{4}$, we arrive at $\rho^{2}+r^{2}=\varphi(x)+\varphi(a-x)$.

Properties of the function $\varphi(x): 1) \varphi(x) \uparrow$ on the interval $[R ; a-R]$
2) $\varphi^{\prime}(x)=\frac{2(\sqrt{x+R}-\sqrt{R})^{3}}{\sqrt{x+R}}>0, x \in[R ; a-R]$

$\varphi^{\prime}(x) \uparrow$ on the interval $[R ; a-R]$.
3) $\varphi^{\prime \prime}=\frac{(\sqrt{x+R}-\sqrt{R})^{2}(2 \sqrt{x+R}+\sqrt{R})}{(\sqrt{x+R})^{3}}>0, x \in[R ; a-R]$, i.e., the function

Critical points of the function:

$\left(\rho^{2}+r^{2}\right)^{\prime}=\varphi^{\prime}(x)-\varphi^{\prime}(a-x)=0 \rightarrow \varphi^{\prime}(x)=\varphi^{\prime}(a-x) \rightarrow x=a-x \rightarrow x^{*}=\frac{a}{2}$

At the endpoints of the interval, the values of the function $\rho^{2}+r^{2}=\varphi(x)+\varphi(a-x)$ are the same and equal to

$\varphi(R)+\varphi(a-R)=(\sqrt{2 R}-\sqrt{R})^{4}+(\sqrt{a}-\sqrt{R})^{4}=\frac{b^{2}}{4}(\sqrt{2}-1)^{4}+\frac{1}{4}(\sqrt{2 a}-\sqrt{b})^{4}$. This value is the greatest for $\rho^{2}+r^{2}$ when $b \leq a \leq 2 b$. Under the same conditions, the minimum value of $\rho^{2}+r^{2}$ is achieved at $x=a / 2$ and it is equal to $2 \varphi\left(\frac{a}{2}\right)=\frac{(\sqrt{a+b}-\sqrt{b})^{4}}{2}$. Then the maximum value of the sum of the areas of the circles

$$
S_{\max }=\pi\left(R^{2}+\rho^{2}+r^{2}\right)_{\max }=\frac{\pi}{4}\left(6(3-2 \sqrt{2}) b^{2}+(\sqrt{2 a}-\sqrt{b})^{4}\right)
$$

and the minimum value

$$
S_{\min }=\pi\left(R^{2}+\rho^{2}+r^{2}\right)=\pi\left(\frac{b^{2}}{4}+\frac{(\sqrt{a+b}-\sqrt{b})^{4}}{2}\right)(* *)
$$