3. In the plane, there are two identical circles with a radius of 1, which have an external tangency. Consider a rectangle containing both circles, each side of which touches at least one of them. Determine the largest and smallest possible area of such a rectangle.
SOLUTION. Let $k_{1}, k_{2}$ be two identical circles with radius $r=1$ and centers $O_{1}, O_{2}$, respectively, which are externally tangent. Let $A B C D$ be a rectangle satisfying the conditions of the problem (sides $A B$ and $B C$ of the rectangle are tangents to circle $k_{1}$, and sides $C D$ and $D A$ are tangents to circle $k_{2}$). From the problem statement, it is clear that the centers of both circles lie inside the sought rectangle. Let $P$ be the intersection of the lines passing through the center $O_{1}$ parallel to side $A B$ and through the center $O_{2}$ parallel to side $B C$ (Fig. 2). Let $\varphi$ be the measure of angle $\mathrm{PO}_{1} \mathrm{O}_{2}$. Considering the symmetry of the problem, it suffices to consider only values $\varphi \in\left\langle 0 ; \frac{1}{4} \pi\right\rangle$.

![](https://cdn.mathpix.com/cropped/2024_04_17_bd202734538ab0b74df6g-04.jpg?height=783&width=1117&top_left_y=1413&top_left_x=481)

Fig. 2

For the area $S$ of the rectangle $A B C D$ according to Fig. 2, we have

$$
S=|A B| \cdot|B C|=(2 r+2 r \cos \varphi)(2 r+2 r \sin \varphi)=4(1+\sin \varphi)(1+\cos \varphi)
$$

(substituting the value $r=1$). Now we determine the minimum and maximum values of the area $S$ with the variable angle $\varphi \in\left\langle 0 ; \frac{1}{4} \pi\right\rangle$. It suffices to determine the minimum and maximum values of the expression (function)

$$
V(\varphi)=(1+\sin \varphi)(1+\cos \varphi) \quad \text { for } \varphi \in\left\langle 0 ; \frac{1}{4} \pi\right\rangle
$$

We first modify this function expression as follows:

$$
V(\varphi)=1+\sin \varphi+\cos \varphi+\sin \varphi \cos \varphi=\frac{1}{2}+(\sin \varphi+\cos \varphi)+\frac{1}{2}(\sin \varphi+\cos \varphi)^{2}
$$

Everything thus depends on the value of $u=\sin \varphi+\cos \varphi$. We will show that for any $\varphi \in\left\langle 0 ; \frac{1}{4} \pi\right\rangle$, $1 \leqq u \leqq \sqrt{2}$. Since for such $\varphi$ the sum $\sin \varphi+\cos \varphi$ is clearly positive, we can instead of finding the minimum and maximum values of the expression $u$ find the minimum and maximum values of the expression $u^{2}$ (on the same interval). We have

$$
u^{2}=\sin ^{2} \varphi+2 \sin \varphi \cos \varphi+\cos ^{2} \varphi=1+\sin 2 \varphi \text {. }
$$

From this, considering the inequality $0 \leqq \sin 2 \varphi \leqq 1$, we get

$$
1 \leqq u^{2} \leqq 2, \quad \text { hence } \quad 1 \leqq u \leqq \sqrt{2}
$$

where $u=1$ if and only if $\varphi=0$, and $u=\sqrt{2}$ if and only if $\varphi=\frac{1}{4} \pi$.

Since the quadratic function $V(u)=\frac{1}{2}+u+\frac{1}{2} u^{2}=\frac{1}{2}(u+1)^{2}$ is increasing on the interval $\langle 1 ; \sqrt{2}\rangle$, for any such $u$ the inequalities hold

$$
2=V(1) \leqq V(u) \leqq V(\sqrt{2})=\frac{3+2 \sqrt{2}}{2}
$$

which already lead to the necessary conclusions about the area $S=4 V(\varphi)$:

$$
8 \leqq S \leqq 6+4 \sqrt{2}
$$

The minimum value of $S$ in relation (1) is achieved precisely when $\varphi=0$, i.e., in the case where the sides $A B$ and $C D$ of the sought rectangle $A B C D$ are tangents to both given circles $k_{1}$ and $k_{2}$, and the sides $B C$ and $D A$ are tangents to circles $k_{1}$ and $k_{2}$, respectively. This is the case where the sought "circumscribed" rectangle (according to the conditions of the problem) is a rectangle with sides 4 and 2, whose area is indeed 8. Similarly, the maximum value of $S$ in relation (1) is achieved precisely when $\varphi=\frac{1}{4} \pi$, i.e., in the case where the sought "circumscribed" rectangle is a square $A B C D$ with area $6+4 \sqrt{2}$, on whose diagonal $B D$ the centers of both given circles lie.

Answer. The minimum possible area is 8, the maximum is $6+4 \sqrt{2}$.