
3. Two externally tangent unit circles are given in the plane. Consider any rectangle (or a square) containing both the circles such that each side of the rectangle is tangent to at least one circle. Find the largest and the smallest possible area of such a rectangle.

(Jaroslav Švrček)


Solution. Denote the circles by $k_{1}, k_{2}$, their radius by $r=1$, and their centers by $O_{1}, O_{2}$, respectively. Let $A B C D$ be one such rectangle (or a square) and without loss of generality assume that the sides $A B, B C$ are tangent to $k_{1}$ while the sides $C D$, $D A$ are tangent to $k_{2}$. Let $P$ be the intersection of a line through $O_{1}$ parallel to $A B$ and a line through $O_{2}$ parallel to $B C$. Finally, let $\phi=\angle P O_{1} O_{2}\left(\phi \in\left[0, \frac{1}{4} \pi\right]\right.$, Fig. 2).

Then

$$
[A B C D]=A B \cdot B C=(2 r+2 r \cos \phi)(2 r+2 r \sin \phi)=4(1+\sin \phi)(1+\cos \phi) \text {. }
$$

It remains to analyze the expression $V(\phi)=(1+\sin \phi)(1+\cos \phi)$ for $\phi \in\left[0, \frac{1}{4} \pi\right]$. Multiplying out, this rewrites as

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

![](https://cdn.mathpix.com/cropped/2024_04_17_c0cc9d17594ecfc50dfag-05.jpg?height=786&width=1125&top_left_y=281&top_left_x=431)

Fig. 2

and it remains to analyze $u=\sin \phi+\cos \phi$. Squaring and using the formula $\sin 2 \phi=$ $2 \sin \phi \cos \phi$, we obtain $1 \leqslant u \leqslant \sqrt{2}$. Since the function $V(\phi)=\frac{1}{2}+u+\frac{1}{2} u^{2}$ is increasing on interval $[1, \sqrt{2}]$, we get $2 \leqslant \phi \leqslant \frac{3}{2}+\sqrt{2}$ and finally

$$
8 \leqslant[A B C D] \leqslant 6+4 \sqrt{2} .
$$

The first inequality is sharp for $\phi=0$, that is if both $A B$ and $C D$ are tangent to both the circles. The second inequality is sharp for $\phi=\frac{1}{4} \pi$, that is if $A B C D$ is a square.
