In a parallelogram, we placed two unit circles such that they touch each other and three of the parallelogram's sides. One vertex of the parallelogram is $\sqrt{3}$ units away from the nearest point of tangency.

What is the area of the parallelogram?
Since both circles touch the 3-3 sides of the parallelogram, there are at least 2 sides that both circles touch. These 2 sides are common tangents of the two circles, and indeed, they are common external tangents because both circles are inside the parallelogram. The radii of the circles are equal, so the two tangents are parallel, meaning the two sides are opposite sides of the parallelogram. The circles can only touch each other externally, so the two circles can only be arranged in the parallelogram as shown in the diagram.

$$
\text { Figure 1 }
$$

Using the notation in the diagram. Let $D D_{C}=\sqrt{3}$, the hypotenuse of the right triangle $D D_{C} O_{1}$, according to the Pythagorean theorem, is $\sqrt{(\sqrt{3})^{2}+1^{2}}=2$, so this triangle is congruent to half of an equilateral triangle, i.e., $O_{1} D D_{C} \angle=30^{\circ}$. The line $D O_{1}$ is the angle bisector of $\angle A D C$, so $\angle A D C=60^{\circ}$. The sum of the adjacent angles of the parallelogram is $180^{\circ}$, so $\angle D C B=180^{\circ}-\angle A D C=120^{\circ}$. The line $O_{2} C$ bisects $\angle D C B$, so the angle $\angle O_{2} C C_{D}$ in the right triangle $O_{2} C C_{D}$ is $60^{\circ}$, making this triangle also congruent to half of an equilateral triangle. Therefore, $C_{D} C=\frac{\sqrt{3}}{3}$. The quadrilateral $D_{C} O_{1} O_{2} C_{D}$ is a rectangle, so $D_{C} C_{D}=O_{1} O_{2}=2$, and thus the length of the side $D C$ of the parallelogram is

$$
D C=D D_{C}+D_{C} C_{D}+C_{D} C=\sqrt{3}+2+\frac{\sqrt{3}}{3}=\frac{6+4 \sqrt{3}}{3}
$$

The height corresponding to the side $D C$ of the parallelogram is $D_{C} A_{B}=2$, so the desired area is $\frac{6+4 \sqrt{3}}{3} \cdot 2=\frac{12+8 \sqrt{3}}{3} \approx$ 8.62 square units.

Based on the work of Gábor Braun (Budapest, Szent István Gymnasium, 1st year)

![](https://cdn.mathpix.com/cropped/2024_05_02_99e562434f0f449bd697g-1.jpg?height=355&width=830&top_left_y=1096&top_left_x=626)