5. Four circles with radius 1 and centers at $A, B, C, D$ are arranged on a plane such that each circle touches exactly two others. A fifth circle touches two of these circles externally and passes through the centers of the other two. Find all possible values of the area of quadrilateral ABCD.
5. Answer: 4 or 3. Solution: Let's assume that the circle centered at $B$ touches the circles centered at $\mathrm{A}$ and $\mathrm{C}$, and the circle centered at $\mathrm{C}$ touches the circles centered at $\mathrm{B}$ and $\mathrm{D}$. It is easy to see that $A B C D$ is a rhombus with side length 2. Let $\mathrm{O}$ be the center of the fifth circle. There are two fundamentally different cases. The fifth circle passes through two adjacent vertices of this rhombus (for example, A and B) or through the vertices of two opposite vertices (for example, A and C). In the first case, $\mathrm{OA}=\mathrm{OB}$ and $\mathrm{OC=OD}$. This means that the perpendicular bisectors of the sides $A B$ and $C D$ intersect. But then $A B C D$ is a square with area 4. In the second case, $O$ is the point of intersection of the diagonals of the rhombus. If the radius of the fifth circle is $\mathrm{x}$, then by the Pythagorean theorem $x^{2}+(1+x)^{2}=4$, from which $x=\frac{\sqrt{7}-1}{2}, 1+x=\frac{\sqrt{7}+1}{2}, S=3$.

Criteria: full solution - 7 points. Only the square case is considered (with an explanation of why it is a square) - 2 points. Only the second case is considered, and the square case is lost - 4 points. Everything is considered correctly, but there is an error in the area calculation - 5 points. In all other cases, 0 points.