Three circles each of radius 1, touch one another externally and they lie between two parallel lines. The minimum possible distance between the lines is

Options

A. `2+sqrt(3)`
B. `3+sqrt(3)`
C. 4
D. `2+(1)/(sqrt(3))`

Explanation

To solve the problem of finding the minimum possible distance between two parallel lines that enclose three externally touching circles of radius 1, we can follow these steps:

### Step 1: Understand the Configuration
We have three circles, each with a radius of 1, arranged such that they touch each other externally. The centers of these circles form an equilateral triangle.

### Step 2: Determine the Distance Between the Centers
Since the circles touch each other externally, the distance between the centers of any two circles is equal to the sum of their radii. Therefore, the distance between the centers of any two circles is:
\[ d = r_1 + r_2 = 1 + 1 = 2 \]
Thus, the distance between the centers of the three circles is 2 units.

### Step 3: Calculate the Height of the Triangle Formed by the Centers
The centers of the three circles form an equilateral triangle with each side measuring 2. To find the height (h) of this triangle, we can use the formula for the height of an equilateral triangle:
\[ h = \frac{\sqrt{3}}{2} \times \text{side length} \]
Substituting the side length (2):
\[ h = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \]

### Step 4: Determine the Total Distance Between the Parallel Lines
The total distance between the two parallel lines is the height of the triangle formed by the centers of the circles plus the radius of the circles at the top and bottom. Since the radius of each circle is 1, we need to add the radius twice (once for each line):
\[ \text{Total distance} = \text{height} + \text{radius} + \text{radius} = \sqrt{3} + 1 + 1 = \sqrt{3} + 2 \]

### Final Answer
Thus, the minimum possible distance between the two parallel lines is:
\[ \sqrt{3} + 2 \]