
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
The graphs of the functions \( y=2a+\frac{1}{x-b} \) and \( y=2c+\frac{1}{x-d} \) are centrally symmetric with respect to the point with coordinates \( \left(\frac{b+d}{2}, a+c\right) \) and, therefore, have exactly one common point if and only if

\[ 2a + \frac{1}{x-b} = 2c + \frac{1}{x-d} = a+c \]

when \( x = \frac{b+d}{2} \). This condition is equivalent to the equality \( (a-c) \times (b-d) = 2 \).

Similarly, it can be shown that this equality is also equivalent to the condition that the graphs of the functions \( y=2b + \frac{1}{x-a} \) and \( y=2d + \frac{1}{x-c} \), which are centrally symmetric with respect to the point with coordinates \( \left(\frac{a+c}{2}, b+d\right) \), have exactly one common point.

Present the answer in LaTex format: \boxed{Your answer}
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