In triangle ABC, line joining the circumcenter and orthocenter is parallel to side AC, then the value of tan A tan C is equal to

Options

A. `sqrt3`
B. 3
C. `3sqrt3`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given information and derive the required value of \( \tan A \tan C \).

### Step 1: Understand the Given Information
We have a triangle \( ABC \) where the line joining the circumcenter \( O \) and orthocenter \( H \) is parallel to side \( AC \). This implies a specific relationship between the angles of the triangle.

### Step 2: Use the Distance Formulas
The distance from the circumcenter \( O \) to side \( AC \) can be expressed as:
\[ d(O, AC) = r \cos B \]
where \( r \) is the circumradius of triangle \( ABC \).

Similarly, the distance from the orthocenter \( H \) to side \( AC \) is given by:
\[ d(H, AC) = 2r \cos A \cos C \]

### Step 3: Set the Distances Equal
Since the line \( OH \) is parallel to side \( AC \), we can equate the two distances:
\[ r \cos B = 2r \cos A \cos C \]

### Step 4: Simplify the Equation
Assuming \( r \neq 0 \) (which is valid since we are dealing with a triangle), we can divide both sides by \( r \):
\[ \cos B = 2 \cos A \cos C \]

### Step 5: Use the Cosine Angle Sum Identity
Using the identity \( \cos(A + C) = \cos A \cos C - \sin A \sin C \), we can rewrite the equation:
\[ \cos B = 2 \cos A \cos C \]
This can be rearranged to:
\[ \cos A + \cos C = 2 \cos A \cos C \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ \cos A + \cos C - 2 \cos A \cos C = 0 \]

### Step 7: Use the Sine Product Identity
Using the identity \( \sin A \sin C = \frac{1}{2} (\cos(A - C) - \cos(A + C)) \), we can relate the angles:
\[ \sin A \sin C = \frac{1}{3} \cos A \cos C \]

### Step 8: Divide by \( \cos A \cos C \)
Dividing both sides by \( \cos A \cos C \) yields:
\[ \tan A \tan C = 3 \]

### Conclusion
Thus, the value of \( \tan A \tan C \) is:
\[ \boxed{3} \]