Vertices of a variable acute angled triangle ABC lies on a fixed circle. Also a, b, c and A, B, C are lengths of sides and angles of triangle ABC, respectively. If `x_(1),x_(2) and x_(3)` are distances of orthocentre from A, B and C, respectively, then the maximum value of `((dx_(1))/(da)+(dx_(2))/(db)+(dx_(3))/(dc))` is

Options

A. `-sqrt3`
B. `-3sqrt3`
C. `sqrt3`
D. `3sqrt3`

Explanation

To solve the problem, we need to find the maximum value of the expression \(\frac{dx_1}{da} + \frac{dx_2}{db} + \frac{dx_3}{dc}\), where \(x_1\), \(x_2\), and \(x_3\) are the distances from the orthocenter of triangle \(ABC\) to the vertices \(A\), \(B\), and \(C\) respectively.

### Step-by-Step Solution:

1. **Identify the distances from the orthocenter**:
The distances from the orthocenter \(H\) to the vertices \(A\), \(B\), and \(C\) can be expressed as:
\[ x_1 = 2R \cos A, \quad x_2 = 2R \cos B, \quad x_3 = 2R \cos C \]
where \(R\) is the circumradius of triangle \(ABC\).

2. **Differentiate \(x_1\), \(x_2\), and \(x_3\)**:
We need to differentiate these distances with respect to the angles \(A\), \(B\), and \(C\):
\[ \frac{dx_1}{da} = \frac{d(2R \cos A)}{dA} = -2R \sin A \]
\[ \frac{dx_2}{db} = \frac{d(2R \cos B)}{dB} = -2R \sin B \]
\[ \frac{dx_3}{dc} = \frac{d(2R \cos C)}{dC} = -2R \sin C \]

3. **Relate the sides to the angles**:
The sides \(a\), \(b\), and \(c\) of triangle \(ABC\) are related to the angles by the law of sines:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
Therefore, we can differentiate these expressions with respect to the angles:
\[ \frac{da}{dA} = 2R \cos A, \quad \frac{db}{dB} = 2R \cos B, \quad \frac{dc}{dC} = 2R \cos C \]

4. **Apply the chain rule**:
Using the chain rule, we can express \(\frac{dx_1}{da}\), \(\frac{dx_2}{db}\), and \(\frac{dx_3}{dc}\):
\[ \frac{dx_1}{da} = \frac{dx_1}{dA} \cdot \frac{dA}{da} = \frac{-2R \sin A}{2R \cos A} = -\tan A \]
\[ \frac{dx_2}{db} = \frac{dx_2}{dB} \cdot \frac{dB}{db} = \frac{-2R \sin B}{2R \cos B} = -\tan B \]
\[ \frac{dx_3}{dc} = \frac{dx_3}{dC} \cdot \frac{dC}{dc} = \frac{-2R \sin C}{2R \cos C} = -\tan C \]

5. **Combine the results**:
Now we can combine these results:
\[ \frac{dx_1}{da} + \frac{dx_2}{db} + \frac{dx_3}{dc} = -\tan A - \tan B - \tan C \]

6. **Use the inequality for the sum of tangents**:
For an acute triangle, we know:
\[ \tan A + \tan B + \tan C \geq 3\sqrt{3} \]
Therefore:
\[ -(\tan A + \tan B + \tan C) \leq -3\sqrt{3} \]

7. **Conclusion**:
Thus, the maximum value of \(\frac{dx_1}{da} + \frac{dx_2}{db} + \frac{dx_3}{dc}\) is:
\[ \boxed{-3\sqrt{3}} \]