If H is the orthocentre of triangle ABC, R = circumradius and `P = AH + BH + CH`, then

Options

A. `P = 2 (R + r)`
B. max. of P is 3R
C. min. of P is 3R
D. `P = 2 (R -r)`

Explanation

To solve the problem, we need to analyze the relationship between the orthocenter \( H \), circumradius \( R \), and the sum \( P = AH + BH + CH \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Orthocenter**:
The orthocenter \( H \) of triangle \( ABC \) is the point where the three altitudes of the triangle intersect.

2. **Using the Relationship of Distances**:
The distances from the vertices \( A \), \( B \), and \( C \) to the orthocenter \( H \) can be expressed in terms of the circumradius \( R \) and the angles of the triangle:
\[ AH = 2R \cos A, \quad BH = 2R \cos B, \quad CH = 2R \cos C \]

3. **Calculating \( P \)**:
Now we can substitute these expressions into our equation for \( P \):
\[ P = AH + BH + CH = 2R \cos A + 2R \cos B + 2R \cos C \]
Factoring out \( 2R \):
\[ P = 2R (\cos A + \cos B + \cos C) \]

4. **Using the Cosine Sum Formula**:
We know from trigonometric identities that:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]
where \( r \) is the inradius of triangle \( ABC \).

5. **Substituting Back**:
Substituting this identity back into our expression for \( P \):
\[ P = 2R \left(1 + \frac{r}{R}\right) = 2R + 2r \]

6. **Finding the Maximum and Minimum Values**:
We know that \( r \leq \frac{R}{2} \) (the inradius is less than or equal to half the circumradius). Thus:
\[ P \leq 2R + 2 \left(\frac{R}{2}\right) = 3R \]

7. **Conclusion**:
Therefore, we conclude that:
\[ P < 3R \]
and the maximum value of \( P \) is \( 3R \).

### Final Result:
Thus, the final result is:
\[ P < 3R \]