A triangle is inscribed in a circle of radius 1. The distance between the orthocentre and the circumcentre of the triangle cannot be

Options

A. 1
B. 2
C. `3//2`
D. 4

Explanation

To solve the problem, we need to determine the possible distances between the orthocenter (H) and the circumcenter (O) of a triangle inscribed in a circle of radius 1.

### Step-by-Step Solution:

1. **Understanding the Triangle and Circle**:
- We have a triangle inscribed in a circle (circumcircle) with a radius of 1. The circumcenter (O) of this triangle is at the center of the circle, which is the origin (0, 0).

2. **Vertices of the Triangle**:
- The vertices of the triangle can be represented as \( A(\cos \theta_1, \sin \theta_1) \), \( B(\cos \theta_2, \sin \theta_2) \), and \( C(\cos \theta_3, \sin \theta_3) \).

3. **Finding the Orthocenter**:
- The orthocenter (H) of the triangle can be calculated as:
\[ H = \left( \cos \theta_1 + \cos \theta_2 + \cos \theta_3, \sin \theta_1 + \sin \theta_2 + \sin \theta_3 \right) \]

4. **Distance Between Orthocenter and Circumcenter**:
- The distance \( d \) between the orthocenter (H) and the circumcenter (O) is given by the distance formula:
\[ d = \sqrt{(H_x - O_x)^2 + (H_y - O_y)^2} \]
- Since \( O \) is at the origin (0, 0), this simplifies to:
\[ d = \sqrt{(\cos \theta_1 + \cos \theta_2 + \cos \theta_3)^2 + (\sin \theta_1 + \sin \theta_2 + \sin \theta_3)^2} \]

5. **Using Trigonometric Identities**:
- We can use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) to simplify the expression:
\[ d^2 = (\cos \theta_1 + \cos \theta_2 + \cos \theta_3)^2 + (\sin \theta_1 + \sin \theta_2 + \sin \theta_3)^2 \]
- Expanding this gives:
\[ d^2 = 3 + 2(\cos \theta_1 \cos \theta_2 + \cos \theta_2 \cos \theta_3 + \cos \theta_3 \cos \theta_1 + \sin \theta_1 \sin \theta_2 + \sin \theta_2 \sin \theta_3 + \sin \theta_3 \sin \theta_1) \]
- This can be rewritten using the cosine of the angles between the vertices.

6. **Finding Maximum and Minimum Values**:
- The maximum value of \( d \) occurs when the triangle is equilateral, and the minimum value occurs when the triangle degenerates into a line segment. The maximum distance between the orthocenter and circumcenter for a triangle inscribed in a circle of radius 1 is less than 3.

7. **Conclusion**:
- The distance \( d \) can take values from 0 to less than 3. Therefore, the distance cannot be 4, as it exceeds the maximum possible distance.

### Final Answer:
The distance between the orthocenter and circumcenter of the triangle cannot be **4**.