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 distance between the orthocenter and the circumcenter of a triangle inscribed in a circle of radius 1, and identify which of the given options cannot be this distance.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- The **circumcenter** of a triangle is the point where the perpendicular bisectors of the sides intersect. For a triangle inscribed in a circle (circumcircle), the circumcenter is at the center of the circle.
- The **orthocenter** is the point where the altitudes of the triangle intersect.

2. **Setting Up the Problem**:
- Let the circumradius \( R \) of the triangle be 1 (since the triangle is inscribed in a circle of radius 1).
- The circumcenter \( O \) is at the origin, \( O(0, 0) \).

3. **Finding the Orthocenter**:
- The coordinates of the orthocenter \( H \) can be expressed in terms of the angles of the triangle. If the angles are \( \theta_1, \theta_2, \theta_3 \), then the coordinates of the orthocenter can be given as:
\[ H = ( \cos \theta_1 + \cos \theta_2 + \cos \theta_3, \sin \theta_1 + \sin \theta_2 + \sin \theta_3 ) \]

4. **Calculating the Distance**:
- The distance \( d \) between the orthocenter \( H \) and the circumcenter \( O \) is given by:
\[ d = \sqrt{(\cos \theta_1 + \cos \theta_2 + \cos \theta_3)^2 + (\sin \theta_1 + \sin \theta_2 + \sin \theta_3)^2} \]

5. **Using the Cauchy-Schwarz Inequality**:
- By applying the Cauchy-Schwarz inequality, we can derive that:
\[ d^2 \leq (1 + 1 + 1)(\cos^2 \theta_1 + \cos^2 \theta_2 + \cos^2 \theta_3 + \sin^2 \theta_1 + \sin^2 \theta_2 + \sin^2 \theta_3) = 3 \]
- Therefore, \( d \leq \sqrt{3} \).

6. **Evaluating the Options**:
- The options given are:
- A) 1
- B) 2
- C) \( \frac{3}{2} \)
- D) 4
- Since the maximum possible distance \( d \) can be \( \sqrt{3} \), any value greater than \( \sqrt{3} \) cannot be the distance between the orthocenter and circumcenter.

7. **Conclusion**:
- The only option that exceeds the maximum possible distance \( \sqrt{3} \) is option D) 4. Therefore, the distance between the orthocenter and the circumcenter of the triangle cannot be 4.

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