Prove that the points `(0, 0), (3, (pi)/(2))` and `(3, (pi)/(6))` are the vertices of an equilateral triangle.

Answer

To prove that the points \( (0, 0) \), \( (3, \frac{\pi}{2}) \), and \( (3, \frac{\pi}{6}) \) are the vertices of an equilateral triangle, we will first convert the polar coordinates to Cartesian coordinates and then calculate the distances between the points.

### Step 1: Convert Polar Coordinates to Cartesian Coordinates

1. **Point A**: \( (0, 0) \)
- Cartesian coordinates: \( (x, y) = (0, 0) \)

2. **Point B**: \( (3, \frac{\pi}{2}) \)
- \( x = r \cos(\theta) = 3 \cos\left(\frac{\pi}{2}\right) = 3 \cdot 0 = 0 \)
- \( y = r \sin(\theta) = 3 \sin\left(\frac{\pi}{2}\right) = 3 \cdot 1 = 3 \)
- Cartesian coordinates: \( (0, 3) \)

3. **Point C**: \( (3, \frac{\pi}{6}) \)
- \( x = r \cos(\theta) = 3 \cos\left(\frac{\pi}{6}\right) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \)
- \( y = r \sin(\theta) = 3 \sin\left(\frac{\pi}{6}\right) = 3 \cdot \frac{1}{2} = \frac{3}{2} \)
- Cartesian coordinates: \( \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) \)

### Step 2: Calculate the Distances Between the Points

1. **Distance AB**:
- \( A(0, 0) \) and \( B(0, 3) \)
- \( d_{AB} = \sqrt{(0 - 0)^2 + (3 - 0)^2} = \sqrt{0 + 3^2} = \sqrt{9} = 3 \)

2. **Distance AC**:
- \( A(0, 0) \) and \( C\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) \)
- \( d_{AC} = \sqrt{\left(\frac{3\sqrt{3}}{2} - 0\right)^2 + \left(\frac{3}{2} - 0\right)^2} \)
- \( = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2} \)
- \( = \sqrt{\frac{27}{4} + \frac{9}{4}} = \sqrt{\frac{36}{4}} = \sqrt{9} = 3 \)

3. **Distance BC**:
- \( B(0, 3) \) and \( C\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) \)
- \( d_{BC} = \sqrt{\left(\frac{3\sqrt{3}}{2} - 0\right)^2 + \left(\frac{3}{2} - 3\right)^2} \)
- \( = \sqrt{\left(\frac{3\sqrt{3}}{2}\right)^2 + \left(-\frac{3}{2}\right)^2} \)
- \( = \sqrt{\frac{27}{4} + \frac{9}{4}} = \sqrt{\frac{36}{4}} = \sqrt{9} = 3 \)

### Conclusion

Since all three distances \( d_{AB} \), \( d_{AC} \), and \( d_{BC} \) are equal to \( 3 \), the points \( (0, 0) \), \( (0, 3) \), and \( \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right) \) form an equilateral triangle.