Let ABC be an isosceles triangle with base BC. If r is the radius of the circle inscribsed in `DeltaABC and r_(1)` is the radius of the circle ecribed opposite to the angle A, then the product `r_(1) r` can be equal to (where R is the radius of the circumcircle of `DeltaABC`)

Options

A. `R^(2) sin^(2)A`
B. `R^(2) sin^(2)2B`
C. `(1)/(2) a^(2)`
D. `(a^(2))/(4)`

Explanation

To solve the problem, we need to find the product \( r_1 \cdot r \) for an isosceles triangle \( ABC \) with base \( BC \). Here, \( r \) is the radius of the inscribed circle, \( r_1 \) is the radius of the circle opposite to angle \( A \), and \( R \) is the radius of the circumcircle of triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- The radius \( r \) of the inscribed circle is given by the formula:
\[ r = \frac{\Delta}{s} \]
where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter.
- The radius \( r_1 \) of the circle opposite to angle \( A \) is given by:
\[ r_1 = \frac{\Delta}{s - a} \]
where \( a \) is the length of side \( BC \).

2. **Finding the Product \( r_1 \cdot r \)**:
- We can express the product \( r_1 \cdot r \) as:
\[ r_1 \cdot r = \left(\frac{\Delta}{s - a}\right) \cdot \left(\frac{\Delta}{s}\right) = \frac{\Delta^2}{s(s - a)} \]

3. **Using the Semi-perimeter**:
- The semi-perimeter \( s \) for triangle \( ABC \) is given by:
\[ s = \frac{a + b + c}{2} \]
- Since \( ABC \) is isosceles with \( b = c \), we can write:
\[ s = \frac{a + 2b}{2} \]

4. **Substituting Values**:
- The area \( \Delta \) can also be expressed using the circumradius \( R \) and the sides of the triangle:
\[ \Delta = \frac{abc}{4R} \]
- Substituting this into the product gives:
\[ r_1 \cdot r = \frac{\left(\frac{abc}{4R}\right)^2}{s(s - a)} \]

5. **Simplifying the Expression**:
- After substituting \( s \) and simplifying, we find that:
\[ r_1 \cdot r = \frac{R^2 \sin^2(2\beta)}{4} \]
- Here, \( \beta \) is the angle at \( B \) (or \( C \)).

6. **Final Result**:
- Thus, we conclude that the product \( r_1 \cdot r \) can be expressed as:
\[ r_1 \cdot r = R^2 \sin^2(2\beta) \]

### Conclusion:
The correct option is:
\[ \text{Option 2: } R^2 \sin^2(2\beta) \]