`rr_1 r_2 r_3 =`

Options

A. `R^2`
B. `S^2`
C. `s^2`
D. none

Answer

To solve the problem \( r \cdot r_1 \cdot r_2 \cdot r_3 \), where \( r \) is the radius of the incircle and \( r_1, r_2, r_3 \) are the radii of the excircles of a triangle, we can follow these steps:

### Step 1: Understand the Definitions
- Let \( r \) be the radius of the incircle of triangle \( ABC \).
- Let \( r_1, r_2, r_3 \) be the radii of the excircles opposite to vertices \( A, B, C \) respectively.

### Step 2: Use the Formulas
The formulas for the inradius \( r \) and the exradii \( r_1, r_2, r_3 \) are:
- \( r = \frac{A}{s} \)
- \( r_1 = \frac{A}{s - a} \)
- \( r_2 = \frac{A}{s - b} \)
- \( r_3 = \frac{A}{s - c} \)

Where:
- \( A \) is the area of the triangle.
- \( s \) is the semi-perimeter of the triangle, given by \( s = \frac{a + b + c}{2} \).

### Step 3: Substitute the Values
Now, substituting the values into the expression \( r \cdot r_1 \cdot r_2 \cdot r_3 \):
\[ r \cdot r_1 \cdot r_2 \cdot r_3 = \left(\frac{A}{s}\right) \cdot \left(\frac{A}{s - a}\right) \cdot \left(\frac{A}{s - b}\right) \cdot \left(\frac{A}{s - c}\right) \]

### Step 4: Combine the Terms
This can be simplified as:
\[ r \cdot r_1 \cdot r_2 \cdot r_3 = \frac{A^4}{s \cdot (s - a) \cdot (s - b) \cdot (s - c)} \]

### Step 5: Use Heron's Formula
Using Heron's formula for the area \( A \):
\[ A = \sqrt{s(s - a)(s - b)(s - c)} \]
Substituting this into the equation gives:
\[ A^2 = s(s - a)(s - b)(s - c) \]
Thus:
\[ A^4 = (s(s - a)(s - b)(s - c))^2 \]

### Step 6: Final Expression
Now substituting back into our expression:
\[ r \cdot r_1 \cdot r_2 \cdot r_3 = \frac{(s(s - a)(s - b)(s - c))^2}{s \cdot (s - a)(s - b)(s - c)} \]
This simplifies to:
\[ r \cdot r_1 \cdot r_2 \cdot r_3 = s \cdot (s - a)(s - b)(s - c) \]

### Conclusion
Thus, the final result is:
\[ r \cdot r_1 \cdot r_2 \cdot r_3 = s \cdot (s - a)(s - b)(s - c) \]