`r_2r_3+r_3r_1+r_1r_2=`

Options

A. `s^2`
B. `S^2`
C. `S^2//r^2`
D. `R^2`

Answer

To solve the expression \( r_2 r_3 + r_3 r_1 + r_1 r_2 \), we will use the formulas for the inradii of a triangle. The inradii \( r_1, r_2, r_3 \) are defined in terms of the area \( S \) of the triangle and its semi-perimeter \( s \).

### Step-by-Step Solution:

1. **Define the Inradii**:
- The inradii \( r_1, r_2, r_3 \) can be expressed as:
\[ r_1 = \frac{S}{s-a}, \quad r_2 = \frac{S}{s-b}, \quad r_3 = \frac{S}{s-c} \]
where \( S \) is the area of the triangle, \( s \) is the semi-perimeter, and \( a, b, c \) are the lengths of the sides opposite to the vertices where the inradii are defined.

2. **Substitute the Inradii into the Expression**:
- We substitute \( r_1, r_2, r_3 \) into the expression:
\[ r_2 r_3 + r_3 r_1 + r_1 r_2 = \left(\frac{S}{s-b}\right)\left(\frac{S}{s-c}\right) + \left(\frac{S}{s-c}\right)\left(\frac{S}{s-a}\right) + \left(\frac{S}{s-a}\right)\left(\frac{S}{s-b}\right) \]

3. **Simplify Each Term**:
- Each term can be simplified:
\[ = \frac{S^2}{(s-b)(s-c)} + \frac{S^2}{(s-c)(s-a)} + \frac{S^2}{(s-a)(s-b)} \]

4. **Combine the Terms**:
- To combine these fractions, we need a common denominator:
\[ = S^2 \left( \frac{(s-a) + (s-b) + (s-c)}{(s-a)(s-b)(s-c)} \right) \]

5. **Simplify the Numerator**:
- The numerator simplifies to:
\[ (s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s \]
- Thus, we have:
\[ = S^2 \cdot \frac{s}{(s-a)(s-b)(s-c)} \]

6. **Final Expression**:
- Therefore, the final expression for \( r_2 r_3 + r_3 r_1 + r_1 r_2 \) is:
\[ \frac{S^2 s}{(s-a)(s-b)(s-c)} \]