`(r_1-r)(r_2+r_3)=?`

Options

A. `a^2`
B. `b^2`
C. `c^2`
D. none

Answer

To solve the expression \((r_1 - r)(r_2 + r_3)\), we need to understand the definitions of \(r_1\), \(r_2\), and \(r\) in the context of triangles.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \(R_1\) be the radius of the excircle opposite to side \(A\).
- Let \(R_2\) and \(R_3\) be the radii of the excircles opposite to sides \(B\) and \(C\) respectively.
- Let \(r\) be the radius of the incircle of the triangle.

2. **Formulas for Excircle and Incircle Radii**:
- The radius of the excircle \(R_1\) is given by:
\[ R_1 = \frac{\Delta}{s - A} \]
- The radius of the incircle \(r\) is given by:
\[ r = \frac{\Delta}{s} \]
- Where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter defined as:
\[ s = \frac{A + B + C}{2} \]

3. **Substituting the Values**:
- Substitute \(R_1\) and \(r\) into the expression:
\[ (R_1 - r) = \left(\frac{\Delta}{s - A} - \frac{\Delta}{s}\right) \]
- To combine these fractions, find a common denominator:
\[ (R_1 - r) = \frac{\Delta s - \Delta (s - A)}{(s - A)s} = \frac{\Delta A}{(s - A)s} \]

4. **Calculating \(R_2 + R_3\)**:
- Similarly, we can express \(R_2\) and \(R_3\):
\[ R_2 = \frac{\Delta}{s - B}, \quad R_3 = \frac{\Delta}{s - C} \]
- Therefore,
\[ R_2 + R_3 = \frac{\Delta}{s - B} + \frac{\Delta}{s - C} \]
- Finding a common denominator:
\[ R_2 + R_3 = \Delta \left(\frac{(s - C) + (s - B)}{(s - B)(s - C)}\right) = \Delta \left(\frac{2s - (B + C)}{(s - B)(s - C)}\right) \]

5. **Final Calculation**:
- Now we can substitute \(R_1 - r\) and \(R_2 + R_3\) back into the original expression:
\[ (R_1 - r)(R_2 + R_3) = \left(\frac{\Delta A}{(s - A)s}\right) \left(\Delta \frac{2s - (B + C)}{(s - B)(s - C)}\right) \]
- This simplifies to:
\[ = \frac{\Delta^2 A (2s - (B + C))}{(s - A)s(s - B)(s - C)} \]

### Final Answer:
\[ (r_1 - r)(r_2 + r_3) = \frac{\Delta^2 A (2s - (B + C))}{(s - A)s(s - B)(s - C)} \]