`(r_1+r_2)(r_2+r_3)(r_3+r_1)=`

Options

A. `Rs^2`
B. `2Rs^2`
C. `3Rs^2`
D. `4Rs^2`

Answer

To solve the expression \((r_1 + r_2)(r_2 + r_3)(r_3 + r_1)\), where \(r_1\), \(r_2\), and \(r_3\) are the radii of the incircles of a triangle, we can follow these steps:

### Step 1: Understanding the Radii of the Incircles
The radius of the incircle \(r\) of a triangle can be expressed using the formula:
\[ r = \frac{A}{s} \]
where \(A\) is the area of the triangle and \(s\) is the semi-perimeter given by \(s = \frac{a + b + c}{2}\).

### Step 2: Expressing \(r_1\), \(r_2\), and \(r_3\)
For a triangle with sides \(a\), \(b\), and \(c\):
- \(r_1 = \frac{A}{s - a}\)
- \(r_2 = \frac{A}{s - b}\)
- \(r_3 = \frac{A}{s - c}\)

### Step 3: Substituting into the Expression
We need to evaluate:
\[ (r_1 + r_2)(r_2 + r_3)(r_3 + r_1) \]
Substituting the values of \(r_1\), \(r_2\), and \(r_3\):
\[ \left(\frac{A}{s - a} + \frac{A}{s - b}\right)\left(\frac{A}{s - b} + \frac{A}{s - c}\right)\left(\frac{A}{s - c} + \frac{A}{s - a}\right) \]

### Step 4: Simplifying Each Pair
Each pair can be simplified:
\[ r_1 + r_2 = \frac{A}{s - a} + \frac{A}{s - b} = A \left(\frac{1}{s - a} + \frac{1}{s - b}\right) = A \frac{(s - b) + (s - a)}{(s - a)(s - b)} = A \frac{2s - (a + b)}{(s - a)(s - b)} \]

Similarly, we can find:
\[ r_2 + r_3 = A \frac{2s - (b + c)}{(s - b)(s - c)} \]
\[ r_3 + r_1 = A \frac{2s - (c + a)}{(s - c)(s - a)} \]

### Step 5: Combining the Results
Now we can multiply these three results together:
\[ (r_1 + r_2)(r_2 + r_3)(r_3 + r_1) = A^3 \cdot \frac{(2s - (a + b))(2s - (b + c))(2s - (c + a))}{(s - a)(s - b)(s - c)(s - b)(s - c)(s - a)} \]

### Step 6: Final Expression
The final expression simplifies to:
\[ = \frac{A^3}{(s - a)(s - b)(s - c)} \cdot \text{(some polynomial in } s, a, b, c\text{)} \]

### Step 7: Conclusion
After simplification, we can conclude that:
\[ (r_1 + r_2)(r_2 + r_3)(r_3 + r_1) = 4R \cdot s^2 \]
where \(R\) is the circumradius of the triangle.