`r_1/(bc)+r_2/(ca)+r_3/(ab)=`

Options

A. `1/(2R) -1/r`
B. `2R-r`
C. `r-2R`
D. `1/r-1/(2R)`

Answer

To solve the problem \( \frac{r_1}{bc} + \frac{r_2}{ca} + \frac{r_3}{ab} \), where \( r_1, r_2, r_3 \) are the inradii corresponding to angles \( A, B, C \) of triangle \( ABC \), we will follow these steps:

### Step 1: Write the expressions for \( r_1, r_2, r_3 \)
The inradii \( r_1, r_2, r_3 \) can be expressed in terms of the semi-perimeter \( s \) and the sides \( a, b, c \) of the triangle:
\[ r_1 = \frac{A}{s}, \quad r_2 = \frac{B}{s}, \quad r_3 = \frac{C}{s} \]
where \( A, B, C \) are the areas of the triangle corresponding to angles \( A, B, C \), and \( s = \frac{a+b+c}{2} \).

### Step 2: Substitute the values of \( r_1, r_2, r_3 \) into the expression
Now substituting these values into the expression:
\[ \frac{r_1}{bc} + \frac{r_2}{ca} + \frac{r_3}{ab} = \frac{A}{s \cdot bc} + \frac{B}{s \cdot ca} + \frac{C}{s \cdot ab} \]
Factoring out \( \frac{1}{s} \):
\[ = \frac{1}{s} \left( \frac{A}{bc} + \frac{B}{ca} + \frac{C}{ab} \right) \]

### Step 3: Use the area formula
The area \( A \) of triangle \( ABC \) can be expressed using the sides and the circumradius \( R \):
\[ A = \frac{abc}{4R} \]
Thus, we can substitute \( A, B, C \) in terms of \( R \):
\[ = \frac{1}{s} \left( \frac{\frac{abc}{4R}}{bc} + \frac{\frac{abc}{4R}}{ca} + \frac{\frac{abc}{4R}}{ab} \right) \]
This simplifies to:
\[ = \frac{1}{s} \cdot \frac{abc}{4R} \left( \frac{1}{b} + \frac{1}{a} + \frac{1}{c} \right) \]

### Step 4: Simplify the expression
Now, we can express \( \frac{1}{b} + \frac{1}{a} + \frac{1}{c} \) in terms of \( s \):
\[ = \frac{1}{s} \cdot \frac{abc}{4R} \cdot \frac{a+b+c}{abc} = \frac{1}{s} \cdot \frac{1}{4R} \]
Thus, we have:
\[ = \frac{1}{4Rs} \]

### Step 5: Final expression
Now, substituting \( s = \frac{a+b+c}{2} \):
\[ = \frac{2}{4R(a+b+c)} = \frac{1}{2R} \]

### Final Answer
The final expression is:
\[ \frac{1}{R} - \frac{1}{2R} \]