`1/(bc)+1/(ca)+1/(ab)=`

Options

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

Answer

To solve the expression \( \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} \), we will follow these steps:

### Step 1: Find a common denominator
The common denominator for the fractions \( \frac{1}{bc} \), \( \frac{1}{ca} \), and \( \frac{1}{ab} \) is \( abc \).

### Step 2: Rewrite each fraction with the common denominator
We rewrite each term:
\[ \frac{1}{bc} = \frac{a}{abc}, \quad \frac{1}{ca} = \frac{b}{abc}, \quad \frac{1}{ab} = \frac{c}{abc} \]

### Step 3: Combine the fractions
Now we can combine the fractions:
\[ \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} = \frac{a + b + c}{abc} \]

### Step 4: Define the semi-perimeter
The semi-perimeter \( s \) of a triangle with sides \( a, b, c \) is given by:
\[ s = \frac{a + b + c}{2} \]
Thus, \( a + b + c = 2s \).

### Step 5: Substitute into the expression
Substituting \( a + b + c \) into our combined fraction gives:
\[ \frac{a + b + c}{abc} = \frac{2s}{abc} \]

### Step 6: Relate \( abc \) to the area of the triangle
Using the formula for the circumradius \( R \) of a triangle, we have:
\[ R = \frac{abc}{4\Delta} \]
where \( \Delta \) is the area of the triangle. Rearranging gives:
\[ abc = 4R\Delta \]

### Step 7: Substitute \( abc \) back into the expression
Now substituting \( abc \) into our expression:
\[ \frac{2s}{abc} = \frac{2s}{4R\Delta} = \frac{s}{2R\Delta} \]

### Step 8: Relate area \( \Delta \) to the semi-perimeter \( s \)
The radius \( r \) of the incircle is given by:
\[ r = \frac{\Delta}{s} \]
Therefore, \( \Delta = rs \).

### Step 9: Substitute \( \Delta \) back into the expression
Substituting \( \Delta \) gives:
\[ \frac{s}{2R\Delta} = \frac{s}{2R(rs)} = \frac{1}{2Rr} \]

### Final Answer
Thus, the final result for the expression \( \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} \) is:
\[ \frac{1}{2Rr} \]

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