Question

In \( \triangle ABC \), if \( r = 3 \) and \( R = 5 \) then \( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \)

• A. \( \frac{1}{30} \)
• B. \( \frac{12}{15} \)
• C. \( \frac{1}{15} \)
• D. \( \frac{5}{36} \) Correct Answer

Hint:

Remember key formulas for properties of triangles: Semi-perimeter: \( s = (a+b+c)/2 \) Area \( \Delta \): \( \Delta = rs \), \( \Delta = \frac{abc}{4R} \) Inradius \(r\), Circumradius \(R\). Combining expressions often leads to simplifications using these formulas.

Answer 1

The Correct Option is A

Step 1: Combine the terms in the expression.
\[ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{c + a + b}{abc} \]
Step 2: Relate the numerator and denominator to known triangle formulas.
The numerator \( a+b+c = 2s \), where \( s \) is the semi-perimeter.
The denominator \( abc \).
We know the area of the triangle \( \Delta = \frac{abc}{4R} \), so \( abc = 4R\Delta \).
Also, the inradius \( r = \frac{\Delta}{s} \), so \( \Delta = rs \).

Step 3: Substitute these into the expression.
\[ \frac{2s}{abc} = \frac{2s}{4R\Delta} \] Substitute \( \Delta = rs \): \[ = \frac{2s}{4R(rs)} \]
Step 4: Simplify the expression.
Assuming \( s \ne 0 \) (which is true for a triangle), we can cancel \( s \): \[ = \frac{2}{4Rr} = \frac{1}{2Rr} \]
Step 5: Substitute the given values of \( r \) and \( R \).
Given \( r = 3 \) and \( R = 5 \).
\[ \frac{1}{2Rr} = \frac{1}{2 \cdot 5 \cdot 3} = \frac{1}{30} \] This matches option (1).