Question

In $ \triangle ABC $, the expression $ \frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} $ is equivalent to:

• A. \( \frac{4R}{r} - 1 \)
• B. \( \frac{R}{r} - 3 \)
• C. \( \frac{2R}{r} - 1 \)
• D. \( \frac{4R}{r} - 2 \)

Hint:

This identity \( \frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} = \frac{4R}{r} - 2 \) is useful in various geometric applications involving the circumradius \( R \) and inradius \( r \) of a triangle. It's derived from the relationship between the sides of a triangle and its incircle and circumcircle.

Answer 1

The Correct Option is D

We are asked to evaluate the expression: \[ \frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} \] where \( a, b, c \) are the side lengths of the triangle \( \triangle ABC \) and \( s \) is the semi-perimeter, defined as: \[ s = \frac{a + b + c}{2} \] This expression involves the semi-perimeter and sides of the triangle. We will now proceed to find the value of this expression in terms of the circumradius \( R \) and inradius \( r \). 
Step 1: Simplifying the expression.
The expression \( \frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} \) is a known identity in triangle geometry. We can derive its value by using the relationship between the sides, the semi-perimeter, and the circumradius and inradius. 
Step 2: Use of known formula.
There is a well-established identity for this expression: \[ \frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} = \frac{4R}{r} - 2 \] where: \( R \) is the circumradius of the triangle,
\( r \) is the inradius of the triangle.
This formula can be derived from the properties of the triangle, but it is typically found in advanced triangle geometry and is an important identity. 
Step 3: Conclusion.
Since the expression \( \frac{a}{s-a} + \frac{b}{s-b} + \frac{c}{s-c} \) simplifies to \( \frac{4R}{r} - 2 \), the correct answer is: \[ \boxed{\frac{4R}{r} - 2} \] Thus, the correct option is Option 4.