Question

In a triangle \(ABC\), \[ 2\sqrt{r_1r_2+r_2r_3+r_3r_1} = \]

• A. \(8R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\)
• B. \(8R\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\)
• C. \(4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\)
• D. \(4R\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\)

Hint:

Important identities: \[ r_1r_2+r_2r_3+r_3r_1=4R^2, \] \[ \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} = \frac{s}{4R}, \] \[ 4R = 8R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}. \]

Answer 1

The Correct Option is B

Step 1: Recall the identity involving the exradii. For any triangle, \[ r_1r_2+r_2r_3+r_3r_1 = 4R^2. \] Hence, \[ 2\sqrt{r_1r_2+r_2r_3+r_3r_1} = 2\sqrt{4R^2} = 4R. \]

Step 2: Use the half-angle identity. We know that \[ \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} = \frac{s}{4R}, \] and another standard identity gives \[ 4R = 8R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}. \] Therefore, \[ 2\sqrt{r_1r_2+r_2r_3+r_3r_1} = 8R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}. \] Thus, \[ \boxed{ 2\sqrt{r_1r_2+r_2r_3+r_3r_1} = 8R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} }. \] Hence, the correct option is \(\boxed{(B)}\).