Question

In \( \triangle ABC \), if \( (r_2 - r_1)(r_3 - r_1) = 2r_2r_3 \), then \( 2(r + R) = \):

• A. \( a + b \)
• B. \( c + a \)
• C. \( 2 \sqrt{2} R \cos \left( \frac{C - A}{2} \right) \)
• D. \( 2 \sqrt{2} R \cos \left( \frac{B - C}{2} \right) \)

Hint:

For equations involving inradius and circumradius, use geometric identities and properties to simplify the given relations. Often, trigonometric functions help relate angles and side lengths.

Answer 1

The Correct Option is D

We are given the equation: \[ (r_2 - r_1)(r_3 - r_1) = 2r_2r_3. \] This is related to the semi-perimeter and other geometrical properties of the triangle. To solve for \( 2(r + R) \), we apply various geometric and trigonometric relationships, leading to the formula: \[ 2(r + R) = 2 \sqrt{2} R \cos \left( \frac{B - C}{2} \right). \] Thus, the correct answer is \( 2 \sqrt{2} R \cos \left( \frac{B - C}{2} \right) \).

Answer 2

Given in \( \triangle ABC \):
\[ (r_2 - r_1)(r_3 - r_1) = 2 r_2 r_3 \] Find \( 2(r + R) \), where:
- \( r \) is the inradius
- \( R \) is the circumradius
- \( r_1, r_2, r_3 \) are the exradii opposite to vertices \( A, B, C \) respectively.

Step 1: Recall the relations:
\[ r_i = \frac{\Delta}{s - a_i} \quad \text{and} \quad r = \frac{\Delta}{s}, \quad R = \frac{abc}{4\Delta} \] where \( s \) is the semiperimeter, \( \Delta \) is the area, and \( a_i \) is side length opposite vertex \( A_i \).

Step 2: Using known identities and the given relation, after algebraic manipulation, it can be shown that:
\[ 2(r + R) = 2 \sqrt{2} R \cos \left( \frac{B - C}{2} \right) \] This connects the sum of inradius and circumradius with the angle difference of \( B \) and \( C \).

Therefore,
\[ \boxed{2(r + R) = 2 \sqrt{2} R \cos \left( \frac{B - C}{2} \right)} \]