Question

In \( \triangle ABC \), if \( r_1 + r_2 = 3R \), \( r_2 + r_3 = 2R \), then what type of triangle is \( \triangle ABC \)?

• A. \( \text{ABC is a right-angled isosceles triangle} \)
• B. \( B = \frac{\pi}{3} \)
• C. \( A = 90^\circ, \quad a \neq b \neq c \)
• D. \( C = 90^\circ, \quad a:b:c = 2:1:\sqrt{3} \)

Hint:

When analyzing triangle properties using circumradius and exradii, remember the relation \( r_1 + r_2 + r_3 = 4R \). This can help in classifying the type of triangle based on given conditions.

Answer 1

The Correct Option is C

We are given the conditions: \[ r_1 + r_2 = 3R, \quad r_2 + r_3 = 2R \] where: - \( r_1, r_2, r_3 \) are the exradii, - \( R \) is the circumradius of \( \triangle ABC \). 

Step 1: Recall the exradius and circumradius relation 
For any triangle, \[ r_1 + r_2 + r_3 = 4R \] From the given conditions, \[ r_1 + r_2 = 3R \] \[ r_2 + r_3 = 2R \] Adding both equations, \[ r_1 + 2r_2 + r_3 = 5R \] Since we know \( r_1 + r_2 + r_3 = 4R \), subtracting gives: \[ r_2 = R \] Substituting \( r_2 = R \) in \( r_1 + r_2 = 3R \): \[ r_1 + R = 3R \Rightarrow r_1 = 2R \] Similarly, from \( r_2 + r_3 = 2R \): \[ R + r_3 = 2R \Rightarrow r_3 = R \] 

Step 2: Identify the type of triangle 
Since \( r_1 = 2R \), \( r_2 = R \), and \( r_3 = R \), we use the identity: \[ r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \] For \( A = 90^\circ \): \[ r = 4R \sin \frac{90^\circ}{2} \sin \frac{B}{2} \sin \frac{C}{2} \] which satisfies the given conditions. This confirms that \( A = 90^\circ \), and the triangle is right-angled but non-isosceles. Thus, the correct answer is: \[ \mathbf{A = 90^\circ, \quad a \neq b \neq c} \]