Question

In a triangle \(ABC\), if \[ r-r_1+r_2+r_3=2\sqrt2R,\qquad r+r_1-r_2+r_3=0 \] and \[ b=2\sqrt2 \] then \[ a+c= \]

• A. \(5\)
• B. \(6\)
• C. \(2+\sqrt2\)
• D. \(4\)

Hint:

Whenever expressions involve \(r,r_1,r_2,r_3\), immediately convert them into area and semiperimeter relations.

Answer 1

The Correct Option is A

Concept: Important triangle identities: \[ r=\frac{\Delta}{s},\qquad r_1=\frac{\Delta}{s-a},\qquad r_2=\frac{\Delta}{s-b},\qquad r_3=\frac{\Delta}{s-c} \] Also \[ \Delta=\frac{abc}{4R} \] These relations connect inradius, exradii and circumradius.

Step 1: Use standard identity involving exradii.
After substituting formulas for \(r,r_1,r_2,r_3\) and simplifying both equations, we obtain side relation \[ a=c \] Hence triangle becomes isosceles.

Step 2: Apply second condition.
Using standard reduction and substituting \[ b=2\sqrt2 \] we obtain \[ a=\frac52 \] Since \[ a=c \] therefore \[ c=\frac52 \]

Step 3: Find required value.
\[ a+c = \frac52+\frac52 \] \[ =5 \] Thus \[ \boxed{5} \]