Question

In a triangle \(ABC\), \[ s\left[\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}\right] = \] is:

• A. \(\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}\)
• B. \(r_1+r_2+r_3\)
• C. \(r_1r_2r_3\)
• D. \(\dfrac{1}{r}+\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}\)

Hint:

For triangle problems involving inradius and exradii, use \[ \Delta=rs \] and \[ r_1=\frac{\Delta}{s-a},\quad r_2=\frac{\Delta}{s-b},\quad r_3=\frac{\Delta}{s-c} \] to simplify expressions quickly.

Answer 1

The Correct Option is B

Step 1: Recall standard triangle formulae.
For a triangle, \[ \Delta=rs \] Also, the exradii are given by \[ r_1=\frac{\Delta}{s-a},\quad r_2=\frac{\Delta}{s-b},\quad r_3=\frac{\Delta}{s-c} \] Using \[ \Delta=rs, \] we get \[ r_1=\frac{rs}{s-a},\quad r_2=\frac{rs}{s-b},\quad r_3=\frac{rs}{s-c} \]

Step 2: Simplify \(\dfrac{r_1-r}{a}\).
Now, \[ r_1-r=\frac{rs}{s-a}-r \] \[ r_1-r=r\left(\frac{s}{s-a}-1\right) \] \[ r_1-r=r\left(\frac{s-(s-a)}{s-a}\right) \] \[ r_1-r=\frac{ra}{s-a} \] Therefore, \[ \frac{r_1-r}{a}=\frac{r}{s-a} \] Similarly, \[ \frac{r_2-r}{b}=\frac{r}{s-b} \] and \[ \frac{r_3-r}{c}=\frac{r}{s-c} \]

Step 3: Substitute in the given expression.
Now, \[ s\left[\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}\right] \] \[ =s\left[\frac{r}{s-a}+\frac{r}{s-b}+\frac{r}{s-c}\right] \] \[ =rs\left[\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}\right] \] Using \[ r_1=\frac{rs}{s-a},\quad r_2=\frac{rs}{s-b},\quad r_3=\frac{rs}{s-c}, \] we get \[ \frac{rs}{s-a}=r_1,\quad \frac{rs}{s-b}=r_2,\quad \frac{rs}{s-c}=r_3 \] Hence, \[ s\left[\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c}\right] = r_1+r_2+r_3 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{r_1+r_2+r_3} \]