Question

In a triangle \(ABC\),
\[ r_1\cot\frac{A}{2}+r_2\cot\frac{B}{2}+r_3\cot\frac{C}{2}= \] where \(r_1,r_2,r_3\) are the exradii and \(s\) is the semi-perimeter.

• A. \(s\)
• B. \(2s\)
• C. \(3s\)
• D. \(\dfrac{s}{2}\)

Hint:

Remember the identities \(\Delta=rs\) and \(\cot\frac{A}{2}=\frac{s-a}{r}\). They simplify many triangle geometry expressions involving inradius and exradii.

Answer 1

The Correct Option is C

Step 1: Recall the exradius formula.
In a triangle, the exradii are given by
\[ r_1=\frac{\Delta}{s-a}, \qquad r_2=\frac{\Delta}{s-b}, \qquad r_3=\frac{\Delta}{s-c} \] where \(\Delta\) is the area of the triangle.
Also, we know that
\[ \tan\frac{A}{2}=\frac{r}{s-a} \] where \(r\) is the inradius.
Hence,
\[ \cot\frac{A}{2}=\frac{s-a}{r} \]
Similarly,
\[ \cot\frac{B}{2}=\frac{s-b}{r}, \qquad \cot\frac{C}{2}=\frac{s-c}{r} \]

Step 2: Compute each term.
First term:
\[ r_1\cot\frac{A}{2} = \frac{\Delta}{s-a}\cdot\frac{s-a}{r} \]
\[ =\frac{\Delta}{r} \]
Since
\[ \Delta=rs, \] we get
\[ r_1\cot\frac{A}{2}=s \]
Similarly,
\[ r_2\cot\frac{B}{2}=s \] and
\[ r_3\cot\frac{C}{2}=s \]

Step 3: Add all the terms.
Therefore,
\[ r_1\cot\frac{A}{2} + r_2\cot\frac{B}{2} + r_3\cot\frac{C}{2} \]
\[ =s+s+s \]
\[ =3s \]

Step 4: Final conclusion.
Hence,
\[ \boxed{3s} \]