Prove that the triangle ABC is equilateral if `cotA+cotB+cotC=sqrt3`

Answer

To prove that triangle ABC is equilateral if \( \cot A + \cot B + \cot C = \sqrt{3} \), we will follow these steps:

### Step 1: Start with the given equation
We have:
\[ \cot A + \cot B + \cot C = \sqrt{3} \]

### Step 2: Square both sides
Squaring both sides gives:
\[ (\cot A + \cot B + \cot C)^2 = (\sqrt{3})^2 \]
This simplifies to:
\[ \cot^2 A + \cot^2 B + \cot^2 C + 2(\cot A \cot B + \cot B \cot C + \cot C \cot A) = 3 \]

### Step 3: Use the angle sum property
In triangle ABC, we know that:
\[ A + B + C = \pi \]
From this, we can express \( A + B \) as:
\[ A + B = \pi - C \]

### Step 4: Apply cotangent to both sides
Taking the cotangent of both sides gives:
\[ \cot(A + B) = \cot(\pi - C) \]
Using the identity \( \cot(\pi - x) = -\cot x \), we have:
\[ \cot(A + B) = -\cot C \]

### Step 5: Use the cotangent addition formula
The cotangent addition formula states:
\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]
Thus, we can write:
\[ \frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C \]

### Step 6: Rearranging the equation
Cross-multiplying gives:
\[ \cot A \cot B - 1 = -\cot C (\cot A + \cot B) \]
Rearranging this leads to:
\[ \cot A \cot B + \cot C \cot A + \cot C \cot B = 1 \]
Let’s denote this as Equation (2).

### Step 7: Substitute Equation (2) into the squared equation
From our squared equation:
\[ \cot^2 A + \cot^2 B + \cot^2 C + 2(\cot A \cot B + \cot B \cot C + \cot C \cot A) = 3 \]
We can express \( 2(\cot A \cot B + \cot B \cot C + \cot C \cot A) \) using Equation (2):
\[ \cot^2 A + \cot^2 B + \cot^2 C + 2 \cdot 1 = 3 \]
This simplifies to:
\[ \cot^2 A + \cot^2 B + \cot^2 C + 2 = 3 \]
Thus:
\[ \cot^2 A + \cot^2 B + \cot^2 C = 1 \]

### Step 8: Use the identity for squares
Using the identity:
\[ x^2 + y^2 + z^2 - xy - xz - yz = \frac{1}{2}((x - y)^2 + (y - z)^2 + (z - x)^2) \]
we can substitute \( x = \cot A, y = \cot B, z = \cot C \):
\[ \cot^2 A + \cot^2 B + \cot^2 C - \cot A \cot B - \cot B \cot C - \cot C \cot A = 0 \]
This implies:
\[ (\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 = 0 \]

### Step 9: Conclude that the angles are equal
Since the sum of squares is zero, each term must be zero:
\[ \cot A - \cot B = 0 \implies \cot A = \cot B \]
Similarly, we find:
\[ \cot B = \cot C \quad \text{and} \quad \cot C = \cot A \]
Thus, we have:
\[ A = B = C \]

### Conclusion
Since all angles are equal, triangle ABC is equilateral.