<s>[INST] How to show that the triangle is equilateral triangle? 
If $\cot A+\cot B+\cot C=\sqrt3$ then prove that the triangle is equilateral triangle.

Trial: I can counter check that this is true as $\cot 60+\cot 60+\cot 60=3 \frac{1}{\sqrt3}=\sqrt3$.Here I also know that $A+B+C=180$. But I am unable to show it. Please help.
 [/INST] Since $A + B + C = \pi$, $\cot C = \cot(\pi - (A + B)) = -\cot(A + B)$. Using the angle addition formula for cotangent:
$$\cot A + \cot B - \frac{\cot A \cot B - 1}{\cot A + \cot B} = \sqrt{3}$$
and if we introduce new variables such that $p = \cot A + \cot B, q = \cot A \cot B$:
$$p + \frac{1 - q}{p} = \sqrt{3}$$
$$p^2 - \sqrt{3}p + (1 - q) = 0$$
$$q = p^2 - \sqrt{3}p + 1$$
By Vieta, the roots of $x^2 - px + q = 0$ are $\cot A, \cot B$. We want these roots to be real, thus we can set a condition on the discriminant:
$$p^2 - 4q ≥ 0$$
$$p^2 - 4(p^2 - \sqrt{3}p + 1) ≥ 0$$
$$-3 \left(p^2 - \frac{4}{\sqrt 3}p + \frac{4}{3} \right) ≥ 0$$
$$-3 \left(p - \frac{2}{\sqrt 3} \right)^2 ≥ 0$$
and for this concave down quadratic, this can only occur when $p = \frac{2}{\sqrt 3}$ and $q = \frac{4}{3} - 2 + 1 = \frac{1}{3}$.
Since we already know that the quadratic is a perfect square, $\cot A, \cot B$ must be the roots of $\left(x - \frac{1}{\sqrt 3} \right)^2 = 0$, which implies $\cot A = \cot B = \frac{1}{\sqrt 3}$ or that $A = B = \frac{\pi}{3}$ as $A + B + C = \pi$.
Thus $\Delta ABC$ must be equilateral.
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