If a triangle is equilateral, then all its angles are 60 degrees, and the tangent of 60 degrees is \(\sqrt{3}\), so the sum of the tangents is \(3\sqrt{3}\).

If \(\tan A + \tan B + \tan C = 3\sqrt{3}\), then the triangle must be equilateral.

Using the identity \(\tan A + \tan B + \tan C = \tan A \tan B \tan C\) for any triangle, if \(\tan A + \tan B + \tan C = 3\sqrt{3}\), then \(\tan A \tan B \tan C = 3\sqrt{3}\).

By AM-GM inequality, \(\frac{\tan A + \tan B + \tan C}{3} \ge \sqrt[3]{\tan A \tan B \tan C}\), with equality if and only if \(\tan A = \tan B = \tan C\).

Combining the identity and the AM-GM inequality, we can show that if \(\tan A + \tan B + \tan C = 3\sqrt{3}\), then \(\tan A = \tan B = \tan C = \sqrt{3}\), which implies \(A = B = C = 60^\circ\).