IMO Hk TST

The minimum value is $3 - 2\sqrt{2}$.
Condition (i) can be rewritten as
$$(\sin A - |\sin B|)(\cos B - |\cos A|) = 0.$$
If $\sin A = |\sin B|$, then we have $\sin^2 A = \sin^2 B = 1 - \cos^2 B$.
If $\cos B = |\cos A|$, then we have $\cos^2 B = \cos^2 A = 1 - \sin^2 A$.
Therefore, in any case, $(\sin A, \cos B)$ is a point on the unit circle on the coordinate plane. Since the steps are reversible, it can be any point on the unit circle.

Next, $(\tan C, \cot C)$ can be any point on the hyperbola $xy = 1$. Therefore, we are asked to find the minimum distance between a point on the unit circle and a point on the hyperbola. In view of the geometry, since the line $y = x$ is an axis of symmetry of both geometric objects, the minimum distance is attained

between points on the line $y = x$. By symmetry, it suffices to consider $x > 0$. Clearly, $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ is a solution to $x^2 + y^2 = 1$ and $y = x$, while $(1, 1)$ is a solution to $xy = 1$ and $y = x$. Therefore, the minimum distance is
$$\left(1 - \frac{1}{\sqrt{2}}\right)^2 + \left(1 - \frac{1}{\sqrt{2}}\right)^2 = 3 - 2\sqrt{2},$$
and we are done.