Given a set of points $S$ and a fraction $\alpha$ I would like to find exactly one hyperplane which divides $S$ such that approximately $\alpha$ points lie on one side and $1-\alpha$ points on the other.
Can this be formulated as a strictly (maybe strongly) convex optimisation problem?
In general I think you can't, this is a typical combinatorial problem. But You can solve it efficiently I guess.
Take any direction $m$ and compute $m^T x_i$ for each $x_i \in S$. This induces an ordering along the direction $m$. Then just take the $\alpha|S|-$th point and compute $d$ such that $m^Tx+d=0$. Do the same for the $\alpha|S|+1$ obtaining a value $d1$. The plane $m^Tx + (d+d1)/2$ is what you are looking for. There might be some special cases to consider.
Otherwise you can formulate the problem as a binary linear program, but unless the continuous relaxation has integer solution, it is going to be hard to solve.
