Convex hull of a subset of the unit circle Let U be the unit circle. Let V be all points on the unit circle that have a rational angle from the positive X axis. Let H be the boundary of the convex hull of V.
Is H equal to either U or V?
 A: Since V is dense in the unit circle, its convex hull H contains every interior point of the unit disk. Hence the boundary of H is all of the unit circle.
A: Let $\mathcal{O}$ represent the origin $(0,0)$, and let $X = (1,0)$.
Let $U$ be the set of points $(u_x,u_y)$ with radius $r \leq 1$ from $\mathcal{O}$.
Let $V$ be the set of points $(v_x,v_y)$ with radius $r \leq 1$ where the angle $v_{\theta} = \angle X \mathcal{O} V_{i}$ is rational for all $V_{i} \in V$.
Let $H$ be the set of points $(h_x,h_y)$ on and within the boundary of the convex hull of $V$.  
It is clear that $V \subset U$, and $V \bigcup I = U$ where $I$ is the set of points in $U$ with irrational angles and $V \bigcap I = \emptyset$.
So the question is: does $H = U$?
@user72694 is certainly correct in that $V$ is dense on $U$ since the rational numbers are dense in the reals.  However, density does not imply equality, that is, density of $V$ in $U$ does not mean that $V = U$.  We may refer to the points in $I$ with radius $r=1$ as the limit set.  $H$ does not include the limit set, which is included in $U$, therefore $H \neq U$ even though it is infinitesimally close to it!  
I can see this being argued though, and there is probably much more that can be said :)  
