Deciding if a point is inside a convex polygon in O(logn) time

Take any vertex of the polygon and use it as an "anchor" in a binary 
search.
First attach the vertex to one roughly n/2 vertices away. Decide if 
your query point is to the left or right of this segment. This eliminates 
half of the polygon. You are still left with a convex polygon.  Repeat, by 
attaching the anchor to a vertex at distance n/4, etc.  In the end, you 
are left with a cone, formed by the anchor and two successive vertices, u 
and v.  Now you can just check in O(1) time if the query point is inside 
the triangle (anchor,u,v) i.e. inside the polygon.

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Computing the two tangents (new hull edges) for a point outside a convex
polygon.    ... in O(logn) time

If you join an edge E from the new point to any vertex V of the polygon, 
you 
get just a few simple cases, depending on what side of E you  will find
edges V-1,V  and  V,V+1.  Only at the tangents will both of these 
consecutive edges be to one side of E.  It is very simple geometry to
decide which direction to proceed on a binary search, in the other cases.

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Updating the hull by computing tangents

You can add the two tangents to a hull in O(logn) time, as mentioned 
above.
  IF  you want to delete the edges adjacent to vertices that all of the 
sudden became 
interior points, it might take you O(n) time for a specific inserted 
point.  But overall you can only delete edges that you constructed,
so there is only a linear number of deletions after inserting n points
to a hull.

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So, for  n  sequential insertions of points, you get the hull at
every iteration, using a total of O(nlogn) time.