Volume of 1/2 using hull of finite point set with diameter 1 It's easy to bound a volume of a half.  For example, the points $(0,0,0),(0,0,1),(0,1,0),(3,0,0)$ can do it.  The problem is harder if no two points can be further than 1 apart. Bound a volume of 1/2 with a diameter $\le 1$ point set.
With infinite points at distance 1/2 from the origin, a volume of $\pi/6 = 0.523599...$ can be bound. But we want a finite point set. What is the minimal number of points?  
(A 99 point set used to be here.  See Answers for a much better 82 point set)
Here's a picture of the hull. Each vertex is numbered. Green vertices have one or more corresponding blue faces with vertices at distance 1. Each blue face has a brown number giving the opposing green vertex. Red vertices and yellow faces lack a face/vertex pairing.

Some may think that Thomson problem solutions might give a better answer. The first diameter 1 Thomson solution with a volume of 1/2 is 121 points with volume .500069. 
These points will not fit in a diameter 1 sphere, but the maximal distance between points is less than 1. Similarly, a unit equilateral triangle will not fit in a diameter 1 circle.
Is 99 points minimal for bounding a volume of 1/2 using a point set with diameter 1?  Or, to phrase it as a hypothesis:
99 Point Hypothesis
99 points of diameter 1 in Euclidean space.
99 points with a volume of a 1/2.
Take one off, move them around (without increasing diameter)
You can't get a volume of 1/2 any more.   
 A: A related question is to ask the volume of the largest $n$-vertex polyhedron which can be inscribed in a sphere of given size. It is important to recognize that this is a different problem, as a polyhedron of diameter $d$ is not generally inscribable in a sphere of diameter $d$. To illustrate, your 110-vertex polyhedron actually contains 70 points outside of the diameter-$1$ sphere centered at the origin (I have not verified whether some other diameter-$1$ sphere not centered at the origin might contain fewer than 70 vertices of your polyhedron in its exterior).
With that said, the question of the largest $n$-vertex polyhedron which can be inscribed in a sphere of a given size is asked here. Its solution is unknown for $n>8$, which hints that your problem might be hard to solve.
The accepted answer to that question links to a 1994 page entitled "Maximal Volume Spherical Codes", which appears to use Thomson problem solutions for their putatively optimal arrangements, as they list $n=121$ as the smallest $n$ which breaches the $1/2$ threshold. 
A: Just for fun, I got down to 162 points and a volume of .5058  by starting with a triangulation of a Icosahedron and subdividing each triangle into 4 smaller triangles twice.
I improved my own first try by using a Fibonacci Sphere for n points I than calculated the volume for 100 points up to 150 poimts. At 128 points, it goes over 0.5 num = 127 volume = 0.49984077982 num = 128 volume = 0.500172211602.
