TITLE: Volume of 1/2 using hull of finite point set with diameter 1
QUESTION [14 upvotes]: 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.

REPLY [6 votes]: (Update)
My current result is $82$ points:
consider this point set:
pts = {
{39331, -1787,  125739},
{-42020,    -78476, 96709},
{97017, -83209, 30835},
{-17033,    70737,  109597},
{-54599,    29504,  115688},
{-69547,    63866,  91701},
{-84862,    -62280, -80052},
{111630,    -49662, -51118},
{110858,    44843,  -58218},
{7570,  -94324, 91248},
{115828,    -36578, 50910},
{-103422,   33617,  73525},
{13903, 130088, -24865},
{-48488,    -30540, -119577},
{13546, 105208, 78574},
{92754, -90941, -22055},
{-87842,    -12726, -97961},
{17890, -95311, -90222},
{-32617,    127358, -17688},
{-83770,    -100939,    6478},
{-67513,    -103415,    -46172},
{-15435,    70574,  -111233},
{42948, 122369, 28253},
{82827, -31757, -98975},
{-8841, 14824,  130515},
{-31918,    -116156,    52485},
{-124638,   33189,  26548},
{46151, -58101, 108697},
{-107711,   76927,  -3256},
{8590,  -131155,    -3832},
{-2349, -45047, 123671},
{-67052,    113066, 17470},
{-49845,    -26471, 118738},
{45038, -56580, -110986},
{124167,    -45279, 903},
{60780, -115738,    12319},
{-109374,   -68092, -27125},
{-40207,    -124921,    2722},
{74952, 40665,  100449},
{88162, -58830, 78010},
{60461, 114907, -29946},
{110136,    -3355,  -73936},
{70896, 79060,  -79787},
{56554, -97875, 67358},
{72446, -84584, -71147},
{30586, 57713,  114256},
{-15936,    -120088,    -52161},
{-480,  -46761, -124154},
{-72908,    103917, -38653},
{-101424,   28721,  -80454},
{-45115,    103290, 68859},
{41881, -117921,    -41667},
{-74575,    -93889, 53049},
{108114,    53390,  54482},
{15266, -123265,    42434},
{40723, -3854,  -126221},
{90334, 94409,  22158},
{96396, 85431,  -32579},
{-63349,    75478,  -88497},
{122169,    52183,  -1811},
{108487,    5280,   74810},
{-88785,    -956,   96779},
{-7851, 14221,  -131625},
{64857, 88850,  73124},
{23713, 102177, -81511},
{129972,    1413,   -27143},
{-119337,   -14421, 52312},
{-88103,    -51438, 82718},
{-10887,    127563, 33645},
{33805, 54367,  -116181},
{-102814,   64657,  -52366},
{-126644,   25744,  -26822},
{-25275,    110536, -68979},
{-112785,   -59627, 30034},
{-129858,   -19908, 289},
{-36740,    -84005, -95750},
{78058, 29755,  -103069},
{-118373,   -22382, -53597},
{-55526,    28946,  -116699},
{-94065,    79056,  48080},
{80742, -15619, 102763},
{129505,    8123,   26059}
}

Then (Mathematica code)
Volume[ConvexHullMesh[pts]]

is $\approx 9.00744\times10^{15}$. 
And Mathematica sketch:
ConvexHullMesh[pts]]

 
Another picture.  If all vertices of a face are at distance one from another vertex, the face is colored blue.

Since all point coordinates are integer, then one can write it directly (with arbitrary small computational errors):
$$Diameter = \sqrt{68\;719\;348\;253} \approx 262\;143.\;754\;938;$$
$$Volume = \dfrac{54\;044\;635\;971\;533\;362}{6} \approx 9\;007\;439\;328\;588\;893.\;666\;667.$$
If multiply all coordinates by $\dfrac{1}{2^{18}}$, then we'll get:
$$Diameter = \frac{\sqrt{68\;719\;348\;253}}{262\;144}  \approx 0.999\;999\;065;$$
$$Volume = \dfrac{54\;044\;635\;971\;533\;362}{2^{54}\times 6} \approx 0.\;500\;013\;326.$$

Note: when add any point (with real coordinates) rather close to (the center of) any face, one will get the set of $83, 84, ...$ points with described property.