[Suppose we are given a disc. Circumscribe it
with a regular hexagon just grazing its
circumference, and circumscribe that in turn
with a circle. I call this the hexagonally
circumscribed circle of the original disc. If
P is any point outside the original disc, we
can make a kind of cap with P as vertex
from the convex hull of P together with the
disc. This convex hull I call the cap
corresponding to P and the disc. The key
property we shall need later on is that its
vertex angle at P is a decreasing function of
the distance of P from the disc. This vertex
angle will be exactly 120o precisely when P
lies on the hexagonally circumscribed circle,
and when this angle is less than 120o the
point lies outside the hexagonally
circumscribed circle. ]

--> Circumscribe A DISC OF RADIUS r
with a regular hexagon [just grazing its
circumference], and circumscribe THE HEXAGON 
with a circle. THIS GIVES WHAT I call [this] the hexagonally
circumscribed circle of the original disc, A CONCENTRIC CIRCLE OF
RADIUS $R=2r/\sqrt{3}$. If
P is any point outside the original disc, 
THE TWO TANGENTS FROM P TO THE DISC
BOUND WHAT I WILL
[can make a kind of cap with P as vertex
from the convex hull of P together with the
disc. This convex hull I] call the cap
corresponding to P and the disc. The key
property we shall need later on is that [its] THE
vertex angle [at P] OF THE CAP is a decreasing function of
the distance of P from the disc. THIS FOLLOWS
FROM ELEMENTARY TRIGONOMETRY, SINCE IF R IS THE
RADIUS OF THE DISC, AND D THE DISTANCE FROM P TO
THE CENTER, THAT ANGLE IS 2 ARCSIN(R/D). This vertex
angle will be exactly 120o precisely when P
lies on the hexagonally circumscribed circle,
(BECAUSE THEN IT IS A VETEX OF A CIRCUMSCRIBED HEXAGON!),
SO [and] when this angle is less than 120o [the
point lies] P MUST LIE outside the hexagonally
circumscribed circle. 

***************

We shall need also another property of these circumscribed
     circles. 

[Suppose two of them intersect, but that the discs
     themselves do not intersect. If P is a point in the intersection
     then it lies at least as close to one of these two discs than to
     any other disc, and if in fact it lies as closely to some third
     disc then these three discs will be in the configuration of
     discs in a hexagonal packing. Because if the two discs are
     this close, there will be a non-empty dead region on the
     bisecting line. If the intersection contains a point in the
     Voronoi cell of a third disc, then the triple point associated to
     these three discs will also lie in this intersection. But each of
     the vertex angles at the capped discs from this triple point
     must then be at least 120o. This can only happen if this angle
     is exactly 120o in the circumstances indicated. ]


{I think the following is more intelligible; but it does introduce
the diameter of the circumscribed circle - which I snuck in above}



 Suppose two of them intersect, but that the discs
     themselves do not intersect. THEN THAT INTERSECTION CAN ONLY
INTERSECT THE VORONOI CELL OF A THIRD DISC IF THE THREE DISCS
ARE in the configuration of THREE
     MUTUALLY TANGENT
     discs in a hexagonal packing. 

HERE IS WHY: ANY POINT IN THE
INTERSECTION IS AT DISTANCE $\leq R=2r/\sqrt{3}$ FROM THE
CENTERS OF THE TWO DISCS SO, AS WE CALCULATED ABOVE, THE VERTEX
ANGLES OF THE CAPS FROM THAT POINT MUST BOTH BE $\geq 120^{\circ}$.
If the intersection contains a point in the
     Voronoi cell of a third disc, THEN BY DEFINITION OF THE VORONOI
CELL, THAT POINT MUST BE AT DISTANCE $\leq R=2r/\sqrt{3}$ FROM
THE THIRD CENTER, AND THE VERTEX ANGLE OF THE CAP ON THE THIRD
CIRCLE FROM THAT POINT MUST ALSO BE $\geq 120^{\circ}$. SINCE 
CAPS FROM ONE POINT TO NON-INTERSECTING DISCS CANNOT INTERSECT,
THE ONLY POSSIBILITY IS FOR ALL THREE OF THESE ANGLES TO EQUAL
EXACTLY $120^{\circ}$. SO THE POINT IS A VERTEX OF THREE
CIRCUMSCRIBED HEXAGONS, I.E. these three discs will be in the 
configuration of THREE
     MUTUALLY TANGENT
     discs in a hexagonal packing.


     In the diagram to the right, this means that the points in the
     yellow parallelogram are never in the Voronoi cell of the
     third disc. 


then change caption to "A triple point ..."

**********************

[Hoe] HOW dense can a disc be in its Voronoi cell?

   We want to show that a disc take up no larger proportion of the area in its Voronoi cell than it does in its
   circumscribing regular hexagon. To show this, we partition the cell into sub-regions of different types. 

   Suppose we now look at a cell in a layout. Draw the circumscribing circle as well, or at least its intersection with
   the cell. Where it extends beyond the cell, the line cut off by the cell boundary will be the [line between two
   triangles attached to neighbouring discs.] THE BISECTOR OF THE
PARALLELOGRAM ASSOCIATED TO TWO DISCS WHOSE HEXAGONALLY CIRCUMSCRIBED
CIRCLES INTERSECT.

{triangle is an unintroduced concept}

We can now divide up the cell into distinct regions of three types: 
    The points in the cell which do not
    belong to the enlarged disc.
                             The points in the cell belonging to a
                             [triangle] PARALLELOGRAM
                              associated to a neighbouring
                             cell. 
                                     The parts of the circumscribing disc not
                                     belonging to one of these [triangles]
                                      PARALLELOGRAMS.

{I think it would read better if these phrases were put below
the diagrams, like captions.}

********************


   The density of the [discs] DISTRIBUTION in the first [region] TYPE is of course 0, since by definition such a region contains no part of a
   disc. The density of the DISTRIBUTION in the third type is just the ratio of the radius of the disc to that of the larger one.
   Both densities are strictly less than the density of the disc in its regular circumscribed hexagon. The crux of the
   argument is now to examine regions of the second type. 


*********************

{Who is Rogers?}

So now we look at one of [those] THE TYPE-TWO regions. 
We [must] WILL show that the ratio of the area inside the disc to that of the
   whole TYPE-TWO region is at most equal to the ratio in the special case 
when the central angle is 60o. This can be done by
   an explicit calculation, and the claim then reduces to the observation that the function sin(x) / x is monotonic
   decreasing in the range between 0 and /2. 


*************


But it can also be demonstrated by a purely geometric argument. [In the] 
THE figure on the left above [we are looking at] SHOWS
   the special case. In the figure on the right the animation generates the 
other cases. As the central angle decreases
   we also generate the image of the left [hexagonally circumscribed circle]
DISC {!!} under the unique linear transformation
   which acts by scaling vertically and horizontally, transforming the original equilateral triangle to the narrower
   one, as indicated in the figure above. [Keep in mind that] SINCE 
linear transformations preserve [ratios along lines, and] the
   ratio of areas on the plane COMMA, THE animation [then] shows that 
[our bound is valid.] THE LARGEST RATIO OF CIRCULAR SECTOR TO TRIANGLE
(I.E. THE LARGEST DENSITY) OCCURS WHEN THE CENTRAL ANGLE IS 60 $^{\circ}$.

*****************

IN SUMMARY, THE
   [The] density will achieve the maximum possible only when there are no 
regions of the first or third type AND WHEN THE CENTRAL ANGLE OF EACH 
TYPE-TWO REGION IS EXACTLY 60 $^{\circ}$. THESE CRITERIA UNIQUELY DETERMINE
THE HEXAGONAL PACKING. [Thus the
   same proof explains the argument about the 6 discs surrounding a single one, since it shows that the hexagon
   circumscribing a disc is the smallest possible cell containing it. ]

{Something wrong here: the circumscribing dodecagon obviously is smaller. ????}