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The two-dimensional version of Kepler's
conjecture asks for the densest packing
of unit disks in the plane. If we
inscribe a disk in each hexagon in the
regular hexagonal tiling of the plane,
the density of the packing is
p/Ö[12] » 0.9069. Thue's
theorem, announced in 1890, affirms
that this is the highest density
possible. There is a common
misconception that the proof of Thue's
theorem is not elementary. The proof
here is based on an idea of Rogers's
and does not require calculus. The
ideal presentation of this proof would
be one that develops interactively on
the computer screen without written
words. Below we combine words and
static images...
Take an arbitrary packing of the plane
by nonoverlapping disks of radius 1. We
will partition the plane into regions,
and will show that each region has
density at most p/Ö[12]. Center
a larger circle of radius 2/Ö3
around each disk (Figure i).
Whenever two of these large circles
intersect, draw the segment between the
two points of intersection, and draw
two congruent isosceles triangles with
this segment as base and vertices at
the centers of the two circles (Figure
ii).
There cannot be a point interior to
three large circles. Indeed, in the
extreme case, three large circles meet
at a point, the circumcenter, if the
centers are the vertices of an
equilateral triangle of edge 2
(Figure iii).
This gives our partition of space:
regions outside all of the larger
circles (white), the isosceles
triangles (cyan), and the part inside
the larger circles but outside all
triangles (purple) - see Figure
iv.
The regions outside all of the larger
circles have density 0, which is
certainly less than p/Ö[12].
The density of the interior of the
larger circles is the square 3/4 of
the ratio (1:2/Ö3) of the
smaller to larger radius, again less
than p/Ö[12]. This inequality
can be seen geometrically by drawing a
hexagon that the small circle inscribes
and the large one circumscribes (Figure
v).
The hexagon has density
p/Ö[12], and this will be
greater than the density inside the
full larger circle.
To calculate the density of an
isosceles triangle, we apply a linear
transformation to the triangle
(preserving ratios of areas and hence
densities) to transform it into an
equilateral triangle with edge
2/Ö3. The transformation scales
along the orthogonal axes through the
base and altitude of the triangle and
fixes the vertex v opposite the base
of the isosceles triangle. The unit
circle is transformed into an ellipse.
(See Figure vi.)
The linear transformation preserves
the lengths of the two equal edges of
the isosceles triangle. Hence the
ellipse cuts those edges at distance 1
from its center v. This identifies
all four points of intersection of two
conic sections, the ellipse and the
unit circle centered at v.
Knowing these points of intersections,
we deduce that the intersection of the
ellipse with the interior of the
equilateral triangle is contained in a
disk of radius 1 centered at v. In
particular, the density is of the
equilateral triangle is increased by
replacing the ellipse with a circle of
radius 1. The equilateral triangles fit
together to form hexagons with
inscribed disks of radius 1. The
density of these pieces is therefore
p/Ö[12]. This completes the
proof of Thue's theorem.
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Figure (i)
Figure (ii)
Figure (iii)
Figure (iv)
Figure (v)
Figure (vi)
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