Now we're in great shape. We have annexed our annulus, and although parts have been snipped away, no two snipped-away parts overlap. Within the sector defined by one of the snipped-away parts, the annexed area is on the order of the original area. (See Figure 20.)
Figure 20: Comparing areas within a snipped sector.
Thus we hope and expect that there is enough room for the tubes to grow. To verify this, let us specify precisely how the tubes are to run.
The region we are supposed to be cutting up is the portion of
G between height h and height h/2
and lying above 
,
the union of the base of the hemisphere and the annexed territory.
Just as we divided the plane up into pieces,
we will divide
into pieces, and treat each such piece separately.
We begin by cutting
along the radii that extend to the
tips and centers of the snips.
(See Figure 21.)
This yields pieces of three kinds: sectors, right triangles, and ``left triangles.'' In discussing how to lay out the tubes we will ignore left triangles, since they are just like right triangles.
For a given piece S,
either a sector or a triangle,
let 
be the portion of the plane at height z
that lies above S,
and let
Our task is to extend the tubes that have run into 
down through the region H until they run into
.
Since the tubes are to remain more or less vertical,
this amounts to specifying a correspondence 
between and for each 
.
Our strategy is to choose these correcpondences so that the
available horizontal cross-section is shared equally among the tubes,
that is,
so that
for all 
.
This condition doesn't determine the 's,
so we require in addition that the 's be nice and smooth,
and take ``radii'' of to ``radii'' of S (z).
To see what this entails,
introduce polar coordinates 
in the (x,y)-plane
so that the point r=0 is the center of the disk D
and the ray 
lies just to the right of S.
(See Figure 22.)
Then H is determined by the inequalities
where
when S is a sector, and
when S is a right triangle. Set
and define
The function 
tells what fraction of the area
of has 
-coordinate between 0 and .
The function 
tells what fraction of the area of the infinitesimal strip between
angles and 
has r-coordinate
between 
and r.
We make the tubes run along the curves 
, 
.
The correspondences associate points in
and that have equal values of u and v.
These correspondences are illustrated in
Figures 23 and 24.
Figure 23 shows a sector;
here the tubes are being deflected radially.
Figure 24 shows
a right and left triangle together;
here in addition to being deflected radially
the tubes are being squeezed from the center of the snip out towards the tips.
Figure 23: Tubes above a sector.
Figure 24: Tubes above a pair of triangles.
Have we succeeded in keeping the tubes more or less vertical? This comes down to checking whether there is a universal upper bound for
where the subscript u,v indicates that u and v are to be held constant. For a sector, this fact is geometrically obvious. For a triangle, this fact is not quite so obvious, but nonetheless true. To see why, consider that the only free parameter is the ratio h/a, and the only way the expression above can fail to be bounded is if something bad happens as h/a goes to 0. So you just have to convince yourself that nothing bad happens as h/a goes to 0. As a last resort, this can be verified by computation.
