All labeling refers to figure 1 above.

 

1. The length of the perpendicular bisector of the largest possible equilateral triangle that fits in the space between the circles, GI is:

GI = 3d/2 * (√3 -1 – 1/√3)

Proof: First, we note that the angle GHI is 60º and angle GHK is 30º, thus GI is 3 times longer than GK because tan(60º)/tan(30º) = 3.

            GI = 3 GK.                  [I]

Next we note that angle DAK is 30º, and calculate

AK = AD/cos(30º) = 2r/√3 = √3 r – r/√3,        [II]

where r = radius AD.   

GK = AK – r                            [III]

From III and II,    

GK = √3 r - r/√3 – r = d/2 * (√3 -1 – 1/√3)     [IV]

where 2r = d (diameter)
Therefore from  [I] and [IV],

GI = 3d/2 * (√3 -1 – 1/√3)                  QED.

 

2. The length of the perpendicular bisector of the entire cusp, GE is:

GE = AE – AG = √3 r – r.

Proof:

GE = AE – AG = AE – r.                     [I]
AE = AB cos(30º) = √3 r.                    [II]

Therefore,

GE = √3 r – r.              QED.

Note, the length produced during a real deposition depends on the deposition conditions and any subsequent annealing.

 

3. The percentage area occupied by the space between the nanosphere mask is: 1 – π/(√3*2) ≈ 9%

Proof:  Since the nanosphere mask can be decomposed into identical unit triangles, ABC, we examine ABC.

Area of triangle ABC = ½ CB * AE     [I]
AE = AB cos(30º) = √3d /2 = √3r       [II]

Where d (sphere diameter) = 2r (radius).

Therefore from [I] and [II],
Area ABC = ½ 2r * √3r = √3 r^2           [III]

This area includes 3 arcs of circles, each 1/6th of a circle in area, thus the area occupied by the circle arcs is

3 * π r^2 / 6 = π r^2 / 2        [IV]

The area unoccupied by the circles is total area – circle area, from [III] and [IV], this is

r^2 (√3 – π/2).                 [V]

As a percentage area, this is [V]/[III]

r^2(√3 – π/2)/ (√3 r^2) =  1 – π/(√3*2) ≈ 9%       QED.

 

4. The % area occupied by the smallest equilateral triangle that fits in the space between the circles HIJ is: 7√3/3 –4 ≈ 4%

Proof: Since ABC and HIJ are both equilateral triangles, the area of HIJ is the square of the ratio between two similar sides.

% area = (GI/AE)2       [I]
AE = AB cos(30º) = √3d /2     [II]
From above (see proof 1), GI = 3d/2 * (√3 -1 – 1/√3). [III]

Therefore, from [I], [II], [III], percentage area

(GI/AE)2 = 3 d/2 * (√3 -1 – 1/√3)2/(√3 d/2 )
= √3 *(√3 -1 – 1/√3)2 = 7√3/3 –4 ≈ 4%          QED.