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See below.
The rotation matrix is
and the reflexion matrix about the
then
and
Attached a plot showing
2)
We will use several facts.
First, note that the volume of a parallelepiped with coterminous edges
Two more facts about the scalar triple product we will use:
Note that the first fact means that any scalar triple product involving the same vector twice will clearly be
Finally, we will use that both the dot product and the cross product distribute over addition. With all that, we will now calculate the volume of the new parallelepiped as the scalar triple product of its coterminous edges:
#=|(vec(a)+vec(b)) * [(vec(a) xx vec(b)) + (vec(a) xx vec(c)) + (vec(c) xx vec(b)) + cancel((vec(c) xx vec(c))]|#
#=|cancel(vec(a) * (vec(a) xx vec(b))) + cancel(vec(a) * (vec(a) xx vec(c)))#
# + vec(a) * (vec(c) xx vec(b)) + cancel(vec(b) * (vec(a) xx vec(b)))#
# + vec(b) * (vec(a) xx vec(c)) + cancel(vec(b) * (vec(c) xx vec(b)))|#
#=|-vec(a) * (vec(b) xx vec(c)) - vec(a) * (vec(b) xx vec(c))|#
#=2|vec(a) * (vec(b) xx vec(c))|#
#=2(40)#
#=80#
The farthest distance that between to points on the circle is the sum of the distance between the centers and the two radii:
Here is a graph of the two circles:
The circles overlap but the larger circle does not contain the other. This could have been deduced by the following process.
We assign
For circle 1:
For circle 2:
Compute the distance,
If the larger circle contained the smaller, then the distance between the centers would be less than difference between the two radii, 1. This is not the case.
The farthest distance that between two points on the circle is the sum of the distance between the centers and the two radii:
Let
Supposing that
Please see the explanation.
To go from
To go from
Therefore,
The length of the semi-major axis, a, is the distance from the center to either vertex:
Let A = the angle of rotation, then:
Rationalizing both denominators:
Here is a reference for An Rotated Ellipse that I not at the origin
Solving for
Force this to contain the point
I used WolframAlpha to do the evaluation:
Here is the final equation:
Here is a graph to prove it:
Warning: long answer...
Synopsis
First, I will show that the corners of three intersecting golden rectangles with width
Second, I will determine the outer radius of such an icosahedron, that is the distance from the centre to each vertex.
Third, I will determine the inner radius of the same icosahedron, that is the distance from the centre to the centre of each face.
Fourth, I will determine the surface area of an icosahedron with edges of length
Fifth, I will use that to write down formulae for the surface area in terms of the inner and outer radii.
Proposition
The following
#(+-1/2, +-1/2varphi, 0)#
#(0, +-1/2, +-1/2varphi)#
#(+-1/2varphi, 0, +-1/2)#
where
Each of the three sets of
Proof
Note that
#varphi^2 = varphi+1#
Due to the symmetrical specification, we just need to check that the distance between a corner of one of the golden rectangles and a selected corner of another is
The distance between
#d = sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#
#color(white)(d) = sqrt((0-1/2)^2+(1/2-1/2varphi)^2+(1/2varphi-0)^2)#
#color(white)(d) = sqrt(1/4+1/4(1-varphi)^2+1/4varphi^2)#
#color(white)(d) = sqrt(1/4+1/4(2-varphi)+1/4(varphi+1))#
#color(white)(d) = sqrt(1)#
#color(white)(d) = 1#
Outer radius of an icosahedron
The outer radius of an icosahedron with edges of length
#sqrt((1/2)^2+(1/2varphi)^2+0^2) = sqrt(1/4+1/4varphi^2)#
#color(white)(sqrt((1/2)^2+(1/2varphi)^2+0^2)) = sqrt(1/4(2+varphi))#
#color(white)(sqrt((1/2)^2+(1/2varphi)^2+0^2)) = 1/2sqrt(2+varphi)#
Inner radius of an icosahedron
One of the faces of the icoshedron described above has corners:
#(1/2, 1/2varphi, 0)# ,#(0, 1/2, 1/2varphi)# ,#(1/2varphi, 0, 1/2)#
The centre of this face is therefore located at:
#(1/6(1+varphi), 1/6(1+varphi), 1/6(1+varphi))#
The distance of this point from the origin is:
#sqrt(3(1/6(1+varphi))^2) = sqrt(1/12(1+varphi)^2) = 1/2sqrt(2/3+varphi)#
This is the inner radius of an icosahedron with edges of length
Area of an equilateral triangle with unit sides
An equilateral triangle with sides of length
Surface area of icosahedron
The surface area of our icosahedron with edges of length
#20*sqrt(3)/4 = 5sqrt(3)#
If we had an icosahedron of inner radius
#(5sqrt(3))/(1/2sqrt(2/3+varphi))^2r^2 = (20sqrt(3))/(2/3+varphi)r^2#
If we had an icosahedron of outer radius
#(5sqrt(3))/(1/2sqrt(2+varphi))^2R^2 = (20sqrt(3))/(2+varphi)R^2#
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