GraphingCalculator 3.5;
Window 112 62 811 1003;
PaneDivider 162;
FontSizes 18 12 8;
BackgroundType 0;
Slider 0 1;
SliderControlValue 99;
U -1 1;
V -1 1;
3D.View 0.767925921570821 0.584512089447441 -0.261983580154276 -0.553321311940251 0.811384946463828 0.18838788285211 0.322684528190167 0.000293159725629645 0.94650652893956;
3D.Speed 0;
Text "=================================
Matemagi® by Ambjörn Naeve ©Dialectica
=================================
The (white) unit sphere is given by";
Color 17;
Opacity 0.7;
Expr x^2+y^2+z^2=1;
Text "
and its (green) tangent plane at the south pole
is given by";
Color 17;
Opacity 0.7;
Expr z=-1;
Text "
The ""north pole"" of the unit sphere is given by the red point";
Color 2;
Grain 0.1333333333333333;
Expr vector(0,0,1);
Text "
A horizontal circle on the sphere
with radius (sin2πc)
is given by the blue circle";
Color 17;
Expr vector(x,y,z)=vector([sin(2*pi*c)]*cos(2*pi*t),[sin(2*pi*c)]*sin(2*pi*t),cos(2*pi*c));
Text "
The cone that is tangent to the sphere
along the blue circle is given by the light-blue surface";
Color 17;
Opacity 0.7;
Expr z=cos(2*pi*c)+[tan(2*pi*c)]*[sin(2*pi*c)]-([tan(2*pi*c)]*[x^2+y^2]^(1/2));
Text "
This tangent cone has its vertex in the point C, where";
Color 17;
Expr C=vector(0,0,cos(2*pi*c)+[tan(2*pi*c)]*[sin(2*pi*c)]);
Text "
Now, ""switch off"" the blue circle
and the light-blue tangent cone
(i.e. make them invisible)
before you proceed further.
Rotating the blue circle by an angle of 2πa around the y-axis
and then by an angle of 2πb around the z-axis
produces a circle with parametric equation (x, y, z) = V(t)
where";
Color 17;
Expr function(V,t)=matrix(3,3,cos(2*pi*b),sin(2*pi*b),0,-sin(2*pi*b),cos(2*pi*b),0,0,0,1)*matrix(3,3,cos(2*pi*a),0,sin(2*pi*a),0,1,0,-sin(2*pi*a),0,cos(2*pi*a))*vector([sin(2*pi*c)]*cos(2*pi*t),[sin(2*pi*c)]*sin(2*pi*t),cos(2*pi*c));
Text "
This rotated (red) circle is displayed by";
Color 17;
Expr vector(x,y,z)=function(V,t);
Text "
The coordinate functions X(t), Y(t), Z(t)
of a point on the red circle are given by";
Color 17;
Expr function(X,t)=dot(function(V,t),vector(1,0,0)),function(Y,t)=dot(function(V,t),vector(0,1,0)),function(Z,t)=dot(function(V,t),vector(0,0,1));
Text "
The projection from the north pole of the sphere
onto the tangent plane at the south pole
of a point on the red circle
is given by the tip of the red vector";
Color 17;
Expr vector(0,0,1),vector(2*function(X,n)/(1-function(Z,n)),2*function(Y,n)/(1-function(Z,n)),-1);
Text "
As the point on the red circle moves,
the tip of the red vector traces out a curve
in the green tangent plane
given by the red curve";
Color 17;
Expr vector(x,y,z)=vector(2*function(X,n*t)/(1-function(Z,n*t)),2*function(Y,n*t)/(1-function(Z,n*t)),-1);
Text "NOTE: This curve is in fact a circle.
This is a remarkable property which requires a proof.
We will now contruct the centre-point of this circle.
Let the point P be the point on the sphere,
which is obtained by rotating the north pole
by an angle of 2πa around the y-axis
and then an angle of 2πb around the z-axis.
The coordinates of P are given by";
Color 17;
Expr P=matrix(3,3,cos(2*pi*b),sin(2*pi*b),0,-sin(2*pi*b),cos(2*pi*b),0,0,0,1)*matrix(3,3,cos(2*pi*a),0,sin(2*pi*a),0,1,0,-sin(2*pi*a),0,cos(2*pi*a))*vector(0,0,1);
Text "
The same two rotations applied to the vertex-point C
gives the point Q, where";
Color 17;
Expr Q=matrix(3,3,cos(2*pi*b),sin(2*pi*b),0,-sin(2*pi*b),cos(2*pi*b),0,0,0,1)*matrix(3,3,cos(2*pi*a),0,sin(2*pi*a),0,1,0,-sin(2*pi*a),0,cos(2*pi*a))*C;
Text "
Hence, the yellow vector";
Color 17;
Expr P,Q;
Text "points out the vertex
of the (correspondingly) rotated light-blue tangent cone.
This rotated tangent cone is given by the yellow surface";
Color 17;
Opacity 0.7;
Expr vector(x,y,z)=matrix(3,3,cos(2*pi*b),sin(2*pi*b),0,-sin(2*pi*b),cos(2*pi*b),0,0,0,1)*matrix(3,3,cos(2*pi*a),0,sin(2*pi*a),0,1,0,-sin(2*pi*a),0,cos(2*pi*a))*vector(u,v,cos(2*pi*c)+[tan(2*pi*c)]*[sin(2*pi*c)]-([tan(2*pi*c)]*[u^2+v^2]^(1/2)));
Text "This surface is tangent to the sphere along the red circle.
The yellow tangent cone has its vertex in the point Q,
whose coordinates (X, Y, Z) are given by";
Color 17;
Expr X=dot(Q,vector(1,0,0)),Y=dot(Q,vector(0,1,0)),Z=dot(Q,vector(0,0,1));
Text "
The projection from the north pole of the sphere
through the vertex Q
of the yellow tangent cone to the red circle
onto the green tangent plane at the south pole
is given by the white vector";
Color 17;
Expr vector(0,0,1),vector(0,0,1)+2/(1-Z)*vector(X,Y,Z-1);
Text "
It can be shown that
the tip of this vector
points out the centre-point
of the projected (red) circle
in the green tangent plane.
Hence, stereographic projection maps circle onto circles
WITHOUT preserving centre-points.
NOTE: It follows that the centre-point of a circle
cannot be constructed by the use of only a ruler.
===================================
The position and size of the red circle on the sphere
can be changed with the following sliders:";
Color 17;
MathPaneSlider 54;
Expr a=slider([0,1]);
Color 17;
MathPaneSlider 43;
Expr b=slider([0,1]);
Color 17;
MathPaneSlider 22;
Expr c=slider([0,1]);