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Text "=================================
Matemagi® by Ambjörn Naeve ©Dialectica
=================================
The proof that stereographic projection preserves angles is given in the text below.
We will now show that the stereographic map is conformal,
i.e. that it conserves angles.
Consider the following three draggable points:
1) The draggable (red) point p";
Color 2;
Expr p=0.921875-(0.71875*i);
Text "
2) The draggable (black) point q";
Color 8;
Expr q=0.1875-(1.171875*i);
Text "
3) The draggable (blue) point s";
Color 3;
Expr s=1.265625+0.28125*i;
Text "
The equation of a circle
with centre p and passing through q
is given (in parametric form)
by (x, y) = (X(t), Y(t)) where";
Color 17;
Expr function(X,t)=Re(p)+abs(q-p)*cos(2*pi*t);
Color 17;
Expr function(Y,t)=Im(p)+abs(q-p)*sin(2*pi*t);
Text "
The graph of this (red) circle is given by";
Color 2;
Expr vector(x,y)=vector(function(X,t),function(Y,t));
Text "
The equation of the tangent line
of the (red) circle at the (black) point q
is given (in parametric form)
by (x, y) = (A(t), B(t)) where";
Color 17;
Expr function(A,t)=Re(q)+100*[t-1/2]*Im([q-p]);
Color 17;
Expr function(B,t)=Im(q)-(100*[t-1/2]*Re([q-p]));
Text "
The equation of a circle
with centre s and passing through q
is given (in parametric form)
by (x, y) = (U(t), V(t)) where";
Color 17;
Expr function(U,t)=Re(s)+abs(q-s)*cos(2*pi*t);
Color 17;
Expr function(V,t)=Im(s)+abs(q-s)*sin(2*pi*t);
Text "
The graph of this (blue) circle is given by";
Color 3;
Expr vector(x,y)=vector(function(U,t),function(V,t));
Text "
The equation of the tangent line
of the (blue) circle at the (black) point q
is given (in parametric form)
by (x, y) = (C(t), D(t)) where";
Color 17;
Expr function(C,t)=Re(q)+100*[t-1/2]*Im([q-s]);
Color 17;
Expr function(D,t)=Im(q)-(100*[t-1/2]*Re([q-s]));
Text "
Now, let us consider a sphere with unit radius
that has its centre at the point (0, 0, 1),
and hence is tangent to the xy-plane at the origin.
This (yellow) sphere is given by the equation ";
Color 6;
Opacity 0.7;
Expr prime(x)^2+prime(y)^2+[prime(z)-1]^2=1;
Text "where the x', y', z' variables
ensure that the sphere is displayed in a separate window to the right.
The xy-plane has the equation z = 0,
and this (green) plane is displayed in the right window by";
Color 4;
Opacity 0.7;
Expr prime(z)=0;
Text "
Now, stereographic projection
maps the points on a sphere (apart from its north pole)
to the points on its tangent plane at the south pole
by projecting the sphere-points from the north pole.
The inverse of this map
maps the points on the tangent plane at the south pole
to the sphere points
by carrying out the north pole projection ""in reverse"".
The (red) circle (x, y ) = (X(t), Y(t)) to the left
corresponds to the (red) circle (x', y', z') = (X(t), Y(t), 0) to the right
whose graph is given by";
Color 2;
Expr vector(prime(x),prime(y),prime(z))=vector(function(X,t),function(Y,t),0);
Text "
This (red) circle in the (green) xy-plane
is stereographically projected
to the following (red) circle on the (yellow) sphere:";
Color 2;
Expr vector(prime(x),prime(y),prime(z))=vector(4*function(X,t)/(function(X,t)^2+function(Y,t)^2+4),4*function(Y,t)/(function(X,t)^2+function(Y,t)^2+4),2*[function(X,t)^2+function(Y,t)^2]/(function(X,t)^2+function(Y,t)^2+4));
Text "
NOTE: The fact that stereographic projection maps circles to circles
is a highly remarkable property that requires a separate proof.
The (blue) circle (x, y ) = (U(t), V(t)) to the left
corresponds to the (blue) circle (x', y', z') = (U(t), V(t), 0) to the right
whose graph is given by";
Color 3;
Expr vector(prime(x),prime(y),prime(z))=vector(function(U,t),function(V,t),0);
Text "
This (blue) circle in the (green) xy-plane
is stereographically projected
to the following (blue) circle on the (yellow) sphere:";
Color 3;
Expr vector(prime(x),prime(y),prime(z))=vector(4*function(U,t)/(function(U,t)^2+function(V,t)^2+4),4*function(V,t)/(function(U,t)^2+function(V,t)^2+4),2*[function(U,t)^2+function(V,t)^2]/(function(U,t)^2+function(V,t)^2+4));
Text "
We will now show that the angle
between the red and the blue curve in the green xy-plane
is the same as the angle between the red and the blue curve
on the yellow sphere.
In order to obtain a proof that is valid for general curves,
we can think of the two circles in the xy-plane
as being tangent to two general curves at the black point q.
As described above, the (purple) tangent to the red curve (circle)
at the (black) point q (to the left) is given by";
Expr vector(x,y)=vector(function(A,t),function(B,t));
Text "
In the window to the right,
this tangent line corresponds to the (purple) line";
Expr vector(prime(x),prime(y),prime(z))=vector(function(A,t),function(B,t),0);
Text "
Analogously, the (light-blue) tangent to the blue curve (circle)
at the (black) point q (to the left) is given by";
Color 5;
Expr vector(x,y)=vector(function(C,t),function(D,t));
Text "
In the window to the right,
this tangent line corresponds to the (light-blue) line";
Color 5;
Expr vector(prime(x),prime(y),prime(z))=vector(function(C,t),function(D,t),0);
Text "
Now, by definition,
the angle between the purple line and the light-blue line
is the same as
the angle between the red curve and the blue curve in the green xy-plane
that they are tangents to.
We will now show that this angle is the same
as the angle between the red and the blue curve on the yellow sphere
at the corresponding point of intersection
The correspondence is given by the (white) projection line";
Color 7;
Expr vector(prime(x),prime(y),prime(z))=vector(0,0,2)*t+vector(Re(q),Im(q),0)*[1-t];
Text "
Now, the purple line
that is tangent to the red circle in the green xy-plane
is stereographically projected
to the following purple circle
which is tangent to the red circle on the yellow sphere,
and which goes through the north pole of the yellow sphere:";
Expr vector(prime(x),prime(y),prime(z))=vector(4*function(A,t)/(function(A,t)^2+function(B,t)^2+4),4*function(B,t)/(function(A,t)^2+function(B,t)^2+4),2*[function(A,t)^2+function(B,t)^2]/(function(A,t)^2+function(B,t)^2+4));
Text "
NOTE: The fact that the projected (purple) curve on the sphere is a circle
follows directly from the fact that it must lie in the plane
that contains the purple line and the white projection line
In the same way, the light-blue line
that is tangent to the blue circle in the green xy-plane
is stereographically projected
to the following light-blue circle
which is tangent to the blue circle on the yellow sphere,
and which goes through the north pole of the yellow sphere:";
Color 5;
Expr vector(prime(x),prime(y),prime(z))=vector(4*function(C,t)/(function(C,t)^2+function(D,t)^2+4),4*function(D,t)/(function(C,t)^2+function(D,t)^2+4),2*[function(C,t)^2+function(D,t)^2]/(function(C,t)^2+function(D,t)^2+4));
Text "
Now, by symmetry, the two purple circle and the light-blue circle
intersect at the same angle at both of their points of intersection.
But since
the purple circle and the light-blue circle
is in the same plane as
the purple line respectively the light-blue line (in the green xy-plane),
and since
the tangent plane to the yellow sphere at its north pole
is parallel to
the tangent plane to the yellow sphere at its south pole,
it follows that
the tangent lines at the north pole
to the purple circle and the light-blue circle
must be parallel
to the purple line respectively the light-blue line (in the green xy-plane).
The tangent line
to the purple circle
at the north pole of the yellow sphere
is given by the purple line";
Expr vector(prime(x),prime(y),prime(z))=vector(function(A,t),function(B,t),0)+vector(-Re(q),-Im(q),2);
Text "
The tangent line
to the light-blue circle
at the north pole of the yellow sphere
is given by the light-blue line";
Color 5;
Expr vector(prime(x),prime(y),prime(z))=vector(function(C,t),function(D,t),0)+vector(-Re(q),-Im(q),2);
Text "
Therefore,
since the two purple lines are parallel
and the two light-blue lines are parallel
the purple line and the light-blue line that meet at the north pole
intersect at the same angle as
the purple line and the light-blue line in the green xy-plane,
and this angle is identical to
the angle under which the purple circle and the light-blue circle meet
at both of their points of intersection.
Hence,
the red and the blue curves meet under the same angle
on the sphere and in the xy-plane,
which shows that
stereographic projection preserves angles,
i.e. that it is conformal.";