Kepler's Idea for Projective Geometry
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Kepler's Ideal Lines and Points 
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Geometers call the set of all lines through a point a 'pencil' of 
concurrent lines.
Johannes Kepler (ca 1600) had the idea of extending this concept to all
lines parallel to a given line as the pencil of lines concurrent at an 
'ideal point'. 
The stars, which for astronomer Kepler were infinitely 
far away, became the physical representations of the ideal points. So we
also call them 'points at infinity'.  Like real (or finite) points, 
ideal points could lie on an 'ideal line' insofar as the lines in 
each pencil are all parallel to the same (real) plane. Modern geometers 
prefer to define new objects in terms of sets of points. 

.Definition.
==================
An ideal point pass:[$ X $] is a set of lines parallel to a line. 
By pass:[$ (\ell X) $] we shall mean that pass:[$ \ell \in X $].
In the plane, the set of all ideal points comprise the ideal line,
pass:[$ \infty $]. Thus every line pass:[$ \ell $] has an ideal
point on itself, pass:[$ X = (\ell \infty) $], and two lines are 
parallel if and only if they meet at infinity: 
pass:[$ \ell || k \equiv (\ell \infty) = (k \infty) $]
==================

.Question.
==================
In space we have many ideal lines, one for each set of all 
mutually parallel planes. The ideal lines comprise one ideal
plane, Kepler's firmament of stars. Given two geometrical objects
taken from the set of points, lines and planes, where do they 
meet? 
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Two non-parallel planes meet along a line. Two parallel planes now meet
along their common ideal line. A plane and a line (that does not
lie in the plane) must meet, either where the line pierces the plane.
Or, if the line is parallel to the plane, they meet "at infinity", i.e.
the ideal point of the line lies on the ideal line on the plane. 

Two lines, however, need not meet at all, either at any real point
or at any ideal point. Such lines are called 'skew'.  
Can you figure out what it means for a real and an ideal line 
to be skew?
pass:[</div></div>]

Kepler's peculiar geometry had already been discovered,
practically speaking, by the Rennaissance artists in 
their study of perspective. The 'horizon',
particularly familiar for residents of Illinois, is an ideal line. In 
perspective, the horizon is drawn as a line in the painter's canvas.

Later, these ideas were adopted by mathematicians as  
'projective geometry'. It differs from the traditional Euclidean 
geometry in many ways. Most notably, there is greater logical 
symmetry between points and lines. 
In projective geometry, any two points are joined by a line ( as they
do in Euclid's geometry) 
and any two lines (in a plane) have a point in common (non-Euclidean).
In projective 3-space, any two points are joined by a line (Euclidean),
and any two planes meet along line (non-Euclidean).

To exercise this expanded way of thinking we re-examine Desargues' 
theorem, but allowing for certain points to be ideal.

Desargues Theorem
-----------------

The figure we usually draw to illustrate Desargues Theorem looks 
like the following. But you should experiment with figures where 
the red and green triangles are in more general positions along the
three lines.

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|pass:[<A href="desargues1.seg"><IMG src="desargues1.png" alt="desargues1.png"></A>]
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.Desargues Theorem.
*************
For &#x025B3;asciimath:[ABC] and &#x025B3;asciimath:[A'B'C'] and asciimath:[D=(A A')(B B')(C C')] we have that asciimath:[C^{**}=(A'B')(AB), B^{**}=(C'A')(CA)], and asciimath:[A^{**}=(B'C')(BC)] are collinear, i.e. asciimath:[(A^{**}B^{**}C^{**})].
*************

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Now let's look at one ideal case, when both pass:[$ D \and A^** $]
are idea, i.e. pass:[$ (A'A) || (B'B) || (C'C) \and (BC) || (B'C') $].

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In this case, we have three parallal lines supporting the two triangles. 
asciimath:[C^{**}=(A'B')(AB), B^{**}=(C'A')(CA)] are real (draw them !) 
but asciimath:[A^{**}=(B'C')(BC)] is ideal. Now what does the conclusion of
Desargues' theorem, that asciimath:[(A^{**}B^{**}C^{**})] 
is a line, mean in purely Euclidean terms? 
It means that asciimath:[A^{**}] is on asciimath:[(B^{**}C^{**})], 
i.e. that asciimath:[(B^{**}C^{**})] is a line parallel to 
(both) asciimath:[(BC)] and asciimath:[(B'C')]. 
In other words, asciimath:[(B^{**}C^{**})] is in the pencil of 
asciimath:[(BC)].

.Exercise.
======================
Easy: Give a similar description of the case that only pass:[$ D $] is ideal.

Challenging: What if only pass:[$ A $] is ideal? 
======================

Next, let's play this game in reverse, and see what happens when ideal
points are made real. Here is a classical theorem about parallel lines
(see Tondeur, Chapter 1).

Pappus Theorem
--------------

.Pappus Theorem.
*************
Let asciimath:[(ACE)] and asciimath:[(BDF)], and the two lines be parallel. 
If asciimath:[(AB)\ ||\ (DE)] and asciimath:[(BC)\ ||\ (EF)] then asciimath:[(CD)\ ||\ (FA)].
*************

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.Question.
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Restate Pappus' theorem in Kepler's terms.
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Consider four points pass:[$ W = (AC)(BD), X=(AB)(DE), Y=(BC)(EF), 
Z = (CD)(FA) $]. The theorem says that if pass:[$ W, X, Y $] are 
ideal, then pass:[$ Z $] must also be ideal. In particular, 
pass:[$ (XYZ) = \infty $]. 
pass:[</div></div>]

.Exercise.
==================
Use KSEG or ruler and compass to investigate the case that 
pass:[$ W $] is a real point. Is it still true? If both pass:[$ X, Y $]
be real too, is it still true that pass:[$ (XYZ) $]. 
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Pascal's Theorem in KSEG
------------------------

Having extended the concept of 'collinearity' to include ideal and 
real points in the plane, you should revisit Pascal's generalization of
Pappus' theorem we mentioned earlier. Pascal was so amazed by the
result he discovered, he gave it a fancy name: 


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.Hexagrammum Mysticum
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If pass:[$ ABCDEF $] is a hexalateral inscribed in a 'conic section' then
pass:[$ G=(AB)(DE) $], pass:[$ H=(BC)(EF) $] and pass:[$ J=(CD)(FA) $] are
collinear. 
*************

The circle is a well-known  conic section, as are two crossing lines. 
In Kepler's sense, two parallel lines are also qualify. They are a section
of a cylinder, and a cylinder is a cone with its vertex at infinity.

The intersections of the three pairs of opposite sides of a regular hexagon 
are ideal, and hence collinear. 
Inscribe a hexalateral in a circle, and see where these 
three points are now? Don't be afraid to have sides of the polygon cross.
Move one vertex of the hexalateral off the circle and see what happens.
Finally, move all the vertices of the polygon off the circle, but carefully
perserving the collinearity you've observed. Any conjectures about where
these vertices must lie?