Pascal   Brianchon


Angles on unit circle (in degrees)
1 (on green, blue):     
2 (on blue, purple):    
3 (on purple, green):   
4 (on green, blue):     
5 (on blue, purple):    
6 (on purple, green):   

                        Hyperbola   Ellipse

If hyperbola:  angle of first point at infinity
If ellipse:  angle of perp. to new line at infinity
                               

If hyperbola:  angle of second point at infinity
If ellipse:  length of perp. to new line at infinity:

Suggestions for Inputs

These are some suggestions for inputs to the above form. They are only suggestions, and you'll have more fun if you play around with the angles yourself. However, all of the suggestions below should produce intelligible pictures, and each one corresponds to a different Euclidean theorem. Can you formulate the theorems corresponding to the pictures?

These inputs were designed for the Pascal option on the form. They will also work with the Brianchon option; however, in that case the diagrams will probably be better if you put the vertices in increasing or decreasing order. 

Suggestions for Angles of Vertices

30  -90 120 -45  90 -135 (non-convex hexagon)
 0   50 100 135 200  -90 (convex hexagon)
-160 135  0  60 -120 -50 (hexagon from movie)
120 -30 180  45 -90  -90 (pentagon)
 0    0 150 150  80 -100 (quadrilateral 1)
20   20 210 150 150  -60 (quadrilateral 2)
105 105  30 30 -135 -135 (triangle)

Hexagons with Unusual Pascal Lines

45  180 -90   0 120 -135 (one inters at inf)
xx   xx  xx  xx  xx   xx (two inters at inf)
 0   60 120  180 240 300 (all inters at inf)
30  -30 120 -150 150  60 (all inters at inf)

Placement of the Line at Infinity

If you use the ellipse option with a final input
greater than 1, your second picture will be an
ellipse.  It probably won't be much more interes-
ting than the circle; however, it may be easier
to see lines and points in the ellipse.  You can
also try these options:

135 180 -90 45  0  0 ellipse 90 1.414
  (one inters at inf)
135 180 -90 45  0  0 ellipse 90 2.414
  (two inters at inf)
180 135  90 45  0 -90 ellipse 90 2.414
  (Pascal line at inf)
180 135  90 45  0 -90 ellipse 0 3.414
  (one pt from Pascal line at inf)

There is no specific parabola option, but you can
get parabolas as a limiting case of either the
ellipse option or the hyperbola option.  For
instance, either "ellipse 135 1" or "hyperbola 135
135" will give you the parabola in which the 135-
degree point has been taken to infinity.  Try any
of the above suggestions using "hyperbola x 1" at
the bottom of the form, where x is a random angle,
the angle of a vertex, or the angle of a repeated
vertex.

Almost anything you do with the hyperbola option
will produce interesting results.  Try this option,
using for your angles:

two random angles
a random angle and the angle of a vertex
a random angle and the angle of a repeated vertex
the angles of two consecutive vertices*
the angles of two almost-consecutive vertices*
the angles of two opposite vertices*
(*=try this when neither, either, or both of the
 vertices are repeated vertices)

As in the elliptical case, it's also interesting
to make some intersections on the line at infinity:

60 -60  150 30 -120 180 hyperbola 0   90
  (one inters at inf)
60 -60 -120 30 150 180 hyperbola  0   90
  (one inters at inf)
60 -60 120 120 180 0 hyperbola -90 90
  (two inters at inf)
60 -60 -120 30 150 180 hyperbola 45 -135
  (Pascal line at inf)