Null points and null lines | Universal Hyperbolic Geometry 12 | NJ Wildberger

Null points and null lines are central in universal hyperbolic geometry. By definition a null point is just a point which lies on its dual line, and dually a null line is just a line which passes through its dual point. We extend the rational parametrization of the unit circle to the projective parametrization of null points and null lines. And we determine the joins of null points and meets of null lines using these coordinates.

Video Content

00:00 Introduction
5:35 Recall: Rational parametrization of unit (null) circle
10:21 Projective parametrization of the unit circle
14:42 Parametrization and linear algebra, chromogeometry
18:11 Join of null points theorem
27:38 Meet of null lines theorem
32:31 A triply nil triangle (ST2)

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[njwildberger]

@EmptySpaceEnterprise Your hard work in making all these CONTENT SUMMARIES is much appreciated, not only by me but also I'm sure by many viewers.

[ChrisDjangoConcerts]

orhic axis: (-61:-30:133)
orthocenter = basecenter = orthostar [-61/133, -30/133]

So the for triply null triangles these seem to coincide.

Also there appears an additional collinearity, which may have a deeper meaning I'm not seeing at the moment, of 3 sets of 4 points
d1, d2, (a1a2), c3
d2, d3, (a2a3), c1
d1, d3, (a3a1), c2
where d_i are the vertices of the dual orthic triangle and c_i meets of the sides of the triangle and its orthic triangle, so they are on the line of perspectivity of the two triangles

[mavincabanilla]

Good day..

Our professor uses your video as our lecture.. and it is really amazing because every time we get confused into the subject then we just review your videos.. you're so amazing.. :D

[njwildberger]

@EmptySpaceEnterprise Do you have pg1-pg3 for this one?

[robharwood]

It looks like to me that the Join of Null Points theorem also works even if the points are *not* distinct (i.e. they are the same point), in which case it directly gives the dual (tangent) line through the single null point on the circle! Nice!

[njwildberger]

Hi goldilocksbamboozles, Remember to get rid of any common factors when expressing points and lines in homogeneous coordinates! Keep up the good work...

[njwildberger]

@EmptySpaceEnterprise Can't see pg1-3..

[AnschelOron]

Hi . you are a very good teacher and these vid are very fun to watch, look at the views count on this one, i think it is becasue the last theorms of 10 aren't explained enough, and that makes 10 and 11 hard ones, maybe you can add 10b?

[floretion]

Awesome series of lectures, thanks! Concerning the proof of "Every null point alpha is of the form ..." @11:32, aren't we missing the other half of the theorem? That being that if alpha is a null point, then a rational "u" and "t" can always be found such that [u*u - t*t : 2*u*t : u*u + t*t] ? Here's my "proof" (which requires taking the square root, something I realize we're trying to avoid here- unsure how to avoid it): Assume alpha = [x : y : z] is a null point, then x*x + y*y - z*z = 0 by definition. With no loss to generality, we can assume z = 1 or equivalently alpha = [x : y : 1] (otherwise just divide by z and relabel x and y). Define u = sqrt((x+1)/2) and t = y/(2*u). That does the trick (check).

[postbodzapism]

27:11 wow, and that's a very good way in factorizing expresssions.

[vilbjrgbroch]

hello, the e(t:u) parametrization is an amazing step!!! In WildTrig 15 you reveal a connection between the e(t) parametrization and rotation with complex numbers : Zs = Zr*Zt; s = (r+t)/(1-r*t) . This simple algorithm is fabulous and I have been using it extensively in sound-programming the past two years, so I have been trying to find something similar for e(t:u). I have been scribbling many pages full but I am not coming closer, Is this solved or is it definitely not solvable???thanks and thanks again for your generous internet postings. Vilbjørg