Midpoints and bisectors | Universal Hyperbolic Geometry 16 | NJ Wildberger

Midpoints of sides may be defined in terms of reflections in points in hyperbolic geometry. Reflections are defined by 2x2 trace zero matrices associated to points. The case of a reflection in a null point is somewhat special. The crucial property of reflection is that it preserves perpendicularity, which then implies that reflections send lines to lines. Midpoints of a side bc can be constructed with a straightedge when they exist, and in general there are two of them! This is a big difference with Euclidean geometry. Bisectors of vertices are defined by duality.

Video Content:
00:00 Introduction
3:28 Definition of reflection of a general point
7:32 Null reflection theorem
9:29 Matrix perpendicularity theorem
14:29 Reflection (preserves) perpendicularity theorem
18:11 Reflection (preserves) lines theorem
21:24 Midpoint between 2 points
24:25 Geometrical construction of midpoints 1a)
28:59 Geometrical construction of midpoints 1b)
31:41 Geometrical construction of midpoints 2
33:34 Not all sides have midpoints; side/vertex midpoints/bisectors

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[pickeyberry]

Video Content

00:00 Introduction
09:29 Matrix perpendicularity theorem
14:29 Reflection (preserves) perpendicularity theorem
18:11 Reflection (preserves) lines theorem
24:25 Geometrical construction 1a)
28:59 Geometrical construction 1b)
31:41 Geometrical construction 2

[tezlaf]

I have to say that your discovery is absolutely beautiful! I feel very fortunate for your open and instructive approach to sharing these hugely important findings. I will be incorporating universal hyperbolic geometry immediately into my own research. -Scott

[njwildberger]

No, you don't need to know the Cartesian coords of the two points in order to construct the midpoints of the side. This is a purely straightedge construction using the null circle.

[njwildberger]

In Universal Hyperbolic Geometry one circle, called the null circle, plays a distinguished role. All constructions ultimately come down somehow to this null circle. The existence of midpoints of a segment is dependent on a quadratic equation, which can have either two solutions or zero (or the special case of one in between). If one point is inside the circle and the other outside, then as you say there is no midpoint.

[jacobolus]

Note that on slide 2, [-54: -18: 63] all have a common factor of 9, so this can be written [-6: -2: 7]. Or it might be easier to start by dividing the matrix ma*mb*ma = [18, -9; 117, -18] by 9 to get the matrix [2, -1; 13, -2].

[leonig]

Sorry for persisting but I think this is important for my understanding. When I draw a line between two points in Hyperbolic Geometry I must specify the cartesian coordinates of the points otherwise there is no way that the midpoint between them can be defined using the null circle. This is not so with Euclidean geometry.

[postbodzapism]

Rewatching the UnivHypGeom playlist and found lots of things I missed 4 years ago...
I think it would be better if more motivations are provided. For example, why is reflection defined as such? Some motivations of the definitions would give a more intuitive idea about the concept.

[leonig]

In Euclidean geometry all lines have midpoints. In hyperbolic geometry my understanding was that the circle was used as an intermediate step to get the midpoints. This is obviously not the case as points one inside and one outside the circle do not have a midpoint as discussed in the video. In other words if I draw a line without the complimentary circle does it not have a reflection of points on it?

[waterheaven]

About the Null Reflection Theorem, the problem rises when b lies on "a-perp" A - a null line, which yields a factor of zero, in the matrix calculation setting. However, pictorially there seems to be no exception wrt the points on that null line. Perhaps these points should be taken extra care of in such an matrix formulation.

[MartinPitti]

I can't figure out the m_𝛼 m_b m_𝛼 = m_𝛼 execise/proof. With pure matrix mult. there is not too much one can do: you cannot multiply with m_𝛼 as that would make both sides zero (as m_𝛼² = 0), and multiplying with m_b doesn't yield something sensible. I tried to do the multiplication with the definition of the matrices, but that just yields four huge element expressions without any common factor. Any hints? Thank you in advance!

[liadorel]

I have just realized, and I hope I am right, that if the two sides of a triangle have rational midpoints, then the third side has it as well.