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I was studying inversion in Olympiad geometry, and they (Evan Chen's book EGMO) mentioned that we can extend the Euclidean plane by adding a point $P_{\infty}$ such that each line passes through it, and no circle passes through it.

The reason for this they said was that now two parallel lines meet at that point only, and the center can go there on inversion.

But now I have a really stupid confusion:

Does this mean that all non parallel lines meet at two points and parallel lines meet at only one?

I am very confused by this part now, I tried looking up some things on Wikipedia but they had defined very different things and it just made me more confused.

I would really appreciate if anyone could clear this really dumb doubt of mine,

Thank you!

Aditya_math
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    Parallel lines are tangent at the point-at-infinity, so they meet at two coincident points. (Consider inverting parallel lines in a circle.) – Blue Jan 01 '23 at 14:10
  • @Blue , oh wow, that actually makes a lot of sense, infact, that made it very clear. thank you. – Aditya_math Jan 03 '23 at 09:46

2 Answers2

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Does this mean that all non parallel lines meet at two points and parallel lines meet at only one?

Yes, for any two distinct lines, that would be the case. You're trading the axiom that "two distinct points determine a line" for a different one where "three distinct points determine a 'cline'" (I thought they called them "lircles" actually...)

You are not doing Euclidean geometry anymore, but have passed to Möbius geometry. I suppose the chapter you are reading teaches you how to transition between the two.

rschwieb
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  • Thank you so much for your answer. btw, the textbook is named: "Euclidean Geometry in Math Olympiad" (EGMO), so it was not an introduction to Möbius geometry, but just teaching the method of inversion in the plane. This answer was still very useful in reinforcing the concept of parallel lines in the extended plane, so Thank you! – Aditya_math Jan 03 '23 at 09:52
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    @Aditya_math I know, I found the chapter you were talking about (available as a sample online) in a matter of a minute or two. It doesn't really matter if he calls it Mobius geometry or not, that's just what it is – rschwieb Jan 03 '23 at 12:37
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You asked

Does this mean that all non parallel lines meet at two points and parallel lines meet at only one?

I think that the Wikipedia article Stereographic projection will make this visually clear. In this projection, all lines and circles in the plane of projection come from circles on the surface of the sphere. All of the circles that pass through the pole project down onto straight lines in the plane while all of the circles that do not pass through the pole project down onto circles in the plane. Any two distinct spherical circles are either disjoint, or are tangent at one point, or else have two points of intersection just as circles do in the plane.

Thus, any two distinct circles that pass through the pole already intersect at the pole. If they are tangent then that is the only intersection point and they project onto two parallel lines, otherwise they intersect at another point and they project down onto two intersecting lines.

Note that the pole on the sphere corresponds to the ideal point $P_\infty$ which is the only point on the sphere which does not project down to a point in the plane.

Somos
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  • "Any two distinct great circles are either disjoint, or are tangent at one point, or else have two points of intersection"

    This is true of circles on the sphere, not great circles, all of which intersect in exactly two points, right? E.g. $\mathbb{CP}^1$ is often introduced as the space of lines through the origin in $\mathbb{C}$, which correspond to great circles passing through $N$.

     – Mo Pol Bol Jan 01 '23 at 22:06
  • @MoPolBol Oops! Thanks for that comment! I have fixed it! – Somos Jan 01 '23 at 22:45
  • No worries @Somos, just realised there’s a mistake in my reply too: $\mathbb{RP}^1$ is the space of lines through $O$ in $\mathbb{R}^2$. $\mathbb{CP}^1$ is the space of (complex) lines through $O$ in $\mathbb{C}^2$. Ofc Euclidean lines through $O$ in the plane still correspond to great circles through $N$ under stereographic projection. – Mo Pol Bol Jan 02 '23 at 00:00
  • Thank you so much for this answer, thinking about inversion in the sense of a stereographic projection really helped make it more intuitive. Thank you! – Aditya_math Jan 03 '23 at 09:53