I would have asked this on a math community but I couldn’t find an active one.

In a spherical geometry, great circles are “straight lines”. As such, a triangle can have two or even three right angles to it.

But what if you go the long way around the back of the sphere? Is that still a triangle?

(Edit:) I guess it’s a triangle! Fair enough; I can’t think of what else you would call it. Thanks, everyone.

    • ByteJunk@lemmy.world
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      2 months ago

      Why the down votes? Bro asking a question and being legit curious, don’t be hating on someone that’s looking to challenge what they know just because it’s trivial to you.

      • towerful@programming.dev
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        2 months ago

        I feel my comment adds to the discussion and wants more details.
        But it was too simply phrased.
        I guess the details of such a question should be obvious. And if you need the details, the question doesn’t actually add the the discussion… It just seems idiotic!

        I felt like there might be a really cool scenario where a vertice isn’t considered a vertice.
        Like, there actually might be some case on a 2d plane “where actually” applies.
        I’m fine being wrong

    • RightHandOfIkaros@lemmy.world
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      2 months ago

      If a shape has 3 corners and 4 edges, it is incomplete or open and therefore not a shape yet but a collection of edges (or possibly, two triangles that share an edge).

      A shape with 4 corners and 3 edges is not possible. An edge cannot have a corner in the middle of it, that would make it two edges.

      • towerful@programming.dev
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        2 months ago

        I felt like adding something about the specific case of 180° between edges and a vertice.
        Makes sense.
        And I guess too many vertices means an open set of edges (ie not close, this not a shape).
        I was kinda hoping for a strange edge case, like a mobius strip or Klein bottle.

        I guess a mobius strip is a 2d representation of a 1d paradigm. And a klein bottle is a 3d representation of a 2d paradigm.
        It would be too much to ask of a 1d representation of a ??d paradigm.