this post was submitted on 15 May 2025
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[–] SuperNovaStar@lemmy.blahaj.zone 9 points 2 weeks ago

You're all thinking too two-dimensionally. Clearly the people are being instructed to arrange themselves into a tetrahedron.

[–] LovableSidekick@lemmy.world 12 points 2 weeks ago (1 children)

Tetrahedrons man, tetrahedrons.

[–] intensely_human@lemm.ee 3 points 2 weeks ago (1 children)

So this makes me wonder if one could force a move into a higher dimension by somehow constraining a set of connected distances in this way.

Sort of like protein folding as a way to bootstrap a dimensional jump.

[–] LovableSidekick@lemmy.world 2 points 2 weeks ago* (last edited 2 weeks ago)

You might like And He Built a Crooked House by Robert Heinlein - the story of a tesseract-shaped house that folds itself into a real tesseract during an earthquake.

https://en.wikipedia.org/wiki/And_He_Built_a_Crooked_House

[–] regdog@lemmy.world 9 points 2 weeks ago

Middle one should be the square root of 4.5 meters, or 2.12 meters

[–] Randelung@lemmy.world 16 points 2 weeks ago (1 children)
[–] Mistic@lemmy.world 10 points 2 weeks ago* (last edited 2 weeks ago) (2 children)

Equal sides in a triangle are only possible if the corners are equal. So, 60⁰ each.

But its height cannot be half of base because of the same Pythagorean theorem

(1,5)²+(1,5/2)²=2,8125

sqrt(2,8125) ≈ 1,677, which is half of a diagonal

So, we get 4 sides that are 1,5 in a parallelogram, but diagonals are 1,5 and 3,354, as opposed to both being 1,5 as shown on the picture

TL;DR: Won't work because Pythagorean theorem

[–] Obi@sopuli.xyz 4 points 2 weeks ago (2 children)

It's funny how we say "because of such and such theorem" as if if some greek dude didn't come up with his little story, the height could totally be half of base.

[–] bss03@infosec.pub 2 points 2 weeks ago

Decolonialize Maths!

We do need short names, but they don't all have to be wyt guys. Pre-globalization, I'm sure many true maths statements were independently discovered by many people.

[–] rowinxavier@lemmy.world 3 points 2 weeks ago (2 children)
[–] kryptonianCodeMonkey@lemmy.world 3 points 2 weeks ago (1 children)

Yes, it is possible with a 3-sided pyramid, i.e. tetrahedron. If we dont look at all 4 points as being on the same plane but 2 opposite corners being offset above or below the other two, this could totally be a tatrahedron.

[–] ptu@lemm.ee 1 points 2 weeks ago

So those two darker green symbols would represent someone shorter or taller. Totally plausible.

[–] Mistic@lemmy.world 1 points 2 weeks ago (1 children)
[–] TheButter_ItSeeps@lemmy.world 8 points 2 weeks ago (2 children)

They could each be on the vertices of a tetrahedron for all we know...

[–] bss03@infosec.pub 3 points 2 weeks ago (1 children)

We do know that, with those measurements, they aren't confined to a single (Euclidean) plane.

[–] LovableSidekick@lemmy.world 2 points 2 weeks ago* (last edited 2 weeks ago)

Exactly! I regret that I have but one upvote to give.

[–] Mistic@lemmy.world 2 points 2 weeks ago

I was thinking of plane surfaces, but if their altitudes are different, I guess it'd be possible.

[–] xia@lemmy.sdf.org 36 points 2 weeks ago (1 children)
[–] bampop@lemmy.world 11 points 2 weeks ago

ikr? It's like some people don't even recognize a tetrahedron

[–] Danitos@reddthat.com 13 points 2 weeks ago

Funnily enough, this is valid under Chebyshev metric, same that kings in chess follow.

[–] untakenusername@sh.itjust.works 13 points 2 weeks ago (1 children)

if the people were aranged in 3d in the shape of a tetrahedron (triangular pyramid) this would work out fine

[–] Uranus_Hz@lemm.ee -4 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

No

Unless the measurement is from the corner to where the lines cross (peak of the pyramid), but that is not at all clear from how the diagram is drawn.

[–] davidagain@lemmy.world 7 points 2 weeks ago (1 children)

It's not a square based pyramid, it's a triangular based pyramid. Imagine the top right hand one floating up onto the air and moving to hover above the centre of the other three (which move to make an equalateral triangle). The distances work but the layout changes.

[–] Uranus_Hz@lemm.ee 2 points 2 weeks ago (1 children)
[–] Luffy879@lemmy.ml 2 points 2 weeks ago (1 children)

Explanation:

So, Theres the sentence of Pythagoras. It says that c^2^ = a^2^ + b^2^ when the triangle has a 90° corner

Since a square is just 2 triangles, it applies. That means c (the distance from Person a to Person c) should be √(2×1.90^2^). But that is 2.7m.

[–] Knock_Knock_Lemmy_In@lemmy.world 1 points 2 weeks ago (1 children)

These instructions are in 3d.

[–] Luffy879@lemmy.ml 1 points 2 weeks ago (1 children)

Even if we would take it from the head of person a to the toe of c, its still a triangle

It's the plan view of 4 equilateral triangles arranged into a tetrahedron. There are no right angles so Pythagoras doesn't apply.

[–] mumblerfish@lemmy.world 8 points 2 weeks ago (1 children)

What? Everytime I meet other people we always arange ourselves in the shape of a simplex of the appropriate dimension. Doesn't everyone?

[–] davidagain@lemmy.world 5 points 2 weeks ago* (last edited 2 weeks ago) (2 children)

So the fifth person to arrive moves to the centre of the tetrahedron and shifts roughly 1.299m into the past or future.

I have a few questions.

  1. How do you attain time offset?
  2. Doesn't that make conversation difficult?
  3. What even is the fifth dimension?
  4. How do you convert a distance in metres into a distance in time? You would surely then have a universal m/s? Oh, wait, there is a universal speed, it's the speed of light, which means 1.299m is equivalent to about 4.3 billionths of a second, which is considerably less impressive for question 1 and just not at all problematic for question 2.
  5. If you're using very fast motion for your time offset, doesn't that make conversation even more difficult? How fast would you need to be going to dilate time for a few billionths of a second? Doesn't Heisenberg uncertainty start to have an impact here? How can you be sure you got it right?
[–] bss03@infosec.pub 2 points 2 weeks ago (1 children)
  1. So, the diagram doesn't represent it well, but the 1.5m is a minimum. So, I just delay myself by half a heartbeat which is well over 4.3^e-9s.
[–] davidagain@lemmy.world 2 points 2 weeks ago (1 children)

Yes, but if they're just minimums, there's no need for even using the third dimension, let alone the fourth.

[–] bss03@infosec.pub 1 points 2 weeks ago (1 children)

Oh, I may have violated distancing protocols then. My personal delay device doesn't have sub-microsecond accuracy. Should I will have gotten a test for time-invariant COVID ?

[–] davidagain@lemmy.world 2 points 2 weeks ago

I won't tell if you won't tell.

[–] ivanafterall@lemmy.world 4 points 2 weeks ago (1 children)

If you have to ask, you wouldn't understand.

[–] davidagain@lemmy.world 3 points 2 weeks ago

If I understood, I wouldn't have to ask.

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