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Where am I?

If I'm stuck in the middle, but can look around as much as I'd like; I can see 120, 24 times.

These 24 are actually arranged into groups of 3 across 8 complete imaginary ●'s of view.


However, as soon as I get cornered; whatever corner that may be; I can see straight again. In any such case, 3 of the 120's are no longer visible to me, since I am standing in them; resulting in only 3 120's which I can still count.

These 3 are again the parts of 1 full ● of view.

Suggested reading

An entry in the Official Dictionary of Dubiously-used HTML Characters:

"●", origin: abstract.

A ● is the solid version of an o, and is a big version of an •. The Ancient Greeks considered the ● to be more perfect than a 0 or a □.

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  • $\begingroup$ Welcome to Puzzling SE! Please be more clear in your question, at this moment it's incomprehensible! For example, what does it mean "I can see 120, 24 times"?. Also, what's that dot? $\endgroup$ – leoll2 Apr 19 '15 at 12:34
  • $\begingroup$ Is this a riddle? Are you sure the calculation-puzzle and the visual tags are appropriate? $\endgroup$ – leoll2 Apr 19 '15 at 12:43
  • $\begingroup$ @leoll2 Thank you for your welcome. Visual, I guess is applicable; even though it might be unneeded, since the tag geometry is already used. Calculation-puzzle is also applicable, since solving the riddle will most likely require e.g. a very simple division. $\endgroup$ – Vincent Mia Edie Verheyen Apr 19 '15 at 12:46
  • $\begingroup$ I've just edited the tags: removed calculation-puzzle as we have differents standards for those puzzles; added riddle. $\endgroup$ – leoll2 Apr 19 '15 at 12:49
  • $\begingroup$ @leoll2 Thanks for that mate. Hey, your answer was very interested as well. I think it also fits the riddle to a large degree, or perhaps even completely. Perhaps re-add it? I am already interested in it, solely to discover new geometric objects. If I could up-vote it, I would certainly do that. $\endgroup$ – Vincent Mia Edie Verheyen Apr 19 '15 at 13:16
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The answer is

Cube

enter image description here

If the point of view is in the middle of the cube, you can see all the edges forming 120° angles between them. It's easy to see, just look at the corner of your room where two walls and the ceiling meet, it's a perspective matter!

When the point of view is in a vertex, you can only see three 120° in the opposite corner, while the three you're standing in are hidden by you!

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  • $\begingroup$ Very nice! This was the mathematical object which I initially had in my mind; via abstraction by staring at the corner of my room. Very nice explanation. I haven't heard of the objects described in Bob's answers yet, but find them really adept as well. I would love to give you both the credits, but perhaps I would give them to the one lower in score? Or perhaps I will accept your extented answer, which corresponds with my initial thought. ||| Anyway, thank you for an excellent answer. What was the mathematical object of your earlier answer please? I would love to leave it here as a reference. $\endgroup$ – Vincent Mia Edie Verheyen Apr 19 '15 at 13:40
  • $\begingroup$ Just in-case you don't know how "credits" work in SE: "up"/"down" vote can be assigned to multiple answers, while "best answer" can only be assigned once per question by the author. By the way, my previous answer was octahedron, but it was incorrect as I confused 60° and 120° eheh! $\endgroup$ – leoll2 Apr 19 '15 at 13:46
  • $\begingroup$ Yeah perspective makes it all so clear. As long as you look directly at the opposite corner. Definitely don't look along an edge! $\endgroup$ – Bob Apr 19 '15 at 13:47
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    $\begingroup$ Thanks for playing both; I enjoyed reading the answers. For simplicity, I'll accept the cube as "the answer"; since this was my original thought & the answer is pedagogically extended. ||| The 4 cornered spherical polyhedron is very ingenious as well, however; one can not "see straight" lines when standing in at an edge, I guess. ||| I very much like the idea of a truncated hexahedron as well, which is a very neat answer. Credits & thanks to Bob for them both. $\endgroup$ – Vincent Mia Edie Verheyen Apr 19 '15 at 15:05
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I think I have a better answer than my last...

...a 4 cornered spherical polyhedron, essentially a cube inflated like a ball...

...because when inflated like a ball the angles become $1/3$ of a circle that magic $120^\circ$ and this time there are $24$ of them.

It looks like this :http://upload.wikimedia.org/wikipedia/commons/7/7c/Uniform_tiling_432-t0.png

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I'm guessing (because I'm not sure how this answer fits with part of the puzzle):

Inside a truncated hexahedron.

The ●s are corners each have 3 vertices.

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  • $\begingroup$ Hexahedron=cube? Or is this an irregular hexahedron? $\endgroup$ – leoll2 Apr 19 '15 at 13:02
  • $\begingroup$ @Bob Very good, I'm impressed. The OP was actually thinking of the most famous h_x_h_d_o_. Could you explain why you think other h_x_h_d_o_s might also fit the riddle? I would be interested. | Also: the ● can very well be considered to be a corner. However, there is a geometric object which is more appropriate to the description of ●, as given in the Official Dictionary of Dubiously-used HTML Characters. If you could find out something about the Ancient Greeks, you might be able to guess it? =) But, again: the interpretation of ● is dubious for the first part. Yours is fine. $\endgroup$ – Vincent Mia Edie Verheyen Apr 19 '15 at 13:09
  • $\begingroup$ I chose that type because it has corners which are triangles, which fits the part where you say there are groups of three. I realise now I've looked at octagonal faces, thought they were hexagons and linked that with the 120 as 120 degree angles. Of course that obviously wrong so I'm going to delete that bit. I'm even less sure now. $\endgroup$ – Bob Apr 19 '15 at 13:11
  • $\begingroup$ @Bob Well, actually I think your answer: "truncated hexahedron" is a very good answer as well; of which I hadn't thought yet. I wouldn't delete it? $\endgroup$ – Vincent Mia Edie Verheyen Apr 19 '15 at 13:12
  • $\begingroup$ I'll leave the answer. But I've cut the obvious nonsense. $\endgroup$ – Bob Apr 19 '15 at 13:17

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