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What is the maximum number of faces of totally convex solid that one can "see" from a point?

...and, more importantly, how can I ask this question better? (I'm a college student with little experience in asking well formed questions, much less answering them.)

By "see" I mean something like this: you point a camera from a point at the solid, and look at the picture. How many of the faces of the solid look like faces and not just lines? Let's assume that the lens is just a point in space (no lenses wider than the solid itself) and that the camera is a finite distance from the solid. I know this is a crude definition... if you have any ideas for a more rigorous definition, this would be awesome, then maybe there's ways to prove the answer to my question mathematically.

For example, in the picture of this cube, you can see 3 faces. This is the maximum you can see for a cube. How can that be proved?

What methods might you use to prove this for a convex solid of any size and shape? Are there ways to do so using only relatively basic math (Multivariable calc, linear algebra, high school geometry)?

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    $\begingroup$ The answer will be different or every solid. Viewed from above the apex you can see every face but one of a pyramid with $n$ faces. $\endgroup$ – almagest Apr 19 '16 at 14:24
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    $\begingroup$ Proof for a cube: you can't see two parallel faces simultaneously. $\endgroup$ – Abstraction Apr 19 '16 at 14:26
  • $\begingroup$ I know that it'll be different for every solid. If you have a dodecahedron, it'll be 6. If you have a cube it's 3. If you have a tetrahedron it'll be 3... I'd like to know if there is a good way to prove each of these? @Abstraction, I'm not sure that constitutes a proof...? $\endgroup$ – jhch Apr 19 '16 at 14:28
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    $\begingroup$ Let say that the solid is convex "enough" so that each face can radiate without hindrance in all the outward directions. Then you can consider the centre of each face and compute if a ray from it is reaching the camera (with a sufficient angle, if you exclude seeing lines). $\endgroup$ – G Cab Apr 19 '16 at 14:33
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    $\begingroup$ @JohnHughes What I meant was, out of all the possible polyhedra, you cannot ever see more than $n-1$ faces, but you can see exactly $n-1$ in a few special cases. See my answer. $\endgroup$ – shardulc says Reinstate Monica Apr 19 '16 at 14:57
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You can always find a place from which to see at least half the faces.

To see why, start by considering a polyhedron with central symmetry. Imagine a viewpoint from which you don't see any lines as points or faces as lines (i.e. general position) and far enough away so that you can see all the faces whose normal points into your side of the half plane perpendicular to your line of sight. Then think about what you see from far enough away in the opposite direction. You can see all the faces from one side or the other and no face from both sides, so the symmetry says you see half each time.

Four of the five regular polyhedra have a center of symmetry. The tetrahedron does not: there's no place to put the origin that allows invariance under the map $x \to -x$.

Even without central symmetry, you see all the faces from one side or the other, so you see at least half from at least one side. Pyramids represent an extreme case. You can see all but one face from one direction and just one from the other, as @almagest points out in a comment.

Since the polyhedron has only finitely many faces, "far enough away" in the preceding proof does not have to be at infinity (though it may be pretty far). As @JohhHughes comments, if you put your camera close enough to any face that's the only face you'll see.

Note: the arguments work in all dimensions. They are particularly easy to visualize in the plane. (On the line they're trivial.)

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    $\begingroup$ Maybe I'm wrong here... doesn't a tetrahedron have a center of symmetry? But you can see 3 out of 4 sides, not 2 out of 4. $\endgroup$ – jhch Apr 19 '16 at 14:40
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    $\begingroup$ No, a tetrahedron does not have a center of symmetry. There's no way to place it in space so that the map $x \to -x$ maps the object to itself. $\endgroup$ – Ethan Bolker Apr 19 '16 at 14:40
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    $\begingroup$ @GCab: The Question is about the maximum number of faces that are visible, so if a closer point of view reduces them, don't do that. $\endgroup$ – hardmath Apr 19 '16 at 14:43
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    $\begingroup$ Rather than "a center of symmetry", a better choice might be saying the polyhedron has "central symmetry" (or "point symmetry"). $\endgroup$ – hardmath Apr 19 '16 at 14:45
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    $\begingroup$ With any convex solid I think you can place the point of view close enough to reduce the number of faces seen to exactly 1. (if I'm wrong, that is super cool and please tell me more.) $\endgroup$ – jhch Apr 19 '16 at 14:46
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As @almagest has pointed out, the absolute maximum number of faces you can see of a polyhedron with $n$ faces is $n-1$. This is achieved in the case of a right pyramid with a base and $n-1$ sides; if you view the pyramid from above the apex, you can see all the sides except the base. This is perhaps true for non-right pyramids and other shapes as well.

The absolute minimum number of faces you can see is 1, as you said: just place yourself (or the camera) arbitrarily close to any one face. As the polyhedron is convex, none of the faces will 'tower over' any one shape and you will see only one shape.

Both these bounds, however, are quite obvious and useless. As in the answer above, half the faces of a regular polyhedron with a center of symmetry can be seen from sufficiently far away. I would extend this to say that roughly half of the faces of a roughly regular polyhedron can be seen from sufficiently far away, where 'roughly' is an appropriate tolerance constant. I think it is illustrative to think of the sphere that completely circumscribes the polyhedron, of which you can obviously see exactly half. Maybe that half can be 'mapped' on to the polyhedron's faces.

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  • $\begingroup$ Can we say that for any given polyhedron with number of faces > 4 the maximum you can see purely by changing viewpoint is n/2? Are there any special cases other than a tetrahedron? $\endgroup$ – jhch Apr 19 '16 at 14:59
  • $\begingroup$ @JohnHughes What if I give you a pyramid? $\endgroup$ – shardulc says Reinstate Monica Apr 19 '16 at 15:01
  • $\begingroup$ I think a pyramid has 4 faces, right? For the pyramid, it's n – 1. For other polyhedrons, is it n/2 or might there be other exceptions? $\endgroup$ – jhch Apr 19 '16 at 15:09
  • $\begingroup$ @JohnHughes There is an $n-2$ case: consider a pyramid whose base isn't completely flat. Instead, it has an outwards crease in the middle, so that there are in fact 2 faces that you don't see from the top. This can be generalized to $n-k$ for any $2 \leq k < n - 1$ I suppose. $\endgroup$ – shardulc says Reinstate Monica Apr 19 '16 at 16:43
  • $\begingroup$ I'm having trouble imagining this pyramid whose (triangular) base isn't completely flat, for which you couldn't just change the viewpoint to see at least one of those two faces on the base. $\endgroup$ – jhch Apr 19 '16 at 19:29
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Perhaps this paper which deals with the situation in a rather abstract way, and for higher dimensional polytopes might interest some: http://www-math.mit.edu/~rstan/transparencies/vis.pdf

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Not only for a pyramid, but also for a "hemi-spheric diamond" cut with any number of flat faces, when looked from far enough, will show $n-1$ faces.
If you consider regular polyhedron only, than the answer may vary.
Coming back to a general method, for a general convex polyhedron and a "common" camera, then consider the camera visual cone (something less than $ {2\pi }$ steradians), with the vertex placed at the camera position. It will collect all the rays within that cone. Now consider the polyhedron as light emitting. You may consider that each face radiate from its centre in all the directions radiating outwards. If the polyhedron is convex, each face should freely radiated in the outward hemisphere. Exclude rays at $90°$ from vertical if you do not account for face=line. Check if there is a radiated ray that can reach the camera, and if it is within its view cone. Or invert the ray path.
With a proper math description of the polyhedron, the SW task is not too difficult.

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  • $\begingroup$ Can you clarify what a "hemi-spheric diamond" is? $\endgroup$ – jhch Apr 19 '16 at 15:46
  • $\begingroup$ @JohnHughes I mean a polyhedron, with a flat (large) face and with all vertices lying on a dome over that, perimeter (rim) included. View point on the vertical axis, as far as it can view the most inclined faces along the perimeter. $\endgroup$ – G Cab Apr 19 '16 at 19:14
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Here are some definitions (in dimention 3, but you can easily generalize):

Definition : Given a finite number of points with coordinates $ P_1 = (x_1,y_1,z_1), .., P_n = (x_n,y_n,z_n) $, a convex solid is the convex hull of these points, i.e. all the points of the space defined as $\sum_{1\leq i \leq n} \lambda_i P_i $ with $\forall i \in [1,n], \lambda_i \in \mathbb{R}, 0 \leq \lambda_i \leq 1, \sum_{1 \leq i \leq n } \lambda_i = 1 $

For instance, the points (0,0,0); (1,0,0); (0,1,0); (1,1,0); (0,0,1); (1,0,1); (0,1,1); (1,1,1) define a cube.

Rq : If I add the point (0.5,0.5,0.5), I define the same cube.

Definition : An oriented plane is 4 real numbers $a, b, c$ and $d$.

Definition : An oriented plane is a face of a solid, iff at least three points defining the solid verify $ax+by+cz = d$ (are in the plane), and all points of the solid verify $ax+by+cz \leq d$.

For instance, the plane defined by $a=1, b=0, c=0, d=1$ in the cube described above is a face. The one defined $a=1, b=1, c=0, d=1$ is not, since it fails the second condition of the definition. The one defined by $a=-1, b=0, c=0, d=-1$ is not either. Indeed, the definition specifies $ax+by+cz \leq d$, and it is not the case here.

The definition of a face given here enables to control on which "side" of the plane the solid is. Putting a "-" sign in front of all the coefficients defining a face would mean the solid is on the other side of the plane.

Definition : A view direction is a vector $ (x_v,y_v,z_v) $ with $x_v^2+y_v^2+z_v^2=1$.

The direction can be identified at the way you look at. Here, it is assumed you look from a point at an infinite distance.

Definition : A face $(a,b,c,d)$ of a solid is visible from a direction $ (x_v,y_v,z_v) $ iff $ax_v+by_v+cz_v < 0 $.

Here is the proof (short version) for the cube (with the points described earlier).

All faces of this cube are defined by the following quadruples (and no others) :

  • (1,0,0,1)

  • (0,1,0,1)

  • (0,0,1,1)

  • (-1,0,0,0)

  • (0,-1,0,0)

  • (0,0,-1,0)

So, with any viewing direction, one sees at most three faces (the ones corresponding to the sign of your viewing direction). (you see less if one of the viewing direction coordinates is 0, it would mean we see only some of its edges).

For a different kind of solid, you need its defining points, and then the proof would basically go this way

  • first find all faces of this solid

  • assume there is another face and discover it is one you already have (so you have proven that you have all of them)

  • Assume you have a viewing direction. Then, a lot of cases to look at, like "if my direction is such that $x_v + ... < ...$"

  • find, in all the cases, the maximal number of faces

If you think there is a leaner/more conventional way to define the problem, do not hesitate to mention it.

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