Metal Ball Cage Template Cardinality: A Brilliantly Lazy PROOF

N.B. - I'm looking for the simplest way to ascertain the number of templates $T$ (see below) comprising the structure from just one angle alone; that is, I'm sitting down looking up at this thing, and I want a way to compute its cardinality based on the simplest methods, but perhaps relying on some underlying abstract concept, in particular, graph theory, geodesics, topology, algebra, or something totally new.

My working theory is that we need only the equation

$$V-E+F=2,$$

and the Inclusion-Exclusion Principle. In fact, we might also need the fact that the star is a fraction of the whole, and some other symmetry to rely on to create a system of equations. That is

$$\frac{V}{n}-\frac{E}{n}+\frac{F}{n}=2$$

and

$$\frac{V}{m}-\frac{E}{m}+\frac{F}{m}=2,$$

where we know the relationship between $m$ and $n$. Actually, that makes no sense... Hmm...

UPDATE: I have verified, with sufficient effort, Mr. Narain's proposal that the structure is a snub dodecahedron: There are indeed 60 pieces, however, I'd still like a lazy method using the ideas I've alluded to all throughout this post...

I'm at a fancy restaurant, and I saw these template ball lights: http://tinypic.com/r/2co6b9w/5

I'm trying to figure out the number of template pieces, call them $T$; they look like this: http://tinypic.com/r/29cs8cw/5

Here is my approach: Count the number of things that look like this: http://tinypic.com/r/wmnssn/5

Now notice that for each $T$ coming out from the center there are four legs which the rest are connected to, two of which are connected to adjacent $T$'s which are coming out from the same center mentioned before. I feel this problem is one of algebra. My friend here thinks that if you measured the shape of $T$, then you could find it easily with the surface area of a sphere, but she can't seem to work out how to get the number of $T$'s. If you need more photos, let me know.

Just in case you don't see it: This appears to be a snub dodecahedron--as was pointed out by one of the commenters--and can be seen in an overlay here: Look at this: I believe these two can be related in a system of equations via the Euler Characteristic--perhaps I'd need a third distinct shape...

Here are some statistics: Based on these statistics, here is the template--just in case you want to make one for yourself: If you pay me $50 I'll make a larger one for you out of balsa wood. ^_^ THE BIG IDEA: Assume I am a tree--maybe I'm an African Baobab, and my Baobab friend next to me has this thing dangling motionlessly from her branches. Me being an Baobab, I don't know about snub dodecahedrons, but--for some genetically mutative reason--I know a bit of mathematics. So, now, I'm looking at this thing wondering if I can count how many$T$'s there are (see above) just by noting how the arms of the$T$'s are connected. What is the least amount of data that I need from my single, grounded point of view to ascertain the number of the$T$'s comprising this object? • What theory should be applied here? How do I define "ballness," as it were? – Trancot Nov 6 '13 at 0:10 • Maybe we could be more Tao Te Ching and just focus on the empty space. There are six holes in the center, then fifteen holes surrounding the main star, and so on... – Trancot Nov 6 '13 at 0:20 • Or perhaps we should consider the degree of the vertices. There appear to be three types; one has degree five, one has degree four, and one has degree three. – Trancot Nov 6 '13 at 0:22 • A diagram would be very nice if you included it in your analysis. – Trancot Nov 6 '13 at 0:55 • This may sound complicated, but can anything be extracted from the dual graph of sub-region of a map that happens to be a fraction of of a sphere--in this case the star? – Trancot Nov 6 '13 at 2:09 3 Answers$N$stick figures. Belly vertices:$N$. Hands and feet vertices:$4N/3$. Head vertices:$N/5$. Sticks (arms, legs, neck):$5N$. Open spaces:$5N/2$. $$\left(N + \frac{4N}{3} + \frac{N}{5}\right) - (5N) + \left(\frac{5N}{2}\right) = 2.$$ $$\frac{N}{30}=2.$$ So$60$stick figures in total. • That's about what I have here on paper. Good! Maybe you wouldn't mind explaining to the audience, in particular, why is$V$,$E$and$F$this way, and so on. Explain your technique/approach--if you wouldn't mind. – Trancot Nov 6 '13 at 21:05 • Best answer! Bravo! – Trancot Nov 6 '13 at 21:11 One approach is to use the surface area of the sphere. If you wrap a tape measure around the sphere you can get the circumference and hence the radius$R$. The surface area is$4\pi R^2$. Now straighten out in your mind the edges of your shape and measure its area. Divide that into the area of the sphere and you have it. You don't have to be that precise, as you are trying to tell the difference between$8$and$9$, so you need$10\%$accuracy, which isn't hard. Another approach is to try to determine the geometry of the ball. The ring of arms of your group of five looks like a pentagon to me. If the pattern looks the same at each juncture, it would be a dodecahedron, which has$12$pentagons. This pentagon uses$\frac 25$of each template (it uses one arm and shares two other arms with the neighboring pentagon), so you should need$\frac 52 \cdot 12=30$templates. I suspect that there are other styles of vertex than three pentagons. If you can define what meets at each vertex, you can get the number of arms. The Euler characteristic formula can help here. A more practical approach is to put a white dot at the center of each piece and count them. • Yes, which is the initial conclusion I came to vis a vis your last comment; however, I'm lazy, remember? Therefore, I have no way of practically computing such things. I'm looking for an answer that is so brilliant it could be explained to a child. Much in the spirit of Einstein, I'd like a sort of "If you can't explain it to a six year old" kind of approach to the problem using algebra and... well... something else. – Trancot Nov 6 '13 at 0:48 • All the six year olds I know can count to$40$. That many Post-its they can understand. – Ross Millikan Nov 6 '13 at 3:45 • I'm lost... Post-its? – Trancot Nov 6 '13 at 4:16 • I would use Post-it notes to identify the centers, then take them off and count. – Ross Millikan Nov 6 '13 at 14:07 Let$R$be the ball's radius. Let's go really lazy and just use the picture. The apparent diameter of the piece with the green outline is roughly$2/3$of the apparent diameter of the ball. So is its radius, it's$2R/3$. It's apparent area of the piece is the n$4\pi R^2/9$(approximated as a flat disk) and the apparent area of the spherical ball is$4\pi R^2\$, dividing the spherical area with the piece's area is exactly 9. However, the piece is actually a spherical cap so it's area is larger than what's apparent but not by much. Therefore the answer should be less than 9. We get 8 such pieces.

• What I'm really looking for is an answer that uses the vertices, their respective degrees, and the architecture of how they're connected, to come to an even quicker answer. I'm not looking for a geometrically approximating approach, but something more reliant on topology, graph theory, and algebra--and perhaps something I've never even heard of. – Trancot Nov 6 '13 at 1:20
• Are you sure you are not moving the goal-posts? Your comment on Ross's answer speaks of a brilliant answer that could be explained to a child. On my answer you wrote you want something more reliant on topology, graph theory, etc, or even something you've never heard of. You are making the constraints extremely tight. – Alexander Vlasev Nov 6 '13 at 1:28
• Is there not a simple way to count the concentric holes that are generated outward from the center? Namely, the middle has 5, then the next ring has 15, and so on. – Trancot Nov 6 '13 at 1:28
• Any such counting will lead you to something more complicated than just counting the number of pieces. Your friend's approach with measuring the area and all that is more practical and simpler. It's more so in the spirit of Fermi problems in general where you easily estimate some quantity to get a ball-park figure. In this case it gives you the right answer within a few lines of reasoning – Alexander Vlasev Nov 6 '13 at 1:32
• Exactly correct, in my opinion topology and graph theory are just labels for very human ways of interpreting things. Children understand topology and graph theory at an intuitive level. The theory might, to us, seem abstract at first, but in my experience topological things are easier to explain to a child--of course, with sufficient thought. You can tell them there are 3 things coming out of this point, and there are 4 things coming out here, and they are all connected like this, so... Does that make sense? Have I won you over? Are you on my side now? – Trancot Nov 6 '13 at 1:32