Category Archives: MathStream

So I had zeroed in on the proof of the Archimedean volume relationship as the source of my giveaway for G4G13. But how to create an interesting variation?
The first thing to notice is that the function of h that appears in the proof in the cross-sectional area of of the cone, namely h², could be any function f(h) whatsoever, and the proof would be just as valid — solids where the cross-sectional areas went as f(h) and 1 – f(h) would have volumes that add up to that of a cylinder.

So, in some sense the simplest example would be solids with cross-sectional area going as h and 1-h. What would they look like? Well, they would correspond to radii at height h of √h and √(1-h). In other words, they would be solids of revolution about the z-axis of the graphs of √h and √(1-h), like so:

First, these two objects look identical (just one is upside down as compared to the other). But that’s no surprise: as h goes from 0 to 1, 1-h goes from 1 to 0, and at the same rate. But that means that each one must have half the volume of the corresponding cylinder. And further, since the graph of √h is a parabola (on its side), these objects are paraboloids. So we’ve established

the paraboloid of unit height and radius has half the volume of the cylinder with unit height and radius,

which is kind of cool already. So should this be the giveaway?

It was time to figure out what the giveaway items would actually be. Since the principle was about volumes, they needed to be something which was related to volume, but I didn’t want to create another hourglass. So the obvious answer was that they should be measuring cups. And that decision suggested that I shouldn’t use f(h) = h, since two identical measuring cups would be boring.

Since I’d now looked at h and h², the natural next choice would be h³. Here’s what the corresponding solids look like:

Those look pretty interesting. So if I used these shapes, what would the volumes be? Now it’s by far easiest to go ahead and use that late-17th-century invention, calculus. The volume of the “pointy” shape should be $\int_0^1 \pi h^3 dh = \pi \left.h^4/4 \right|_0^1 = \pi/4$. In other words, if the cylinder were scaled to have a volume of one cup, say, then the pointy shape would have a volume of 1/4 cup and the rounded one would have a volume of 3/4 cup. That seemed pretty good: since most measuring cup sets do not include one with volume 3/4 cup, the resulting set might actually be useful.

But there’s another place you can vary the construction in the proof: the shape of the cross-sections. There’s no requirement that they be circles; you can scale any shape, and the overall volume will scale by the area of the largest cross section. (That’s the $\pi$ in the above formulas, namely the area of the top circle of the pointy shape or the bottom circle of the rounded one.) So I decided to use this flexibility to create one more link to G4G13: make the cross section of the “cylinder” be a regular 13-gon, the 3/4-cup measure a regular octagon, and the quarter-cup measure be a regular pentagon. These shapes echo the Fibonacci decomposition 13 = 8 + 5 where the mathematical free association began.

So although that brings us to the shapes that actually went into the G4G13 giveaway, this exploration wouldn’t be complete without emphasizing that there’s no reason f(h) has to be a power of h at all. We’re literally free to use any function at all, so long as it takes values between 0 and 1. In particular, why not try the most famous pair of functions that sum to 1 for the cross-sectional areas of the two cups, namely cos²(h) and sin²(h)? That would produce these shapes:

As in the first case, the shapes are identical, although in this case because of the identity sin(x) = cos(π/2 – x). So again, this shape has half the volume of the corresponding cylinder, even though it is clearly not a paraboloid (as you can see, for example, by the fact that it comes to a point at its apex; a paraboloid is smooth there). We could spend forever exploring different combinations of functions that would give Archimedes-like decompositions of a cylinder, but it’s time to turn attention to fabricating the G4G13 giveaway.

So it became time to decide on Studio Infinity’s giveaway at the 13th Gathering for Gardner (G4G13). By tradition, at least, it’s considered a plus for giveaways to connect with the number of the conference — 13 in this case. So this mathematical free association starts with the number 13.

What thoughts does 13 evoke? First and foremost, it’s a Fibonacci number, 13 = 8 + 5. So something about breaking something down by addition? Addition is so simple, what could an interesting giveaway to do with addition be like? What if it’s not just addition of numbers, but addition of something more involved? Like addition of volumes? Is there an interesting instance of volumes adding up nicely?

Well, in fact there is: the so-called Archimedes hourglass, which I had seen people make and show at previous G4Gs, such as this beautiful example by Rod Bogart:

The reason this works is that Archimedes determined that the volume of a cylinder of unit radius and two units height is equal to the sum of the volumes of a sphere of unit radius and a cone of unit base radius and two units height. Is there some variation on that which could be interesting? To try to find something like that, maybe we should look at how the proof of that fact goes.

But first, those occurrences of “two” seem a bit out of place. It seems a much more natural statement to say that the volume of a cylinder of unit radius and unit height is equal to the sum of of the volumes of a hemisphere of unit radius and a cone of unit base radius and unit height. And indeed, that statement has a simple and natural justification. Archimedes looked at it differently, but the 5th century Chinese mathematician Zu Gengzhi elucidated a principle (not rediscovered in Western mathematics until Bonaventura Cavalieri in the 17th century, and so sometimes called the “Cavalieri Principle”) that makes this straightforward, the Gengzhi Principle:

If the cross-sectional areas of two solids along every plane parallel to a fixed plane are equal, the solids have equal volume.

How does this apply to the cylinder, cone, and hemisphere? Let’s place them side-by-side this way:

The yellow plane is at height h above the x–y plane (light blue), which is the common base of all three volumes (well, actually, we’ve placed the apex of the cone at z=0). The yellow plane cuts the cylinder in a circle with radius 1, the cone in a circle with radius r, and the hemisphere in a circle with radius s. Moreover, since the width of the cone increases linearly from 0 at the blue plane to 1 at a height of 1, r = h. And since the equation of the red semicircle (which is the vertical cross-section of the hemisphere) is h² + s² = 1, we have that s = √(1-h²).

Therefore, the area of the horizontal cross section (in the yellow plane) of the cylinder is π·1² = π, while the areas of the cross sections of the other two solids are πh² for the cone and π(√(1-h²))² = π(1-h²). The latter two obviously sum to the former, since h² + (1-h²) = 1. Since this holds for any plane parallel to the light blue plane, we conclude by the Gengzhi Principle that the volume of the cylinder is the sum of the other two, just as Archimedes established. (Living six centuries earlier than and half a world away from Gengzhi, Archimedes of course had to rely on other methods, considerably more complicated, to justify this relationship.)

But now, as we will see in the next post, there is a great deal of flexibility in this proof of the Archimedes relationship — plenty to create an interesting G4G13 giveaway.

This post just takes care of some of the calculations used in planning the Golden Gate STEM Fair event. First, the plan was to make the structure shown at the right: a regular tetrahedron on top of a regular octahedron. Moreover, the resulting construction was intended to be five meters tall. Hence, the question arises: How long should each truss of the structure be to achieve the desired height?

It’s easier to work in the other direction. Assume that each edge of the structure has length s; how tall is the overall structure? Let’s start with the tetrahedron part. The apex lies directly over the center of the equilateral triangular base of the tetrahedron. We know the side length of the that triangle is s so we use the formula relating the radius r of a circle and the side of an inscribed regular polygon: s = 2rsin τ/2n, where n is the number of sides of the regular polygon, in this case three (and τ is the full circle constant, equal to 2π). So we get sin τ/6 = sin 60°, which is √3/2, from which we conclude that s = √3 r or r = s/√3.

Now we have a right triangle from the base of the tetrahedron to its apex, from which we conclude that h = √(s²-s²/3) = s√2/√3.

That tells us the height of the tetrahedron on top, but what about the octahedral base? We could go through a similar sort of calculation, albeit more involved, but we can also save ourselves a lot of trouble by noticing that if we lay the whole structure down on its side, we get a tetrahedron nestled up against the side of an octahedron, which makes it clear that the regular octahedron and regular tetrahedron are the same height. Thus, the tower overall has height s2√2/√3. Solving this for s when the height is 500 centimeters yields about 306 cm. And since we’re using boxes measured in inches and a foot is about 30cm, it’s easiest to think of that as 10 feet. So the moral of the story is that we need each truss to be ten feet long.

How many boxes will that take? Well, the boxes are 6 inches on a side, but they lie on the trusses along a face diagonal, which is 6√2 or about 8.5″. So we need the trusses to be 120/8.5 = 14.1 boxes long, or say 15 boxes long to be on the safe side. Extending a truss by one unit typically takes four boxes (one on the top and bottom and two in the middle), so that’s roughly 60 boxes per truss. With 15 trusses in the whole structure (just by counting in the diagram at the top), we can estimate about 900 boxes in the entire structure. And we know that’s an overestimate, because the trusses overlap each other where they join. So it’s a safe number to use for planning.

In the last MathStream post, we concluded that if you took spheres with holes at the points indicated by black dots in the diagram below, you could connect them with struts to form a lattice composed of alternating octahedra and tetrahedra.

But for building large-scale constructions, we’d like something comprised of components that are a little easier to make or obtain. The goal of this post is to show how that same oct-tet lattice can be constructed simply from cubical boxes.

The first step is to notice that we could shrink the struts to be as short as we like, or even do away with them altogether. If we had a whole lot of spheres just touching (or rather, glued together) at the black spots, they would form an oct-tet lattice.

But spheres are a bit hard to work with. Instead of thinking of them as points on a sphere, instead connect those same twelve points with straight lines and planar faces:

That produces a shape called a cuboctahedron, and what we have established is that cuboctahedra joined vertex-to-vertex form an oct-tet lattice.

But where do cuboctahedra come from? If we read much of the way down the Wikipedia page, we see the statement that “a cuboctahedron is a rectified cube.” Unraveling that word “rectified,” this statement just means that if you start with a cube, take the midpoint of every edge, and then connect the new points when the edges they correspond to connect, you get a cuboctahedron. Or we can “undo” the rectfication and re-create the cube from the cuboctahedron we have.

Therefore, connecting cuboctahedra vertex-to-vertex is the same as connecting cubes edge-to-edge, so we have established that cubes joined edge-to-edge form an oct-tet lattice. We’ve taken enough steps that this statement may now seem a bit mysterious, but hopefully this final image will help tie the whole thing together; notice that each strut in the oct-tet lattice passes through the midpoint of an edge of the cube.

The ease with which we could draw an equilateral triangle on the sphere naturally leads to wondering whether other constructions work out so nicely. For example, can we construct the square of points that would be needed to locate the holes for building a regular octahedron from styrofoam balls and sticks?

Since the central angle between any two adjacent vertices of the square is again τ/6 = 60°, the sides of the square are the same length as the triangle’s sides in the previous construction. So let’s just look up the geometric construction of a square and use that on the sphere. Unfortunately, you immediately hit a snag: the first step is to “extend the line segment PQ.” And I don’t know of a really practical way to extend line segments on a sphere. You could try a flexible ruler, but it’s hard to get it lined up and to draw with it in place. Or you can make three equilateral triangles that all share the same vertex; then the side of the third one extends the original one. That’s fine in theory, but in practice it’s a lot of work and it’s easy for small errors to accumulate, leaving the sum of the three angles meeting at the vertex a bit different from 180° = τ/2.

So to make things much easier, we will use a special property of this particular square. As you can see in the picture, the regular octahedron can also be viewed as a square bipyramid. That tells us that the central angle between points A and C diagonally opposite on the square we want is τ/4 = 90°. And that in turn means that the diagonal of the square on the surface of the sphere is exactly 1.5 times as long as the side of the square. Contrast this to the situation of a square in the plane, where the diagonal is the much more computationally difficult √2 times the side of the square, but beware! This relationship does not hold for all squares on the surface of a sphere. Indeed, the ratio of diagonal to side length of spherical squares ranges from √2 to 2, and in fact, you can determine the side length of a spherical square from just that ratio (and the radius of the sphere).

But for the particular square we want, we do have this lovely relationship that the diagonal is 1.5 times the length of the side, and so we can find the other two corners just as we did for the equilateral triangle, simply using a thread of length d for one arc and 1.5d for the other arc. So building a regular octahedron should also be quite feasible.

We can go a bit further than this. If you play around with constructing these squares and triangles, you will find that alternating them completely covers the sphere.

In other words, there are just twelve points on the surface of the sphere so that each one is a vertex of two equilateral triangles and two squares, alternating. That’s a pretty special structure, and it allows the existence of a very strong framework filling space called the oct-tet lattice consisting of alternating octahedra and tetrahedra.

For this project, I needed to figure out (a) where should the holes be in spheres to connect them by straight lines to form a regular tetrahedron, and (b) how to locate those points on a physical sphere. The diagram makes part (a) fairly straightforward. We can see that the angle between any two holes (as viewed from the center of one of the spheres) should be 60°, the angle at each vertex of an equilateral triangle. And since 60° is one-sixth of a full circle (or an angle of τ/6, as the tauists point out), we can find the required distance d between any two holes on the surface of a sphere of radius r to be d = rτ/6, or approximately d ≈ 1.047r.

Now, how should we actually locate the points? As you can see from the diagram, they form a sort of “spherical equilateral triangle,” each point the same distance from each of the other two. And as you may recall, it’s pretty easy to construct an equilateral triangle in the plane. Fortunately, exactly the same procedure works on a sphere: First, select any two points a distance d apart on the sphere. Then using each of the two points as center, trace out a spherical circle with radius d on the surface of the sphere. (In practice, you only need a small section of each circle in the vicinity of where they’re going to intersect.) Each of the two points at which those two circles intersect represents one of the two possible locations for the third vertex of the desired triangle.
Moreover, it’s easy to perform this construction in practice. And mathematical curiosity makes us wonder: if it works for triangles, will it work for other constructions as well?

When I set the goal of creating a new tensegrity structure for the Storm King Art Center workshop, I decided that I wanted to create a highly regular, symmetric structure, to contrast with the more free-form, organic style of Snelson’s Free Ride Home at the Center. (Snelson also created many very regular tensegrity sculptures himself.)

As is often the case, the creative process began with an existing structure that shared many of the characteristics I wanted, namely the classic six-strut tensegrity structure. In a previous article, we saw how that structure is based on a regular icosahedron; in fact, the compression members are just six diagonals of an icosahedron that are equivalent under the symmetries of the icosahedron. Moreover, the end points of the diagonals exhaust the twelve vertices of the icosahedron.

Thus, it seemed that I might obtain a different interesting structure if I started with a different polyhedron that shared some key properties with the icosahedron, and then looked for an appropriate set of diagonals. So the first question was, what are those key properties of the icosahedron? Well, we want all of the edges of the shape to be the “same” in some sense so that when we fabricate them from tension members with the same tension, they will balance each other uniformly. And what does “same” mean in this context? Well, a natural answer is that there should be some symmetry of the shape — in other words, some shape-preserving transformation, which in this case will be a rotation — which will match any edge up with any other edge. When the symmetries of an object completely scramble some component of that object, mathematicians call the object “transitive” for that component. So I could find a list of candidate shapes by an internet search for “edge-transitive polyhedron.” That search turned up a fancy synonym for edge-transitive, namely “isotoxal,” and it also turned up this Wikipedia page, of which the first table is of particular interest.

The rhombic triacontahedron immediately caught my eye. However, it includes two different categories of vertices, the ones where five edges meet and the ones where three edges meet; and what’s worse, there are twelve of the former but twenty of the latter. That means there will be no way to make sixteen equivalent diagonals, since some of them will not terminate at a five-edge vertex at either end, while others will.

So instead, I set my sights on it’s left-hand neighbor in that table, the icosidodecahedron. It has 30 identical vertices (consistent with our theme, the symmetries of the icosidodecahedron act transitively on the vertices). So then I searched for a set of fifteen diagonals of the icosidodecahedron, all the same length, to be the struts of the new tensegrity.

I began with what seemed like the simplest sort of diagonal, namely a diagonal between two vertices that are only “two edges apart,” i.e., both adjacent to the same vertex. I was able to draw a set of diagonals that exhausted the vertices, like this:

However, there’s a problem: you can see on the right-hand side of the image that two of the diagonals selected intersect. While that’s OK mathematically, there’s no way to direct struts to pass through each other. And try as I might, I couldn’t avoid two of these diagonals intersecting. From a math point of view, the natural response was to try to prove that any such set of diagonals contains (at least) two that intersect each other. That turns out to be the case, and here’s how the proof goes: Any diagonal that connects two vertices that are adjacent to the same vertex is actually a diagonal of one of the pentagonal faces of the icosidodecahedron. There are only twelve such faces, but we need 15 diagonals. Hence, at least one of the faces will end up with two diagonals. And (as you can demonstrate for yourself by drawing some diagrams) any two diagonals of a regular pentagon intersect.

So these “simplest” diagonals were out. And it also occurred to me that for building a tensegrity structure, we likely want long diagonals that go through the body of the polyhedron, to make it easier for the tension of the rubber bands along the edges of the polyhedron to pull against them. The very longest diagonals of the icosidodecahedron all intersect each other at the center of the polyhedron, so they’re out. Therefore, my next attempt was to use the second-longest diagonals, like this partial effort:

You can see that this selection of diagonals already has intersecting pairs. In fact, I was not able to find any selection of diagonals that fit the requirements above, leading to this Math StackExchange question. So what did I do? I settled for the “near miss” below. The ten diagonals that connect a “polar” vertex to an “equatorial” vertex are transitive, but there’s no way to transform them to one of the other five diagonals (that connect two “mid-latitude” vertices, one above the equator and one below.) In fact, the two types of diagonals are different lengths, with the first type about 20% longer than the second. So I couldn’t expect the resulting tensegrity structure to come out regular, using struts the same length for diagonals that should theoretically be the same length. But I hoped it would at least be close.

So this image became the template for building the new tensegrity structure, which you can read about in the next MakeStream post.

As mentioned at the end of the last MathStream post, the actual shape that the six-strut tensegrity structure takes on is close to, but not quite precisely, a regular icosahedron. And that fact immediately makes you want to build a tensegrity structure that will under ideal circumstances assume the shape of a truly regular icosahedron. What would that entail? Why is the classic tensegrity not a regular icosahedron? One possibility that immediately comes to mind is that not all of the edges of the icosahedron are represented in the same physical way in the model. Namely, as you may recall from this diagram,

there is no rubber band lying along some of the edges of the icosahedron. To test if that’s the reason that the six-strut is not regular, we’d want a different arrangement of elements so that there will be exactly the same tension on a rubber-band connection between every closest pair of endpoints of the six struts, as shown in this diagram.

So, you might immediately start tying to build such a configuration. But that will bounce you right back to a mathematical question: how can we route the rubber bands so that there is exactly the same amount of rubber band between every closest-neighbor pair of endpoints? In fact, one of the wonderful things that happens at Studio Infinity is that many of the things we try to build lead to new mathematical questions, and many of the mathematical discoveries we encounter suggest new things to build. So the MakeStream and MathStream really build on each other.

Getting back to the question at hand, it amounts to finding a route for each of several rubber bands so that the route of each band traverses the same number of edges of the icosahedron we are trying to achieve, and so that every edge is covered. Since the icosahedron has thirty edges, that immediately narrows down the possible number of rubber bands we might use. For example, there might be ten rubber bands each covering three edges, or six rubber bands each covering five edges, or five rubber bands each covering six edges, and so on. And since each rubber band is a loop, the edges covered will have to form a loop on the surface of the icosahedron. For example, there are fairly obvious loops one could make with three edges (in blue) or with five edges (in red) in this diagram.

So you can just start trying to cover the edges with loops of the same length. In fact, I recommend that you give it a try right now, maybe using the six-strut model you’ve made, or the diagram above. See if you can find a way to route rubber bands to cover every edge, so that every loop is the same length, before you read further.

After a while of trying, you may start to get discouraged. You might start to feel that this is an impossible task. And that would not be too surprising, because it is impossible. But when we make a statement like that in mathematics, we have to back it up. If you want to claim that something is impossible, you can’t just list the seventeen things you tried that didn’t work. You’ve got to find reasons why no possible routing of rubber bands, no matter how clever, will cover all of the edges and have every loop be the same length.

And the key to that in this case is even and odd. First count how many edges meet at every vertex: five, an odd number. Next count how many of those edges rubber bands can cover. Well, every rubber band is a loop, so it must arrive at the vertex by some edge, and then leave that vertex by a different edge. It might later come back to that vertex by yet another edge, but if so, it has to leave again by a still further edge, in order to form a single closed loop overall. So in short any one rubber band covers an even number of edges touching the vertex. And since the sum of a collection of even numbers is even, we can be sure that the rubber bands are covering an even number of edges touching the vertex.

And so now we run into the problem. An even number cannot be equal to an odd number, so we can’t possibly have covered all of the edges. So there in fact is no way to route the rubber bands as desired.

So we regretfully conclude that there is no way to construct the regular icosahedron we are looking for, right? Not quite. Another habit in mathematics is, when something doesn’t work out the way you thought it might, to see if there’s a way to change the assumptions so that it does work out. And in this case, there is something we can change: rather than providing the tension between two neighboring endpoints by virtue of a single segment of rubber band, why couldn’t there be exactly two different (but identical in length) segments of rubber band covering each edge? Then there would be ten different segments of rubber band reaching each vertex, which is an even number, and the contradiction underlying our impossibility argument would melt away in a puff of logic.

And now we can very pleasantly notice that there are exactly twelve of the five-edge pentagonal paths similar to the one highlighted in red in the last diagram above, and that each edge lies on exactly two of these pentagonal paths. So now the plan is simple: connect one rubber band along each pentagon, and the icosahedron should simply materialize from the balance of forces. Let’s return to the MakeStream to see if it works…

So maybe you’ve made the classic six-strut tensegrity (or perhaps you’ve just looked at the pictures) and you’re wondering what shape that is, exactly. Naturally, since mathematics is among other things the science of shape and structure, understanding that is going to involve a little math. And in mathematics, sometimes it’s easiest to understand something by breaking it down into the pieces that are there, but other times it’s actually easier to understand something by adding (or pretending to add) pieces that aren’t there.

So take a look again at the structure you built.

Notice that in many cases, two neighboring endpoints of struts are connected by segments of rubber band. However, in other cases, there are endpoints of two struts that are close together but are not connected at all. So let’s imagine what the structure would look like if all of the nearest neighbors were connected.

Now, if we look at the outer surface of the structure, we see a much simpler pattern. Every outermost flat shape of the structure — which we call a face of the structure — is a triangle. There are 20 of these triangles, and at each endpoint of a strut, which makes a sort of “corner” of the structure called a vertex, exactly five of the triangles meet. There’s just one geometric shape fitting those facts, called the icosahedron.

This shape is simple and natural enough to have been discovered at least two and a half millennia ago, as one of the five Platonic Solids of the ancient Greeks. These are the shapes with identical regular polygonal faces and identical vertices, and it is a remarkable and beautiful fact that only five different Platonic solids exist. Yet the icosahedron is also intriguing and pleasing enough that it’s still of interest today.

One of the most important characteristics of the icosahedron is that it has very many symmetries. A symmetry of an object is some transformation, like a rotation or a translation or some combination thereof, which brings the shape back to coincide with itself. For example, if we were to rotate the icosahedron one-fifth of a turn around the axis indicated by the dotted blue line in this diagram, it would again look exactly the same, even though we moved it.

In the idealized world of this Platonic icosahedron, how can we represent the struts? Each one of them connects two of the vertices of the icosahedron, so it is what you might call a diagonal of the icosahedron. Let’s draw all six of them on our diagram.

Note how these diagonals are arranged; each one connects a vertex to another vertex that’s not adjacent, but lies “two edges away” from the starting vertex. They come in three pairs of parallel diagonals. Moreover, all six diagonals are “equivalent” under the symmetries of the icosahedron, which means to say that there’s a symmetry of the icosahedron which moves any of the struts to the position of any one of the other struts. That corresponds with the fact that all of our struts are the same length: in order to get an icosahedron, it’s necessary to use struts that are the same length, because there’s no way to tell one strut from another. They’re all interchangeable by the symmetries of the icosahedron.

We’ll close this tour of the math underlying the six-strut tensegrity with a perhaps inconvenient-seeming observation. If you look closely at the structure you’ve built (or at the pictures above), you will see that the icosahedron is not actually perfectly regular. In the diagram, the dotted sides of the triangles are not the same lengths as the rubber-band sides of the triangles. So our tensegrity structure is not actually a Platonic regular icosahedron; you might instead dub it an “icloseahedron” since it’s really close to, but not quite identical to, the idealized structure we envisioned.

But why isn’t this structure regular? After all, we built it in a very symmetric way. Well, there’s one way in which the physical structure of the six-strut tensegrity differs markedly from a true icosahedron — and we’ll take that up in the next MathStream article.

HERE!! go on about symmetries, show how the struts relate, point out that the tensegrity isn’t precisely a regular icosahedron, end with wondering why, saying that will have to wait for another post.