30 Jul

How does the value of the following improper integral compare to 1? I.e., is it smaller, larger, or exactly equal to 1?

(This problem was proposed to Math Horizons Playground by Mehtaab Sawney of Commack High School. And for all of you $\pi$-ists out there, $\tau$ is of course just the radian measure of a full circle, i.e., $\tau=2\pi$.)

Stumped? You can peek at the solution using the password “strapyb”.

30 Jul

30 Jul

For definiteness, the unlabeled points do in fact trisect each of the sides of the triangle.

30 Jul

30 Jul

What are the next few terms in this sequence of smallish numbers?

# 1 1 13131 131 131 131 1 1321 132131 131 1 1321 1321 13 ? ? ?

(And, of course, what’s the rule generating the sequence?)

Stumped? You can peek at the solution using the password “1p1a2s”.

30 Jul

11 Apr

This post contains the details of the claim made in “More Spherical Construction” that you can determine the side length of a spherical square from the ratio between the lengths of its diagonals. We’ll do this on a sphere of radius one; everything scales by a factor of the radius for a general sphere.

The advantage of working on a sphere of radius one is that then the side length is nothing other than the measure of the central angle (in radians) between two adjacent vertices of the square. Similarly, the diagonal is the central angle between opposite vertices of the square.

The easiest way to construct a square on the unit sphere is to place its center at the intersection of the sphere and the x-axis, and then choose a (positive) angle $\theta \leq \tau/4$ (where $\tau = 2\pi$ is the radian measure of a full circle) and place two vertices at plus or minus $\theta$ (from the square’s center) in the longitudinal direction and two vertices at plus or minus $\theta$ in the latitudinal direction. In spherical coordinates, the center is at $(1,0,\tau/4)$ and the four vertices are at $(1,-\theta,\tau/4), (1,\theta,\tau/4), (1,0,\tau/4-\theta), (1,0,\tau/4+\theta)$. For this square, the diagonal is obviously $2\theta$; it remains only to compute the side length of the square.

Using the great-circle distance formula on the two points $(1,\theta,\tau/4)$ and $(1,0,\tau/4-\theta)$ yields a side length of $\arccos(\cos(\tau/4)\cos(\tau/4-\theta) + \sin(\tau/4)\sin(\tau/4-\theta)\cos(\theta)\;).$

Since $\cos(\tau/4)$ is 0 and $\sin(\tau/4)$ is 1, this expression equals $\arccos(\sin(\tau/4-\theta)\cos(\theta)).$

And further, $\sin(\tau/4-\theta) = \cos(\theta)$, so the side length is just $\arccos(\cos^2 \theta).$

Therefore, the ratio of the diagonal to the side of the square is $2\theta/\arccos(\cos^2 \theta)$. Graphing this on the allowed interval $(0,\tau/4]$ for $\theta$ immediately demonstrates the claims made in the referring post: no value of the ratio is repeated for different values of $\theta$, so the ratio determines the diagonal (and side length) of the square; the maximum value is 2; and any value greater than $\sqrt2$ is achievable.

If you’d rather not rely on graphing software to extract these facts, but rather demonstrate them just from the formulas, here’s how you can proceed. The value 2 at $\theta=\tau/4$ comes from straight substitution. The value at 0 by substitution is the indeterminate 0/0, so we obtain the value by L’Hôpital’s rule: the derivative of the numerator is 2, but the derivative of the denominator is $\frac{2\cos \theta}{\sqrt{1+\cos^2\theta}},$ so the diagonal-side ratio has the limiting value $\sqrt2$ at $\theta =0$. Finally, we need to show that the ratio is monotone increasing on the interval of interest. By the quotient rule, this is equivalent to showing that on this interval $\arccos(\cos^2\theta) > \frac{2\theta\cos\theta}{\sqrt{1+\cos^2\theta}}.$ Both sides are 0 at $\theta=0$, so we can take derivatives one more time; the derivative of the left-hand side is of course $2\cos\theta(1+\cos^2\theta)^{-1/2}$ again, which on this interval is always larger than the derivative of the right-hand side, namely $2\cos\theta(1+\cos^2\theta)^{-1/2} – 2\theta\sin\theta(1+\cos^2\theta)^{-3/2}$. So the left-hand side is also always larger than the right hand side, i.e., the derivative of the diagonal-side ratio is positive on this interval.

10 Apr

To compute the height $H$ of the Gengzhi Goblets so that the 13-gon prism has volume 1 cup ≈ 236 cubic centimeters, and the radius of the pentagon goblet so that its maximum cross section is the same as the 13-gon (and hence its volume is exactly one quarter as large, or 59 cc):

Note that we assume that the height of the 13-gon prism will equal the distance from one vertex to the midpoint of the opposite side. Referring to the Wikipedia page on regular polygons for formulas, that is to say that the height will equal the circumradius plus the apothem. Now the volume $V$ of the 13-gon prism is the height times the area of the 13-gon, which is $\frac{13}2r^2\sin(\tau/13)$. (As usual we use $\tau$ for the radian measure of a full circle, namely $2\pi$, as it makes most formulas clearer in terms of fractions of circles.) Combining these two observations, we get that
$V = \frac{13}2r^2\sin(\tau/13)(r + r\cos(\tau/26)) = r^3(\frac{13}2\sin(\tau/13)(1+\cos(\tau/26))).$
Setting $V$ equal to 236 and solving for $r_{13} = r$, we get $r_{13} = \sqrt[3]{472/(13\sin(\tau/13)(1+\cos(\tau/26))} \approx 3.41$cm. The height is then $H = r_{13}(1+\cos(\tau/26)) \approx 6.72$cm.
Finally, to find the radius $r_5$ such that pentagon with circumradius $r_5$ has the same area as the 13-gon with circumradius $r_{13}$, we set the respective area formulas equal and solve: $r_5 = r_{13}\sqrt{13\sin(\tau/13)/5\sin(\tau/5)} \approx 3.84$cm.

10 Apr

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.

09 Apr

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 xy 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.

05 Apr

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.

05 Apr

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.