For me, geometry for breakfast is not unusual. This morning, though, I’m sprinkling calculus on top before eating it.

It’s important to explain, right up front, that Ronald Reagan was president when I last took calculus. However, I have a new determination to learn the subject. I have a hunch this may go better without the “help” of actually being enrolled in a calculus class, since the way I learn things, and the way most people learn things, aren’t much alike.

My current calculus puzzle started when I noticed that taking the derivative of the volume of a sphere, in terms of the radius, (4/3)πr³, yields the formula for the surface area of a sphere, 4πr². That was both unexpected and exciting, so I tried applying the same idea to another solid: the cube. With edge length e, the volume of a cube is e³, and the derivative of that is 3e² . . . but that’s only half of the surface area of a cube, which is 6e².

Half? What’s going on here? I mentioned this puzzle on Facebook, where I have many on my friends’-list whose mathematical knowledge exceeds my own. It was pointed out to me that I’d made an important and unhelpful change by going from using the radius, for the sphere, to the edge length, for the cube.

So I’ll try this again, but do it in terms of the radius of the cube, rather than the edge length. For a cube, the radius extends from the center to any of the cube’s eight vertices. Both the light and dark blue segments in the diagram below are cube radii.

This radius is sqrt(3)/2 times the cube’s edge length, as can be verified by applying the Pythagorean Theorem twice, first to triangle ABC (which shows that the green face-diagonal is sqrt(2) times the edge length), and then to triangle BCD (which yields sqrt(3) times the edge length for the interior diagonal DC, half of which is the radius).

It then follows that, if r = [sqrt(3)/2]e, that e = [2/sqrt(3)]r, which “cleans up” to e = (2/3)sqrt(3)r, when the denominator is rationalized.

If a cube’s volume is e³, and e = (2/3)sqrt(3)r, it then follows that V = [(2/3)sqrt(3)r]³ = (8/27)(3)sqrt(3)r³ = (24/27)sqrt(3)r³ = [8sqrt(3)/9]r³. If I take the derivative of the last expression, I get [8sqrt(3)/3]r² for the derivative of the volume, which I now need to compare to the surface area of a cube, in terms of its radius, rather than edge length.

So here goes . . . SA = 6e² = 6[(2/3)sqrt(3)r]² = [48(3)/9]r² = 16r², which isn’t what I got for the derivative of the volume, above.

Well, I was using, as the radius, the radius of the cube’s circumscribed sphere. Perhaps I should have used the inscribed sphere, instead? The radius of the cube’s inscribed sphere is the “invisible” segment FM in the diagram above, which I’m going to call “a” (for “apothem,” because this looks like the 3-d version of the apothem of a regular polygon). The length of a is exactly one-half that of e, the cube’s edge length, which means that e = 2a. Therefore, V = e³ = (2a)³ = 8a³, the derivative of which is 24a².

Now to check the surface area, in terms of a: SA = 6e² = 6(2a)² = 24a², and that’s what I got when I took the derivative of the volume, in terms of a.

So this trick works for the cube if you use the radius of the inscribed sphere, but not the circumscribed sphere. This leaves me with three questions to address later:

1. Will this also work for other polyhedra? This is something I intend to explore in future blog-posts, starting with the tetrahedron and the octahedron.
2. Why did this work at all?
3. Why was it necessary to use the radius of the cube’s inscribed sphere, rather than its circumscribed sphere?

If any reader of this post knows the answer(s) to #2 and/or #3, sharing your knowledge in a comment would be very much appreciated.

Working Towards a Solution of the Hemisphere Problem

The hemisphere problem referred to here was described in the previous post. To reword it somewhat, consider this hemisphere, half of a sphere of radius r. The orange cross-section is a circle parallel to the hemisphere’s yellow, circular base.

We are to find the height of the yellow section with the orange circular top (which I shall call x), as a fraction of r, such that the yellow and red sections above have equal volumes.

Since the volume of a hemisphere is (2/3)πr³ (that’s half a sphere’s volume), each of these two sections must have half the hemisphere’s volume, or (1/3)πr³.

Moreover, the top (red) portion is a “spherical cap,” described here on Wikipedia, as was pointed out to me, by a friend, on Facebook. On that Wikipedia page, you can find this diagram, as well as the formula shown below for the volume of the purple spherical cap in the diagram.

Now, as our goal is to find x, as described at the top of this post, it important to remember that r = x + h, where r is the radius of the original sphere (and height of the hemisphere), h is the height of the spherical cap, and x is the height of the hemisphere, after the spherical cap is removed. We now have two expressions for the volume of the spherical cap: (1/3)πr³ (because it is a fourth of the volume of the original sphere), and (1/6)πh(3a² + h²) from the Wikipedia article on the spherical cap (so all of this assumes, then, accuracy in that Wikipedia article). Setting them equal to each other,

(1/3)πr³ = (1/6)πh(3a² + h²)

Next, I’ll clean this up by multiplying left and right by 6/π, to cancel fractions and π from both sides.

2r³ = h(3a² + h²)

A right triangle exists in the blue-and-purple figure above, and the unlabeled leg is x, the problem’s original goal. I’ll add an “x” to this diagram.

Using this right triangle, and the Pythagorean Theorem, it can be seen that a² = r² – x². Also, since r = x + h, it follows that h = r – x. By substitution for a and h, then,

2r³ = h(3a² + h²)

becomes

2r³ = (r – x)[3(r² – x²) + (r – x)²], which then distributes to

2r³ = (r – x)(3r² – 3x² + r² – 2rx + x²), which expands as

2r³ = 3r³ – 3rx² + r³ – 2r²x + rx² -3r²x  + 3x³ – r²x + 2rx² – x³, which simplifies to

0 = 2r³ – 6r²x + 2x³, which becomes, by division:

0 = r³ – 3r²x + x³.

I’m trying to find x, as a fraction of r, meaning that x = kr, and I want k. On that basis, I’ll now substitute kr for each x in the last equation above.

0 = r³ – 3r²(kr) + (kr)³, which then becomes

0 = r³ – 3kr³ + k³r³, or

3kr³ = r³ + k³r³, and then dividing by r³ yields

3k = 1 + k³, which can be rearranged to

3k – k³ = 1, which factors,on the left side, to yield

(k)(3 – k²) = 1.

For k and (3 – k²) to have a product of one, they must be reciprocals. Therefore, 1/k = 3 – k². I can then graph y = 1/k, as well as y = 3 – k², and find the solution by seeing where the graphed functions intersect above the k-axis, with a “k” value between zero and one, since no value of k less than zero or greater than one would make sense, as a solution to the original problem. (I’ll be using k coordinates, and a k-axis, in place of the usual x coordinates and x-axis.) Here’s the initial graph:

The only intersection in the specified range of zero to one is between point A and point B, so I brought them together, as closely as I could get Geometer’s Sketchpad to let me.

With points A and B almost on top of each other, k = 0.34958 by one equation, and k = 0.34820 by the other. To two significant figures, then, I can conclude that the horizontal cut in the original problem should be made 35% of the way from the base of the hemisphere to the hemisphere’s top.

Using a graphing calculator, a more precise answer of 0.34729636 was obtained. I’d still like to have an exact answer, but this will do for now.

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Later addition: a helpful reader led me to a Wolfram Alpha site where I could get an exact answer, as well as a decimal approximation with a greater degree of precision. In the pic below, I have omitted the two solutions of the third-order polynomial which are not the one solution of interest. Here’s the one which is:

Now, however, I have another mystery: how can an exact answer, with all those imaginary units in it, have a real-only approximation? To this question, at least for now, I don’t even have the beginnings of an answer.

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Even later post-script: I have been assured by friends on Facebook that the imaginary units in the above exact solution somehow cancel, although I must concede that I still do not see how, myself. I’ve also been shown another way to express the solution, for 2sin(π/18) is also ≈ 0.3472963553338606977034333. This surprised me, due the the lack of any explicit appearance by a π/18 (= 10°) angle in the original problem, and the fact that no trigonometric functions were used to solve it.

The Hemisphere Problem (See Next Post for the Solution)

A hemisphere rests with its circular base on a horizontal, level surface, and is to be cut into two pieces of equal volume. If the hemisphere’s radius is r, at what fraction of r above the floor should the horizontal cut be made?

An Unsolved Problem Involving the Icosahedron and the Dodecahedron, and Their Circumscribed Spheres

This is apparently a problem, posed by Gregory Galperin, which went unsolved at the Bay Area Math Olympiad in 2005. I haven’t solved it yet, but I’m going to try, as I work on this blog-post. My 2010 source is a paper about Zome which may be read, as a .pdf, at bact.mathcircles.org/files/Summer2010/zomes-6-2010.pdf. The problem involves a dodecahedron and an icosahedron, each inscribed inside the same sphere of radius r, and asks which has the greater volume. At the time the authors wrote this paper, they knew of no solution, and I know of none now, but I do like a challenge.

My idea for solving this begins with Zome (info on Zome:  see http://www.zometool.com, as well as other sites you can find by googling “Zome”). In the Zome geometry system, using B1 struts for the edges of both a dodecahedron and an icosahedron, R1 struts are the radii of the circumscribed sphere for the icosahedron,  and Y2 struts are the radii for the circumscribed sphere of the dodecahedron. Since volume formula for polyhedra are generally given in term of edge-length, I need to find B1 in terms of R1 for the icosahedron, and find B1 in terms of Y2 for the dodecahedron.

Icosahedron:  find B1, in terms of R1.

There exists a right triangle which can be built in Zome which has a hypotenuse equal to 2R1, and legs epqual to B1 and B2. B2 = φB1, so, by the Pythagorean Theorem, (2R1)^2 = (B1)^ + φ²(B1)², which simplifies to 4(R1)^2 = (1 + φ²)(B1)^2, which can then be solved for B1 as B1 = sqrt[4(R1)^2/(1 + φ²)]. B1 here is the icosahedron’s edge-length, while R1 is the radius of its circumscribed sphere.

Dodecahedron:  find B1, in terms of Y2.

In the Zome system, Y2 = φY1, and Y1 = [sqrt(3)/2]B1. Rearrangement of the first of these equations yields Y1 = Y2/φ, and substitution then yields [sqrt(3)/2]B1 = Y2/φ, which then can be rearranged to yield B1 = 2Y2/[φsqrt(3)]. B1 here is the dodecahedron’s edge-length, while Y2 is the radius of its circumscribed sphere.

Next, find the volume of the icosahedron inscribed inside a sphere, in terms of that sphere’s radius.

According to http://mathworld.wolfram.com/Icosahedron.html, the volume of an icosahedron is given by V = (5/12)[3 + sqrt(5)]a³, where a is the edge length, or B1 in the first indented section, between the two images, above.  Then, by substitution, V = (5/12)[3 + sqrt(5)]{sqrt[4(R1)^2/(1 + φ²)]}³, which then becomes (with “r” being the radius of the circumscribed sphere) V = (5/12)[3 + sqrt(5)][2r/sqrt(1 + φ²)]³ = (40/12)[3 + sqrt(5)][1/sqrt(1 + φ²)]³r³ = (10/3)[3 + sqrt(5)][1/sqrt(1 + φ²)]³r³. Then, using the identity φ² = φ + 1, this can be further simplified to V = (10/3)[3 + sqrt(5)][1/sqrt(2 + φ)]³r³.

Next, find the volume of the dodecahedron inscribed inside the same sphere, in terms of that sphere’s radius, r.

According to https://en.wikipedia.org/wiki/Dodecahedron, the volume of an icosahedron is given by V = (1/4)[15 + 7sqrt(5)]a³, where a is the edge length, or B1 in the second indented section, below the second image, above.  Then, by substitution, V = (1/4)[15 + 7sqrt(5)]{2Y2/[φsqrt(3)]}³, which then becomes (with “r” being the radius of the circumscribed sphere) V = (8/4)[15 + 7sqrt(5)]{1/[φsqrt(3)]}³r³ = 2[15 + 7sqrt(5)]{1/[3sqrt(3)]}(1/φ³)r³ = (2/3)[15 + 7sqrt(5)][sqrt(3)/3](1/φ³)r³  = [2sqrt(3)/9][15 + 7sqrt(5)](1/φ³)r³.

So, with the “r” in each case being the same, the icosahedron is larger than the dodecahedron iff (10/3)[3 + sqrt(5)][1/sqrt(2 + φ)]³ > [2sqrt(3)/9][15 + 7sqrt(5)](1/φ³), which simplifies to (5)[3 + sqrt(5)][1/sqrt(2 + φ)]³ > [2sqrt(3)/3][15 + 7sqrt(5)](1/φ³), which simplifies further to {5/[sqrt(2 + φ)]³}[3 + sqrt(5)] > [2sqrt(3)/3φ³][15 + 7sqrt(5)], which is, as a decimal approximation, is (0.726542528)(5.2360679774998) > (3.464101615/12.708203932)(30.6524758), or 3.804226 > 8.355492, which is false, meaning that the dodecahedron is larger, not the icosahedron.

Now for the bad part:  I think I’m wrong, but I don’t know where the error lies. I’m also tired. If any of you see the mistake, please point it out in a comment, and I’ll try to fix this after I’ve rested.

Update:  if the websites http://rechneronline.de/pi/icosahedron.php and http://rechneronline.de/pi/dodecahedron.php work correctly, then the dodecahedron is larger. Evidence:

This does not, however, mean that I did the problem correctly. I merely stumbled upon the correct answer. How do I know this? Simple:  the ratio I obtained was too far off. Therefore, I would still welcome help clearing up the mystery of where my error(s) is/are, in the calculations shown above.

A 240-Atom Fullerene, and Related Polyhedra

The most well-known fullerene has the shape of a truncated icosahedron, best-known outside the world of geometry as the “futbol” / “football” / “soccer ball” shape — twenty hexagons and twelve pentagons, all regular. The formula for this molecule is C60. However, there are also many other fullerenes, both larger and smaller. One of my favorites is C240, simply because I sometimes make class projects out of building fullerene models with Zome (available at www.zometool.com), and the 240-atom fullerene is the largest one which can be built using Zome. Here’s what it looks like, as molecular models are traditionally colored.

This polyhedron still has twelve pentagons, like its smaller “cousin,” the truncated icosahedron, but far more hexagons. What’s more, these hexagons do not have exactly the same shape. If this is re-colored in the traditional style of a polyhedron, rather than a molecule, it looks like this. In this image, also, the different shapes of hexagons each have their own color.

Like other polyhedra, a compound can be made from this polyhedron and its dual. In this case, the dual’s faces are shown, below, as red triangles. The original fullerene-shape is in purple for the pentagonal faces, and orange for the hexagons.

In the base/dual compound above, it can be difficult to tell exactly what this dual is, but that can be clarified by removing the original fullerene. What’s left is called a geodesic sphere — or, quite informally, a ball made of many triangles. The larger a fullerene is, the more hexagonal rings/faces it will have, and the more triangles will be found on the geodesic sphere which is its dual. For the 240-atom fullerene shown repeatedly, above, here is the dual, by itself, with different colors indicating slightly different triangle-shapes. (An exception is the yellow and green triangles, which are congruent, but have different colors for aesthetic reasons.)

I made these four rotating images using Stella 4d:  Polyhedron Navigator. To try this program for yourself, simply visit www.software3d.com/Stella.php. At that site, there is a free trial download available.

On Sportsball, As Viewed By One Aspie

Image

Since I live in the American South, I hear a lot of talk about about sportsball. I have a hard time, though, telling exactly which variety of sportsball is being discussed. I don’t find sportsball interesting, and so I’m not fluent in any variant of sportsball jargon. For that reason, it can be difficult for me to tell which sportsballspeak-dialect is being used.

So, sometimes, just to try to make friendly conversation (while still being myself), I ask sportsball-fans questions, in order to find out which version they’re so intently discussing. (Figuring out why people obsessively talk about sportsball so much, I think, is a mystery I’ll never solve. Understanding the strange behavior of non-Aspies is much more difficult than the types of problems I enjoy trying to solve. As Albert Einstein said, when declining the presidency of Israel, “I have no head for human problems.”)

Here’s an example of one such question: “Are you talking about the type of sportsball often played inside, with a bunch of people chasing an orange sphere around on a wooden rectangle, and trying to get the sphere to pass through a metallic, elevated circle of slightly larger diameter than the sphere itself?”

Now, if someone mentions the sportsball game most people call “football,” there’s an obvious follow-up question that needs to be asked . . . so, of course, I ask it:  “Which one?”

Replies to that question usually go something like this: “Whaddya mean, which one? Football! We’re talking about football, ya nerd!”

“But there are at least two games called by that name, which confuses me. Do you mean the sportsball-version where the players chase a prolate spheroid, or a rounded version of a truncated icosahedron?”

If they don’t understand that question, I attempt clarification: “You know, both those versions of sportsball are played on rectangles covered with grass . . . but the one with the prolate spheroid has two giant tuning forks at opposite sides end of the grassy rectangle, is usually played in the USA, and has a far higher rate of injuries, even fatal ones. The one that uses a truncated icosahedron doesn’t have tuning forks, is called ‘football’ by far more people than that American game, and isn’t nearly as dangerous. I think it’s at least a little more interesting than that other game people call ‘football,’ because of the Archimedean solid they chase around, since I like polyhedra. Which one are you discussing?”

If they tell me they’re talking about American football, I usually follow-up with a brief rant, for that sportsball-variant’s name confuses me. “Why do people call it that, anyway? I’ve seen it being play a few times — not for a full game, of course, but I can stand to watch it for a few minutes. That’s long enough to tell that the players only rarely use their feet to kick the prolate spheroid, and usually carry or throw it instead, using, of course, their hands. They usually use their feet just to run around chasing each other. Calling that version of sportsball by the term ‘football’ doesn’t make sense at all. In the game the rest of the world calls ‘football,’ the players kick the ball all the time, so I can understand why it has that name, but that prolate-spheroid version really should be called something else! Also, why are the games sometimes called ‘bowl games?’ They still play on a rectangle, and chase a prolate spheroid — there’s no actual bowl involved, is there?”

On occasion, they aren’t talking about any of these three varieties, though, but yet another form of sportsball. (Why are there so many?)

“Oh! You must mean the one played on a ninety-degree sector of a circle, with a square (confusingly called a ‘diamond,’ for some reason) in its interior, positioned such that one of its vertices is at the circle’s center. At that vertex, there’s a convex-but-still-irregular pentagon on the ground, while the other three vertices of the large, grass-covered square have much smaller squares on the ground, instead of a pentagon. The guy standing at the pentagon is always trying to hit a red-and-white sphere with a wooden or aluminum stick, but he usually misses. The guy who throws the sphere toward the region above that pentagon usually scratches himself, and spits — a lot. He must be important in some way, for he’s provided with a small hill to stand on, literally placing him above the rest of the players. Have I got it now?”

Sometimes, people try to get me to stop calling these strange activities “sportsball,” by bringing up hockey as an objection. “You can’t call all sports ‘sportsball!’ What about hockey? It doesn’t even have ball! It’s got a puck!”

I’m always ready for this objection, though. “You mean the one with the short black cylinder that slides across ice? That’s a sport? I thought it was just an excuse to have fights!”