This is a slight thing that crossed my reading yesterday. You might enjoy. The question is a silly one: what’s the “optimal” way to slice banana onto a peanut-butter-and-banana sandwich?
Here’s Ethan Rosenthal’s answer. The specific problem this is put to is silly. The optimal peanut butter and banana sandwich is the one that satisfies your desire for a peanut butter and banana sandwich. However, the approach to the problem demonstrates good mathematics, and numerical mathematics, practices. Particularly it demonstrates defining just what your problem is, and what you mean by “optimal”, and how you can test that. And then developing a numerical model which can optimize it.
And the specific question, how much of the sandwich can you cover with banana slices, one of actual interest. A good number of ideas in analysis involve thinking of cover sets: what is the smallest collection of these things which will completely cover this other thing? Concepts like this give us an idea of how to define area, also, as the smallest number of standard reference shapes which will cover the thing we’re interested in. The basic problem is practical too: if we wish to provide something, and have units like this which can cover some area, how can we arrange them so as to miss as little as possible? Or use as few of the units as possible?
Robert Austin, of the RobertLovesPi blog, got to thinking about one of those interesting mathematics problems. It starts with the equations that describe the volume and the surface area of a sphere.
If the sphere has radius r, then the surface area of the sphere is 4πr2. And the volume is (4/3)πr3. What’s interesting about this is that there’s a relationship between these two expressions. The first is the derivative of the second. The derivative is one of the earliest things one learns in calculus. It describes how much a quantity changes with a tiny change in something it depends on.
And this got him to thinking about the surface area of a cube. Call the length of a cube’s side s. Its surface is six squares, each of them with a side of length s. So the surface area of each of the six squares is s2, which is obvious when you remember we call raising something to the second power “squaring”. Its total surface area then is 6s2. But its volume is is s3. This is why we even call raising something to the third power “cubing”. And the derivative of s3 is 3s2. (If you don’t know calculus, but you suspect you see a pattern here, you’re learning calculus. If you’re not sure about the pattern, let me tell you that the derivative of s4 would be 4s3, and the derivative of (1/3)s2 would be (2/3)s.)
There’s an obvious flaw there, and Austin’s aware of it. But it got him pondering different ways to characterize how big a cube is. He can find one that makes the relationship between volume and surface area work out like he expects. But the question remains, why that? And what about other shapes?
I think that’s an interesting discussion to have, and mean to think about it some more myself. And I wanted to point people who’d be interested over there to join in.
Before painting a room you should spackle the walls. This fills up small holes and cracks. My father is notorious for using enough spackle to appreciably diminish the room’s volume. (So says my mother. My father disagrees.) I put spackle on as if I were paying for it myself, using so little my father has sometimes asked when I’m going to put any on. I’ll get to mathematics in the next paragraph.
One of the natural things to wonder about a set — a collection of things — is how big it is. The “measure” of a set is how we describe how big a set is. If we’re looking at a set that’s a line segment within a longer line, the measure pretty much matches our idea of length. If we’re looking at a shape on the plane, the measure matches our idea of area. A solid in space we expect has a measure that’s like the volume.
We might say the cracks and holes in a wall are as big as the amount of spackle it takes to fill them. Specifically, we mean it’s the least bit of spackle needed to fill them. And similarly we describe the measure of a set in terms of how much it takes to cover it. We even call this “covering”.
We use the tool of “cover sets”. These are sets with a measure — a length, a volume, a hypervolume, whatever — that we know. If we look at regular old normal space, these cover sets are typically circles or spheres or similar nice, round sets. They’re familiar. They’re easy to work with. We don’t have to worry about how to orient them, the way we might if we had square or triangular covering sets. These covering sets can be as small or as large as you need. And we suppose that we have some standard reference. This is a covering set with measure 1, this with measure 1/2, this with measure 24, this with measure 1/72.04, and so on. (If you want to know what units these measures are in, they’re “units of measure”. What we’re interested in is unchanged whether we measure in “inches” or “square kilometers” or “cubic parsecs” or something else. It’s just longer to say.)
You can imagine this as a game. I give you a set; you try to cover it. You can cover it with circles (or spheres, or whatever fits the space we’re in) that are big, or small, or whatever size you like. You can use as many as you like. You can cover more than just the things in the set I gave you. The only absolute rule is you must not miss anything, even one point, in the set I give you. Find the smallest total area of the covering circles you use. That smallest total area that covers the whole set is the measure of that set.
Generally, measure matches pretty well the intuitive feel we might have for length or area or volume. And the idea extends to things that don’t really have areas. For example, we can study the probability of events by thinking of the space of all possible outcomes of an experiment, like all the ways twenty coins might come up. We find the measure of the set of outcomes we’re interested in, like all the sets that have ten tails. The probability of the outcome we’re interested in is the measure of the set we’re interested in divided by the measure of the set of all possible outcomes. (There’s more work to do to make this quite true. In an advanced probability course we do this work. Please trust me that we could do it if we had to. Also you see why we stride briskly past the discussion of units. What unit would make sense for measuring “the space of all possible outcomes of an experiment” anyway?)
But there are surprises. For example, there’s the Cantor set. The easiest way to make the Cantor set is to start with a line of length 1 — of measure 1 — and take out the middle third. This produces two line segments of length, measure, 1/3 each. Take out the middle third of each of those segments. This leaves four segments each of length 1/9. Take out the middle third of each of those four segments, producing eight segments, and so on. If you do this infinitely many times you’ll create a set that has no measure; it fills no volume, it has no length. And yet you can prove there are just as many points in this set as there are in a real normal space. Somehow merely having a lot of points doesn’t mean they fill space.
Measure is useful not just because it can give us paradoxes like that. We often want to say how big sets, or subsets, of whatever we’re interested in are. And using measure lets us adapt things like calculus to become more powerful. We’re able to say what the integral is for functions that are much more discontinuous, more chopped up, than ones that high school or freshman calculus can treat, for example. The idea of measure takes length and area and such and makes it more abstract, giving it great power and applicability.
Figure 1. A triangle (meant to be a right triangle) with an inscribed circle of radius r. Three radial lines, perpendicular to the bases they touch, are included.
So here’s another geometry-based approach to finding what the radius of the circle that just fits inside the triangle has to be. We started off with the right triangle, and sides a and b and c; and there’s a circle inscribed in it. This is the biggest circle that’ll fit within the triangle. The circle has some radius, and we’ll just be a little daring and original and use the symbol “r” to stand for that radius. We can draw a line from the center of the circle to the point where the circle touches each of the legs, and that line is going to be of length r, because that’s the way circles work. My drawing, Figure 1, looks a little bit off because I was sketching this out on my iPad and being more exact about all this was just so, so much work.
The next step is to add three more lines to the figure, and this is going to make it easier to see what we want. What we’re adding are liens that go from the center of the circle to each of the corners of the original triangle. This divides the original triangle into three smaller ones, which I’ve lightly colored in as amber (on the upper left), green (on the upper right), and blue (on the bottom). The coloring is just to highlight the new triangles. I know the figure is looking even sketchier; take it up with how there’s no good mathematics-diagram sketching programs for a first-generation iPad, okay?
Figure 2. A triangle (meant to be a right triangle) with an inscribed circle of radius r. The triangle is divided into three smaller triangles meeting at the center of the inscribed circle.
If we can accept my drawings for what they are already, then, there’s the question of why I did all this subdividing, anyway? The good answer is: looking at this Figure 2, do you see what the areas of the amber, green, and blue triangles have to be? Well, the area of a triangle generally is half its base times its height. A base is the line connecting two of the vertices, and the height is the perpendicular distane between the third vertex and that base. So, for the amber triangle, “a” is obviously a base, and … say, now, isn’t “r” the height?
It is: the radius line is perpendicular to the triangle leg. That’s how inscribed circles work. You can prove this, although you might convince yourself of it more quickly by taking the lid of, say, a mayonnaise jar and a couple of straws. Try laying down the straws so they just touch the jar at one point, and so they cross one another (forming a triangle), and try to form a triangle where the straw isn’t perpendicular to the lid’s radius. That’s not proof, but, it’ll probably leave you confident it could be proven.
So coming back to this: the area of the amber triangle has to be one-half times a times r. And the area of the green triangle has to be one-half times b times r. The area of the blue triangle, yeah, one-half times c times r. This is great except that we have no idea what r is.
But we do know this: the amber triangle, green triangle, and blue triangle together make up the original triangle we started with. So the areas of the amber, green, and blue triangles added together have to equal the area of the original triangle, and we know that. Well, we can calculate that anyway. Call that area “A”. So we have this equation:
Where a, b, and c we know because those are the legs of the triangle, and A we may not have offhand but we can calculate it right away. The radius has to be twice the area of the original triangle divided by the sum of a, b, and c. If it strikes you that this is twice the area of the circle divided by its perimeter, yeah, that it is.
Incidentally, we haven’t actually used the fact that this is a right triangle. All the reasoning done would work if the original triangle were anything — equilateral, isosceles, scalene, whatever you like. If the triangle is a right angle, the area is easy to work out — it’s one-half times a times b — but Heron’s Formula tells us the area of a triangle knowing nothing but the lengths of its three legs. So we have this:
(Right triangle)
(Arbitrary triangle)
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Since we started out with a Pythagorean right triangle, with sides 5, 12, and 13, then: a = 5, b = 12, c = 13; a times b is 60; a plus b plus c is 30; and therefore the radius of the inscribed circle is 60 divided by 30, or, 2.
The set of posts about the area of a trapezoid seems to form a nearly coherent enough whole that it seems worthwhile to make a convenient reference point so that people searching for “how do you find the area of a trapezoid in the most convoluted and over-explained way possible?” have convenient access to it all. So, this is the path of that whole discussion.
[ I figure this to be the last important bit of Trapezoid Week. ]
I have one last bit of proving the area formula for the trapezoid, and figure this should wrap up all the exposition I mean to do here. I’m starting again with a trapezoid, with the two parallel bases AB and DE drawn so they’re horizontal. I’m making no assumptions about the legs AD and BE; they might be parallel, they might not be. This makes a proof that’s improved from the difference-of-triangles proof, since it works for more shapes. It’ll work naturally for parallelograms and rectangles, whether you want them to be trapezoids or not. It’s also improved from the twin-trapezoid proof, because it requires so much less extra work.
[ More of Trapezoid Week! Here we make finding the area simpler by doubling the number of trapezoids on the screen. ]
Figuring out the area of a trapezoid based on making it the difference between two triangles works all right. “All right” carries with it a sense of inadequacy. The complaints against it are pretty basic. The first is that it doesn’t work for everything which might be called a trapezoid. Maybe we don’t want to consider parallelograms and rectangles to be particular kinds of trapezoids, but, why rule them out if we don’t have to? The second point is the proof is a little convoluted, requiring us to break out of thinking about trapezoids to remember details of similar triangles. It’d be nice if we had a more direct way of proving things.
(It strikes me, this might just as well be Trapezoid Week here. )
Since I did work out the area of a trapezoid starting from the area formula for triangles, and since I was embarrassed to have not seen it sooner, I decide to share it here, where it may do someone some good, particularly if it’s me for next time I teach a class like this. The punch line is known far ahead of time. The trapezoid is a four-sided figure with two sides parallel. The parallel sides have lengths b1 and b2; they’re considered bases. The two bases are an altitude a apart. The area of the trapezoid then is a * (b1 + b2)/2.
I haven’t got done listing kinds of trapezoids, of course. Arguably I’d never be able to finish, since, after all, couldn’t any possible length of the two bases — the parallel lines — and of different lengths of the diagonal legs be imagined? Well, perhaps, although a lot of those kinds are going to look the same. An isoceles trapezoid where the long base is 10 and the short base 8 looks a lot like one where the long base is 11 and the short base 7.5, at least if the bases are the same distance apart. But there are more cases imaginable.
When I lecture I like to improvise. I prepare notes, of course, the more detailed the more precise I need to be, but my performing instincts are most satisfied when I just go in front of the class with some key points to hit and maybe a few key lines worked out ahead of time. But I did recently make an iconic mistake, repeating the mathematics instructor’s equivalent of the lawyer asking in court a question without already knowing what the answer will be. Improvisation has to be carefully prepared.
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nebusresearch 