Geometry Puzzle

11 October, 2010

We’re thinking about solar power over at the Azimuth Project, so Graham Jones wrote a page on solar radiation. This led him to a nice little geometry puzzle.

If you float in space near the Earth, and measure the power density of solar radiation, you’ll get 1366 watts per square meter. But because this radiation hits the Earth at an angle, and not at all during the night, the average global solar power density is a lot less: about 341.5 watts per square meter at the top of the atmosphere. And of this, only about 156 watts/meter2 makes it down the Earth’s surface. From 1366 down to 156 — that’s almost an order of magnitude! This is why some people like the idea of space-based solar power.

But when I said “a lot less”, I was concealing a cute and simple fact: the average global solar power density is one quarter of the power density in outer space near Earth’s orbit:

1366/4 = 341.5

Why? Because the area of a sphere is four times the area of its circular shadow!

Anyone who remembers their high-school math can see why this is true. The area of a circle is πr2, where r is the radius of the circle The surface area of a sphere is 4πr2, where r is the radius of the sphere. That’s where the factor of 4 comes from.

Cute and simple. But Graham Jones posed a nice followup puzzle: What’s the easiest way to understand this factor of 4? Maybe there’s a way that doesn’t require calculus — just geometry. Maybe with a little work we could just see that factor of 4. That would be really satisfying.

But I don’t know how to “just see it”. So this is not the sort of puzzle where I smile in a superior sort of way and chuckle to myself as you folks struggle to solve it. This the sort where I’d really like to know the best answer.

But here’s something I do know: we can derive this factor of 4 from a nice but even less obvious fact which I believe was proved by Archimedes.

Take a sphere and slice it with a bunch of parallel planes, like chopping an apple with a cleaver. If two slices have the same thickness, they also have the same surface area!

(When I say “surface area” here, I’m only counting the red skin of the apple slices.)

There’s an interesting cancellation at work here. A slice from near the top or bottom of the sphere will be smaller, but it’s also more “sloped”. The magic fact is that these effects exactly cancel when we compute its surface area.

If you think a bit, you can see this is equivalent to another nice fact:

The surface area of any slice of a sphere matches the surface area of the corresponding slice of the cylinder with the same radius. If you don’t get what I mean, see the picture at Wolfram Mathworld.

And this in turn implies that the surface area of the sphere equals the surface area of the cylinder, not including top and bottom. But that’s the cylinder’s circumference times its height. So we get

2πr × 2r = 4πr2

So we get that factor of 4 we wanted.

In fact, Archimedes was so proud of discovering this fact that he put it on his tomb! Cicero later saw this tomb and helped save it from obscurity. He wrote:

But from Dionysius’s own city of Syracuse I will summon up from the dust—where his measuring rod once traced its lines—an obscure little man who lived many years later, Archimedes. When I was questor in Sicily [in 75 BC, 137 years after the death of Archimedes] I managed to track down his grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns. I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave. And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder. I immediately said to the Syracusans, some of whose leading citizens were with me at the time, that I believed this was the very object I had been looking for. Men were sent in with sickles to clear the site, and when a path to the monument had been opened we walked right up to it. And the verses were still visible, though approximately the second half of each line had been worn away.

I don’t know what the verses said.

It’s been said that the Roman contributions to mathematics were so puny that the biggest was Cicero’s discovery of this tomb. But Archimedes’ result doesn’t by itself give an easy intuitive way to see where that factor of 4 is coming from! You may or may not find it to be a useful clue.

(In fact, “equal surface areas for slices of equal thickness” is a special case of a principle called “Duistermaat-Heckman localization”. For that, try page 23 of chapter 2 of the book by Ginzburg, Guillemin and Karshon. But that’s much fancier stuff than I’m wondering about here, I think.)

Probability Puzzles (Part 2)

5 September, 2010

Sometimes places become famous not because of what’s there, but because of the good times people have there.

There’s a somewhat historic bar in Singapore called the Colbar. Apparently that’s short for “Colonial Bar”. It’s nothing to look at: pretty primitive, basically a large shed with no air conditioning and a roofed-over patio made of concrete. Its main charm is that it’s “locked in a time warp”. It used to be set in the British army barracks, but it was moved in 2003. According to a food blog:

Thanks to the petitions of Colbar regulars and the subsequent intervention of the Jurong Town Council (JTC), who wanted to preserve its colourful history, Colbar was replicated and relocated just a stone’s throw away from the old site. Built brick by brick and copied to close exact, Colbar reopened its doors last year looking no different from what it used to be.

It’s now in one of the few remaining forested patches of Singapore. The Chinese couple who run it are apparently pretty well-off; they’ve been at it since the place opened in 1953, even before Singapore became a country.

Every Friday, a bunch of philosophers go there to drink beer, play chess, strum guitars and talk. Since my wife teaches in the philosophy department at NUS, we became part of this tradition, and it’s a lot of fun.

Anyway, the last time we went there, one of the philosophers posed this puzzle:

You know a woman who has two children. One day you see her walking by with one. You notice it’s a boy. What’s the probability that both her children are boys?

Of course I instantly thought of the probability puzzles we’ve discussed here. It’s not exactly any of the versions we have already talked about. So I thought you folks might enjoy it.

What’s the answer?

Probability Puzzles (Part 1)

24 August, 2010

Today Greg Egan mailed me two puzzles in probability theory: a “simple” one, and a more complicated one that compares Bayesian and frequentist interpretations of probability theory.

Try your hand at the simple one first. Egan wrote:

A few months ago I read about a very simple but fun probability puzzle. Someone tells you:

“I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”

Please give it a try before moving on. Or at least figure out what this is:

Of course, your first reaction should be “it’s irrelevant the boy was born on a Tuesday“. At least that was my first reaction. So I said:

I’d intuitively assume that the day Tuesday is not relevant, so I’d ignore that information – or else look at some hospital statistics to see if it is relevant. I’d also assume that boy/girl births act just like independently distributed fair coin flips — which is surely false, but I’m guessing the puzzle wants us to assume it’s true. And then I’d say there are 4 equally likely options: BB, BG, GB and GG.

If you tell me “one is a boy”, it’s very different from “the first one is a boy”. If one is a boy, we’re down to 3 equally likely options: BB, BG, and GB. So, the probability of two boys is 1/3.

But that’s not the answer Egan gives:

The usual answer to this puzzle — after people get over an initial intuitive sense that the “Tuesday” can’t possibly be relevant — is that the probability of having two sons is 13/27. If someone has two children, for each there are 14 possibilities as to boy/girl and weekday of birth, so if at least one child is a son born on a Tuesday there are 14 + 14 – 1 = 27 possibilities (subtracting 1 for the doubly-counted intersection, where both children are sons born on a Tuesday), of which 7 + 7 – 1 = 13 involve two sons.

If you find that answer unbelievable, read his essay! He does a good job of making it more intuitive:

• Greg Egan, Some thoughts on Tuesday’s child.

But then comes his deeper puzzle, or question:

That’s fine, but as a frequentist if someone asks me to take this probability seriously and start making bets, I will only do so if I can imagine some repetition of the experiment. Suppose someone offered me $81 if the parent had two sons, but I had to pay $54 if they had a son and a daughter. The expected gain from that bet for P(two sons)=13/27 would be $11.

If I took up that bet, I would then resolve that in the future I’d only take the same bet again if the person each time had two children and at least one son born specifically on a TUESDAY. In fact, I’d insist on asking the parent myself “Do you have at least one son born on a Tuesday?” rather than having them volunteer the information (since someone with two sons born on different days might not mention the one born on a Tuesday). That way, I’d be sampling a subset of parents all meeting exactly the same conditions, and I’d be satisfied that my long-term expectation of gain really would be $11 per bet.

But I’m curious as to how a Bayesian, who is happier to think of a probability applying to a single event in isolation, would respond to the same situation. It seems to me (perhaps naively) that a Bayesian ought to be happy to take this bet any time, and then forget about what they did in the past — which ought to make them willing to take the bet on future offers even when the day of the week when the son was born changes. After all, P(two sons)=13/27 whatever day is substituted for Tuesday.

However, anyone who agreed to keep taking the bet regardless of the day of the week would lose money! Without pinning down the day to a particular choice, you’re betting on a sample of parents who simply have two children, at least one of whom is a son. That gives P(two sons)=1/3, and the expectation for the $81/$54 bet becomes a $9 loss.

Now, I understand how the difference between P(two sons)=13/27 and P(two sons)=1/3 arises, despite the perfect symmetry between the weekdays; the subsets with “at least one son born on day X” are not disjoint, so even though they are isomorphic, their union will have a different proportion of two-son families than the individual subsets.

What’s puzzling me is this: how does a Bayesian reason about the thought experiment I’ve described, in such a way that they don’t end up taking the bet every time and losing money?