As the story goes, somebody once posed a problem to

Johnny von Neumann at a cocktail party:

A gnat flies back and forth between the front wheels of two bicycles that are approaching one another at respective speeds of 10 mph. The gnat flies at 20 mph and turns around instantly when it reaches either wheel. If the bicycles are initially 1 mile apart, then how far has the gnat traveled by the time it gets crushed?

If you visualize the problem, or anyhow an idealized version of it, then you will see that the gnat makes infinitely many trips back and forth, each trip shorter than the last. You could figure out the distance of each trip and sum an infinite series to find the answer. For the numbers given in the problem, this turns out to look like

^{2}⁄_{3}(1 + ^{1}⁄_{3} + ^{1}⁄_{9} + ^{1}⁄_{27} + ... ) miles.

The sum of an infinite series of the form 1 + *r* + *r*^{2} + ... (for *r* < 1) is ^{1}⁄_{(1 − r)}. Putting *r* = ^{1}⁄_{3}, we find the gnat's total distance as

^{2/3}⁄_{(1 − 1/3)} = 1 mile.

The fast way to solve the problem is to reason that since the initial 1 mile separation between the bicycles is closing at a rate of 20 miles per hour, it will take ^{1}⁄_{20}th of an hour for the bicycles to collide; and the gnat is flying 20 mph for that whole time, so it covers a total distance given by speed × time = 20 ^{mi}⁄_{hr} × ^{1}⁄_{20} hr = 1 mile.

Now the story goes that von Neumann, presented with the problem, quickly answered "one mile." "Ah ha," said his friend, "you discovered the trick." "Trick?" said von Neumann. "It's not a very difficult series to sum."

As brilliant as von Neumann was, I have a hard time believing the story. But it's part of the lore—I heard the story back when I was a graduate student, and there are dozens of variations floating around. If the events actually did happen, then my interpretation of them isn't the usual one. Rather than concluding, as I think most people do, that von Neumann had instantly solved the problem the long way in his head—I just don't think that's possible, even for the person Los Alamos used as a human supercomputer—and rather than concluding that he was trying to suggest that he had (let's assume he wouldn't do something that lame), I think von Neumann was making a joke. The joke is that he had instantly solved the problem the long way in his head. (If this joke seems too meta, consider that von Neumann was the co-inventor of game theory.)

Once I was faced with a similar problem. I wanted to calculate the cooling effect in an expanding ideal gas directly from the motion of the gas particles. Instead of a single gnat, I was analyzing the motion of an enormous number of them, not all of them moving at the same speed, as they bounced back and forth between a moving piston and a container wall. Another important difference from the gnat problem was that the particle speeds weren't constant: the distribution of particle speeds changes as particles rebound from a receding piston. My

results are published in the

*American Journal of Physics*. The answer takes a very complicated form, otherwise I'd suspect you could find it somewhere in an old notebook of Boltzmann, Maxwell, or Gibbs.

I was reminded of all this the other day when I was looking at a triangle shaped like this blue one:

The area can be found by simply calculating one-half the base times the height. The result is 10.5. Now, to prove that "one-half base times height" works for a triangle like this, we usually show a picture like so:

Suppose we've previously established the principle that the area of a *right* triangle is one-half base times height. By this principle, the large right triangle has area ^{1}⁄_{2} × 5 × 7 = 17.5. Meanwhile, the small right triangle has area ^{1}⁄_{2} × 2 × 7 = 7. So the area of the triangle we want, the blue triangle, is 17.5 − 7 = 10.5.

This is how it's usually done;

here is Salman Khan, for example (skip to the 7-minute mark). But the other day, I was suddenly dissatisfied with a method of calculating the area of the blue triangle that forces me to contemplate the area of shapes that lie

*outside* the blue triangle.

If all we want is the area inside the blue triangle, then can't we find it by calculating only areas of shapes that are inside the blue triangle?

In the Borges short story

"Funes the Memorious," the character of Ireneo Funes invents an absurd number system in order to resolve his dissatisfaction with having to use more than one number symbol to represent the number of gauchos in an episode from Uruguayan history. The method I used to calculate the area of the blue triangle is no less absurd—but perhaps similarly amusing. It goes like this.

Draw infinitely many horizontal and vertical segments to carve the blue triangle into infinitely many right triangles.

Use similarity to determine the lengths of the horizontal and vertical segments, then calculate the areas of the right triangles using "one-half base times height." I'll let you verify that this gives an infinite series,

area = ^{441}⁄_{50} (1 + ^{4}⁄_{25} + ^{16}⁄_{625} + ^{256}⁄_{15625} + ...).

As the saying goes, it's not a very difficult series to sum, and the result is

area = ^{441/50}⁄_{(1 − 4/25)} = 10.5

as expected.

One can do the problem symbolically, replacing 3 by *b*, 2 by *x*, and 7 by *h*. Then we get

area = \(\frac{1}{2}b^2h\frac{b+2x}{(b+x)^2}\sum_{n=0}^\infty{\left(\frac{x}{b+x}\right)^{2n}}\).

It is straightforward to verify that this gives

^{1}⁄_{2}*bh*, as expected. Thus, the area is one-half base times height.

Part of the "fun" of this method is to conceive of the area of a triangle as an infinite series. But just as one can derive 1 +

*r* +

*r*^{2} + ... =

^{1}⁄

_{(1 − r)} recursively by noticing that 1 +

*r* +

*r*^{2} + ... = 1 +

*r*(1 +

*r* +

*r*^{2} + ...), one can find the triangle area recursively as well, thus avoiding the "dot-dot-dot." To do this, denote by

*A* the area of the blue triangle. Then

*A* is the sum of the largest right-triangle area (found by similarity as

^{63}⁄_{10}), the next-largest right-triangle area (found by similarity as

^{126}⁄_{50}), and the area of a triangle that is similar to the blue triangle by a scale factor

^{2}⁄_{5}. So we have

*A* = ^{63}⁄_{10} + ^{126}⁄_{50} + ^{4}⁄_{25}*A*.

Solve this linear equation to find

*A* = 10.5.

Borges himself had occasion to write down an infinite series in his essay

"Avatars of the Tortoise." The number of thinkers who appear in this essay is vast (Zeno, Aristotle, Hui Tzu, Thomas Aquinas, Lewis Carroll, and Williams James, among others), and I don't entirely follow the reasoning. But the series in the essay makes a fitting end to today's post:

10 + 1 + ^{1}⁄_{10} + ^{1}⁄_{100} + ^{1}⁄_{1000} + ^{1}⁄_{10,000} + ....

Needless to say, the sum of this series is 11.111\(\cdots\).