In 2014 I described Descartes’ theorem, which shows how to find a fourth circle that’s tangent to three “kissing circles.”

Descartes’ equation refers to the “curvature” of each circle: this is just the reciprocal of the radius, so a circle with radius 1/3 would have a curvature of 3. (This makes sense intuitively — a circle with a small radius “curves more” than a larger one.)

Remarkably, if the four starting circles all have integer curvature, then so will every circle we pack into the figure, each kissing the three around it. In the limit the figure becomes a fractal containing an infinite number of circles. It’s called an Apollonian gasket.

07/29/2016 UPDATE: By coincidence, the U.S. quarter, nickel, and dime make three of the four generating circles in an integral packing — see the caption accompanying the first figure on this page. (Thanks, Trevor.)

Here’s an isosceles triangle. Sides AB and AC are equal, and this means that the angles opposite those sides are equal as well.

That’s intuitively reasonable, but proving it is tricky. Teachers of Euclid’s Elements came to call it the pons asinorum, or bridge of donkeys, because it was the first challenge that separated quick students from slow.

The simplest proof, attributed to Pappus of Alexandria, requires no additional construction at all. We know that two triangles are congruent if two sides and the included angle of one triangle are congruent to their corresponding parts in the other (the “side-angle-side” postulate). So Pappus suggested simply picking up the triangle above, flipping it over, and putting it down again, to produce a second triangle ACB. Now if we compare the respective parts of the two triangles, we find that angle A is equal to itself, AB = AC, and AC = AB. Thus the “left-hand” angles of the two triangles are congruent, and it follows that the base angles of the original triangle above are equal.

In his 1879 book Euclid and his Modern Rivals, Lewis Carroll accepts the proof but remarks that it “reminds one a little too vividly of the man who walked down his own throat.”

What’s the most effective strategy for loading an airplane? Most airlines tend to work from the back to the front, accepting first the passengers who will sit in high-numbered rows (say, rows 25-30), waiting for them to find their seats, and then accepting the next five rows, and so on. Both the airline and the passengers would be glad to know that this is the most effective strategy. Is it?

In 2005, computer scientist Eitan Bachmat of Ben-Gurion University decided to find out. He devised a model that considers parameters of the aircraft cabin, the boarding method, the passengers, and their behavior, and found that the most important variable is a combination of three parameters: the length of the aisle blocked by a standing passenger, multiplied by the number of seats in a row, divided by the distance between rows. If rows are 80 centimeters apart, there are six seats in a row, and a standing passenger and his hand luggage take up 40 centimeters of the aisle, then the passengers headed for a single row will block the aisle space of three rows while they’re waiting to reach their seats.

This quickly backs things up. Even if the airline admits only passengers with row numbers 25-30, half the aisle will be completely blocked and most passengers will have to wait until everyone in front of them has sat down before they reach their seats. The time it takes to fill the cabin grows in proportion to the number of passengers.

A better policy would be to call up the passengers in rows 30, 27, and 24; then those in 29, 26, and 23; and so on (perhaps using color-coded boarding passes). These combinations of passengers would not block one another in the aisles.

An even better policy, Bachmat found, would be to dispense with seat assignments altogether and let passengers board the plane and pick their seats as they please. “With this method, or lack of a method,” writes George Szpiro, “the time required to get people on board and into their seats would only be proportional to the square root of the number of passengers.”

(Eitan Bachmat et al., “Analysis of Airplane Boarding Times,” Operations Research 57:2 [2009]: 499-513 and George S. Szpiro, A Mathematical Medley, 2010. See All Aboard.)

In January 2007, inspired by this article by computer scientist Scott Aaronson, philosophers Agustín Rayo of MIT and Adam Elga of Princeton joined in the “large number duel” to come up with the largest finite number ever written on an ordinary-sized chalkboard.

The rules were simple. The two would take turns writing down expressions denoting natural numbers, and whoever could name the largest number would win the duel. No primitive semantic vocabulary was allowed (so that it would be illegal simply to write the phrase “the smallest number bigger than any number named by a human so far”), and the two agreed not to build on one another’s contributions (so neither could simply write “the previous entry plus one”).

Elga went first, writing the number 1. Rayo countered with a string of 1s:

111111111111111111111111111111111111111111111111111111111111

and Elga erased a line through the base of half this string to produce a factorial:

1111111111111111111111111111!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The two began defining their own functions, and toward the end Rayo wrote this phrase:

The smallest number bigger than any number that can be named by an expression in the language of first-order set theory with less than a googol (10¹⁰⁰) symbols.

With some tweaking, this became the winning entry, now enshrined as “Rayo’s number.”

“It was a great game,” Elga said after the match. “Heated at times, but nevertheless, a really great game.”

The use of philosophy was “crucial,” Rayo said. “The limit of math ability was reached at the end. Knowing a bit of philosophy, that was the key.”

Asked whether he thought his entry had set the Guinness world record, “It’s hard to be sure,” Rayo said, “but the number is bigger than any number I have ever seen.”

(Thanks, Erik.)

Here’s the item I mentioned in Episode 99 of the podcast — New York City engineer John Waterhouse published it in July 1899. It’s not a proof of the Pythagorean theorem, as I’d thought, but rather a related curiosity. It made a splash at the time — the Proceedings of the American Society of Civil Engineers said it “interested instructors of geometry all over the country, bringing many letters of commendation to him from prominent teachers.” Listener Colin Beveridge has been immensely helpful in devising the diagram above and making sense of Waterhouse’s proof as it appears on page 252 of Elisha Scott Loomis’ 1940 book The Pythagorean Proposition. Click the diagram to enlarge it a bit further.

1. Red squares BN = AI + CE — Pythagoras’s theorem
2. Blue triangles AEH, CDN, BMI are all equal in area to ABC, reasoning via X and Y and base sides.
3. Green angles GHI and IBM are equal and green triangle GHI is congruent to IBM (side angle side), so IG = IK = IM. IH′K is congruent to IHK as angle HIK = angle HIG and the adjacent sides correspond. This means G and K are the same distance from the line HH′, so GK is parallel to HI. Similarly, DE is parallel to PF and MN is parallel to LO.
4. GK = 4HI, because TU=HI, TG = AH (HTG congruent to EAH) and UK = UG (symmetry). Similarly, PF = 4DE. Dark blue triangles IVK and LWM are equal, so WM = VK. Similarly, OX = QD (dark green triangles PQD and NXO are congruent). Also, WX=MJ and XN=NJ, so M and N are the midpoints of WJ and XJ. That makes WX=2MN, so LO = 4MN.
5. Each of the trapezia we just looked at (HIKG, OLMN and PFED) have five times the area of ABC.
6. The areas of orange squares MK and NP are together five times EG. This is because:
• the square on MI is (the square on MY) + (the square on IY) = (AC²) + (2AB)² = 4AB² + AC².
• the square on ND is (the square on NZ) + (the square on DZ) = (AB²) + (2AC)² = 4AC² + AB²
• the sum of these is 5(AB² + AC²) = 5BC², and BC = HE.
7. A′S = A′T, so A′SAT is a square and the bisector of angle B′A′C′ passes through A. However, the bisectors of angle A′B′C′ and A′C′B′ do not pass through B and C (resp.) [Colin says Waterhouse’s reasoning for this is not immediately clear.]
8. Square LO = square GK + square FP, as LO = 4AC, GK = 4AB and FP = 4BC.
9. [We’re not quite sure what Waterhouse means by “etc. etc.” — perhaps that one could continue to build squares and triangles outward forever.]