Four straight roads cross a plain. No two are parallel, and no three meet in a point. On each road is a traveler who moves at some constant speed. If Blue and Red meet each other at their crossroad, and each of them meets Yellow and Green at their respective crossroads, will Yellow and Green necessarily meet at their own crossroad?

Suprisingly, the answer is yes. Rob Fatland offers a beautiful solution at the CTK Exchange:

Imagine the whole scenario unfolding from Blue’s rest frame; that is, regard Blue as unmoving and consider the movements of the other travelers relative to him. What does he see? Objectively we know that each traveler moves along a straight line at a constant speed, eventually encounters Blue, and moves on, so from Blue’s perspective each of them moves directly toward him on a straight line, passes through his position, and continues.

Very well, let Red do that. But we know that Red also encounters Yellow and Green without deviating from his own path. And we know that (from Blue’s perspective) Yellow and Green are also traveling straight lines that intersect Blue’s position. This can only mean that Red, Yellow, and Green are all traveling along the same straight line from Blue’s point of view. And this means that Yellow and Green must meet one another.

(It might be objected that two points traveling the same line needn’t meet if they’re going in the same direction at the same speed. But here this would mean that two of the roads are parallel, and that possibility is excluded by the conditions of the problem.)

05/27/2016 UPDATE: Reader Derek Christie recalls a three-dimensional solution: “Make a time axis perpendicular to the plane. Then each traveller moves in a straight line trajectory through this 3D space. Blue and Red meet at a particular place and time, so their two trajectories must meet at a point, and these two trajectories define a plane. Both Yellow and Green trajectories meet the Blue and Red trajectories and so also lie in this same plane. So Yellow and Green must also meet somewhere in that plane.”

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.)