An inquisitive ant sets out from point A at a bottom vertex of the 1 × 1 × 2 box shown above. Of all the possible destinations it might seek in a direct route along the surface of the box, which one requires the longest journey?

Intuitively we might think it’s point B, the farthest vertex on the box roof. But Japanese mathematician Yoshiyuki Kotani discovered that the longest journey actually ends one-fourth of the way along the rooftop diagonal that ends at point B.

This can be seen by “unfolding” the box into a flat diagram, where four different paths can be traced from A to that point. The Pythagorean theorem shows that all four paths have the same length, 2.850…, which is about 0.022 longer than the shortest path to B.

Data scientist Shiro Matsumoto provides some animations here.

(Martin Gardner, “The Ant on a 1 × 1 × 2,” Math Horizons 3:3 [February 1996], 8-9.)

Make an inverted triangle of hexagonal cells with side length 3ⁿ + 1, and color the cells in the top row randomly in three colors. Now color the cells in the second row according to these rules:

1. If the neighboring cells immediately above are of the same color, assign that color.
2. If they’re of different colors, assign the third color.

When you’ve finished the second row, continue through the succeeding ones, applying the same rules. Pleasingly, no matter how large the triangle, the color of the last cell can be predicted at the start: Just apply our two guiding rules to the endmost cells in the top row. If those two cells are both red, the last cell will be red. If one is red and one is yellow (as in the figure above), the bottom cell will be blue.

The principle was discovered by Newcastle University mathematician Steve Humble in 2012. Gary Antonick gives more background here, and see the paper below for a mathematical discussion by Humble and Ehrhard Behrends.

(Ehrhard Behrends and Steve Humble, “Triangle Mysteries,” Mathematical Intelligencer 35:2 [June 2013], 10-15.)

In a family of convex sets in the Euclidean plane, if the intersection of each triple is non-empty, then the intersection of the whole collection is non-empty.

Discovered by Austrian mathematician Eduard Helly in 1913.

 s = 1 + 2 + 4 + 8 + 16 + ...

   = 1 + 2(1 + 2 + 4 + 8 + ...)

   = 1 + 2s,

so

-1 = 1 + 2 + 4 + 8 + 16 + ...

(Regular summation gives a sum of infinity, but the idea above does make a certain sense.)

(Thanks, Murtadha.)

One other oddity concerning π: If you add up the first three sextads in the decimal expansion, you get 1588419:

141592 + 653589 + 793238 = 1588419

That’s a little prophecy: If you now skip ahead 15 places you arrive at the string 88419:

3.1415926535897932384626433832795028841971693 …

(Communicated by P. Olivera.)

This graph has 5 edges, and we’ve managed to label its vertices in a remarkable way: Each vertex bears some integer from 0 to 5, no two receive the same integer, and each edge is now uniquely identified by the absolute difference between its endpoints, such that this magnitude lies between 1 and 5 inclusive. Such a labeling is called graceful.

In 2008 Donald E. Knuth made a graph representing the contiguous 48 states and the District of Columbia in which each pair of states are connected if they’re joined by at least one drivable road. It turns out that this graph can be labeled gracefully.

And, amazingly, in 2020 T. Rokicki discovered that if you undertake an imaginary journey on Knuth’s map, starting in California and going up the Pacific coast and then along the Canadian border, you’ll visit successive vertices labeled 31, 41, 59, 26, 53, 58, 97, 93, 23, 84, 62, 64, 33, 83, and 27. These are the first 30 decimal digits of π!

Knuth called this a “graceful miracle.”

When two or more buses are scheduled at regular intervals on the same route, planners may expect that each will make the same progress, pausing at each stop for the same interval (1). But if Bus B is delayed by traffic congestion (2), it incurs a penalty: Because it arrives late to the next stop, it will pick up some passengers who’d planned to take Bus C (3). Accommodating these passengers delays Bus B even longer, putting it even further behind schedule. Meanwhile, Bus C begins to make unusually good progress (4), as it now arrives at each stop to find a smaller crowd than expected.

As the workload piles up on the foremost bus and the one behind it catches up, eventually the result (5) is that the two buses run in a platoon, arriving together at each stop. Sometimes Bus C even overtakes Bus B.

What to do? Planners can set minimum and maximum amounts of time to be spent at each stop, and buses might even be told to skip certain stops during crowded runs. Passengers might be encouraged to wait for a following bus, with the inducement that it’s less crowded. Northern Arizona University improved its service by abandoning the idea of a schedule altogether and delaying buses at certain stops in order to maintain even spacing. One thing that doesn’t work: adding vehicles to the route — which might, at first blush, have seemed the obvious solution.

(Via Robert A. Carman, “Mathematical Misteaks,” Mathematics Teacher 64:2 [February 1971], 109-115.)

This worrying result was first published by German mathematician Oskar Schlömilch in 1868. (The discrepancy is explained by minute gaps in the diagonals, as explained here.)

Charles Dodgson (Lewis Carroll) seems to have been taken with the paradox — his papers show that between 1890 and 1893 he was working to determine all the squares that might similarly be converted into rectangles with a “gain” of one unit of area, apparently unaware that V. Schlegel had carried out the same task much earlier.

(Warren Weaver, “Lewis Carroll and a Geometrical Paradox,” American Mathematical Monthly 45:4 [April 1938], 234-236.)