On December 30 William Walkington sent this greeting to a circle of magic-square enthusiasts — it’s a traditional magic square (each row, column, and diagonal sums to 15), but the geometric area of each cell corresponds to its number.

He added, “The areas are approximate, and I don’t know if it is possible to obtain the correct areas with 2 vertically slanted straight lines through the square. Perhaps someone will be able to work this out in 2017?”

It’s only January 19, and the answer is already yes — Walter Trump has produced a “third-order linear area magic square” using the numbers 5-13:

There are many further developments, which have opened new questions and challenges, as these discoveries tend to do — see William’s blog post for more information.

(Thanks, William.)

Aachen University physicist Jörg Pretz has devised a binary clock in the shape of a triangular array of 15 lamps. Here’s how to read it:

• When lit, the top lamp denotes 6 hours.
• Each lamp on on the second row denotes 2 hours.
• Each lamp on on the third row denotes 30 minutes.
• Each lamp on on the fourth row denotes 6 minutes.
• Each lamp on on the fifth row denotes 1 minute.

So the clock above shows 6 hours + (2 × 2 hours) + (2 × 30 minutes) + (3 × 1 minute) = 11:03. The lamps’ color, red, shows that it’s after noon, or 11:03 p.m. The same array displayed in green would mean 11:03 a.m. A few more examples:

The time value assigned to each lamp is the total time value of the row below if that row contained one additional lamp.

On each row the lamps light up from left to right, so a row with n lamps can display n + 1 states (all lamps off to all lamps on). So for a triangular array with n lamps on the bottom row, the total number of states is

(n + 1) × ((n – 1) + 1) × ((n – 2) + 1) × · · · × (1 + 1) = (n + 1)!

That is, it’s a factorial of a natural number. And by a happy coincidence, the total number of minutes in 12 hours is such a factorial (720 = 6!).

“Thus the whole concept works because our system of time divisions is based on a sexagesimal system, dating back to the Babylonians, rather than a decimal system, as proposed during the French Revolution.”

There’s more info in Pretz’s article, and you can play with the clock using this applet.

(Jörg Pretz, “The Triangular Binary Clock,” Recreational Mathematics Magazine, March 2016.)

Reader Joe Antognini sent this in: Brazilian mathematician Inder Taneja has found a way to render every number from 1 to 11,111 by starting with either of these strings:

1 2 3 4 5 6 7 8 9

9 8 7 6 5 4 3 2 1

and applying any of the operations addition, subtraction, multiplication, division, and exponentiation. Brackets are permitted. For example:

6439 = 1 + 2 × (3⁴ × 5 × 6 + 789)

and

6439 = 9 × (8 + 7 + 6) + 5⁴ × (3² + 1)

Intriguingly, there’s one hole: There doesn’t seem to be a way to render 10958 from the increasing sequence.

Taneja’s paper is here. (Thanks, Joe.)

01/29/2017 UPDATE: Taneja tells me that, while it can’t be calculated using only basic operations, 10958 can be reached using factorials or square roots. Here are two factorials:

10958 = 1 + 2 + 3!! + (-4 + 5! + 6 – 7) × 89

10958 = 1 × 2 × (3!! – 4! × (5 + 6) + 7! – 8 – 9)

Towns A and B are connected by two roads. Suppose that two cars connected by a rope of length 2r can travel from A to B without breaking the rope. How can we prove that two circular wagons of radius r, moving along these roads in opposite directions, will necessarily collide?

This can be solved neatly by creating a configuration space. Map each road onto a unit segment, and set these up as two sides of a square. The northern car’s progress is reflected by a point moving up the left side of the square, and the southern car’s by a point moving from left to right along the bottom. Now the motion of the two cars from A to B is represented by a continuous curve connecting (0,0) and (1,1).

The wagons start from opposite towns, so their motion is represented by a curve from (0,1) to (1,0), and it’s immediately clear that the two curves must intersect. The intersection point corresponds to the collision of the wagons.

This example, by N. Konstantinov, is reportedly common in Russian mathematical folklore; I found it in Serge Tabachnikov’s 2005 book Geometry and Billiards (of all places).

Mathematicians Steve Humble and Yutaka Nishiyama invented this game to highlight a surprising result in probability, based on a principle discovered by Walter Penney.

Two players play the game using an ordinary deck of cards. The cards will be dealt out in a row, one after another. Before the dealing begins, each player claims an ordered sequence of colors that might turn up, for example “red black red” (RBR) or “black red red” (BRR). As the cards are dealt, if three successive cards turn up in one of these sequences, the player who claimed it gets to collect those three cards as a trick, and the dealing continues. When all 52 cards have been dealt, the player who has collected the most tricks wins. (In a typical game, 7 to 9 tricks are won.)

This sounds like a perfectly even game, but in fact the second player has a strategy that will given him a significant advantage. When the first player has chosen his sequence (say, RRB), the second player changes the middle color, adds it to the start of the sequence, discards the last color, and claims the resulting sequence (in this case, BRR). This gives a decided advantage to the second player no matter which sequence his opponent has chosen. In a computer simulation of 1,000 games, Humble and Nishiyama got these results:

BBB vs RBB – RBB wins 995 times, 4 draws, BBB wins once
BBR vs RBB – RBB wins 930 times, 40 draws, BBR wins 30 times
BRB vs BBR – BBR wins 805 times, 79 draws, RBR wins 116 times
RBB vs RRB – RRB wins 890 times, 68 draws, RBB wins 42 times
BRR vs BBR – BBR wins 872 times, 65 draws, BRR wins 63 times
RBR vs RRB – RRB wins 792 times, 85 draws, RBR wins 123 times
RRB vs BRR – BRR wins 922 times, 51 draws, RRB wins 27 times
RRR vs BRR – BRR wins 988 times, 6 draws, RRR wins 6 times

Penney’s original game uses coin flips; cards are preferable because no record-keeping is required and because the finite number of cards in a deck increases the second player’s chances.

(Steve Humble and Yutaka Nishiyama, “Humble-Nishiyama Randomness Game — A New Variation on Penney’s Coin Game,” Mathematics Today, August 2010.)

Maybe this was inevitable: A team of mathematicians have worked out the most efficient pub crawl in the United Kingdom, connecting 24,727 pubs in the shortest possible closed loop, 45,495,239 meters, or about 28,269 miles. Because it’s a loop, a determined crawler can start at any point and eventually find himself back home.

Despite the pickled application, this represents a serious achievement in computational mathematics, an advance in the so-called traveling salesman problem (TSP), which asks for the shortest route that passes through each of a set of points once and once only. The pub crawl includes more than 100 times the previous record number of stops in a road-distance TSP.

“We, of course, did not have in mind to bring everything mathematics has to bear in order to improve the lot of a wandering pub aficionado,” wrote lead researcher William Cook of the University of Waterloo. “The world has limited resources and the aim of the applied mathematics fields of mathematical optimisation and operations research is to create tools to help us to use these resources as efficiently as possible.”

(Thanks, Danesh.)

Reader Alex Freuman passed this along — a simple method of establishing any row in Pascal’s triangle, attributed to Edric Cane. To establish, for example, the seventh row (after the initial solitary 1), create a row of fractions in which the numerators are 7, 6, 5, 4, 3, 2, 1 and the denominators are 1, 2, 3, 4, 5, 6, 7:

Now multiply these in sequence, cumulatively, to get the numbers for the seventh row of the triangle:

These are the coefficients for

.

Cane writes, “It couldn’t be easier to remember or to implement.” Another example — row 10: