# Review: Mathematical T-shirt

I was recently lucky enough to pick up a free mathematical T-shirt from the University of Essex. Following the success earlier this year of my mathematical socks review, they will surely be happy to see their T-shirt under the spotlight this time.

## Mathematical content

#### 1. Calculator

A strong first image: pi is displayed to 11 decimal places, and correctly rounded at the end. But given there is no pi button, someone has had to type this in themselves. Is this a worthy use of time? Also note the lack of a square root button, a function common on even the weakest calculator models. Solid memory functionality as well.

However, the elephant in the room is the large number of digits on the screen compared to the limited button set. Of course, standard calculators only come with eight digits and this is hard to ignore.

Maths grade: B. Good idea but poorly executed.

#### 2. Equation

$x = -2$.

Maths grade: C. Possibly challenging for a Year 7 student.

#### 3. Delta del-squared f(x)

Finally some harder mathematics! The Laplacian can be represented by either $\Delta$ or $\nabla^2$ and the choice to do both here is a little unorthodox. Nonetheless, this double Laplacian
$\Delta\nabla^2 f(\boldsymbol{x}) = \nabla^4 f(\boldsymbol{x}),$
set equal to zero, can be found haunting the work of many an applied maths PhD student.

Naturally the $x$ really ought to be a vector $\boldsymbol{x}$, and the f is really too big, but otherwise…

Maths grade: A. The conical nature of the $\nabla$ suggests cylindrical coordinates which makes this a good challenge.

#### 4. Abacus

Do you know how to use an abacus? Well that’s OK, neither do they. On the plus side, there are ten beads per line which is pretty standard. But beads in the middle of the lines?! Lunacy! Perhaps even more damning here is the lost potential for a good 5318008 joke.

Maths grade: D. Even the ancients would be stumped.

#### 5. Graph and compasses

Perhaps this pair of compasses has been used to draw this graph. Actually, this is unlikely because the legs of the compass are different lengths, which you can tell below when the red, left leg line is shorter than the green, right leg line. One would have considerable trouble trying to use this apparatus.

Looking at the graph, traditionally the numbers, on the y-axis, say, line up with the little markings, but here the approach is a little more relaxed. What this graph is of is a little hard to say. The axes are probably irrelevant. It looks a bit like the first two terms of a Fourier transform of something, but that first peak seems a little too pointy to be a smooth function.

Maths grade: C. Perhaps the compasses and graph are not related at all.

#### 6. Ruler

It is a matter of some debate whether the natural numbers, $\mathbb{N}$, includes 0 or not. What is not up for debate is that rulers really ought to start with 0. Also can you name a unit which is commonly split into ninths? (Count the little lines…)

#### 7. Rocks

Jesus Christ, Marie, they’re minerals.

## Fashion

Grey T-shirts have never gone out of style. Pair with some colourful shorts for a well-balanced look.

## Fit

Only comes in one size (large), which will not please at least half the market, and is not particularly flattering if you want to show off your beach body. That said, this bagginess allows for the ‘mathematics’ to be more easily read.

## Utility

Lovely horrible Camden High Street. Image: public domain

Surprisingly light, making it ideal for hot weather. I took the T-shirt out for a walk up and down the fashionable Camden High Street. It was an uncomfortable journey, mostly because this area is always full of tourists who think that Camden is as trendy as it was in the nineties, vying with large amounts of traffic trying to leave London by the north. However, the inexpensive, thin fabric kept me cool and well aerated.

## Overall review

Light and inoffensive on first view. But if you are going to take it out for a walk, expect ridicule for its poor mathematical content, so probably best for light housework during the summer months. If you’re looking for a more mathematical T-shirt, you can order one from Chalkdust.

# Winning the Chalkdust coin game

Go is a two-player strategy game. Players score points for surrounding territory and capturing opponents’ pieces. In 2016, Google challenged the world’s top Go player, Lee Sedol, to a five-game match against their AlphaGo program and won 4-1. The program was successful because it learned to play Go through a machine learning algorithm (specifically deep learning) trained on 30 million moves from games played by human experts (you can read more about AlphaGo here).

Write down a quadratic—any quadratic you like, but let’s say it should have integer coefficients between 0 and 20. What is the probability that it factorises?

What I really mean is will it factorise ‘over the integers’. So
$x^2 + 5x + 6 = (x+2)(x+3)$is in, but
$x^2 + 2x + 2 = (x + [1-\mathrm{i}]) (x + [1 + \mathrm{i}]) \quad \text{and} \quad x^2 – 2 = (x-\sqrt{2})(x+\sqrt{2})$are out.

To make it simpler, we will look for quadratics of the form
$x^2 + bx + c$where $b$ and $c$ are both positive. Try extending it to negative coefficients yourself afterwards!

Let’s plot a graph of $c$ against $b$, and colour in the values where $x^2+bx+c$ factorises. We’re going to colour these in with a 1×1 box where the bottom-left corner is at the relevant coordinate.

Quadratics of the form $x^2 + bx + c$ which factorise, for $b,c \leq 20$.

# Crossnumbers and cross products

Once upon a time, there was a Chalkdust crossnumber clue:

27D The number of straight lines that go through at least two points of a 10 × 10 grid of points. (4)

Obviously, I could have looked that up on the OEIS, but that is not how I roll with the crossnumber. That’s tantamount to letting Humbug Scroggs win.

Instead, I wrote some code to do it. And in the process, stumbled across something a bit special. Or, as my MathsJam talk on the matter had it, a bit… Specials.

## You’ve done too much, too much C1

What I needed was a way to give each line a unique reference, so I could avoid double-counting them. Since my previous student had asked me why we would ever write a line in the form $ax + by + c = 0$ (answer: beats me, but it’s as good as any other), that was on my mind: what if I simply indexed each line as a vector $(a,b,c)$?

That’s good, but it’s not quite good enough: any multiple of $(a,b,c)$ gives the same straight line, but there are several possible fixes for that. The simplest fix is to rescale it as a unit vector: this is what I did in my code. (Alternatives include scaling the vector to make the $z$-component 1, which requires special handling if $c=0$ and who can be bothered with that, at least at this stage? and—because we’re dealing with integers here, just making sure $a$, $b$ and $c$ are coprime. Even that runs into sign problems if you’re as lazy as I am, though.)

Once I’d merrily filled in 4492 in the grid, instead of getting on with the important business of doing the rest of the clues and cursing at Scroggs, something caught in my mind: if lines can be written as vectors… where does that lead?

## A message to you, 2D

My first question was: given two lines, $ax + by + c = 0$ and $px + qy + r = 0$, can I use the vectors $(a,b,c)$ and $(p,q,r)$ to find where the lines cross?

In my talk, I employed the well-known proof by MathsJam technique: "obviously it’s valid, or else I wouldn’t be talking about it." However, I know of no such result for proof by Chalkdust, and I can hardly ask you to take me at my word when I say "yes, you can: to find out where the lines cross, you find the cross product (and scale the result so the $z$-coordinate is 1)."

Can’t I at least leave this as an exercise for the re… no? Sheesh, tough editors. Fine.

Doing the cross product,$\begin{pmatrix}a \\ b \\ c\end{pmatrix} \times \begin{pmatrix} p \\ q \\ r\end{pmatrix} = \begin{pmatrix} br-cq \\ cp-ar \\ aq-bp\end{pmatrix},$and scaling it down gives $\left(\frac{br-cq}{aq-bp}, \frac{cp-ar}{aq-bp}, 1\right).$

Multiplying the equation of the first line by $p$ and the equation of the second by $a$ gives $apx + bpy + cp = 0$ and $apx +qay + ar = 0$, which leads to $bpy + cp = qay + ar$ and $(aq-bp)y = cp—ar$, so $y = \frac{cp-ar}{aq-bp}$. A similar process gives $x = \frac{br-cq}{aq-bp}$, as required. $\blacksquare$.

## It’s working for the rat race

I know I’ve proved it, but it’s worth showing that it works as well. Suppose we’ve got the following question, involving two rats, Zeke and Monty:

Using a graph to find where Zeke catches Monty.

• Zeke scampers at 10m/s
• Monty scampers at 8m/s
• Zeke gives Monty a 10m head start
• How long does it take Zeke to catch up?

We can set that up as a pair of equations: $x-10t=0$ and $x-8t-10=0$, and convert them to the vectors $(1,-10,0)$ and $(1,-8,-10)$. Their vector product is $(100,10,2)$, which scales down to $(50,5,1)$; after 5 seconds, Zeke catches up with Monty, and the fact that this happens 50m from where Zeke starts is a freebie.

## The line through C & A

C&A, as seen in the UK pre 2001. Image: Stuart Caie, CC BY 2.0.

We’re only starting to scratch the surface, though. If $(a,b,c)$ corresponds to a line and $(x,y,1)$ corresponds to a point, what happens if you take the cross product of the vectors corresponding to two points?

Surprise! You get a vector corresponding to the line connecting them. For example, if point $A$ is at (3,4) and point $C$ at (5,1), we’d ordinarily figure out a gradient and plug it into a formula leading us to $3x+2y-17=0$.

However, if we take the cross product of $(3,4,1)$ and $(5,1,1)$, we get $(3,2,-17)$, which corresponds—as one would hope—to the same line.

## Abstract jungle

So, what’s going on here?

One way of looking at it is to consider what’s happening in three dimensions—and in particular, where certain lines and planes intersect the $z=1$ plane.

We’ve glossed over, with a great deal of hand-waving, the idea that we can simply take the cross product of the vectors corresponding to two lines and scale the result to have $z=1$. Let’s firm that up a little bit: a point in our system corresponds to a line through the origin; it is represented in 2D coordinates by the $x$ and $y$ values where the line crosses the $z=1$ plane.

A line, meanwhile, is represented by a plane in this system. Its 2D representation is the intersection between the plane and $z=1$.

Now things start to drop out: any two points in the $z=1$ plane lie on a unique plane that also contains the origin. The normal vector of this plane is perpendicular to the vector from the origin to either of the points, so their cross-product will immediately give the normal of the plane, which corresponds to the line in the $z=1$ connecting them.

Similarly, any line in the $z=1$ plane lies on a unique plane through the origin; the line of intersection of two such plane is a line through the origin perpendicular to both normals, which again comes down to the cross product. Where that intersection line crosses the $z=1$ plane is the point of intersection of the original lines.

## The dawning of a new error

More astute readers will have spotted a hole in this: what happens if you take the cross product of two parallel lines? Something interesting, that’s what.

Take, for example, the parallel lines corresponding to $(3,2,-17)$ and $(3,2,1)$. Their cross product is $(36,-54,0)$—and it’s clearly not possible to scale that to meet the $z=1$ plane. If you’re a strict Euclidean, you dust your hands and say "They don’t intersect!"

If you’re more open-minded, you can say "there’s nothing wrong with the line through the origin and $(36,-54,0)$—why can’t it correspond to a point?" Answer: it can! We can associate each direction with a point at infinity—which opens up yet another tool for us to exploit.

For instance, if you know you have (in regular 2D) a line through (6,7) with a gradient of $-\frac{2}{3}$, you can find its equation by taking the cross product of $(6,7,1)$ and $(1,-\frac{2}{3},0)$ (the point at infinity corresponding to that gradient), giving you $(\frac{2}{3},1,-11)$, or $2x+3y-33=0$.

The points at infinity, of course, have lines corresponding to them: they turn out to be exactly the lines through the origin. Indeed, there’s even a line that goes through all of the points at infinity; it corresponds to the vector $(0,0,1)$.

## It’s up to you

Readers, I have to break it to you: we have been dabbling in projective geometry—and in particular, in the principle of duality, neither of which I’ll go into in detail here (you have a Google, correct? Excellent.)

There are all manner of projective peculiarities you can explore—you might like to consider a point/line as a sort of spinning-top shape on a unit sphere, or to consider higher dimensions if you want to take it one step beyond—but that way lies madness.

If this post has inspired you to attempt the Chalkdust crossnumber, you can find the latest one here. You can win a collection of prizes from Maths Gear if you send in a correct answer before 22 July 2017.

# Top 10 emoji for use in mathematics

Maths loves symbols. Everyone loves emoji. It’s 2017 and time we brought the two together. To get you started, here are our top ten emoji for use in mathematics!

## 10.

📐

Don’t leave home without one: it’s the nifty 45° set square. What better reminder is there that
$$\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \qquad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \qquad \tan\left(\frac{\pi}{4}\right) = 1$$ 📐

## 9.

🏹

Perfect for popping over a letter to make it a vector, it’s the bow and arrow:

## Papercupter competition

Finding the best design for a papercupter is not easy as there are many designs, for instance, with longer or shorter flaps, curvier or straight, only a few flaps or as many as you can get from your paper cup. The number of different papercupters is infinite and so finding the one that spins the most or the one that stays in the air for the longest time is impossible.

There are some papercupter designs which clearly won’t work, for instance, one with flaps so small that it does not make the papercupter spin as it falls down, or one with so many flaps that they become thin strips of paper with no air resistance.

Last week we were able to play a papercupter competition (during the 2017 De Morgan Dinner) and more than 50 different designs competed against each other. The papercupters which made it to the final round had, in general, only a small number of long flaps.

Rosalba, winner of the 2017 Papercupter competition and her design for the best papercupter.