Here's a nice problem I found on math stack exchange (question 1182471).

ACE is a straight line and triangles ABC and CDE are equilateral. Prove that CP = CQ.

If you have a solution do put in the comments. I imagine there are a number of different solutions. I'll post mine this Monday coming.

Packing problems ask questions like 'What is the optimal way to pack copies of shape A in shape B?' For example we might be trying to pack equilateral triangles in a square, and the best known arrangement is this. 

We can think of the problem in two equivalent ways

• Choose a size for the square and make the triangles as big as possible
• Choose a size for the triangles and make the square as small as possible

The example above, which I drew in GeoGebra, has been known since 1996 and was discovered by Erich Friedman. It's pleasingly symmetric and you might expect symmetry is the norm. But it absolutely isn't. As increases every new number has its own idiosyncratic pattern, typically somewhat chaotic but with patches of local orderliness. For instance here is the best known arrangement for .

It was discover this month (May 2026) by Emerson Connelly and I found it on the legendary site Erich's Packing Center, maintained by Erich Friedman for the last 30 years. Erich's site has pages for dozens of different combinations. (Pentagons in Dominoes is one I rather like.)

I've known about this site for ages but when I revisited it yesterday I was astonished to find we have just entered a sort of Golden Age of Packing Problems. Of the 45 arrangements of equilateral triangles in squares found on the site, stretching back to 1996, 27 (60%) have been discovered in the last two months, April and May 2026! That's about one every other day.

And it's not just triangles in squares. Across the many different combinations of shapes hosted on Erich's site 46 have seen updated records in 2026, nearly all in that same two months. 

What accounts for this huge upsurge? I asked Copilot and it trawled the internet and proposed a combination of several factors

1. Faster and more sophisticated search algorithms
2. Dramatically greater computing power
3. Widespread parallel experimentation
4. Increasing use of AI/heuristics
5. Rapid online collaboration
6. A problem structure that rewards brute-force discovery

In some ways the problem resembles the search for bigger and bigger prime numbers. Packing problems have practical importance in manufacturing and transport (and prime numbers have practical importance in cryptography) but the search for new records and the application of such extraordinary human ingenuity and massive computing power is not motivated from practical considerations at all. I guess it comes to the spark of human curiosity and the attraction of a challenge.

And perhaps these records are a little like athletic ones; there is scope for a potentially endless series of small improvements and we always strive to reach them.

In 1999 Zoltán Füredi and John E. Wetzel, two covering problems meisters, found a triangle with a remarkable property.^[1]

It can cover^[2] each and every triangle of perimeter 2. It is the smallest region (not just the smallest triangle) that can do this and it is unique. I made a drawing of it using GeoGebra and fitted some sample triangles with perimeter 2 inside it

The length of is , , the length of is and the perimeter of is about .

[1] The smallest convex cover for triangles of perimeter two, Geometriae Dedicata, 2000

[2] To be precise, it can cover a congruent copy of any such triangle.

Can you match each Old English name to the right Dinosaur?

Remember Thorn þ is a 'th' sound, as in 'thin'.

See Comment for information on the Od English versions.

On forums I sign myself like this

Rich 🙂

... but the platform button often suggests a different emoji

🤑 

I was puzzled by this when I first saw it but eventually realised 'rich' is the link; the emoji is called 'money-mouth face', a symbol of wealth (and greed). 

'But rich is a different word', I thought to myself, 'and nothing to do with my name.'

Except... I was wrong. It has, it's fundamentally the same word, but has reached us by two different routes, and thus ended up as two words with different meanings. 

The name Richard means something like 'strong ruler', from Germanic words ric, 'ruler' and hard, 'strong'.  We see the ric element in other names, such as Eric, 'ever ruler' and Wulfric, 'ruler of wolves'. It also survives in the sense of a domain, as in bishopric.

The ultimate origin is the Proto Indo-European (PIE) stem *reg-, which had the sense of being direct and then from that of imposing order, ruling, reigning over, leading, and so on. 

The same PIE stem *reg- gave the Celtic languages rix, 'king', and modern Gaelic still has ri or righ (pronounced 'ree'). The ancient Gauls who Julius Caesar conquered spoke a Celtic language, Gaulish, and their version was rix. This was an early borrowing into Germanic and came down to us as rich.

There may have been influence from French riche (itself a borrowing from Frankish) but from early Middle English on its meaning was widened to magnificence (think 'richly dressed') and nowadays the predominant meaning is wealthy of course. 

And now we come to the freedom fighter. Although familiar with the history I literally had no idea of the etymological connection until I started writing this post

The most famous of the Gauls was Vercingetorix, something like 'great king over fighters', who led a revolt against the Roman rule imposed by Julius Caesar but in the end was forced to surrender. In the modern era he has become a symbol of French spirit and resistance to foreign invasion. Here's an iconic statue to him. See the Comments for more details.

One of the answers in my crossword today was KLEIN BOTTLE which set me wondering where Felix first described his famous mathematical object. It turns out the answer is (as far as we can tell) a set his handwritten lecture notes 1882 and after some searching I was able to find a facsimile of the relevant page at kleinbottle.com. 

But along the way I found that Klein never called it a bottle. He called it a Fläche, 'surface', related to English ply and pleat, but it got misread as Fläsche, 'bottle' related to English flask and flagon.

Why? The obvious answer is they couldn't read his handwriting, but I'm not convinced. The word Fläche appears four times on the relevant page and is written in quite a clear and consistent way.

So maybe someone simply thought 'bottle' worked better than 'surface' (it does) or maybe it was a flash of inspired humour. Who can tell? But below I have pasted an image of the page (courtesy of kleinbottle.com.)

With a steer from Copilot I was able pick out and highlight Fläche in four places and you can judge for yourself whether it is legible. And as a bonus you get to see Klein's sketches where he is showing that if you just join the end of a cylinder in the obvious way you get an anchor-ring (or torus), but if you join them so the end have opposite orientation you get a one-sided surface.

Here's a well known story whose connection will become apparent later.

One day the famous mathematician David Hilbert gave a lecture and finished by saying, 'And the rest of the proof is obvious!'

But then he had a sudden doubt and paused. After thinking for a while he said, 'Perhaps it is not obvious after all. I will think about it and let you know the answer in next week's lecture.'

When next week came the students waited eagerly. Hilbery arrived a little late and spoke to them. 'Students', he said, 'I have been thinking about the problem the whole week long. But it was only a few minutes ago, while I was on my way here, that it finally came to me.

It is obvious.'

A few days ago I ran across a problem I had not seen before. It was posted on the YouTube channel Mind Your Decisions as a hard problem and it goes like this (my description).

In the interior of an equilateral triangle we choose an arbitrary point and from it draw line segments to the triangle's corners. Suppose these have lengths , and and the angles they make around the chosen point are , and .

Now construct a triangle with sides , and . What will its angles be? 

I saw the answer must treat the angles symmetrically and it must take an angle sum of 360° (around a point) down to 180° (in a triangle), and so formed the conjecture that the angles of the new triangle are simply , and .

But could I prove it?

After several days I found a proof but I wasn't happy with it. It was too long. Besides, I thought that because the result was so simple the proof should be obvious. So I went to bed and in the night I had a sort of dream that showed me the solution. And it is obvious. Here is a look-and-see proof using a tiling pattern.

The tiling consists of multiple copies of our equilateral triangle with an interior point. The white triangle are copies of the blue ones rotated 60°. All the interior points are in the same relative position and as well as being joined to the corners of the triangle they lie in they have been joined to the interior points of the three adjacent blue or white triangles.

The lattice this creates contains multiple copies of what I shall refer to as the triangle of interest, ans of equilateral triangles of three different side lengths, equal to the three sides of the triangle of interest. There are many noteworthy things we can find out about the lattice but for our proof we just need to focus on the triangle I have emphasized.

You can see three copies of the triangle of interest meet at the interior point, with each of its three different angles being represented exactly once. Moreover each of the angles we previously labelled  , and the line segments to the corners make at the interior point consist of one angle from the triangle of interest and one 60° angle from an adjacent equilateral triangle. It follows that the angles of the triangle of interest are , and , as conjectured.

Another photo from my brother. A Hummingbird Hawk Moth in his garden. These insects are notoriously hard to photograph because they flit quickly from one blossom to the next and it's hard to anticipate their next move.

If you look carefully you can see it\s using its extended  proboscis to drink nectar from the flower.

Amazingly these insects are summer visitors who migrate here from Southern Europe. They seem to be becoming commoner in the UK and it's thought they might be starting to become resident here as a result of climate change.

I saw this problem on Mind Your Decisions.

At left is an equilateral triangle. From an arbitrary point in its interior we draw line segments to its vertices, making angles , and as shown. If now we construct a second triangle (right) whose sides are equal in length to these three line segments, as indicated by the tick marks—What will the angles of the new triangle be?

I have never seen this before and have not viewed the solution. But I have worked out what the answer must be, just not proved it yet. And it's a truly beautiful result.

I wonder if it can be generalised to non-equalateral triangle? Or to a square?

I thank the marvellous Cut-The-Knot problem collection for this one.

Given a shape consisting of a rectangle with a rectangular hole, as shown, divide it into two parts equal in area with a single straight cut.

Solution tomorrow.

Here's a question I've seen here and there in various forms. This is my version

Double me, you get a square. Triple me you get a cube. I am the smallest such. What number am I?

Sometimes it's captions '90% of people can't solve this' or similar. But it's actually not too hard.

If the number we seek is then we want ('double me') to be a square and this can be achieved by making twice a square, say . Then a square as required.

But we also want ('triple me') to be a cube and the smallest value of that makes this work is , and .

And sure enough, and , a square and a cube exactly as we want.

By why stop there? Can the idea be extended? We can't extend the pattern to (I'll put a proof in the Comments tomorrow) but we can make it work with . To see how to do this let's look at the prime factors of our previous example, .

When we multiply this by we get and both exponents are even so this is a square.

When we multiply it by we get and both exponents are multiples of so this is a cube.

Using this idea but now with , and (after a fair bit of working) we find the exponents work in the way we want if we take

You can see that if we multiply this by all the exponents will be divisible by , if by they all divisible by , and if by divisible by , just as we want. This is the smallest number that meets our goal but doesn't look small, here it is all 31 digits, a big jump from the 2 digits of !

6810125783203125000000000000000

We can continue in this way as long as we like: adding gives 233 digits

150462810922326152710290228433686961530697356776074449373600141938371053848189980134027578261857302770024765419887333164323078738017254430529707573248000000000000000000000000000000000000000000000000000000000000000000000000000000000000

and the numbers just continue growing in a super-exponential way. I suppose there must something we could say about the long term behaviour but it's beyond my technical capabilities.

Still I might be able to write up a description and get it into the Online Encyclopedia of Integer Sequences (OEIS). I'll give it a try.