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The London Eye, by Image Creator (2024)

https://commons.wikimedia.org/wiki/File:Crouton_skibidis_next_to_an_ice_cube.jpg

'Crouton' and 'crystal' both ultimately stem from an ancient root that was something like kreus and meant 'freezing over' or 'crusting'. Crystals were thought in Ancient Greece (not unreasonably) to resemble frost, kryos, and from that was derived krystallos.

Related words are crust, crustacean, a whole series of modern words coined from cryo-, and, surprisingly, custard.

“My job is studying social insects”, said Tom observantly.

From a French word meaning a bag, from Latin and before that from Gaulish. It derives ultimately from a PIE root meaning something like to swell. Many other words have the same root; for example, ball, balloon, bale, bulge and surprisingly, ballot. Apparently, balls were once used for casting votes.

I thought I’d try shoplifting. But they were too heavy for me.

I found this striking fungus in my garden today. I believe it is Armillaria tabescens, the ringless honey fungus, but it's hard to be certain. Rather beautiful.

In one of the Inspector Morse books, Lewis repeats to himself the mnemonic "Five 'esses' in 'possess'". I can't trace the quote, but I think Lewis must have needed to use the word in a report and felt nervous in case he spelt it wrongly. (The first book in the series was published in 1975, pre-spellcheckers). 

I thought I would test an AI on this, so I asked Copilot. It said four esses. So I tried to correct it and it said rather patronisingly:

• • P-O-S-S-E-S-S-E-S

1. 1. P-O-S-S-E-S-S-E-S
   2. P-O-S-S-E-S-S-E-S
   3. P-O-S-S-E-S-S-E-S
   4. P-O-S-S-E-S-S-E-S

• • Assesses
  • Assassins
  • Assessors
  • Possesses
  • Sassiness
  • ...

Something that has fascinate me for many years is the fact that, if a disordered systems is larges enough, it will tend to contain pockets of unavoidable regularity. Order will emerge from disorder.

For example, imagine we colour every point of the 2D plane in one of two colours, red or blue say. Then there will inevitably be equilateral triangles all three of whose vertices are the same colour; either all red or all blue. Here's how we can prove this.

Imagine we set out to colour the plane in a way that avoids a monochromatic equilateral triangle. We shall find that this is impossible. Consider the following steps in our attempt.

We start with two points which are coloured red (Step 1).This must be possible; if not, there would no more than one red point in the entire plane, so either all equilateral triangles would be blue monochromatic, or all the ones that avoided the single red point would be. Either way our attempt to avoid monochromatic equilateral triangles would have failed spectacularly.

At Step 2 both the ringed points will have to be coloured blue, otherwise one or both the triangles marked will be red monochromatic.

At Step 3 the ringed point will have to be coloured red, otherwise the triangle marked will be blue monochromatic.

At Step 4 the ringed point will have to be coloured blue, otherwise the triangle marked will be red monochromatic.

At Step 5 a monochromatic triangle can no longer be avoided; whether we choose red or blue for the uncoloured point a monochromatic triangle with be created.

There really is such a word, it's in the the Oxford English Dictionary.

First recorded in 1881, it is the only words I have come across in the OED that lists a citation from Twitter:

Do you know a Hrothgar? I bet you do…

Hrothgar was a semi-mythical Danish king who appears in several places in Old English and Old Norse literature. His name is dithematic – it consists of two elements hroth + gar, meaning ‘famous’ and ‘spear’.

Many Germanic names follow this dithematic pattern and there are hundreds, possibly thousands of examples attested. Many survive in modified form to the present day and are very common and familiar given names, for instance

William - will + helm = ‘wish helmet’

Mathilda -  maht + hild = ‘mighty battle’

Rosamund - hros + mund = ‘horse guardian

Robert -  hroth +  beraht = ‘famous bright’

Several of these elements are easy to recognise – will, maht, hros, beraht - and helm survives as a poetical word for helmet. mund was still found in Middle English but is now obsolete, and hild seems to have disappeared earlier.

As for Hrothgar, it has become Roger, a fact I only realised a couple of days ago. Hroth is an element in several other names: Wikipedia quotes Rudolph, Roderick, and Roland; but seems extinct now except in names.

But gar = ‘pike’ i.e.a long, pointed spear, hangs on. This is a garfish, courtesy of Wikipedia.

My daughter sent me a list of "groaners" she found somewhere on the web. 

Here are three I rather liked:

I felt it my duty to contribute to this genre, so here are some I came up with:

There is an Only Fools and Horses episode in which the road sweeper Trigger announces that he has won an award for keeping the same broom for 20 years, and adds that the broom has had 17 new heads and 14 new handles.

I was thinking about this and remembered that all the atoms in our bodies get replaced on a surprisingly small time scale. Reading up on it, I found that, according to Time magazine, the turnover is about 98% per year. To put it differently, only about 1 in 50 of the atoms in your body will still be there in a year's time.

In 20 years that would be 1 in 100,000 billion billion billion, which is much more than the estimated number of atoms in a human body. So more or less all the matter that went to make up your body will have been replaced, and you will resemble Trigger's broom in this respect.

I've always been intrigued by this paradox; if all the constituent parts of something are replace, is its identity still the same? It's an important philosophical question, but until now I didn't realise it goes by the name of Theseus's Paradox. 

Plutarch wrote

"The ship wherein Theseus and the youth of Athens returned from Crete [...] was preserved by the Athenians [...], for they took away the old planks as they decayed, putting in new and strong timber in their places."

I found this geometry problem posted a few days ago maths YouTuber Michael Penn. It was originally published in 2002 in the Mathematical Gazette. It caught my attention because Penn's solution and the one originally published seemed to use trigonometry and calculation of distances involving square roots, and I felt it should be possible to use a purely geometric approach. 

After a lot of thought, by stepping outside the problem I think I've managed to solve it. Below left is the problem and at right the idea that set me on the right road. Suppose I drew in a square and an equilateral triangle to provide a context for the 45° and 30° angles, would that provide any insight?

This was nearly right, but then after more thought I saw  a better approach was to use a square and an equilateral triangle of equal side length, like this.

We can successively compute the angles marked in the diagram is three stages.

1 (green angles marked by single arcs). At C we an internal angle of the equilateral triangle, which is 60 degrees, and two angles which are each half of a right angle, so they are 45 degrees. The line BG passes through H, the midpoint of CF and from the symmetry of the equilateral triangle we see that it makes a right angle with CF, and that the angle CBF is 30 degrees.

2 (blue angles marked by double arcs). The angles at C must sum to 180 degrees, and so angle DCF must be 30 degrees.

CD = CF, because the square and the equilateral triangle have equal side length, and so triangle CDF is isosceles. Hence the two angles marked at D and C are equal and must both be 75 degrees because the angles of CDF must add up to 180 degrees.

3 (red angle marked triple arcs). From consideration of the angles in triangle CGF the angle marked at G must be 30 degrees.

Moreover, triangle CGE has two angles of 75 degrees and so it is isosceles. Because the line through BH bisects the base of this triangle at right angles, the triangle is symmetrical about the line, and the line must pass through point G.

And now look at the kite AGDC. Can you see? - it is symmetrical about the diagonal CG, and therefore θ, the angle we are asked to find, is equal to angle CGD = 30 degrees. Problem solved.