A friend of my daughter's snapped this Brocken Spectre near Abergavenny, in the Bannau Brycheiniog (Brecon Beacons).

When the sun shines from behind an observer onto a bank of mist, the observer's shadow may be cast on the mist and depending on the elevation of the sun the shadow may look like a giant figure, giving rise to the legend of the 'Brocken Spectre', named after the Brocken, a peak in the Harz mountains of northern Germany, where the condition are often favourable for seeing the phenomenon.

In this example the shadow appears tiny but we can clearly see the other feature of a Brocken Spectre: the observer's head is surrounded by a 'glory' – a ring of light like a halo. The ring has colours, in the same order as a rainbow, although distributed somewhat differently, but a glory is much smaller that a rainbow and caused by different mechanisms.

The water droplets in the mist are much smaller than the raindrops responsible for a rainbow, by a factor of about 100. The incoming light is back scattered and the colours are caused by the light waves interfering. But how does this back scattering happen?

Theory 1

The water droplets refract the light back. Unfortunately the water does not bend the light enough for this to be the explanation.

Theory 2

The physics of electromagnetic waves predicts the light can become a wave that travels round the surface of the raindrop and then emerges as a back scattered wave. This seems to happen but does not account for the amount of light actually back scattered.

Theory 3

More recent calculations have shown that energy can tunnel into a water droplet from light waves close by and after travelling round inside the droplet emerge as back scattered light, and that this effect can explain the observed brightness of glories. This is all explained in more detail in the excellent 2012 Scientific American article The Science of the Glory by H. Moysés Nussenzveig.

In this section of the Bayeux Tapestry, which shows Harold landing in Ponthieu, we can see the ship is steered by an oar, as was common at the time. Steering oars did not get replaced by rudders until a couple of centuries later. The steering oar was on the right-side of the ship facing forward,  so that side of the ship was called in Old English steorbord, 'steer side', bord being the term used for the ship's side. I guess if you were a left-handed steerer you just had to live with it.

This has direct cognates in Germanic languages e.g. German Steuerbord, Old Norse stýriborð, and because of the influence these peoples had as seafarers the word was borrowed into Romabce languages, so we have e.g. French tribord.

Tracing the Proto-Germanic *steuro further back seems inconclusive, with more than one explanation appearing in the literature.

Turning now to larboard, the side opposite starboard was at first 'backboard', bæcbord in Old English, but by Middle English we have ladde-borde, which seems to mean the side the ship was loaded or unloaded; think of the modern word laden. Naturally this would be the opposite side from the steering oar.

However having starboard and larboard, although a pleasing rhyming pair, is a recipe for confusion give the similar sounds, and so larboard was been displaced by port, with similar connotations of being the side that comes alongside the quay, so now larboard is rare and archaic, and 'port and starboard' is the normal expression.

Picture credit: Wikimedia, Section of the Bayeux Embroidery showing Harold arriving in Ponthieu, Public Domain.

This iconic picture shows a minister gliding serenely across the ice, but how is thos possib;e. Why is ice slippery enouhgh?

At school we were taught the reason is the pressure of the skates melts the ice and the resulting water act as a lubricant. This is a wonderful explanation, put forward by James Thomson in 1849 based on the fact that pressure lowers the melting point of ice, and it seemed so satisfying that everybody took it for granted that it was correct and it was long taken as received wisdom.

But it can't be quite right; even if the pressure is sufficient to melt the ice when we are talking about ice skating, it is nothing like enough to explain how skis glide along (or why I fall over on icy pavements for that matter!)

So rival theories have been advanced. One is that friction could melt the ice. It's possible this could have an effect but we'd expect melt water from friction to trail behind the sliding object and so not provide much lubrication going forward.

A third idea, based on an observation by Michael Faraday, is premelting. According to this theory the water molecules in the interior of the ice are tightly locked into a crystal lattice that restrict their movement but ones on or close to the surface are bonded to fewer close neighbours, allowing them more mobility, so they can act rather like water. Computer simulations support this as a plausible reason why the surface of ice should act as though it is wet.

It's conceivable all three of the mechanisms outlined above play a part. But very recently a fourth idea has emerged: the skate or ski (or the sole of my shoe) disrupts the crystal lattice of the ice and creates a layer of disorder molecule that are not fixed in a lattice and yet not quite liquid water, but which can act a lubricant. So far this has only been investigated in computer simulations and at present other researchers are not sure how it should be interpreted. 

So we have not finally answered the question 'Why is ice slippery?' but this new research does represent an important advance, and we have edged closer to understanding exactly what is happening.

To read more about this see Quanta Magazine here.

For a good video about the new research see Anton Petrov, 'For 200 Years We Were Wrong About Why Water Ice Is Slippery',

Picture by Sir Henry Raeburn - Reverend Robert Walker (1755 - 1808) Skating on Duddingston Loch. Public Domain.

The American comedian George Carlin once did a routine in which he said a new integer, called bleen, had been found between five and six.

The Nobel Prize in mathematics has been awarded to a California professor who has discovered a new number. The number is “bleen,” which he claims belongs between six and seven. ^[1]

This idea was taken up in a New York Times puzzle column. There they wrote

Suppose that a “new integer” is “discovered” between 5 and 6 called bleen, written B. 

This is a little misleading and doesn't quite make sense; at bottom, we are not being asked to imagine any new integers have been discovered, but that we are going to represent the same old integers in a different number base, specifically eleven, so that for example whereas 42 normally stands for 4 tens and two units, if taken to be in the new system it would stand for 4 elevens and two units.

Most people nowadays will have met different bases in school and hexadecimal — base sixteen — is widely used in computing. In that base we write digits — we need sixteen of them now, so we co-opt some letters — .

And the Dozenal Society of Great Britain has long campaigned for us to use a system of counting in dozens, on the grounds that a dozen has more divisors than ten. Dozenal uses the digits .

Notice that in both these real-life examples, we add symbols representing the extra digits needed at the end of the list, and that would be the normal convention. 

The genius of Bleen is that we break this convention and insert the extra digit, B for Bleen, between 5 and 6. So now we have . This quixotic decision makes the new system quite challenging.

Frank Potter, who submitted the puzzle, then invited us to work these out in the new system

B + 3
B + B
10 x B
B²

Solutions at bottom of page.

And here are two harder ones.

B³
42 - 6

[1] CARLIN ON CAMPUS (1984) – Transcript

The OED has this to say about the etymology of awkward

< awk adj. + ‑ward suffix, i.e. ‘in an awk direction’; compare forward, backward. 

Northward, southward, upward, downward, etc., are all familiar directions, but awk seems to take us somewhere different from any of those. perhaps into another dimension.

It seems to be a borrowing from Old Norse afug, 'turned the wrong way, back foremost', similar to the colloquial expression 'arse about face', and in Icelandic for example the related word öfugur means 'inside out'.

So awkward at first meant 'in a contrary direction' but has shifted semantically and now usually means clumsy or social inept, embarrassed ('An awkward silence fell'), uncooperative etc.; the original meaning is obsolete. 

So where did afug come from? Etymonline suggests it is from a PIE root *apo-, 'from, away', surviving in Modern Greek with the sense 'from', e.g. Apo pou isse?, 'Where are you from?' (a useful conversational opener). It might also be the origin of Latin ab-, seen in words like abscond, absence, abstain, etc.

Awk once had derived words; awky, awkly and awkness, al sadly obsolete now, but easy enough to understand.

There's a decent amount of linguistic evidence that swan and song both come from an ancient root *swen-, '(make a) sound'. In Old English the bird was called swan, from a Germanic root  *swanaz, 'singing bird', related to Old English geswin, 'song' and swinsain, 'to make a musical sound'.

This root *swen- is probably also the origin of sound, Latin sonus, Sanskrit svan, Irish seinn, 'play music'.

But, I hear you cry, the kind of swan we are most familiar with is the Mute Swan, which as its name implies doesn't make a very conspicuous or musical sound. How can calling it the singing bird make sense?

One theory runs like this. When the swan was dubbed 'singing bird' the people that named it were more familiar with a different kind of swan, such a Whooper Swan, which does produce a very noticeable sound.

Then as people and the Indo-European languages migrated south they met swans that didn't sing. But people were still vaguely aware that swan was something to do with singing, which produced a cognitive dissonance, which was resolved by inventing a myth.

The swan is silent all its life but then at the approach of death sings for the first and last time, its Swansong.

And so a meme was born.

Sunrise 24 September 2026, captured by my brother.

Here is a Giant Redwood (AKA as a Giant Sequoia) in a village near where. I often pass it and always stop to admire it.  I couldn't get a good angle to take the photo so a bit of another tree is visible at upper left but I hope you can see the Redwood is very impressive.

In their native range, the Sierra Nevada in California, the species is threatened: a combination of historical logging, suboptimal fire policies that led to build-up of combustible material causing the fires that finally occurred to be far more damaging; drought which made trees susceptible to bark beetles; and climate change has reduced the number of trees to around 80k.

However in Victorian times seeds were brought to Europe and in the UK particularly owners of large gardens grew them enthusiastically. And the trees liked it here and have thrived, so now there are about 5k notable specimens and about 500k younger redwood trees. A small minority are Coast Redwoods and Dawn Redwood but the bulk are Giant Redwoods. 

Of course most of these are young trees, as Redwoods go. The oldest is a mere 150 years old, compared with the 3,000 years some wild trees have reached. They grow fast here though; the tallest documented is 55 m, compared with around 95 m for wild trees, although the oldest wild specimens have a relatively greater volume.

But paradoxically Giant Redwoods may be commoner in the UK than they are in California (it's not an open and shut case though, because the 80k in California are large trees and we don't have a count of how many smaller trees have survived recent devastating fires and beetle attack). It's good that these magnificent trees have been able to grow so successful on another continent from the one where they evolved and let's hope some of the ones growing here today — perhaps the one picured — will live for 3,000 years.

I found ths on 'Mind Your Decisions' who got it from 'CueMath' and I don't know where 'CueMath' got it. I hadn't seen it before but as is often the case I'd just been missing out, because a search shows it's in lots of places on the web. Anyway, here is my solution, I haven't looked at other answers because I wanted to solve it the way Euclid might have. It's a really nice problem.

We are given a square in which is inscribed a quarter-circle and inside that a semi-crcle tangent to the quarter-circle, as shown. Length PQ = 4. What is the ratio of the portion shaded light purple (the semi-circle) to the area shaded green (the area of the quarter-circle not also in the semi-circle)?

Scroll down for solution

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The Line of Centres Theorem If two circles are tangent to one another, the line passing through their centres passes the the point at which they are tangent. ^[1]

So we can draw in line PQR passing through these points.

Thales's theorem  If the hypotenuse of a right-angles triangle is the diameter of a circle, 

Since ST is a diameter of the semi-circle and P a right angle, P must lie on the circle through R, S and T. Therefore PQ is a radius of that circle and equal in length to QR which is also a radius. This makes PR, a radius of the quarter-circle, equal to 2QR, a radius of the semi-circle.

So radius of quarter-circle : radius of semi-circle = 2:1

Area of of quarter-circle : area of semi-circle = : = 1 : = 2 : 1.

So the quarter-circle as a whole is exactly twice the semi-circle, making the light purple and green areas the same. The length 4 is irrelevant, a pure red herring!

References

[1] Euclid, Elements, Book III, Proposition 11.

[2] Euclid, Elements, Book III, Proposition 31.

I found this problem on Cut The Knot, who got it from ^[1].

You are given seven distinct positive integers that sum to 100. Prove that some three of them must add up to at 50 or more.

I'll put my solution in the comments later this evening.

[1] Andreescu, T. and Răzvan, G. Mathematical Olympiad Challeges(Burkhäuser, 2004, p 60).

You may not have met this remarkable palindromic number before.

The Pigeonhole Principle says, of course, that if there are more pigeons than pigeonholes then we must be able to find a pigeonhole with more than one occupant.

The Cut The Problem ask for a proof (using the Pigeonhole Principle) that there is a power of whose last three digits are , which sounds quite surprising.

But let's imagine we have pigeonholes numbered . Calculate distinct power of , divide each by and find the remainder, then put that power in the pigeonhole with the same number as the remainder.

There is one more 'pigeons' than pigeonholes, so there must be a pigeonhole with two occupants, that it, two powers that leave the same remainder on division by .

Suppose these are and , being the smaller. Because they leave the same remainder when divided by , must be a multiple of .

can't divide a power of , so it must divide . This means when worked out ends in three zeros , which in turn means ends in .

We can run a computer search quite easily and we find that in fact

fits the bill.

This result can be generalised of course and we can prove that for example there must a power of that ends in a trillion zeros followed by , although given that even the zeros would take up nearly GB the browser is too small to display it.