I used to take a magazine called Games & Puzzles which ran for 1972 until 1981 and it was there (I believe) that I read about an amazing way to construct a sentence in which arbitrarily many verbs appeared consecutively.
I recalled this recently but I'd forgotten the construction and for a while wasn't very successful when I searched online, but I've found it now. It's called center‑embedding and I thought it would be interesting to apply it to the nursery rhyme "This Is The House That Jack Built". So I did that, which was quite tricky, and then asked Copilot if the result was a legitimate sentence. Copilot said it was but commented
This type of structure is famous in linguistics: it’s grammatical in principle, but humans struggle to process more than one or two levels of embedding. Your sentence pushes it to the extreme for rhetorical effect (and fun).
Here is my sentence, with 9 consecutive verbs. Get ready.
The rat that the cat that the dog that the cow with the crumpled horn that the maiden all forlorn that man all tattered and torn that the priest all shaven and shorn that the cock that crowed in the morn that the farmer sowing his corn kept woke married kissed milked tossed worried killed ate the malt.
On its own initiative Copilot (very helpfully I thought) drew this diagram to clarify the structure.
Edited by Richard Walker, Friday 2 January 2026 at 22:29
On Stack Overflow a Python beginner asked for help in writing a program to find a 4-digit number ABCD such when multiplied by 4 it reverses its digits, that is 4 x ABCD is DCBA. The kindly folks on Stack Overflow duly provided some help and the unique answer is 2178: 4 x 2178 is 8712.
This was new to me; although searching on "2178" threw up ~52,000,000 hits, this puzzle must have passed me by, until now.
Seeing the solution made me wonder if it could be generalised. There is quite a wide field of possible variations, but the first I looked was whether there is a find a 5-digit number that reversed itself when you multiply it by 4.
And there is: it's 21978. That looked familiar somehow and then I searched for a 6-digit counterpart. 219978 was the answer.
You can probably spot the pattern: we are just adding 9s in the middle. Will it work however many of them we insert? Yes it will; here's why.
Imagine we are doing a multiplication, 219.. lots of 9s...978 x 4.
You can see that for each 9 in the middle we shall have carried 3 and we do 4x9 = 36, plus the carry is 39, so we put down 9 and carry 3 again. This will continue as long as the digit being multiplied by 4 is a 9. When reach the 1 we will still have a carry of 3, and we do 4x1 = 4 plus the carry is 7, and then finally 4x2 = 8. So we can have as many 9s as we like sandwiched between 21 an 78 and multiplying by 4 will always reverse the number's digits, which is really cool.
Python deals with very long numbers really well, so just for fun
Edited by Richard Walker, Thursday 1 January 2026 at 16:27
This is something we can (very!) roughly imagine a speaker of the reconstructed Indo-European language saying at the start of a new year about six or seven thousand years ago.
kob is probably the ancient root of Old English hap, "good) fortune", and modern happy and happen (and mishap) derived from it. That k should have become h may seem odd but there are a range of words where we know for certain the same consonant shift happened, e.g. Latin centum (the c is hard) corresponds to hundred, canis to hound,cor to heart.
newo is found in closely related form in a whole raft of languages, from Hittite newa, through Sanskrit nava, Persian nau, Ancient Greek neos, Latin novus, German neu, Old Englis niewe.
So there you have it: Happy New Year!
yeris seen in Ancient Greek oros, Avestan yare, German Jahr, and even related to hour.
Everyone will understand the solution to this competition question
Monday 29 December 2025 at 23:26
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Edited by Richard Walker, Tuesday 30 December 2025 at 23:28
This question was B1 in the December 2025 Putnam competition, which is the top competition for math undergrads at US universities. It's nice because no specialised mathematical knowledge whatsoever is needed to grasp the question and the solution; both just involve simple ideas and the solution needs just a short chain of reasoning.
I'll give the question first and then clarify it a bit.
Suppose that each point in the plane is colored either red or green, subject to the following condition: For every three noncollinear points A,B,C of the same color, the center of the circle
I'm going to change the colour scheme to red and blue, which will work better for anyone red-green colourblind. Here are a couple of sketches to help me explain.
The question is asking us to suppose we have found a scheme that assigns to each point in the plane a colour, either red or blue, in a way that meets the conditions illustrated in the sketches above, i.e.
If three red points lie on a circle, the centre of that circle will always be red, we will never find three red points on a circle with a blue centre.
If three blue points lie on a circle, the centre of that circle will always be blue, we will never find three blue points on a circle with a red centre.
You might say don't any three points lie of a circle, but no, they might all lie on the the same straight line, so we want to rule that out.)
We are asked to prove that these conditions cannot be met unless every point in the plane is the same colour as every other point; every point is red or else every point is blue.
So here is my proof. The idea to imagine trying to colour the points of the plane, always keeping to the conditions bulleted above, using both both red ands blue. We start with two points, one red and one blue, and then try to extend the configuration, still keeping to the rules. We shall find that we must inevitably reach a position for which it is impossible to extend the colouring further without breaking the rules. So no such colouring can exist.
Consider the following diagram, where I have supposed we have two points of opposite colours. It make no difference which is which, so I have made A red and B blue.
From there we can infer colours for points C, D, E, F, G, in alphabetical order, by applying our rules, as follows. To avoid a whole sequence of diagrams each little different from the last I have coloured them in advance but you need to to imagine the colours of the points are not known until we determine each.
First observe that C and D must be different colours; for if both were red C, A and D would be three red points on a circle whos centre B is blue. If both were blue C, B and D would be three blue points on a circle whose centre A is red. Which of C and D is red which blue makes no material difference so we can make C red and D blue as shown.
Now the circle through C, B, D whose centre A is red has two blue points B and D on it already and another blue point on that circle would break our rules. This forces E and F to be red as shown.
Next the circle through C, A, D whose centre B is blue has two red points A and C on it already and another red point on that circle would break our rules. This forces G to be blue as shown.
Now consider the point H which lies on the intersection of a small circle whose centre C is red and a larger circle whose centre B is blue. What colour should H be? Well the small circle with a red centre already has two blue points on it, so a third blue point is ruled out. On the other hand the larger circle with a blue centre already has two red points on it, so a third red point is ruled out. So H cannot be blue and it cannot be red either and it is impossible to colour it in a way consistent with our rules.
So we have found that if any two points are different colours it is impossible to extend that colouring to the whole plane and this the only colouring that meets the conditions is if every point is coloured red or every point is coloured blue.
Musical time values—where did they get their names?
Saturday 27 December 2025 at 22:50
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Edited by Richard Walker, Sunday 28 December 2025 at 18:22
The longest musical note you're likely to meet nowadays is the breve.
In medieval music this was classed as a short note, hence its name, from Latin brevis, "short"; basically the same word as brief. The Latin word is probably, by a regular process of consonant change, derived from a PIE root *mréǵʰus, also meaning short; and surprisingly this may also be the origin of merry. Perhaps the association between something pleasant and something short is the same idea as that expressed by the well-known line
They are not long, the days of wine and roses [1]
In modern music the breve is pretty rare, appearing only occasionally; for example in Brahms's A German Requiem. The longest note in common use is the semibreve, whose meaning is self-explanatory.
Next we have the minim, one half of a semibreve, and called a mimin because it was the minimum, the shortest note. Minimum is from Latin minimus, smallest, thought to be from PIE *mei, "small".
Of course the minim is not the shortest by a long chalk Half a minim is a crochet
The name is taken, presumably in reference to its shape, from a French word crochet, "hook", the origin also of the words crocheting, from the hook used in that craft, and crochety, "irritable, cranky, having eccentric fancies", although I have no idea what the connection is here.
Half a crochet is a quaver. The only explanation I have found for the word's origin is that a quaver is a short note that might get repeated rapidly, so it would sound quavery (possible related to quake?).
But now it's plain sailing: half a quaver is a semiquaver (semi from Latin for half); half a semiquaver is a demisemiquaver (demi from French for half); half a demisemiquaver is a hemidemisemiquaver (hemi from Greek for half).
And it doesn't have to stop there; we can pile on more of the little "flags", as they are called, and repeat the semi-demi-hemi sequence, so this note, of a breve, is hemidemisemihemidemisemiquaver.
of course this is mainly theoretical; the slowest tempo we are likely to meet in reality would make a breve last about 8 seconds, so we'd have to play 128 notes per second to fit them all in, and a frequency that high would become a musical pitch in its own right.
[1] From Vitae Summa Brevis by Ernest Dowson, 1896
Edited by Richard Walker, Wednesday 24 December 2025 at 11:11
I'm trying to teach myself a bit of LaTeX, which if you don't know is a specialised language that allows a user to enter expressions as plain text and have them rendered as mathematical notation.
For a while I've felt not knowing LaTeX is a gap in my toolkit but the immediate prompt is that I'm writing accessibility guides for a couple of modules and I'm looking how someone who is blind or has low vision can read and write mathematical notation, and LaTeX is one route, especially for anyone planning to go on to study more mathematics. So as an exercise I decided to write a post about a neat mathematical trick I learned recently.
My Exercise
Suppose we're asked to solve the following quadratic equation by factorisation
This is not too hard. We just look for a pair of numbers that multiply to give and add to . These are and add to . So the equation factorises to
and the solutions are and .
But what if the coefficient of is greater than ? For example we might have
Now we have to consider all the factors of and of and figure out which combination is the correct one and it all gets a bit messy.
Here's a neat trick that makes this equation as easy to solve as the first example. We say (first coefficient times last), set the first coefficient to and form a new equation (but remember that , we will need it later!)
This is easy to factorise and we get , so the solutions are and . Of course these are not solutions of the original equation but now we bring that back into play, and divide the two solutions we have just found by it (makes a kind of sense you see, first we multiplied by , now we divide by it).
and
and, surprise, surprise, these are the solutions to the original equation!
Why does this work?
Multiplying the original equation by will give
or .
Now replace by and we get the second equation and the roots of this must be times those of the original equation.
In the UK we call them sweets, but in the US the same thing is called candy. Most UK speakers of English will recognise candy but are likely to feel it's an Americanism and even perhaps regard it as a relatively modern transatlantic coinage.
However something prompted me to look it up and it seems candy has a long history, has been borrowed from one language to another and across language multiple times, and has remained surprisingly stable for at least four thousand years. Although the OED is a bit reticent, Wiktionary and Etymology Online broadly agree its ancestry is probably as follows. The language groups are Indo-European IE, Semitic S and Dravidian D.
Modern English candy IE
↑
Middle English candy IE
↑
Old French candi IE
↑
Arabic qand S
↑
Persian qand IE
↑
Sanskrit khanda IE
↑
Proto-Dravidian *kantu D
There is some guesswork in the last step, since PD is a reconstructed language anyway, but Tamil, a modern Dravidian language common in India does have a word kantu, "candy", although of course this might be a back borrowing. But even so candy is a word with a long and distinguished pedigree.
Fun Fact: pedigree is from French pe-de-grue,"crane's foot", because in family trees a branching line of descent looks a bit like crane's foot.
Edited by Richard Walker, Friday 19 December 2025 at 23:58
Suppose we choose 10 digits 0-9 at random, one after another. For example, we might get
3, 0, 5, 2, 1, 5, 3, 1, 9, 2
How many distinct digits appear? In the example there were 6 –
0, 1, 2, 3, 5,
– but the answer could be anything from 1 (all the digits are the same), up to 10 (they are different).
Can we work out the expected number of distinct digits that appear; how many different digits show up on average?
Consider the probability that 0 does not appear. The probability of the first digit not being 0 is , because 9 out of 10 digits aren't 0. The probability that the second digit is not 0 is the same, and so on, up to the tenth digit.
The probability the none of the digits is 0, i.e. 0 does not appear, is therefore .
What then is the probability that 0 does appear. Well, this is easy. It either appears or it doesn't, and the two probabilities must therefore add up to 1. Hence the probability of 0 being present in our sequence is
, or about 0.651.
This is equivalent to saying that the expected number of times 0 will come up is 0.651. But there was nothing special about 0, the calculation would be the same for the other digits.
Now imagine we keep a score of what distinct digits appear, counting 1 if a digit appears and 0 if it doesn't. The probability that a digit appears is 0.651, in which case it will contribute 1 to the total. and 1 - 0.651 = 0.349 that it does not turn up and contributes 0. So it will make a net contribute of 0.651 x 1 + 0.349 x 0 = 0.651.
Now here is a piece of utter magic. There is a theorem, "Linearity of Expectation" which says that to get the overall score across all the digits all we need to do is add up the individual scores! This is actually quite surprising but it's true.
So we can just work out 10 x 0.651 = 6.51 and that is the average number of distinct digits we expect to see.
I always like to simulate these answers, in case I have something horribly wrong, so here is a short Python program
for i in range(trials): count = count + len(set(random.choices(digits, k=10)))
print('mean number of different digits =', round(count/trials, 2))
And here is the output
mean number of different digits = 6.51
But there is more. I started thinking what if we did the same thing in base 16, with 16 digits 0 - 9, A, B, C, D, E, F? Or used the whole alphabet of 26 letters, or went beyond that? Suppose we used a n-character alphabet?
This smelt like a situation where Euler's famous number e ≈ 2.71828 might crop and when I investigated, there it was, right on cue. The beautiful result is that with an n-character alphabet, for large n the expected number of distinct characters in a run of n random choices approaches
For our original case n = 10 ths would give 6.32, so the value of 6.51 that we found is already quite close.
pal is a really interesting word. It is recorded from 1770 and derives from a Romany word phal, "brother, comrade", from Sanskrit bhratr, sharing the same PIE root *bhrater- as English brother, Latin frater, and Ancient Greek phrater, "fellow clansman". There are cognates right across the IE family of languages.
Edited by Richard Walker, Saturday 13 December 2025 at 00:12
I got this rather nice problem from Cut the Knot, and it was originally from the Leningrad Maths Olympiad.
Can we visit each of the 8 vertices of a cube exactly once and return to the starting vertex by following a path made up of 8 straight line segments, each connecting a vertex of the cube to another of the cube's vertices?
The answer is yes and below we see one possible solution, with the dotted path meeting the these conditions.
However you will notice that in this tour 3 of the segments coincide with edges of the cube. Is it possible to find such a path where none of the segments coincide with an edge?
Prove that the answer is no, no such path can exist.
Edited by Richard Walker, Thursday 11 December 2025 at 13:33
Not a firm of Dickensian solicitors, but three unique rhymes.
List of words claimed not to rhyme with any other English word often include silver, bulb and purple, but these do have rhymes, and although (obscure and possible obsolete) they are in the Oxford English Dictionary.
Silver rhymes with Chilver, a young female lamb.
Bulb rhymes with Culb, the item of alchemistic glassware on the bench to the left in this picture from the Welcome Collection¶. AKA a retort.
Purple rhymes with hirple. a Scot dialect word meaning to walk with a pronounced limp.
I found these words when searching for unique rhymes—pairs of words that rhyme only with each other, and with no other word. So we can say for example that the only word that rhymes with silver is chilver and vice versa (although verifying this with certainty would be challenging—what dictionaries do you consult, what counts as a rhyme, how do you even search in a language where urp cam rhyme with irp and erp and earp (as in Wyatt)?)
¶ Woodcut of alchemist trying to transmute metals, circa 1503. Wellcome L0012391
Edited by Richard Walker, Saturday 6 December 2025 at 23:08
Not a lurid headline but a crossword clue, and the "calculating Swiss" is EULER, one of my heroes, pictured below in iconic headgear.
Leonhard Euler (1707-1783)
I was gratified to see that Euler has made it across into the realms of General Knowledge. There aren't many mathematicians famous enough for their name to be the answer to a Times Quick Cryptic Crossword clue.
The picture above, a pastel by Handmann painted in 1753, hangs in theKunstmuseum Basel. This is probably the best known portrait of Euler.
Crossword Clue: "English" = E + "monarch" = RULER which "beheaded" becomes ULER, giving EULER (pro. "Oyler").
Edited by Richard Walker, Tuesday 2 December 2025 at 22:51
This was one of the first jokes I ever heard, in my infant school days.
When is a door not a door?
When it's ajar!
Yesterday I suddenly wondered, don't know why, where "ajar" comes from and what other things , if any, can be ajar (apart from a jar obviously).
The OED gives the definition "
Of a door, gate, or window: so as to be slightly or partially open.
It can apply to anything that closes off a space, so long as it's got hinges. The last bit is important, at least historically, because the original form was on char, where char means something like "turn". In Old English it was spelt in variety of ways, e.g. cyrre, chere, and referred at first to time "coming round". The meaning was extended to something physically turning; the OED gives this example from 1510
The dure on chare it stude.¶
So there we have our door. The meaning was then extended further to something like a "turn of work" — compare ways turn is used; take turns, it's your turn, etc. — and changed its spelling to chore, a piece of work that needs to be done. And so ajar and chore are related words, rather surprisingly.
But chare has left a fossil trace in the language, charwoman.
You probably need a bit of background. The Euler line is a line that passes through three of the four best-known triangle centres Although these centres were known to the mathematicians of Ancient Greece, this line was only discovered in 1765 by Leonhard Euler. My rough sketches below illustrate the three centres concerned and the Euler line.
Our solution to the problem will use the following facts about the circumcentre and the centroid..
The circumcentre is the centre of the unique circle, the circumcircle, that passes through the triangle's vertices.
The centroid is the point where the triangle's medians meet. A median is a line joining a vertex of the triangle to the midpoint of the opposite side. The distance from a vertex to the centroid is twice the distance from the centroid to the corresponding midpoint.
In an equilateral triangle the circumcentre and the centroid coincide,
Consider the following diagram.
Triangle ACB is our 60° triangle. E is its circumcentre and F its centroid. Triangle ADB is an equilateral triangle inscribed in the circumcircle and on the same base AB. We can inscribe this triangle in the circle because any 60° angle subtended by AB must lie on the same circle. Because this triangle is equilateral the centre E is also its centroid, and GD a median.
Line HEFI passed through E the circumcentre of ACB and F its centroid, so it is the Euler line of ABC.
We note also that AD subtends angles ACD and ABD, so they both equal, and we know ABD is 60° because it is an angle of the equilateral triangle. Hence ACD is 60°.
Now the properties of medians come into play. DG is a median of the equilateral triangle ADB and E its centroid. Because of the properties of medians we have DG = 3EG.
Similarly CD is a median of our 60° triangle ACB and F its centroid and again using the properties of medians CG = 3FG
Now consider segment DC. Because DG = 3EG and CG = 3FG it follows that DC is parallel to the Euler line HEFI. Angles DCH and CHF are so-called "Z angles" on these parallel line, so they are equal, and we know already that DCH = DCA is 60°. So CHF = 60° also.
Now recall our problem. HEFI is the Euler line and we want to show it cuts off an equilateral triangle. HCI is 60° by assumption, and we have just shown a second angle CHI in the triangle that is cut off is also 60°. But that means all three angles of triangle CHI must be 60° and it is equilateral as claimed.
PS I've not read the solution(s) given on Cut The Know because I wanted to find my own proof but I'm going to take a look now.
Edited by Richard Walker, Sunday 7 December 2025 at 11:35
The Language-Lover's Lexipedia, by Joshua Blackburn. I learned of this wonderful book via the podcast and YouTube channel Words Unravelled, and it is an absolute joy.
The breadth and depth of Mr Backburn's research and scholarship are simply stunning, but at the same time he manages to maintain a light and playful tone. To give a flavour of the book here's a potted version of just one of the 200+ entries, from Abracadabra to Zyzzyva.
We've all heard of the Bee's Knees and the Cat's Pyjamas I suppose, and perhaps also the Duck's Quack, although that's less familiar. But until now I never real stopped to wonder where they came from. Turns out they were Flapper slang.
When I think about Flappers I have a hazy vision of the Roaring Twenties and bobbed hair style and the Charleston and all that sort of thing. Merriam-Webster's more informed definition is "a young woman predominantly of the 1910s and 1920s who showed freedom from conventions (as in style of dress and conduct)".
And they also had their own Flapper-Speak which included a craze for phrases consisting of the name of an animal followed by something mildly surprising or incongruous (the Duck' Quack is an outlier here). Mr Blackburn has dug up a whole bunch of examples, forgotten now apart from the three I gave above. Here are some of my favourites:
The Bullfrog's Beard
The Clam's Garters
The Eel's Ankle
The Hen's Eyebrows
The Kipper's Knickers
And if you fancy a Bee's Knees cocktail Mr Blackburn has found the recipe.
Edited by Richard Walker, Friday 21 November 2025 at 23:55
A million thanks to New Scientist magazine (Feedback 22 October 2025) for alerting me to the (serious, I promise) scientific paper by Paul Silvia and Meriel Burnett entitled,
What’s brown and sticky? Peering into the ineluctable comedic mystery of dad humor with a handful of machine learning models, hundreds of humans, and tens of thousands of dad jokes *
As well as the paper, the website hosts a downloadable list of, as promised, 32,000 jokes. The download is Joke Database.zip, just over 50 MB. When expanded it holds two datasets, one with the full 32k jokes (all gathered from Reddit) and the other a subset of 501 jokes the researchers put to a panel of human joke evaluators.
I was able to open the big dataset in Excel ¶and sure enough it is a treasure house of (slightly over) 32k (mostly) family-friendly dad jokes. Just imagine, a dad joke a day for almost 90 years.
The database contains a lot of data beyond the actual jokes and would be fertile ground for further research or so I imagine. But for now I thought I would come up with a program to print random dad jokes on request.§ Here's a sample run, as you'll see the jokes are of variable quality.
See a random joke? y/n y A burglar broke into our house last night and stole all of the light bulbs I should be upset, but I'm delighted. See a random joke? y/n y I was really mad when they locked me in a room with nothing but the remnants of a dead chicken, but I got out. Good thing I had a bone to pick. See a random joke? y/n y What do you call someone who does not like physics A physicist . See a random joke? y/n y My Llama roommate yelled at me, "I've had enough of your dad jokes! Leave now!" "Fine," I replied, "Alpaca my bags and leave." See a random joke? y/n y Why are almost all (capital) Greek letters old? Because there is only one Greek letter that is new (N/v). &#; See a random joke? y/n y The cashier at the grocery store asked "paper or plastic?" I said "paper" thinking she meant "cash or charge." She started bagging my stuff in a paper bag so I yelled: "No! Plastic." Then she screamed at me: "Hey! we don't...take...Credit Cards." See a random joke? y/n y what do the eiffel tower and a tick have in common? they are both "parisites" lol See a random joke? y/n n
* Published on OSF Oct 29, 2025, 12:27 PM.
¶ Be warned, the spreadsheet has a lot of columns.
§ Python code below:
import csv, random
jokes = []
with open("Reddit_Dad_Jokes_OSF.csv", mode='r', encoding='utf-8') as file: reader = csv.reader(file) for row in reader: jokes.append(row[1])
quit = False while not quit: resp = input("See a random joke? y/n ") if resp == "y": print(random.choice(jokes)) print() else: quit = True
Edited by Richard Walker, Wednesday 19 November 2025 at 00:27
A phonestheme is when a large number of words that start with the same sound (most typically a consonatn pair) to have related meanings. The phenomenon was first noted by J. R. Firth in 1930. A classic example is fl-, which I remember being fascinated by when first reading about it. Here are some words beginning with this consonant pair fl-.
What all these have in common is some notion of movement, especially sudden or repetitive movement.
The Wikpedia article gives a number of other examples, such as cl-, e,g.clop, clap, cling, clasp, clutch, all to do with moving an object; and sn-, e.g. snort, snitch, snot, snout, snivel, snitch, sneeze, sniff, sneer, snore, snorkel, all nose-related.
So is there some real effect here, is there something inherently fluttery about fl-? If so, is there some deep psychological connect? Or it it mimetic; does "flutter sound like something fluttering? Or is it somehow to do with how we make a "fl" sound, do we flap our tongue when we pronounce it? (probably not). And then we know from historical evidence that words can influence other similar sounding words, so that could be at work too.
Is there even enough statistical evidence to support the idea that fl- carries a semantic load? Perhaps we are just (as people do) see in a pattern because we always look for them?
We might be able to provide some evidence by doing some clever statistical analysis, but it's slippery (there's another example of a phonestheme, sl-; slip, sled. slop, slime slither, slobber, sludge). But what criteria do we use to decide of a word belongs in the list? Should slow be counted? What about slag? Slant? Sleet? Slick?
I tried some rough experiments with the help of Copilot. It says
About 8–10% of common English words starting with “fl” relate to repeated or sudden movement.
Is that evidence. Well maybe. We'd have to come up with some list of representative and comparable (how to define what these mean?) semantic fields and see if 8-10% was unusually high.
Another of the examples in the Wikipedia articles is gl-, which is connected with shining or gleaming etc. I discussed this in a previous post A Little "Eggymology" and it is very likely to be a real connection but one that is simple to explain. The words probably all share an etymology and are descended from a PIE root *ghel-, "shining". So no profound mystery. This time Copilot tells me 6-7%. If the proportion for fl- words sharing the same semantic field is higher than that for gl- where (we think) an effect is known to exist what does that tell us? (But I haven't checjed Copilots findings.)
Another example quoted is stand, stable, state, stasis, but from memory this too is a group of words thought to have a common ancestry, a PIE root *sta-, "stand" or something like that.
So is a phonestheme a real thing? It's an intriguing concept, but the jury is still out. There seems to be a school of thought that believes it is just an illusion, but it seems too tat there are also serious scholars and ongoing research in support of there being something real in the idea. Either way I find it fascinating.
Edited by Richard Walker, Friday 14 November 2025 at 23:09
In this proof of the puzzle discussed here we drop some perpendiculars, which is often a good start in Euclidean style geometry problems.
Given a square ABCD, if a diagonal be drawn from B to D and another line drawn from A to the midpoint E of CD, so as to intersect the diagonal at F, the triangle formed by the two lines and the base AB of the square is one third of the square.
Proof:
We can take the side of the square to be 1, so its area will be 1. Construct perpendiculars EJ from E to BD, FH from F to AB, and FI from F to DA. Let the distance from A to H be d.
Then the angles of AIFH are all right angles, so it is a rectangle, and side FI which is opposite to AH must also have length d.
The point F lies on the diagonal BD, which bisects the angle at D, and therefore the perpendicular distances FI and FG are equal, and FG has length d also.
Triangles AJI and AHF are similar, because they share a common angle FAH and both have a corresponding right angle. The distance JI is 1 and the distance AJ is .§
Therefore the ratio of EJ to AJ is 2:1 and that of FH to AH the same. The length of AH is d and there FH must be 2d.
Then we have HG = 2d + d = 3d, and HG = 1 so FH = 2d = .
The area of the shaded triangle AFB is base x height = x x 1 = , as claimed.
§ We should probably prove this, which is easy enough but clutters uo the proof a bit.
What fraction of the square is shaded? (The line from the bottom left corner needs the top side of the square at its midpoint).
This question attracted a lot of media attention. As the Daily Mail put it, “Tricky maths problem sweeping social media leaves many scratching their heads”. However it passed me by at the time, and I only saw it last weekend, so here are my two takes at it. I haven't looked at any of the many solutions on the internet, but I my guess is the reasons why it has been so popular are
The problem is easy to understand
It's not trivial to solve
But there are lots of different solutions.
Take #1
My first idea is shown in the sketches below.
Suppose the three triangles have areas A, B and C as shown in Figure 2. Then the three "look and see" dissections in Figures 3-5 let us write down three equations. (Figure 3 is a bit ropey but hopefully you get the idea.)
We have three equations in three unknowns and it's fairly easy to solve them.
From Figure 3, B = ½ - A.
From Figure 5, C = ¼ A.
Plug these into the equation illustrated by Figure 4.
½ - A + ¼ A = ¼, multiplying by 4 gives 2 - 4A + A = 1, 1 = 3A and so A is
Take #2
But then I though, if A is made up of 4 small triangles, what if we superimpose a triangular grid on the whole square? I drew this in Geogebra.
Eureka! See what happens if we shift the leftmost triangle by half the length of the square's side to the right.
The area has not changed and we can now see that it is equal to that of 3 rows of 4 = 12 of the small triangles. The area the puzzle asks for is 4 of these small triangles, so it is = of the area of the square.
egg is from Middle English, from Old Norse, replacing an earlier English word ey. Both are of Germanic origin but probably go back to a PIE root *owyo-, which also gave us Latin ovum and Ancient Greek oon, as well as cognates in other languages. Intriguingly there is even a hint of a connection with *awi- "bird" and perhaps even further back to a word that meant "clothed" (birds are clothed in feathers , you see).
yolk is from yellow, OE geolu + a suffix which might be -ca, a diminutive (so "small yellow") or -ock, or something else: there seems no general agreement. And yellow? It is probably from a PIE root *ghel-, "shining", and so perhaps connected with gleam and glisten and a number of other words starting with gl- and semantically related to brightness or shining.
Here is an interesting Old English passage from Aelfric's Catholic Homilies, about 990 CE, in which he uses the analogy of an egg to how how two distinct things can be part of a single whole.
Sceawa nu on anum æge, hu þæt hwite ne bið gemengd to ðam geolcan, & bið hwæðere an æg.
"Look now at an egg, how the white is not mixed with the yolk, and yet it is one egg." (Chat GPT 5)
Edited by Richard Walker, Wednesday 5 November 2025 at 22:02
"There is a tavern in the town, in the town And there my dear love sits him down, sits him down And drinks his wine 'mid laughter free, And never, never thinks of me."
When I first visited Greece and went to a taverna I thought that name might be a borrowing from English. After all Greek has a word "bar". But in fact both languages have borrowed from Latin, Greek directly and English via French (and now again from Greek, so taverna is now an English word too).
The Latin taverna seems to have been something like a shop/stall/shed/tent, with connotations of being moveable. A related word is tabernacle, used in the English translation of the Bible to mean a moveable sanctuary or temple, and deriving from the diminutive form tabernaculum, "tentlet" (the suffix is the same as in pendulum, "little hanging thing").
Other related words are architrave, where the -trave element is from an Italian words for beam and the unusual word contubernal, "living in the same tent". More surprisingly there may just be an ancient connect with tree, which may originally meant "beam" or "dwelling". You can see the semantic connection.
Edited by Richard Walker, Saturday 1 November 2025 at 17:22
Imagine three fair spinners.
If we compare A and B there are 9 equally likely outcomes. The ones where A wins are highlighted.
2-1 2-6 2-8 4-1 4-6 4-8 9-19-69-8
A beats B 5/9 or about 56% of the time.
if we compare B with C there are 9 equally likely outcomes. The ones where B wins are highlighted.
1-3 1-5 1-7 6-3 6-5 6-7 8-3 8-58-7
B beats C 5/9 or about 56% of the the time.
So A beats B most of the time and B beats C most of the time. How will A fare against C? Intuitively we might reason A beats B and B beats C, so A will beat C most of the time too. But not so! C beats A the majority (56% again) of the time. Here's C versus A, C's wins highlighted.
3-2 3-4 3-9 5-25-4 5-9 7-27-4 7-9
This is a simplified version I made up of something calledintransitive dice (Wikipedia), where three dice are numbered in such a way that A will beat B the majority of the time, B will beat C the majority of the time, but when A is pitted against C it is C that has the greater chance of winning. It's also possible to have a longer chain, of four or more dice.
Interesting
If we arrange the numbers 1 through 9 into a 3x3 square and note which numbers are on spinners A, B and C, we find we hey a "Latin square"; each letter appears in each row and each column,
Also note the three numbers on any spinner always sum to 15.
Just for fun I wrote and ran a Python simulation that spins the three spinners a million times and prints out the percentage of the time A beats B, that B beat C and that C beats A, and sure enough I got
A>B 55 % B>C 56 % C>A 56 %
Program below
import random
A = [2, 4, 9] B = [1, 6, 8] C = [3, 5, 7]
abeatsb = 0 bbeatsc = 0 cbeatsa = 0
trials = 1000000
for go in range(trials): a = random.choices(A,k=1)[0] b = random.choices(B,k=1)[0] c = random.choices(C,k=1)[0] if a > b: abeatsb = abeatsb + 1 if b > c: bbeatsc = bbeatsc + 1 if c > a: cbeatsa = cbeatsa + 1
Everywhere I looked agreed it is from Latin tiara, which was in turn a borrowing from Greek, but it's not clear if we can trace it back further. The OED describes it as being "of unknown origin" and Wiktionary likewise, but Etymonline suggests the Greek word is "of Oriental origin". Geoffrey Munn, quoted in Wikipedia, suggested the word is from Persian and was applied to the crown worn by Persian kings. It was indeed, but I'm not sure we can infer from this that the word was Persian.
My doubt arise from the fact that Munn also mentions diadems as being wrapped around said tiaras.
This is also true, but diadem is pure Greek; it has a Greek meaning, something like 'around binding'. The OED explains its etymology in this way, and describes it as "the regal fillet of Persian kings" (which I rather like) and states that Alexander the Great adopted it in the course of his conquests. So the headgear was indeed Oriental but the name used for it by Greek writers was a Greek one.
So we have no real evidence for "tiara" being of Persian origin that I can see; the Greek may have borrowed it from somewhere else, or it might have been pre-Greek, or it might have Greek etymology now lost to us.
So my quest for the origin of "tiara"only took me so far and I think the OED's conclusion that the Greek word τιάρα is of unknown origin is sound.
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Our solution to the problem will use the following facts about the circumcentre and the centroid..
We've all heard of the Bee's Knees and the Cat's Pyjamas I suppose, and perhaps also the Duck's Quack, although that's less familiar. But until now I never real stopped to wonder where they came from. Turns out they were Flapper slang.
