Personal Blogs

In 1685 the mathematician John Wallis recounted that Prince Rupert had a bet that it is possible to pass a cube through a hole in one of smaller side length. Wallis showed how this could be done and out ot curiosity I tracked dow the source in the Internet Archive. Here it is

I only know odd words of Latin, so I asked Copilot what this was (that's all I asked, I gave no clues apart from the image) and quite surprisingly it not only read the text and translated it but also identified what it was and meant.

Septentrionate means "To point North", ultimately derived from Latin septentrio, "North". You have probably spotted the first element sept- is connected to the number seven. the second part, trio, is not to do with three, but apparently mean "plough ox". So what were these seven plough oxen? Well here they are:

A familiar constellation in the Norther sky, Ursa Major and often also called The Big Dipper/The Plough/Charles' Wain.

Latin also had Boreas, the North wind originally and I think borrowed from Greek (think Aurora Borealis). Modern Greek for North is still βορράς, voras, so it hasn't changed much, except that the letter beta has become a "b" sound to a "v" sound in Modern Greek.

I leaned that Septenrio was Latin for North from the excellent Words Unravelled podcast and YouTube channel, which also featured the surprising fact that Romance languages have names for North, South, East and West that are essentially the same as the English words, e.g French nord, sud, est, and ouest. This is as a result of borrowing, most likely from Old English into Old French and from there to the other languages, supplanting the names inherited from Latin.

It's not at all clear why this replacement happened. When I heard about it I assumed that the geographical closeness of France to England and of Spain, Portugal, Italy to France would have been a factor, but to my astonishment I found Romania shares the borrowing too, even though it's much further off. So it's a mystery.

And where did North come from anyway?  Well not everyone agrees with it, but the strongest theory is that the word and its cognates in other Germanic languages come from a root that meant "left" (or perhaps "down" or both). Why would North be left? Because most ancient societies, would probably have been East oriented (literally), rather than North because that is where the Sun rises, and when we look East, where is North then?

photo credit

This is the solution to the puzzle at https://learn1.open.ac.uk/mod/oublog/viewpost.php?post=312323

Given ABC is a square of area 100 sq. units and M and N the midpoints of the sides on which they lie, to find the areas 1, 2, 3 and 4.

Solution

The small triangle (3) can be got from triangle (4) by a rotation of 90° followed by a scaling. The hypotenuse of triangle (4) is equal to the side length of the square, and that of triangle (3) is half the side length of the square, so the dimensions of (3) are half those of (4) and its area one-quarter the area of (4).

But (3) and (4) between them make up one-quarter of the square, which represents an area of 25 sq. units and given their areas are in the ratio 1:4 the area of (3) must be 5 sq. units and that of (4) 20 sq. units.

The combined areas of regions (2) and (3) is also one-quarter of the square, hence the area of region (2) must also be 20 sq. units.

Finally the areas of these three regions add up to (20 + 5 + 20) sq. units, which leaves 55 sq. units for region 1.

So finally we have

Region 1: 55 sq. units

Region 2: 20 sq. units

Region 3: 5  sq. units

Region 4: 20 sq. units

I saw this on a YouTube channel "Knights & Knaves" but this or similar questions seem to have been asked elsewhere too.

In a circle, if we draw 10 random chords, what is the expected number of times two of the chords will intersect? 

(What we mean by a random chord needs defining but for the purposes of this question we take it to be that each chord joins two points on the circumference with a uniform probability distribution, so the probability of the point chosen lying in a given arc is the length of that arc as a proportion of the circumference.)

One solution begins with the case of two chords and reasons along these lines. Suppose we choose four random points and consider the point A. If we join it randomly to one of the other points there are three cases, as shown below, and they are all equally likely, with probability 1/3.

Cases 1 and 3 have no intersections, Case 2 has 1. So the expected number of intersections is 

(1/3)x0 + (1/3)x1 + (1/3)x0 = 0 + 1/3 + 0 = 1/3

This seems to be the generally accepted argument, but I felt it was rather glib, so I got Copilot to write a simulation in Python. You can find the program at the end of this post. I added a bit of code at the end to run it multiple times and after 10,000,000 trials the overall proportion of times it found an intersection was 33.34%, so that seems to support the expectation being 1/3.

The argument continues as follows. We know the expectation for one pair of chords is 1/3 and if there are 10 chords there will be "10 choose 2" which is 10x(10-1)/2 = 45. There is a beautiful fact, called Linearity of Expectation, which tells us if the know the individual expectations of a set of random variables we can find their combined expectation just by adding all the individual expectations together, and this is valid even if the random variables are not independant of one another.

So in the present case we just add 45 lots of 1/3, one for each pair of chords, and find the expected number of intersections is 15. In the general case of n chords we have "n choose 2" pairs, and the expected number of intersections is nx(n-1)/2 lots of 1/3, or nx(n-1)/6.

I think this is the standard line of argument and there is no need to consider how many pairs there are. Here is a simpler approach that occurred to me.

Suppose we have two chords c1 and c2 and we add a third, c3. The expectation that c1 intersects c2 is 1/3 and the expectations that c3 intellects c1 and c2 are 1/3 and 1/3 respectively. By the Linearity of Expectation we can just add all these, so we get 1/3 + 1/3 + 1/3 = 1.

Now if we add a fourth chord c4, the expectation of it intersecting c1, c2 and c3 respectively will all be 1/3 so now the total expectation is 1 + 3x(1/3) = 2. When add a fifth chord the expectation will be 2 + 4x(1/3) = 10/3, and so on. After we have added the n-th chord the total expectation will be

(1 + 2 + 3 +... + (n-1)) x (1/3) = ((n-1)xn/2) x (1/3) = nx(n-1)/6

and voilà, we have the same answer as before!

Here is the Python program generated by Copilot

import math
import random

def generate_random_point_on_circle():
    angle = random.uniform(0, 2 * math.pi)
    return (math.cos(angle), math.sin(angle))

def ccw(A, B, C):
    """Check if three points are listed in a counter-clockwise order."""
    return (C[1] - A[1]) * (B[0] - A[0]) > (B[1] - A[1]) * (C[0] - A[0])

def segments_intersect(A, B, C, D):
    """Return True if line segments AB and CD intersect."""
    return ccw(A, C, D) != ccw(B, C, D) and ccw(A, B, C) != ccw(A, B, D)

def random_chords_intersect():
    # Generate two chords
    A = generate_random_point_on_circle()
    B = generate_random_point_on_circle()
    C = generate_random_point_on_circle()
    D = generate_random_point_on_circle()
    
    # Check if they intersect
    return segments_intersect(A, B, C, D)

count = 0
trials = 10000000
for trial in range(trials):
    if random_chords_intersect():
        count = count + 1
print((count/trials)*100)

A customer walks into a pet shop and asks to buy a giraffe. "Hey, that's a tall order", says the shopkeeper.

30 September, 2025

Anglesey has what may be the finest display of cyclamens in the country, rivalled only by the RHS gardens at Wisley.

The name cyclamen goes back to Greek kyklos, "circle", and we have related words like circus and cycle etc. These words may even be related to ring, the common PIE root being something like *sker-, "turn", with kyklos being a reduplicated form skersker. "ring" is from the same PIE root, so I suppose "circus ring" could be consider a triplication.

But what has this to do with cyclamens? Well the plant grows from a round tuber and the plant is named for that, the theory goes. This seems a bit tenuous to me; many plants have round tubers or bulb or corms, so why are cyclamens special? 

I saw this on "Andymath" but I haven't looked at the solution there. The problem is to find the marked angle.

We have two equilateral triangles, not necessarily of equal size, arranged as shown in the diagram, but no information about the orientation of the upper triangle or its relative size.

One reason geometry problems are so interesting is that often the solution requires some divergent thinking, "out of the box" as the saying goes. This can often involve adding auxiliary elements such as extra lines or circles to the given diagram, and seeing what to add can require considerable creativity.

But experience plays a role too, and one rule of thumb is that it is usually worth drawing in any standout perpendiculars and seeing if they do anything for us. So let's add line CD.

Now we notice that angles CGD and CBD are both interior angles of equilateral triangles, so they are equal. They both stand on CD and the converse of the angles in the same segment theorem tells us CD must be a chord of a circle passing through BG and B. Let's add that circle.

But now we see DB is a chord in that circle with angles DCB and DGB being angles in the same segment and therefore equal. DCB is obviously 30 degrees and so the angle DGB that we want is 30 degrees also. Solved!

A different approach is to say that we were not told the orientation of the upper triangle relative to the lower one and yet expected to solve the puzzle. So we conclude the orientation does not affect the answer and we can choose it to be anything we like. Very well, let's position the upper triangle directly above the lower, like this. 

Now it's self-evident the angle sought is 30 degrees and from the argument just given we see it must have this value irrespective of the orientation of the upper triangle. Another example where the absence of what seems a crucial piece of information turns out to be the key that unlocks the puzzle!

Here are two views of a puzzle object I bought on Amazon.

How can this be? I promise you there is only one object and the viewpoint is all that has changed.

According to Wolfram MathWorld "Any triangle can be positioned such that its shadow under an orthogonal projection is equilateral. "

You can try this for yourself.

I cut a triangle out of printer paper and taped it to a cocktail stick. Then we put a sheet of paper under the kitchen light and a friend experimented with positioning the triangle until we got a shadow that was a pretty fair equilateral triangle. Here's the photo.

A is the paper triangle (the shadow behind is my friend's hand) and B is the triangle's shadow, which you can see is more or less equilateral. It took two or three minutes of trial and error to get the position right but we were pleased with the eventual result!

In Greek mythology the Muses* were the daughters of Zeus and Mnemosyne, whose name is connect with mnēmē "memory", which gives us mnemonic. This probably, although not all authorities agree, comes from a PIE root  *men-, which is to do with thinking and mental (!) capacity in general, and is connected with many words, some quite surprising, for example Ahura Mazda, comment, mandarin, mantra, monster and mosaic — and of course Muse.

*The Muses

Wikipedia gives the following list of Muses and the art or science associated with each, although as you would different authors, classical and later, did not always agree with this list and even with how many the Muses were.

Calliope (epic poetry)
Clio (history)
Polyhymnia (hymn and mime)
Euterpe (flute)
Terpsichore (chorus and dance)
Erato (lyric choral poetry)
Melpomene (tragedy)
Thalia (light verse and comedy)
Urania (astronomy and astrology)

Acknowledgement

Copilot generated the image 20 Sep 2025

A friend in Devon sent this photo yesterday.

After some Googling I found it is a Portuguese Man o' War and quickly messaged my friend warning her to give it a wide berth, because the tentacles you can se in the photo are highly venomous. and in fact the Portuguese Man o' War is seriously dangerous.

It has been quite windy and stormy there and the animal must unfortunately have been blown off course in the Atlantic and become stranded on the beach.

Actually, from my reading, I found it is not an animal but a siphonophore, a whole colony of animals referred to as zooids, each multicellular, and all genetically identical, but having specialised into different roles. One (or maybe some, I am not sure) forms a gas-filled bladder which acts as float for the whole colony. Other zooids form the venous tentacle which incapacitate other animals and presumably draw them back to the colony. A third type of zooid there digest them, and presumably share the nutrients, I don't know how.

Finally a fourth zooid is responsible for sexual reproduction.

In life the colony might have look a bit like my sketch.

There are many other surprising facts about this strange life-form. One I particularly like is the small fish that has evolved immunity to the venom and now lives with impunity amongst the tentacles feeding on any left-overs.

Problem 1
I found on this puzzle 'Watermelon Paradox' on the Mind Your Decisions YouTube channel which cites the original source as the 2010 Indian KVPY exam, but it is probably a classic problem; I found 3,570 Google hits for it. It runs like this (my paraphrase)

I have 100 kg of watermelons which are initially 99% water. After a few days they have lost some moisture and are now only 98% water. How much do they weigh at this point?

Problem 2
This one is very common, with 169,000 Google hits, and I don't know where it originated. It popped up earlier this year and the Mirror wrote of it 'Incredibly hard 1% Club question leaves Brits arguing over the answer' ^[1]. Here is how this one goes (my paraphrase again).

In a room are 100 people, of whom 99% are left-handed. How many left-handers must leave the room to make the proportion of left-handers drop to 98%?

These two problems are the same really, only the way they are framed is different. In each there is an answer we might immediately jump to, but which on reflection turns out wrong. This is quite a good example of the System 1 thinking, as described in the 2011 book Thinking Fast and Slow by Kahneman ^[2]. The book contrasts two modes of thinking

System 1 is fast and intuitive, a snap answer that takes little effort to arrive at and which we don't really have to think about.

System 2 is slow and takes effort, we have to think about how to solve the problem and work the answer out methodically.

Both these systems have their place.  We often must make quick judgements and much of daily life is governed by small familiar decisions taken unconsciously and here System 1 is needed. But other decisions may concern problems that are complex and unfamiliar, with answers that are unintuitive. Here System 2 is required.

Both the problems I have given above trick us into System 1 thinking, so we are tempted to say 99 kg, or 1 person. The correct answers are highly unintuitive; 50 kg and 50 people, which is quite surprising. Especially with Problem 2 I find the answer hard to believe even though I know it's right.

Here is my System 2 analysis of the two problems side by side, so you can see how they parallel one another.

[1] https://www.mirror.co.uk/news/weird-news/incredibly-hard-1-club-question-34387903

[2] https://en.wikipedia.org/wiki/Thinking,_Fast_and_Slow

Wiktionary defines an antiproverb as: " A humorous adaptation of one or more existing proverbs."

There are many forms but a common one starts with one proverb, then switches in midstream to another. For example, the daftly incongruous

"Every dog has a silver lining"

Or the sardonic

"No news is the mother of invention."

Here are more examples from [1]

Don’t count your chickens in midstream
You can lead a horse to water, but you can’t have it both ways.
Too many cooks are better than one.
An apple a day is worth two in the bush.

The word antiproverb was coined by Wolfgang Mieder and there is quite a literature about antiproverbs ^[2][3]. People who study proverbs are paremiologists, a word new to me but paremiology is in the OED and attested from 1861, derived from Latin paroemia, Greek παροιμία.

I thought I would try to generate some antiproverbs of my own, so I got a list of just under 1,000 proverbs and generated many random pairs, looking for good combinations. It turned out harder than I was expecting, but here are some that I feel show definite promise

An apple a day is better than no bread.
Don't count your chickens while the sun shines.
Many a true word is sauce for the gander.
Don’t change horses till the fat lady sings.
It’s an ill wind that never boils.
It’s easy to be wise after a free lunch.

[1]  https://wordsbybob.wordpress.com/2014/01/20/antiproverbs-say-what/

[2] https://www.degruyterbrill.com/document/doi/10.2478/9783110410167.15/html

[3] https://www.researchgate.net/publication/370487535_ENGLISH_ANTI-PROVERBS_AS_STYLISTIC_DEVICES