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.

'And the mill‑stream sang of wheat and rye,
Of the farmer’s care and pain;
Of the golden sheaves and the harvest home,
And the grinding of the grain.'

Juliana Horatia Ewing — The Mill‑Stream

'Never shall I forget the sensations of awe, horror, and admiration with which I gazed about me. The boat appeared to be hanging, as if by magic, midway down, upon the interior surface of a funnel prodigious in circumference, immeasurable in depth, and whose perfectly smooth sides might have been mistaken for ebony, but for the bewildering rapidity with which they spun around, and for the gleaming and ghastly radiance they shot forth, as the rays of the full moon, from that circular rift amid the clouds which I have already described, streamed in a flood of golden glory along the black walls, and far away down into the inmost recesses of the abyss.'

Edgar Allan Poe — A Descent into the Maelström

Two very different images evoked by words — millstream and its evil twin maelström — that are doublets, sharing the same ultimate roots, in fact doubly doublets. The first element in each is from a PIE root *mele-, 'grind' or 'crush', and the second from another PIE root *sreu-, 'flow'. 

The first is seen in many other modern words to do with grinding and crushing; along with mill, we have molar, mallet and malleable; and with what gets ground; emmer, meal (as in oatmeal) and millet. An emolument might have originally been what the miller got paid. 

sreu- is the root of not just stream but has given us catarrh, rheumatic, rhythm and many medical words incorporating Greek rheo (ρεω), 'flow'.

So starting from common beginnings, these words have travelled along different linguistic paths and ending retaining a strong semantic connection but with vastly different connotations. The millstream is water channelled and tamed, pressed into human service to provide us with our bread; but the maelström a vast, violent and uncontrollable elemental force that would destroy us.

What did their respective journeys look like?

Mill apparently entered the Germanic languages fairly early as a borrowing from Latin molina (as in French moulin) and has cognates in other Germanic languages, e.g. Dutch molen

Stream has a more direct history. It seems there was a Proto-Germanic word strauma- that inherited directly from the PIE root. with an added 't' it would interesting to explore in another post.

So we find mylestream already in Old English. Here's an interesting snippet I found in the OED and couldn't resist, from a land charter, quite a few of which have survived (and which mention landmarks that can sometimes still be recognised today).

Of hlippen ham in to þam mylestreame, of þam mylestreame innan þa norð lange dic

AI Overview translates this as

'From the leaping-meadow into the millstream, from the millstream into the long north ditch.'

I don't know what a leaping meadow is, do you?

The compound maelström has a more striking history. It is from Dutch and was used by Dutch explorers and mapmakers in reference to a tidal whirlpool off the coast of Norway, which reputedly sucking in and destroy (grind to pieces?) any vessel ill-advised  enough to approach too closely (although there may have been some exaggeration involved). It featured in an Old Norse tale of sisters Fenja and Menja, giantesses who turned an enormous millstone Grotti, the 'sea mill'. to grind out gold, peace and prosperity, which sounds handy. I'd never met this legend until researching this piece but it looks worth following up.

But it was Poe's story that made the Maelström famous. Published in 1841 (marked) its impact can be seen in this Google ngram.

I'll leave you with a brace of crossword clues you should find easy to solve.

La mer, most disturbed here? (9)

Times Quick Cryptic 1641 by Pedro

A little malingerer, Everyman will watch online channel (10)

Everyman 3903

It's well know that Easter is named after Eostre.,an Old English pagan goddess of spring.

Except maybe not. We only know of a single reference to this goddess. Bede, writing in the early 8c, had this to say (Bede wrote in Latin, this is Copilot's translation)

“Eostur-monath, which is now interpreted as the Paschal month, once took its name from their goddess who was called Eostre, in whose honour they held

festivals in that month. From her name they now call the Paschal season by the old observance’s customary name, calling the joys of the new solemnity by the title of the ancient rite.”

 

ᛖᚩᛋᛏᚱᛖ 

Claim: In any triangle ABC, if we draw a line from each vertex to the point along on the side opposite, the inner triangle so formed has area equal to of the area of the orginal triangle.

I posted a proof of this on 28 March but since then I've thought of a simpler and far nicer proof.

Figure 1

In Figure 1 triangle ABC, R, S and T are trisection points of the sides they lie on; and K, L and M the midpoints of the sides. XYZ is the Feynman "one-seventh" triangle formed by lines AT, BS and CR.

In Figure 2 we have rotated triangles XBL, YCM and ZAK through about the midpoints L, M and N respectively.

Figure 2

Figure 2 shows triangle ABC can be dissected into a figure consisting of seven congruent triangles, all of equal area, and that the Feynman triangle XYZ, numbered 7 in the diagram, is one of them. It follows immediately that its area is the area of triangle ABC.

Of course, like the proof I gave previously, this must have already been discovered by many before me, but I worked it out for myself and the 'εὕρηκα' moment was very pleasing.

In any triangle, if we join each vertex to the point one-third along the side opposite, the area of the triangle this creates has one-seventh the area of the triangle we started with.

Figure 1

The story goes that the Nobel Prize winning physicist Richard Feynman was introduced to this theorem at a dinner following a talk he gave at Cornell University. Feynman was apparently disbelieving at first, and even sought to disprove it, because the combination of numbers 3 and 7 seemed too unlikely be true. After a time he accepted it and then spent the rest of the evening finding a proof.

I read somewhere that Feynman wrote about the difference between physics and mathematics and maybe this story illustrates a difference between practitioner of the two disciplines. I think the typical mathematician would see the surprising combination of 3 and 7 as being so neat that it's difficult to conceive of the theorem not being true!

Coming across the theorem recently I knew I'd seen it before but couldn't remember the proof or even if I'd ever known one. So I thought I would come up with my own. I wanted a proof that didn't use any coordinate geometry (or vectors or complex numbers) or trigonometry or a long chain geometrical reasoning, or any moving parts.

I knew that sometimes with problems of this kind we can prove what we want by using multiple copies of a figure to build up repeating tiling pattern, a tessellation. so I experiments with various possibilities but without joy.

I left it a while and when I went back bingo! I saw how to draw the 'look and see' proof in Figure 2.

Figure 2

This uses 12 additional triangles congruent to the small inner triangle of Figure 1. They are grouped by fours into three parallelograms, and each side of the original triangle exactly bisects a parallelogram. The original triangle is made up from half of each of the three parallelograms plus the small inner triangle. If we let represent the area of the small triangle we have

So the small inner triangle has one-seventh the area of the original one, as claimed.

Doubtless this proof is not new—many people must have discovered it before me—but it was new to me and I was pleased to have worked it out.

A friend captured last night's sunset over Ullswater.

Google 'Married Puzzle' and you'll get around 7,000,000, so it is quite famous. If you've not seen it, it usually goes like this

Jack, who is married, is looks at Anne, but Anne is looking at George, who is unmarried. Is a married person looking at an unmarried person?

1. Yes
2. No
3. Not enough information

It's often captioned '80% Get It Wrong' and indeed Alex Besos wrote about it in the Guardian and ran an online survey that attracted 200,000 responses of which just over 70% were wrong. 

I first saw the puzzle some time ago but encountering it again today reminded me what a well constructed question it is, skilfully written to make it seem more puzzling than it actually is. Here's a different question

I flipped a coin three times. The first time it came up H and the third time it came up T. Did the pair HT appear in the sequence?

I hope it's not too hard to see the answer must be 'Yes'. Given the description we might think about the sequence of coin flips and the only possibilities are HHT and HTT, both of which contain the pair HT.

But the two questions have exactly the same underlying structure; we have just replaced 'married' with H , 'unmarried' with T, and 'is looking at' with 'was followed by'.

However we've lost the distracting baggage the original came with, a storyline that hints at the intriguing possibility of a love triangle (often the question includes a picture of the three people) and perhaps makes us feel vaguely that being married or not might have something to do with the solution. I've set my version in a context that nudges us towards think in a more analytical way and makes it easier to see the answer.

The Guardian article drew on an article in Scientific American that mention the original source of the puzzle as being Hector Levesque, a well-known researcher in artificial intelligence at the University of Toronto, who devised it for a research project. I've tried to find the original publication but haven't to date succeeded in identifying which of his publications it appears in.

In some ways it reminds me of the equally famous Linda Problem from the book Thinking, Fast and Slow, by Daniel Kahneman, which contrasts two Systems of thought: System 1, which is fast, intuitive and economises on mental computation, versus System 2, which is slow, methodical and computationally intensive.

System 1 excels when something is routine or demands rapid and instinctive response, whereas System 2 is needed in more complicated situations that require slow and careful logical thought to arrive at a correct solution.

Here is the Linda Problem

Linda is 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.

Which is more probable?

1. Linda is a bank teller.

2. Linda is a bank teller and is active in the feminist movement.

Most people choose Option 2 but that cannot be correct. There are more bank tellers than there are back tellers who are active in the feminist movement, just like there are more people who live in the UK than there are people who live in the UK and collect stamps. So the probability of 2 being true is less than the probability of 2 being true.

I love this example! The amazing thing to me about it is that when I first saw it I got the right answer but it felt wrong and for all the times I have explained it to someone or written about it, it still feels that 2 is correct. It stands as an example of how wrong an instinctive answer can sometimes be.

I think the link between the two problems, the Married Puzzle and the Linda Problem, is that the majority (and wrong) response in both cases is the System 1 response, but both scenarios require a System 2 approach if we are to solve them correctly.

Being as how today is ' day' (date is 3/14 if you put the month number first), I thought I would present a way of estimating by experiment, rather than by calculating it using one of the many methods.

There are several ways to estimate by experiment. Two of the best known are Buffon's needle and Monte Carlo Integration, and before we had computers people did do physical experiments, to demonstrate these techniques really do give an estimate of . Today it is easier to simulate the experiments using a computer. Although these approaches only converge very slowly and are not of practical use for finding , they are of considerable interest from a recreational mathematics standpoint.

The idea I'm going to describe uses the concept of a random wall on the number line, in the guise of the Drunkard's Walk'.

Initially the drunkard is clinging to the lamp-post at 0 but he sets off, staggering randomly along the number line. At each stage he moved to an adjacent number, either the one to the left or the one to the right ,with equal probability. So when is at his next position will be or , each with probability .

Where does come in? Well, rather surprisingly, after a large number of such moves his expected (absolute) distance from the origin is close to . So if we simulate a large number of long walks and take the average of the distances each walk end up from the origin we can reverse the formula to estimate . Specifically if the walks are moves and the average we find is the estimate for is ,

With the aid of a Python program you can find at the end of this post I simulated 1,000,000 random walks of 10,000 moves, and got 3.142857142857143, an error about 0.04%. That's not spectacularly close but it does support the idea that the simulation is converging to |(\pi\) as expected.

Of course you could make it a practical experiment if you wanted — just flip a coin and keep track of the number of heads vs tails. but it would be a bit tedious to do 10,000 flips

Fun Fact

Our drunkard will return to the origin with probability 1, so return is certain, but it is not possible to set a finite value for the expected number of steps this will take.

Credit to Copilot for the drunkard and lamp-post picture.

Code

import random
import math

N = 10000
moves = [-1,1]
total = 0

for trial in range (1000000):
    x = 0
    for step in range(N):
        x = x + random.choice(moves)
    total = total + abs(x)
avg = total/1000000
pi_estimate = 2*N/avg**2
print(pi_estimate)

Last week I ran a marathon. It's not an easy thing to organise.