Personal Blogs

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)?

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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.

I expect you know the word kibosh; putting the kibosh on something means it has been brought to a sudden stop or been thwarted etc.

I've never thought about its origins until today but tacitly assumed it was from Yiddish — it looks a bit like kibitzer, an onlooker who proffers unwanted advice, which the OED says cautiously is 'perhaps modelled on a Yiddish lexical item'.

The trouble is, it turns out, no-one has found a Yiddish word that kibosh could be derived from, making the Yiddish origin theory unlikely. But plenty of other ideas have been proposed.

From the OED

• From Arabic kirbāš, a judicial whip.
• From Irish Gaelic aidhp bháis, 'death cap', which might have to do with death sentences or executions or a form of torture or the last garment put on a body before interment... but OED says there is no direct Irish evidence to support this.

Wordorigins.org mentions (but doesn't seem convinced by) another possibility

•  It is from Hebrew chi + bosh, 'eighteen + pence'. But why on earth would 'eighteen pence' come to men putting a stop to something?

Michael Quinion

• In his fascinating book Posh Out, Starboard Home (Spoiler: this is not where POSH) Quinion muses that it could be linked with bosh, 'rubbish'. But the change in meaning seems inexplicable.

OUPBlog

In a 2013 post 'Three recent theories of “kibosh”' we find further candidates. 

• It is from a French word meaning to behead a stag, which was borrowed into English as cabosh.
• It is from a iron tool once used by clogmakers to soften leather.
• It was a terms used by builders and sculptors etc. for Portland cement.

Wiktionary

• In a 1904 quotation 'He was proud to be the Husband of the Lady Ki-Bosh of the Local Knickerbockers' it seems to mean in the (possibly excessive) height of fashion. But that doesn't cast light on the possible orgin.

I guess this list is not exhaustive but what stands out is we don't really know where the word came from. It might be from none of them or it might have come from two or more being confused and ending up merged into one. Or someone might just have coined it with humorous intentions and it caught on. This happened with e.g. discombobulate.

I hadn't thought about this until reading something last night with parallel texts in English and German and noticed the German equivalent was KI. Interesting, I thought.

It's short for Künstliche Intelligenz, which is what's called a calque, when individual parts of a compound word are translated literally into the host language.

For example German Regenwald became English rainforest, from Regen, 'rain' and Wald, 'forest'.

Exploring what 'artificial intelligence' translates to in a few other languages, with Gemini doing the heavy lifting, all those I looked into calqued the English phrase, but the abbreviations are overwhelming 'AI', whatever the actual word in a language. For example in Chinese (according to Gemini) artificial intelligence translates as 人工智能 but you could informally write 这个 AI 工具很好用, 'This AI tool is very useful'.

The only exception I found were the Romance languages, which generally put the adjective after the noun, so logically enough we have e.g Spanish inteligencia artificial, IA; Dutch (sometimes) and some Scandinavian languages use KI, like German — but not Swedish where artificial intelligence is artificiell intelligens, AI, more or less the same English).

This is a solution to the problem "A Quick Puzzle — Find x" posted 1/9/26.

Comment Trisecting an angle with straightedge and compasses was a problem posed in antiquity and which many mathematicians attempted to solve over the ages, until Wantzel proved it was impossible in 1837.

So have we achieved the impossible? Well, no. The original problem assumed the use of an unmarked straight edge, and our construction requires one we can mark. Or we could use three rods of equal length.

Our construction formed the basis of an angle trisection method given by Étienne Pascal, father of Blaise, he of the triangle which you will recognise

    1 1

   1 2 1

  1 3 3 1

 1 4 6 4 1

However what was essentially the same method was already known to Archimedes over a millennium earlier.

There is an old joke allegedly from a real exam script

My brother snapped these Mares' Tails, high altitude (20,000 feet and counting) ice crystal clouds. Their meteorological  name is Cirrus uncinus, which apparently means 'curly hooks'. 

It's interesting that different kinds of cloud have double-barrelled 'binomial' names with a genus and a species, a form generally kept for living things. But (thanks Gemini!) I found that viruses have binomial names too, although whether viruses are alive is debatable.

Karl Linnaeus, whose influence is pne of the main reasons binomial names are used in biology. even explored a similar naming scheme for minerals, considered by him a third 'kingdom' (and we still speak of 'Animal, Vegetable and Mineral') but ultimately it was not adopted by the scientific world.

Clouds are an exception amongst non-animate entities: Luke Howard who at the beginning of the 19th century famously pioneered assigning systematic names to cloud forms imitated Linnaeus in using binomial names and his scheme has survived.

Isn't that neat? It was posted on Quora by Alexander Ferrugia, along with a proof that the pattern goes on forever. His proof was quite long though, and I found one which is a bit shorter.

Something that jumped out at me was the resemblance to recurring decimals. , and and I thought there had to be a connection. So I took the second line (the first doesn't fully show what the pattern is, because 6 doesn't appear) and played with it in the Python Shell, trying to get 16, 50 and 33, the numbers on the left hand side ...

100/6, 100/2, 100/3
(16.666666666666668, 50.0, 33.333333333333336)

Of course that's not right, we only want the whole number part. So let's subtract the fractional parts...

100/6 - 2/3, 100/2, 100/3 - 1/3
(16.0, 50.0, 33.0)

... and now it works. What about the right hand side? If we break it up we see it represents .

To get the equivalent expressions for the -th line we can just replace by everywhere and now we get 

, with the left-hand side becoming

We want to prove these two expressions are equal and we can do that very easily. First make them more general by replacing by , giving 

and

It is not hard to see that if these were expanded we would obtain cubic polynomials of the form . How can we prove they are the same? We might use paper and pencil to expand them and collect terms in the good old fashioned way. Or we could enter them into reliable computer algebra system and let it do the heavy lifting.

Or much more elegantly we can use this fact

Why is this true? Because four values let us set up four equations and solve them to find the four constants .

But we have the requisite four values! This post began with them and so we have proved the two expressions are equal for any value of and for any value of in particular.

Imagine 5 unsociable people live on the surface of a tiny spherical world. Being unsociable they want to arrange themselves so the smallest distance between any two them is as big as possible. How can they do this?

If there were 6 of them they could position themselves at the vertices of a regular octahedron, as in this model ^[1] of a Sulphur Hexafluoride molecule, where 6 Fluorine atoms surround a single Sulphur atom. Electrostatic repulsion causes the Fluorine atoms to take up this octagonal configuration.

We can measure the distance between two points that lie on a sphere by giving the angle they make at the centre and you can see that in this case all the angles are 90°.

But now suppose one of them jumps in a spaceship and flies off the some other tiny world. Now the remaining 5 have a little more space to spread themselves out more.

Except they don't! 5 points can't do any better than 6 could and the best minimum distance that can be achieved is still 90°. Here's my attempt at a proof.

Here I showed that

If there are five points on the surface of a sphere, then no matter how they are arranged at least 4 of them lie in the same hemisphere.

I think this result may have first been published by Martin Gardner, although I don't have the reference.

You can think of this as being like the Northern hemisphere, including the Equator and the North Pole, and 2 or more points must lie on the Equator.

If all 4 points lie on the Equator the greatest possible spacing is 90°, achieved when they are situated at the corners of a square inscribed in the Equator.

Otherwise we must have at least 2 on the Equator and at least 1 in the same hemisphere but not on the Equator. The greatest possible separation between this point and the 2 on the Equator is 90°, achieved when that point lies at the North Pole. So, either way, there must be a pair of points separated by 90° or less.

We'd like the configuration to be as symmetrical as possible, so we can put 3 points at the vertices of an equilateral triangle inscribed in the Equator and the other 2 at the North and South Poles.

At this point you might wonder if there are molecules with 5 Fluorine atoms surrounding a single atom of some other element, and if so are the Fluorines in the arrangement described above. The answers are Yes and Yes.  Here's a picture of a Phosphorus Pentafluoride molecule ^[2]

This problem of optimally spacing points on a sphere, which has always fascinated me, is known as the Tammes problem and the best configuration for 2-14 points and 24 points were found and proved between 1943 and 2015 [3]. It is a hard problem, in that it has to be solved for each number individually. I imagine we can find estimates for the general case but finding exact solutions is challenging.

[1] Wikimedia Commons public domain.
[2] Wikimedia Commons public domain.
[3] Wikipedia Tammes problem.

I suppose this must have been solved thousands of times but I wanted to find my own solution,.

I tried with a few 4-digit numbers at first to see if I could get any intuition but without success. I did think of writing a program to do a brute force search for a solution but that felt like cheating. Knowing what I know now it's a good thing I gave up the idea, as you'll see. 

After a lot of thinking I came up with this.

Suppose the number is , where is the last digit and the value represented by the remaining digits. For example if the number was 2026 we would have and . Suppose our number has digits.

Then after has been moved to the front its place value will be . The remaining digits will have shifted one place right, so now represent . For example if we apply the process to 2026 we get .

Since the new number is double the original one we can write

which gives . This means the left-hand side must be divisible by 19 and since is a single digit it cannot be a multiple of 19, so must be.

It's not hard to find a suitable value of by hand but I was lazy and wrote a short Python program to do the heavy lifting.

d = 1
while not (10**d - 20) % 19 == 0:
    d = d + 1
print(d)

This tells us the smallest is 18 so and dividing by 19 gives .

So what is ? If we try setting we get a = 52631578947368421, which doesn't work. Next we try which gives and . Now we have a solution, and it is the smallest possible!

Postscript

I wondered how long the brute force search I contemplated at first would have taken. I worked out a rough estimate and with the standard Python I useit would have been ≅ 1000 years. So I'm glad I didn't try it!

Surprisingly there are hyper-optimised versions of Python that are many orders of magnitude faster and if I had deployed one of these on my Silicon M4 (a technically complex endeavour mind you), it seems I could have found the answer in just under 2 days.

A well known puzzle asks what English number word describes its own length and it's not hard to see the answer is four. This is a fixed-point number: if you count hpw many letters in the word and write that as a word you get back to four.

Less obviously it is also a black hole number: if you begin with any number whatsoever and apply this process repeatedly you always end up at four. Always.

For example

two thousand and twenty six --> twenty three --> eleven --> six --> three -->  five -- > four

I wondered what happens in other languages. Is there always a fixed-point?  If so is it always a black hole. Here's what I found (hope it's all correct, it was a bit triccky making sure I didn;t forget numbers.)

In English four is a fixed point and a black hole.

In Spanish cinco is a fixed point but not a black hole. It drags in nueve and siete, but other numbers get dragged into a 2-cycle, cuatro --> seis and back again.

In French we have a 4-cycle trois --> cinq --> quatre --> six and back again a nd this is an attractor for all other numbers (I think). For example

deux mille seize --> quatorze --> huit --> quatre

Finally, in Italian tre is a fixed point and also a black hole.

We can carry out the same investigation in any language that uses an alphabet (at least 2000 of them) but I think these four give a good feel for the sorts of behaviour that can occur.