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Richard Walker

A Lovely Geometry Theorem

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Suppose equilateral triangles ABE, ACF and ADG share a common vertex A as shown. Then the midpointss of the segments EC, FD and GB form the vertices of an equilateral triangle.

The special case where points BECFDG lie on a circe centred at A was proved by Alon Amit (April 5 2020) in response to a question posed on Quora:

A hexagon is inscribed in a unit circle such that three alternate sides are of unit length with the other three arbitrary. Connect the midpoints of the three arbitrary length sides. Can you prove those points form an equilateral triangle?

Amit's clever proof, using complex numbers, does not in fact depend on the points lying on the unit circle and holds as long as we have the configuration in the figure above, and this more general case is even more surprising than the original version.


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Richard Walker

Pet Lovers 🐈 🐕

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I've just helped towards a Kitty's expensive dental bill. I was happy to do it, because I'm an Ailurophile. The word is relatively modern (1914) and comes from Ancient Greek ἀίλουρος "ailouros" = cat, which is thought to originally mean something like "wavy tail", rather poetic to my mind. A somewhat earlier term for a cat lover was Philofelist, dating from 1843.

A dog lover is a Cynophilist, or Cynophile, from Greek κυνος "kynos" = dog, from Proto-Indo-European kwon, which is also the root of words such as hound, kennel, and canine.

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Richard Walker

Daffynitions

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Edited by Richard Walker, Monday, 7 Apr 2025, 21:46

Condescending: prison inmate using rope to escape.

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Richard Walker

Apple Blossom Time

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Richard Walker

My Latest Startup

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I’ve stated a business that checks Christmas cribs are up to standard. It’s called “QA in a Manger”.

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Richard Walker

Too Little Information

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Edited by Richard Walker, Tuesday, 1 Apr 2025, 23:53
I love puzzles where there seems to be too little information. I found this one on the Mind Your Decisions YouTube channel.


If the radius of the circle is 8 units, find x.

How can we solve this when we don't know where on the diameter point P lies? Well, the mere fact that we aren't told where but are still expected to solve the problem, supplies the extra information we need. The length x must be the same wherever P is.

Very well then, we can choose anywhere we like. So let's pick the centre of the circle.

Now the angle marked in the triangle is 60 degrees, by subtraction from 180, and the two sides adjacent are radii of the circle, so their length is 8 units. Because of the 60 angle the triangle must be equilateral, and so the length of its third side is also 8.

So the answer is x = 8 units.
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Richard Walker

Household Chores

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Richard Walker

Solution to "Five Points of a Sphere"

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See https://learn1.open.ac.uk/mod/oublog/viewpost.php?post=295536

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

Proof: Pick any two of the points. These two points, taken with the centre of the sphere, define a plane that cuts the sphere into two hemispheres, both containing the points we picked.

Three points remain, and of these at least two must lie in the same hemisphere, which will then contain the required four points.

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Richard Walker

Tom Swifty

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"Thank you so much for the lovely sofa and matching armchairs", said Tom sweetly.
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Richard Walker

Five Points of a Sphere Puzzle

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Suppose we mark five points of the surface of a sphere.

Can you prove that however the points are distributed it is always possible to draw a hemisphere that include at least four of the five points in its interior or on its boundary?

This puzzle appears in many places and was included in a maths competition as recently as the early 2000s. but I think it must go back further and may have first been published by Martin Gardner, although I don't have the reference.
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Richard Walker

The Royal Order of Adjectives

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Do you like my grammar hardback red new fat big lovely book?



Most native speakers of English will instinctively agree the order of the adjectives in my sentence above is wrong—weird even—without necessarily being able to describe exactly why; it flouts some rules that we all know but are not normally conscious of. These rules are difficult to write down precisely but roughly speaking follow a sort of semantic spectrum. The Cambridge Dictionary gives this order

opinion - size - physical nature - shape - age - colour - origin - material - type - purpose

If we rewrite my initial sentence using this order we get

my lovely big fat new red hardback grammar book

which sounds perfectly normal, albeit a bit wordy perhaps.

Mu new dictionary—the one pictured—calls the rules "Adjectival order", although "Royal Order of Adjectives" is a more colourful name. 
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Richard Walker

Etymology of "coleslaw"

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Eating my coleslaw on Monday, I wondered about the etymology. The first element is recognisable as meaning "cabbage", as in kale, Old English cole, German kohl, and Italian Cavalo, etc. all stemming (pun intended) from Latin Caulus, stalk or cabbage.

But "slaw" seemed a bit mysterious. 

It turns out to be from Dutch, sla, slaw, a contraction of "salad", and we borrowed "coleslaw" directly from Dutch koolsla,"cabbage salad".
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Richard Walker

Daffynition

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punctual:  I’ll fight everyone 
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Richard Walker

Reedbeds at Fowlmere

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Richard Walker

Answer to "Another Classic Problem"

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See here for the problem I posed.

The answer is the expected proportion of Heads and Tails is 50:50.

A plausible intuition is that the strategy described in the problem with result in more Heads being observed than Tails, but this intuition is misleading

Each coin flip is independent and over a large number of trials Heads and Tails will each turn up with about the same frequency, and no strategy we might decide to follow can change this ratio.

I wrote a Python program to simulate the problem and sure enough if we run it Heads and Tails each appear in with about the same frequency.

import random
H = 0
for i in range(1000000):
throw = random.choice(['H','T'])
    if throw != 'T':
        H = H + 1
    while throw != 'T':
        throw = random.choice(['H','T'])
        if throw  != 'T':
            H = H + 1
print('Heads per 1000000 Tails', H)

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Richard Walker

One Liner

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I was going to buy a cul-de-sac, but I decided it was a dead end.
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Richard Walker

Another Classic Problem

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Suppose 1,000 people take part in the following experiment.

Each participant is given a fair coin, which they proceed to flip one or more times, recording the outcome. As long as their coin turns up heads they keep flipping it. As soon as a tail comes up (but not before) they immediately stop flipping their coin.

What is the expected proportion of heads recorded versus the number of tails?

(This problem is usually framed a bit differently but I think the version given above is preferable to the traditional one.)
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Richard Walker

"Turkey Tail"

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We found this striking Turkey Tail fungus Trametes versicolor growing on a dead log in our garden.


From Wikipedia, it is found world-wide and lives by digesting lignin, a key structural component in plants= tissues. The fungus has been used in traditional Chinese medicine and a range of beneficial properties have also been claimed for it in the setting of conventional medicine. For a summary see this article. I haven't read enough to assess how strong the evidence is, and I think opinion is probably divided.

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Richard Walker

Tom Swifty

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Edited by Richard Walker, Sunday, 2 Mar 2025, 13:43

"I'm counting how many pots of tea you make", said Tom brutally.

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Richard Walker

A classic puzzle from Martin Gardner

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Edited by Richard Walker, Sunday, 2 Mar 2025, 22:44

This puzzle can be found in many places in the internet but was first published by Martin Gardner in his 'Mathematical Recreations' column in Scientific American. It goes like this (this wording is mine, not the original which I don't have to hand.)

Two mathematicians are sitting outside a café, and one say to the other, “I have three children, and the product of their ages is 36. Can you work out their ages?”

“No”, say the second mathematician, “I don’t have enough information”.

“Then what if I tell you the sum of their ages is equal to the number of that house directly opposite?” the first mathematician replies.

“Still not enough information”, answers the second.

“Well what if I tell you the oldest is called Bill?”, say the first.

“Aha, now I know the ages!” says the second mathematician.

1. Can YOU work out what the ages are? (Solution appears further down.)

2. Can you find a number different for 36 but which would still let the second mathematician deduce the ages if they knew the sum of the ages and the fact that the oldest child is called Bill?



PS The second triple is wrong see comments.

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Richard Walker

One Liner

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My crystal ball broke down so I took it back to the shop. They said they’d look into it.

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Richard Walker

Equatorial Plant

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A friend travelling in the Eastern Caribbean sent me some photos of flowers. This lovely plant is paperflower, or lesser bougainvillea, scientific name Bougainvillea glabra.


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Richard Walker

Looking out for Spring

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Richard Walker

"Harvard Interview Question: 90% were eliminated"

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I'm disbelieving of this being a Harvard Interview Question, but that is the title given to a question circulating on social media. It runs as follows:

"7 men have 7 wives. Each man and each wife have 7 children. What's the total number of people?"

It's ambiguous and that's probably why it has gone viral, with people offering a variety of answers and opinions and even the occasional quip, such as "Were they going to St Ives?".

So it's not the maths that is interesting but the impetus to debate, discuss wrangle, argue. disagree and tut about the ambiguity.

Still, to get back to the boring old maths, I thought I would ask:

If 7 is replaced by n, what is the smallest answer that we could put forward a case for, and what is the largest?

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Richard Walker

One Liner

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If Musk got involved in a cover-up, would it be stretching things to call it Elongate?

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