Personal Blogs

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autumn leaves

Another one from Cut the Knot : Probability Riddles by Alexander Bogomolny. This is riddle 3.9, my paraphrase.

Riddle

A standard pack of 52 playing cards contains as we know four aces. If the pack is thoroughly shuffled and we then work our way through it, the first ace we encounter may be at any of the positions . Which of these has the highest probability of being the position where the first ace is found?

Scroll down for solution

My Slightly Simplified Version of the Solution

Let be the probability that the first ace is at position . We count the number of possible ways the aces may be arranged if the first ace is at position .

If the first ace is at position , the remaining must be each be located at one of the positions from to , which can be done in ways, all equally likely to occur, given the pack is in random order.

Now we calculate the total number of ways the fours aces can be arranged altogether, irrespective of where the first ace is. This is .

The probability is therefore . This plainly decreases with and so its maximum is when , and (as we would naturally expect).

This popped nto my head this morning and I believe it is original; at any rate it is not in The Canonical List of Top Swifties, which only lists

80. "Please, Christopher," said Tom crisply.

81. "Let's go get some lettuce," said Tom crisply.

Interestingly, when I went to look at a sample Tom Swift adventure, I discovered the eponymous Tom gives a "crisp" answer near the start of Tom Swift and his House on Wheels, from 1925, which you can find at Project Gutenberg.

Tom Swift, with a negative shake of his head, shoved several papers across the table that separated him from a burly, red-faced man whose eyes narrowly observed the young inventor.

"Then you refuse this contract, Mr. Swift—a contract for constructing over one hundred thousand dollars' worth of machinery on which you can make a handsome profit? You absolutely refuse it?"

The red-faced man in his eagerness was leaning forward now.

"Yes, Mr. Cunningham, I refuse!" was Tom's crisp answer. "The Swift Construction Company does not care to handle it."

Here is the front cover. I think I might try and get an original copy secondhand, if it's not too pricey. Definitely a collector's item.

What's in a name? Here are six Goodies and six Baddies all mixed up. Can you sort out which are the Goodies and which the Baddies?

1 Malzor Kest
2 Elowen Brightmere
3 Tarian Wren
4 Dravik Rane
5 Corin Emberleaf
6 Vexigar Corven
7 Zargath Vorr
8 Alarion Vale
9 Seraphine Lark
10 Krulmar Veyn
11 Lysandra Dawnfall
12 Mordrek Voss

My answers

Credit: I am indebted to Copilot for inventing these, following a set of (unwritten?) conventions for naming fantasy characters.

About 90% of the energy we use comes directly or indirectly from the Sun. Apart from solar panels, we have wind and hydroelectric power, which are ultimately driven by the effect of solar radiation on the climate, and burning of fossil fuels which are the remains of photosynthesis by plants that grew millions of year ago and stored up sunlight. And then there is combustion of wood and biowaste, but these too are stored sunlight.

The bulk of the energy generated in the Sun is from hydrogen atoms fusing together at high pressures and pressures to create helium atoms. In the process some mass is converted to energy according to Einstein's famous equation.

The fusion actually happens in three steps called the Proton-Proton Chain Reaction, with six hydrogens atomic mass 1 being input and one helium atomic mass 4 and two hydrogens emerging at the end. It's a bit more complicated but that is the gist.

Thinking about this I wondered if I could relate something we are all familiar with — powering a light bulb — to actual numbers of helium atoms being created in the sun.

How many helium atoms does it take to power a lightbulb?

Assuming a 10 w bulb, we need to find 10 Joules of energy each second (a Watt is one Joule per second by definition).

The creation of one helium 4 atom releases about Joules, so or around 2,500,000,000,000 helium atoms must be created per second.

Here's a (very) peripherally related joke.

"When Einstein met Pythagoras"

I've just bought a 'Sensitive Plant' or 'Touch-Me-Not', Mimosa pudica.  Mine is quite small at the moment and nowhere near flowering, but I found a good photo on Wikimedia ^[1].

This plant is remarkable in more than one way. Firstly, if you touch it, a kind of impulse runs through its foliage, and the plant immediately folds up nearby leaves. It's this behaviour that causes Linnaeus to dub it pudica, 'shy'. 

Even more remarkably it has a circadian rhythm. During the hours of darkness the plant furls its leaves but begins unfurling them ahead of the dawn; it's got a biological clock.

And it can learn; if its leaves are touched repeatedly without any harm being caused to them, it will come to disregard these stimuli.

Mimosa pudica , a member of the pea and lentil family, came originally from the Caribbean and Central America but is now widespread in parts of the world with warm climates, and a weed in some places. In temperate regions it is a fairly common indoor plant.

For some interesting botanical history of the plant see the article Sensitive Plant (Mimosa pudica) on the Cambridge University Botanic Garden website.

[1] Suyash.dwivedi, Mimosa pudica in September month, Wikimedia, Licensed Under Creative Commons.

There was an Old Man of St Bees,

Who was stung on the Nose by a Wasp.

When they asked, "Does it hurt?"

He replied. "No, not much."

"I'm so glad it wasn't a Hornet." ^[1]

The modern English name of the insect Vespula vulgaris is of course wasp, but in Old English it was wæps or wæsp, and this goes back via a Germanic line of descent to a Proto-Indo-European (PIE) root *wapso-. This is also the origin of Latin vespa, and the latter may have influenced the form of modern wasp.

Interestingly there is a dialect version wopsie, which I've always been aware of and occasionally use, and this is surprisingly close to the ancestral root, so much so that it's possible to imagine that if we could travel 5,000 years back in time we might agree with people living then what the insect is called.

The French name is guêpe and if we recall that the circumflex over the 'e' originated as a scribal abbreviation for an 's', we can see it is not all that far from Latin vespa after all. But it's thought that it is a kind of amalgam of two words, both coming from PIE *wapso-, one from Frankish (so via the Germanic route) wapsu and the other from Latin vespa.

As a final aside, the name of the Roman Emperor Vespasian might derive from vespa, and there are Italian people with the surname Vespa (and also a famous brand of scooter) and Wasp is an English surname, although quite rare.

[1] I have quoted this Limerick as I remember it but checking the original by W.S. Gilber (of Gilbert and Sullivan), poking fun at Edward Lear, I see my version is a little different, but even more ridiculous so I've left it as is.

Hedgehogs – why can't they just share the hedge?

This was the joke that won Dan Antopolski the Funniest Joke of the Fringe award at the 2009 Edinburgh Festival Fringe. ^[1] 

Hedgehogs were not always called that. In old English, it was an igil, cognate with modern German Igel. But in the early part of the Middle English period this was displaced by yrichon, 'urchin', a word of French origin. This was in turn replaced in the 15^th century by heyghoge and hedgehog it has been ever since.

But where did igil and urchin come from? Well Old French herichun is from Latin ericius and the latter is thought to derive from a Proto-Indo-European (PIE) root *ghers-, 'bristly' or 'spiny', probably the origin also of horror, in the sense of your hair standing on end, bristling, you see.

At this point you may be guessing igil is from the same root, but you'd be wrong ; igil is from a different root *heg-, 'hedgehog'. The hedgehog had a PIE root all to itself, hogging it you might say!

This is also the source of Ancient Greek ekhinos, 'hedgehog' or 'sea urchin'. which of course is where we get echinoderm. It seems people have often seen a parallel between between the spiny land animal and the spiny sea creature, and at one time hedgehogs were sometimes called land urchins to distinguish them.

Modern Greek has a different word for hedgehog which I rather like: skandzóchiros, 'prickly pig', which seems a bit more descriptive than the English equivalent.

[1] BBC News. (2009, August 24). Hedgehog joke is Fringe funniest. bbc.co.uk

Leon Tolstoy, as well as being one of the most famous novelists of all time, was an educationist, who founded a school for peasant children in his country estate, and wrote several textbooks for their use, one of which dealt with arithmetic.

I haven't been able to find the original text of Tolstoy's arithmetic book, but one problem that has become quite famous and appeared in a number of places is reproduced in Yakov Perelman's book Algebra Can be Fun. Here is the wording of the problem as it appears there.

A team of haymakers were assigned the task of scything two meadows, one twice the size of the other. Half a day the team worked on the larger meadow. Then it split into two equal groups: the first remained in the larger meadow and finished it by evening; the second group scythed the smaller meadow, but by evening there still remained a portion to do; this portion was scythed the next day by one haymaker in a single day’s work. How many men were there in the team?

Scroll down to see a solution

Solution: There are doubtless many ways to solve the puzzle but this was my take on it.

Suppose there are people in the team. Then in the first half-day they will put in person-days of work. After the team splits, the people remaining in the first field will work for half a day, so putting in a further person-days, for a total of person-days to mow the first meadow.

Over in the second field the people diverted there will work for half a day, making man-days, and on the second day the lone mower will contribute a further person day, for a total of .

But we know the first is twice the size of the second so

, , which implies and that is the answer.

Imagine your calculator is broken. It won't handle numbers beyond 1,000,000,000,000 but you have been asked to find the last two digits of .

Can you still find the answer. What is it?

Scroll down for Solution and Discussion.

Solution

Even with the broken calculator we can start calculating some small powers of and when we get to we discover the last two digits are . This means all end in (01\).

So will end in also and to find the last two digits of we can just multiply by , which tells us the answer is .

In fact it is possible to prove the rather astonishing fact, that for any prime , has the same last two digits as .

And so for example we can be sure that the last two digits of , a number of some 2,000 digits, are .

If we start with a triangle ABC and erect squares on two of its sides, then join the two outer vertices of the squares it creates a new triangle, called a flank triangle.

Theorem:  If we draw a line passing through the midpoint of the outer edge of the flank triangle and the apex of the original triangle, it will be perpendicular to the base of that triangle. In the diagraml ine HCI is at right angles to AB.

Proof: Rotate the flank triangle through 90° about point C.

Consider the triangle ABD'.

From the construction:

ACD' must be a straight line, and AC = CG, so C is the midpoint of AD'.

BD' is the image of GD and H' the image of midpoint H, so H' is the midpoint of BD'.

Hence CH' joins the midpoints of two sides of ABD' and is therefore parallel to the base AB. CH' is perpendicular to CH and hence to HI.

But this means HI must be perpendicular to AB, because CH' and AB are parallel. //

Exercise: Prove the flank triangle and the original triangle have equal areas.

I got this problem from Cut-the-Knot 'Properties of flank triangles' and drew the graphics in GeoGebra.

If you cut a sheet of A4 paper in half you get two sheets of A5, which have half the area but maintain the same proportions of length to width.

Can we do the same in 3D? Yes we can!! Imagine a block with dimensions , and , with being the longest side.

If we "cut along the dotted line" and let be the longest side of the new blocks so formed, we need the proportions to remain the same,so we can write

or . Solving gives

and . For a proper "A" system we need A0 to have a volume of 1 m³, in the same way as an A0 sheet of paper has an area of 1 m². So

m³ and making the dimensions of the A0 block 1260 mm x 1000 mm x 794 mm.

This extends naturally to higher dimensions, even though we can't physically construct the higher-dimensional blocks we can describe them mathematically. In four dimensions, the A0 hyperbox would have dimensions in the ratio and higher dimensions follow the same pattern.