Bent Pent

This is a beautiful and interesting shape, reminiscent of a piece of jewellery:

Pentagons in a ring

I came across it in this tricky little word-puzzle:

Word puzzle using pentagon-ring

Here’s a printable version of the puzzle:

Printable puzzle

Let’s try placing some other regular polygons with s sides around regular polygons with s*2 sides:

Hexagonal ring of triangles

Octagonal ring of squares

Decagonal ring of pentagons

Dodecagonal ring of hexagons

Only regular pentagons fit perfectly, edge-to-edge, around a regular decagon. But all these polygonal-rings can be used to create interesting and beautiful fractals, as I hope to show in a future post.

Sampled (Underfoot)

Some interesting statistics from the American sociologist Elizabeth Wrigley-Field:

Here are three puzzles.

• American fertility fluctuated dramatically in the decades surrounding the Second World War. Parents created the smallest families during the Great Depression, and the largest families during the postwar Baby Boom. Yet children born during the Great Depression came from larger families than those born during the Baby Boom. How can this be?

• About half of the prisoners released in any given year in the United States will end up back in prison within five years. Yet the proportion of prisoners ever released who will ever end up back in prison, over their whole lifetime, is just one third. How can this be?

• People whose cancers are caught early by random screening often live longer than those whose cancers are detected later, after they are symptomatic. Yet those same random screenings might not save any lives. How can this be?

And here is a twist: these are all the same puzzle.

• Answers here: Length-Biased Sampling by Elizabeth Wrigley-Field

Proxi-Performative Post-Scriptum

The title of this post is, of course, a radical reference to core Led Zeppelin track “Trampled Underfoot” (1975).

Paradoxical Puzzle Pair

Two interesting puzzles, one of which looks hard and is in fact easy, while the other looks easy and is in fact hard.

1. Three Cards

The values attached to a deck of bridge cards start with the Two of Clubs as lowest, with Diamonds, Hearts and Ace of Spades as highest.

If you draw three cards at random from the deck, what is the probability that they will be drawn in order of increasing value? (Answer 1)

2. The Hungry Hunter

A hunter, having run out of food, met two shepherds. One of the shepherd had three loaves of bread and the other had five loaves. When the hunter asked for food, the shepherds agreed to divide the eight identical loaves equally between the three of them. The hunter thanked them and gave them $8. How should the shepherds divide the money? (Answer 2)

• Puzzles and answers from Erwin Brecher’s How Do You Survive a Duel? And Other Mathematical Diversions, Puzzles and Brainteasers (Carlton Books 2018)










Answer #1: The puzzle sounds far more complicated than it is. The deck of cards is a red herring. The question reduces to this: Take three cards, say 2, 3 and 4 of clubs, facedown. What is the probability of turning them over in the order 2, 3, 4? There are six possible ways of arranging three cards. Therefore the probability is one-sixth.










Answer #2: It would be wrong to split the money into $3 and $5. Each of the three ended up with 2⅔ loaves. In other words, the first shepherd parted with ⅓ of a loaf and the other shepherd with 2⅓ or 7/3 loaves. The first shepherd should therefore get $1 and the second shepherd $7.

Squooh You

Here’s an interesting little puzzle:

Winnie-the-Pooh and Piglet set out to visit one another. They leave their houses at the same time and walk along the same road. But Piglet is absorbed in counting the birds overhead, and Winnie-the-Pooh is composing a new “hum,” so they pass one another without noticing. One minute after the meeting, Winnie-the-Pooh is at Piglet’s house, and 4 minutes after the meeting Piglet is at Winnie-the-Pooh’s. How long has each of them walked? — “A puzzle by S. Sefibekov” viâ Futility Closet

If you’re good at maths, you should see the answer in an intuitive instant. I’m not good at maths, so it took me much longer, because I didn’t understand what was going on. But I can explain the answer like this. Pooh is obviously walking faster than Piglet. Therefore, Pooh and Piglet can’t have met after one minute, because that would mean Pooh takes one minute to walk the distance walked by Piglet in one minute.

So let’s suppose Pooh and Piglet met after two minutes. If Pooh takes one minute to walk the distance walked by Piglet in two minutes, then Pooh is walking twice as fast as Piglet. Does that work? Yes, because Piglet walks Pooh’s distance in four minutes, which is twice as long as Pooh took. Therefore Piglet is walking twice as slowly as Pooh. It’s symmetrical and we can conclude that they did indeed meet after two minutes. Pooh then walks another minute, for three minutes in total, and Piglet walks another four minutes, for six minutes in total.

But guessing is not a good way to find the answer to the puzzle. Let’s try to reason it through properly. Suppose that Pooh and Piglet meet after one unit of time, during which Piglet has walked one unit of distance and Pooh has walked x units of distance, where x > 1. In other words, Pooh is walking x times faster than Piglet. The distances they walk before meeting are therefore in the ratio:

1 : x

Next, note that Pooh will cover the distance Piglet has already walked in 1 unit / x = 1 minute, while Piglet covers the distance Pooh has already walked in x / 1 = 4 minutes. The times they take are therefore in the ratio:

1 / x : x → 1 : x^2 → 1 : 4

And if 1 : x^2 is 1 : 4 (the ratio of the minutes they walk after meeting), then 1 : x (the ratio of the distances they walk before meeting) = 1 : √(x^2) = 1 : √4 = 1 : 2. Pooh is therefore walking 2x faster than Piglet and Piglet is walking 2x slower than Pooh. If Pooh covers Piglet’s distance in 1 minute, Piglet must have taken 2 minutes to walk that distance. And if Piglet covers Pooh’s distance in 4 minutes, Pooh must have taken 2 minutes to walk that distance.

Therefore, when they meet, each of them has been walking for 2 minutes. Pooh therefore walks 2 + 1 = 3 minutes in total and Piglet walks 2 + 4 = 6 minutes in total.

The result can be generalized for all relative speeds. Suppose that Pooh and Piglet meet after m1 minutes and that Pooh then takes m2 minutes to walk the distance Piglet walked in m1 minutes, while Piglet takes m3 minutes to walk the distance Pooh walked in m1 minutes. The time they walk before they meet, m1 minutes, is therefore supplied by this simple equation:

m1 = √(m3 / m2)

And you can call √(m3 / m2), the square root of m3 / m2, the squooh function:

m1 = √(m3 / m2) = squooh(m2,m3)

Now suppose the distance between Pooh’s and Piglet’s houses houses is 12 units of distance and that Piglet always walks at 1 unit a minute. If Pooh walks at the same speed as Piglet, i.e. 1 unit a minute, then:

After they meet, Pooh walks 6 more min = m2, Piglet walks 6 more min = m3.

How long do they walk before they meet?

m1 = m3 / m2 = 1, √1 * 6 = 6

They meet after 6 min.

Now suppose that after they meet, Pooh walks 2 more min, Piglet walks 8 more min.

Therefore, m3 / m2 = 4, √4 * 2 = 2 * 2 = 4 = m1

Therefore they walk for 4 min before they meet and Pooh walks 2x faster than Piglet.

After they meet, Pooh walks 1 more min, Piglet walks 9 more min (m3 / m2 = 9, √9 * 1 = 3)

Therefore they walk for 3 min before they meet and Pooh walks 3x faster than Piglet.

After they meet, Pooh walks 0·6 more min, Piglet walks 9·6 more min (m3 / m2 = 16, √16 * 0·6 = 4 * 0·6 = 2·4)

Therefore they walk for 2·4 min before they meet and Pooh walks 4x faster than Piglet:

After they meet, Pooh walks 0·4 more min, Piglet walks 10 more min (m3 / m2 = 25, √25 * 0·4 = 5 * 0·4 = 2)

Therefore they walk for 2 min before they meet and Pooh walks 5x faster than Piglet.

And so on.

Pigmental Paradox

From Raymond Smullyan’s Logical Labyrinths (2009):

We now visit another knight/knave island on which, like on the first one, all knights tell the truth and all knaves lie. But now there is another complication! For some reason, the natives refuse to speak to strangers, but they are willing to answer yes/no questions using a secret sign language that works like this:

Each native carries two cards on his person; one is red and the other is black. One of them means yes and the other means no, but you are not told which color means what. If you ask a yes/no question, the native will flash one of the two cards, but unfortunately, you will not know whether the card means yes or no!

Problem 3.1. Abercrombie, who knew the rules of this island, decided to pay it a visit. He met a native and asked him: “Does a red card signify yes?” The native then showed him a red card.

From this, is it possible to deduce what a red card signifies? Is it possible to deduce whether the native was a knight or a knave?

Problem 3.2. Suppose one wishes to find out whether it is a red card or a black card that signifies yes. What simple yes/no question should one ask?

Shareway to Seven

An adaptation of an interesting distribution puzzle from Joseph Degrazia’s Math is Fun (1954):

After a successful year of plunder on the high seas, a pirate ship returns to its island base. The pirate chief, who enjoys practical jokes and has a mathematical bent, hands out heavy bags of gold coins to his seven lieutenants. But when the seven lieutenants open the bags, they discover that each of them has received a different number of coins.

They ask the captain why they don’t have equal shares. The pirate chief laughs and tells them to re-distribute the coins according to the following rule: “At each stage, the lieutenant with most coins must give each of his comrades as many coins as that comrade already possesses.”

The lieutenants follow the rule and each one in turn becomes the lieutenant with most coins. When the seventh distribution is over, all seven of them have 128 coins, the coins are fairly distributed, and the rule no longer applies.

The puzzle is this: How did the pirate captain originally allocate the coins to his lieutenants?

If you start at the beginning and work forward, you’ll have to solve a fiendishly complicated set of simultaneous equations. If you start at the end and work backwards, the puzzle will resolve itself almost like magic.

The puzzle is actually about powers of 2, because 128 = 2^7 and when each of six lieutenants receives as many coins as he already has, he doubles his number of coins. Accordingly, before the seventh and final distribution, six of the lieutenants must have had 64 coins and the seventh must have had 128 + 6 * 64 coins = 512 coins.

At the stage before that, five of the lieutenants must have had 32 coins (so that they will have 64 coins after the sixth distribution), one must have had 256 coins (so that he will have 512 coins after the sixth distribution), and one must have had 64 + 5 * 32 + 256 coins = 480 coins. And so on. This is what the solution looks like:

128, 128, 128, 128, 128, 128, 128
512, 64, 64, 64, 64, 64, 64
256, 480, 32, 32, 32, 32, 32
128, 240, 464, 16, 16, 16, 16
64, 120, 232, 456, 8, 8, 8
32, 60, 116, 228, 452, 4, 4
16, 30, 58, 114, 226, 450, 2
8, 15, 29, 57, 113, 225, 449

So the pirate captain must have originally allocated the coins like this: 8, 15, 29, 57, 113, 225, 449 (note how 8 * 2 – 1 = 15, 15 * 2 – 1 = 29, 29 * 2 – 1 = 57…).

The puzzle can be adapted to other powers. Suppose the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades twice as many coins as that comrade already possesses.” If the pirate captain has six lieutenants, after each distribution each of five will have n + 2n = three times the number of coins that he previously possessed. The six lieutenants each end up with 729 coins = 3^6 coins and the solution looks like this:

13, 37, 109, 325, 973, 2917
39, 111, 327, 975, 2919, 3
117, 333, 981, 2925, 9, 9
351, 999, 2943, 27, 27, 27
1053, 2997, 81, 81, 81, 81
3159, 243, 243, 243, 243, 243
729, 729, 729, 729, 729, 729

For powers of 4, the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades three times as many coins as that comrade already possesses.” With five lieutenants, each of them ends up with 1024 coins = 4^5 coins and the solution looks like this:

16, 61, 241, 961, 3841
64, 244, 964, 3844, 4
256, 976, 3856, 16, 16
1024, 3904, 64, 64, 64
4096, 256, 256, 256, 256
1024, 1024, 1024, 1024, 1024

For powers of 5, the rule runs like this: “At each stage, the lieutenant with most coins must give each of his comrades four times as many coins as that comrade already possesses.” With four lieutenants, each of them ends up with 625 coins = 5^4 coins and the solution looks like this:

17, 81, 401, 2001
85, 405, 2005, 5
425, 2025, 25, 25
2125, 125, 125, 125
625, 625, 625, 625

Shick Shtick

Slightly adapted from Joseph Degrazia’s Math is Fun (1954):

Six Writers in a Railway Car

On their way to Chicago for a conference of authors and journalists, six writers meet in a railway club car. Three of them sit on one side facing the other three. Each of the six has his specialty. One writes short stories, one is a historian, another one writes humorous books, still another writes novels, the fifth is a playwright and the last a poet. Their names are Abbott, Blake, Clark, Duggan, Eccles and Farmer.* Each of them has brought one of his books and given it to one of his colleagues, so that each of the six is deep in a book which one of the other five has written.

Abbott reads a collection of short stories. Clark reads the book written by the colleague sitting just opposite him. Blake sits between the author of the short stories and the humorist. The short-story writer sits opposite the historian. Duggan reads a play. Blake is the brother-in-law of the novelist. Eccles sits next to the playwright. Abbott sits in a corner and is not interested in history. Duggan sits opposite the novelist. Eccles reads a humorous book. Farmer never reads poems.

These facts are sufficient to find each of the six authors’ specialties.

*In the original, the surnames were Blank, Bird, Grelly, George, Pinder and Winch.