First Person

When Elizabeth Thompson married Maj. Sir William Butler in 1877, she was already a respected painter of military subjects. But becoming Lady Butler gave her a unique opportunity: She could now watch maneuvers in person and even stand in front of charging cavalry to study the momentum of the horses.

The startling result, Scotland for Ever, depicts a head-on charge of the Royal Scots Greys, the cavalry regiment that Napoleon had hailed as “those terrible men on grey horses” at Waterloo.

The painting was an enormous success and became a symbol of British military heroism. The scene is a bit exaggerated — in their famous charge the advancing horses had never reached a full gallop due to the broken ground. But then most of the painting’s admirers would never have guessed that the artist had never witnessed a battle.

July 19, 2016 | Art · History

A Separate Peace

After 30 years of searching, acoustic ecologist Gordon Hempton thinks he’s found the “quietest square inch in the United States.” It’s marked by a red pebble that he placed on a log at 47°51’57.5″N, 123°52’13.3″W, in a corner of the Hoh Rainforest in Olympic National Park in western Washington state. The area is actually full of sounds, but the sounds are natural — by quietest, Hempton means that this point is subject to less human-made noise pollution than any other spot in the American wilderness.

Hempton hopes to protect the space by creating a law that would prohibit air traffic overhead. “From a quiet place, you can really feel the impact of even a single jet in the sky,” he told the BBC. “It’s the loudest sound going. The cone of noise it drags behind it expands to fill more than 1,000 square miles. We wanted to see if a point of silence could ripple out in the same way.”

His website, One Square Inch, has more information about his campaign. “Unless something is done,” he told Outside Online, “we’ll see the complete extinction of quiet in the U.S. in our lifetime.”

July 19, 2016July 18, 2016 | Science & Math · Society

Black and White

meyer chess problem

By Heinrich Friedrich Ludwig Meyer. White to mate in two moves.

	Select Click for Answer
1. Bc8! half-pins both black rooks. Now Black has his choice of 23 different rook moves, but each of them allows the white knight to hop into either f4 or g5, giving mate. From the Boy’s Own Paper, 1908.

July 18, 2016July 24, 2016 | Puzzles

Podcast Episode 114: The Desperation of Donald Crowhurst

donald crowhurst

In 1968 British engineer Donald Crowhurst entered a round-the-world yacht race, hoping to use the prize money to save his failing electronics business. Woefully unprepared and falling behind, he resorted to falsifying a journey around the world. In this week’s episode of the Futility Closet podcast we’ll describe the desperate measures that Crowhurst turned to as events spiraled out of his control.

We’ll also get some updates on Japanese fire balloons and puzzle over a computer that turns on the radio.

See full show notes …

July 18, 2016October 5, 2017 | Oddities · Podcast

Road Work

Fed up with endless traffic detours in 1830, London printer Charles Ingrey published a pointed puzzle, Labyrinthus Londoninensis, or The Equestrian Perplexed.

“The object is to find a way from the Strand [lower left] to St. Paul’s [center], without crossing any of the Bars in the Streets supposed to be under repair.”

Mending our Ways, our ways doth oft-times mar,
So thinks the Traveller by Horse or Car,
But he who scans with calm and patient skill
This ‘Labyrinthine Chart of London’, will
One Track discover, open and unbarred,
That leads at length to famed St. Pauls Church Yard.

The image above is a bit too small to navigate, but the British Library has an interactive zoomable version (requires Flash).

I don’t have the solution, but The Court Journal of Dec. 14, 1833, hints that “the farthest way round is the nearest way home.”

07/06/2022 UPDATE: A solution! (Thanks, Paul.)

July 16, 2016July 7, 2022 | Puzzles

Footwork

A poser from Penn State mathematician Mark Levi’s Why Cats Land on Their Feet (2012):

Using only a stopwatch and a sneaker, how can you find an approximate value for $\sqrt{2}$ ?

	Select Click for Answer
Suspend the sneaker by its lace to make a pendulum. Using the stopwatch, record the number of swings it makes in one minute, and call this n₁. Double the lace to halve its length and again record the number of swings in one minute. Call this n₂. Now $\sqrt{2}\approx n_{2}/n_{1}.$ For better precision, record the number of swings over a longer period of time. To estimate $\sqrt{3}$ , use a lace 1/3 as long, and so on.

Click for Answer

July 15, 2016July 24, 2016 | Puzzles

Figures

A foolish young man said to John Wilkes (1725-1797), “Isn’t it strange that I was born on the first of January?”

“Not strange at all,” said Wilkes. “You could only have been conceived on the first of April.”

July 15, 2016July 14, 2016 | Humor

A Square Triangle

https://commons.wikimedia.org/wiki/File:3-Pascal.png — Image: Wikimedia Commons

The product of the six numbers surrounding any interior number in Pascal’s triangle is a perfect square.

July 14, 2016 | Science & Math

“On the Question of Choice”

A leaf was riven from a tree,
“I mean to fall to earth,” said he.

The west wind, rising, made him veer.
“Eastward,” said he, “I now shall steer.”

The east wind rose with greater force.
Said he, “‘Twere wise to change my course.”

With equal power they contend.
He said, “My judgment I suspend.”

Down died the winds; the leaf, elate,
Cried: “I’ve decided to fall straight.”

“First thoughts are best?” That’s not the moral;
Just choose your own and we’ll not quarrel.

Howe’er your choice may chance to fall,
You’ll have no hand in it at all.

— Ambrose Bierce

July 14, 2016July 13, 2016 | Poems

Hat Check

A puzzle from MIT Technology Review, July/August 2008:

Each of three logicians, A, B, and C, wears a hat that displays a positive integer. The number on one of the hats is the sum of the numbers on the other two. They make the following statements:

A: “I don’t know my number.”

B: “My number is 15.”

What numbers appear on hats A and C?

	Select Click for Answer
A is 10 and C is 5. The situation above, in which A doesn’t know his own number but B knows his after learning this, occurs if and only if (1) A and C have the same number or (2) A = 2C and B = 3C. Here’s the proof. If A and C have the same number, x, then B’s number must be 2x. Knowing this, A can conclude only that his own number must be either x or 3x, so he’ll declare that he doesn’t know his own number. B will see that both A and C bear the number x and declare that his own number is 2x. That takes care of (1). For (2), suppose that C’s number is x, A’s number is 2x, and B’s number is 3x. A can conclude only that his own number is either 2x or 4x, so he declares that he doesn’t know his own number. B can see that his own number must be x or 3x, but he realizes that he can eliminate x since, if that were the case, A would have seen two x‘s and could have concluded that his own number was 2x. So that shows that in cases (1) and (2) A and B will make the declarations presented in the puzzle statement. Now we have to show that these are the only cases that produce these two responses. So consider the three possibilities where A, then C, then B is the sum of the other two. (i) Here suppose C’s number is x, A’s is x + y, and B’s is y. If y = x, then A will see this and can conclude that his own number is 2x. If y ≠ x then A can’t know his own number. B knows that his own number must be either y or 2x + y. Neither of these numbers equals x, so both possibilities are consistent with A’s response, and B can’t know his number. (ii) Suppose C’s number is x + y, A’s is x, and B’s is y. A can’t tell whether his own number is x or x + 2y and so responds negatively. B knows that his number is either y or 2x + y. Neither number equals x + y, so B can’t eliminate either possibility and can’t know his own number. (iii) Here C’s number is x, A’s is y, B’s is x + y, and we know that x ≠ y (the case where x = y is covered under (1) above). So there are two possibilities: — If x > y then A responds negatively and B concludes that his own number must be either x + x or x – y. Neither of these possibilities equals x, so B can’t eliminate either one. — If x < y then A responds negatively and B concludes that his own number must be either x + y or y – x. Since x + y ≠ x, he can’t eliminate x + y. But he can eliminate y – x when y – x = x; that is, when y = 2x. This is (2) above, proving the claim. In this particular case we know that (1) is impossible since B claimed the number 15, an odd number. So (2) must be the case, and 3x = 15.

Click for Answer

A is 10 and C is 5. The situation above, in which A doesn’t know his own number but B knows his after learning this, occurs if and only if (1) A and C have the same number or (2) A = 2C and B = 3C.

Here’s the proof. If A and C have the same number, x, then B’s number must be 2x. Knowing this, A can conclude only that his own number must be either x or 3x, so he’ll declare that he doesn’t know his own number. B will see that both A and C bear the number x and declare that his own number is 2x.

That takes care of (1). For (2), suppose that C’s number is x, A’s number is 2x, and B’s number is 3x. A can conclude only that his own number is either 2x or 4x, so he declares that he doesn’t know his own number. B can see that his own number must be x or 3x, but he realizes that he can eliminate x since, if that were the case, A would have seen two x‘s and could have concluded that his own number was 2x.

So that shows that in cases (1) and (2) A and B will make the declarations presented in the puzzle statement. Now we have to show that these are the only cases that produce these two responses. So consider the three possibilities where A, then C, then B is the sum of the other two.

(i) Here suppose C’s number is x, A’s is x + y, and B’s is y. If y = x, then A will see this and can conclude that his own number is 2x. If y ≠ x then A can’t know his own number. B knows that his own number must be either y or 2x + y. Neither of these numbers equals x, so both possibilities are consistent with A’s response, and B can’t know his number.

(ii) Suppose C’s number is x + y, A’s is x, and B’s is y. A can’t tell whether his own number is x or x + 2y and so responds negatively. B knows that his number is either y or 2x + y. Neither number equals x + y, so B can’t eliminate either possibility and can’t know his own number.

(iii) Here C’s number is x, A’s is y, B’s is x + y, and we know that x ≠ y (the case where x = y is covered under (1) above). So there are two possibilities:

— If x > y then A responds negatively and B concludes that his own number must be either x + x or x – y. Neither of these possibilities equals x, so B can’t eliminate either one.

— If x < y then A responds negatively and B concludes that his own number must be either x + y or y – x. Since x + y ≠ x, he can’t eliminate x + y. But he can eliminate y – x when y – x = x; that is, when y = 2x. This is (2) above, proving the claim.

In this particular case we know that (1) is impossible since B claimed the number 15, an odd number. So (2) must be the case, and 3x = 15.

July 13, 2016July 24, 2016 | Puzzles

An idler's miscellany of compendious amusements

Author: Greg Ross