How many watchmen are needed to guard the art gallery at left, so that every part of it is under surveillance? The answer in this case is 4; four guards stationed as shown will be able to watch every part of the gallery.

In 1973 University of Montreal mathematician Václav Chvátal showed that, in a gallery with n vertices, n/3 guards will always be enough to do the job. (If n/3 is not an integer, you can dispense with the fractional guard.) And Bowdoin College mathematician Steve Fisk found a beautifully simple proof of Chvátal’s result.

The figure at right shows another art gallery. Cut its floor plan into triangles, and color the vertices of each triangle with the same three colors. The full area of any triangle is visible from any of its vertices, and that means that the whole gallery can be guarded by stationing watchmen at the points indicated by any of the three colors. Choosing the color with the fewest vertices will give us n/3 guards (again discarding fractional guards).

The Chvátal and Fisk proofs both give an answer that’s sufficient but sometimes not necessary. In this case, the gallery has 12 vertices, and 12/3 guards (say, the four green ones) will certainly do the job, but here as few as two will be enough.

(Steve Fisk, “A Short Proof of Chvátal’s Watchman Theorem,” Journal of Combinatorial Theory, Series B 24:3 [1978], 374.)

Mathematician John Horton Conway invented this game in 1961. A line divides an infinite checkerboard into two territories. An army of soldiers occupies the lower territory, one per cell. They want to deliver a man as far as possible into the upper territory, but they can proceed only as in peg solitaire: One soldier jumps orthogonally over another soldier and lands on an empty square immediately beyond him, whereupon the “jumped” man is removed.

It’s immediately obvious how the soldiers can get a man into the upper territory, and it’s fairly clear how they can get one as far as the fourth row above the line. But, surprisingly, Conway proved that that’s the limit: No matter how they arrange their efforts, the soldiers cannot get a man beyond that row in a finite number of moves.

Christopher, the 15-year-old hero of Mark Haddon’s 2003 novel The Curious Incident of the Dog in the Night-Time, says that Conway’s Soldiers is “a good maths problem to do in your head when you don’t want to think about something else because you can make it as complicated as you need to fill your brain by making the board as big as you want and the moves as complicated as you want.”

You can find any number of proofs online, but the most convincing way to see that the task is impossible is to try it yourself.

The Coudersport Ice Mine is a cave near Sweden Township, Pa., that bears icicles in spring and summer but not in winter. A shaft about 12 feet long is located at the base of a steep hill. In winter, when the shaft is relatively dry, it fills with cold air. In the spring, snow begins to melt, and water accumulates at the bottom of the shaft. At the same time, cold air descends through rock crevices from higher in the hill, which focus it on this spot and freeze the water. By September this fund of cold air has been depleted, the ice melts, and the shaft is dry again when cold weather arrives.

“The general skepticism regarding the existence of this phenomenon has been illustrated many times of late and has furnished the people of Coudersport with an endless source of amusement,” noted the Popular Science Monthly in 1913.

In 1911 a Detroit man offered to bet anyone $100 or more that the story was true. “A millionaire ice manufacturer took the bet and eight other business men of Detroit followed suit. Two newspaper men were selected as stake-holders to decide the bets. They visited the mine and, of course, verified the newspaper story, much to the disgust of the nine losers.”

Patient H.M. went through experimental brain surgery in the 1950s to address a severe epileptic disorder. He emerged with a curiously compromised sense of smell: He could detect the presence and intensity of an odor, but he couldn’t consciously identify odors or remember them. He was unable to say whether two scents were the same or different, or to match one given scent to another. When asked to make conscious choices, he confused an odor’s quality with its intensity. And although he could name common objects using visual or tactile cues, he couldn’t identify them by smell.

“He can describe what he smells in some detail, but the descriptions do not correlate with the stimulus,” wrote chemist Thomas Hellman Morton, who examined and tested H.M. “Descriptions of the same odor vary widely from one presentation to another, and show no obvious trend when compared to his descriptions of different odors.”

Morton calls this “odor deafness,” by analogy with the “word deafness” found in some stroke victims, who can read, write, and hear but can’t recognize spoken words.

This raises an interesting philosophical question: Does H.M. have a sense of smell? If he can detect the presence of a scent and its intensity but can’t recognize it or distinguish it from others, is he smelling it?

(Thomas Hellman Morton, “Archiving Odors,” in Nalini Bhushan and Stuart Rosenfeld, Of Minds and Molecules, 2000.)

In 1972, as Charles Duke was training to visit the moon with Apollo 16, he regretted spending so much time away from his wife and sons. “So just to get the kids excited about what dad was going to do, I said, ‘Would y’all like to go to the moon with me?'” he told Business Insider. “We can take a picture of the family and so the whole family can go to the moon.”

“I talked with Dotty and the boys about it and they were delighted about having a picture of the Duke family on the Moon,” he wrote in his autobiography, Moonwalker. “So one day, Ludy Benjamin, a NASA photographer and good friend, came over to our house in Lago and took a picture of the four of us. On the back of the picture I wrote, ‘This is the family of astronaut Charles Duke of planet Earth, who landed on the moon on the twentieth of April 1972.’ Then we all signed it and put our thumbprints on the back.”

On April 23 Duke and John Young went exploring with the lunar rover in the Descartes Highlands, and he dropped the photo, wrapped in plastic, onto the surface and photographed it with his Hasselblad camera.

He left it there. “After 43 years, the temperature of the moon every month goes up to 400 degrees [Fahrenheit] in our landing area, and at night it drops almost absolute zero,” he said in 2015. “Shrink wrap doesn’t turn out too well in those temperatures. It looked OK when I dropped it, but I never looked at it again and I would imagine it’s all faded out by now.”

(Thanks, Bill.)

Answer these questions:

1. A bat and a ball cost $1.10 in total. The bat costs $1.00 more than the ball. How much does the ball cost? _____ cents
2. If it takes 5 machines 5 minutes to make 5 widgets, how long would it take 100 machines to make 100 widgets? _____ minutes
3. In a lake, there is a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? _____ days

The correct answers are 5 cents, 5 minutes, and 47 days, but each question also invites a quick, intuitive response that’s wrong. In order to succeed, you have to suppress your “gut” response and reflect on your own cognition deeply enough to see the error. Psychologist Shane Frederick devised the three-question test in 2005 to illustrate these two modes of thought, unreflective and reflective, which he called System 1 and System 2.

Scores on the CRT correlate with various measures of intelligence, patience, and deliberation, but cognitive ability alone isn’t strongly correlated with CRT scores: If you’re not disposed to answer impulsively then the problems aren’t hard, and if you do answer impulsively then cognitive ability won’t help you. A sample of students at MIT averaged 2.18 correct answers, Princeton 1.63, Carnegie Mellon 1.51, Harvard 1.43; see the link below for more.

(Shane Frederick, “Cognitive Reflection and Decision Making,” Journal of Economic Perspectives 19:4 [2005], 25-42.) (Thanks, Drake.)

So many more men seem to say that they may soon try to stay at home so as to see or hear the same one man try to meet the team on the moon as he has at the other ten tests.

This ungainly but grammatical 41-word sentence was constructed by Anton Pavlis of Guelph, Ontario, in 1983. It’s an alphametic: If each letter is replaced with a digit (EOMSYHNART = 0123456789), then you get a valid equation:

   SO     31
 MANY   2764
 MORE   2180
  MEN    206
 SEEM   3002
   TO     91
  SAY	 374
 THAT   9579
 THEY   9504
  MAY    274
 SOON   3116
  TRY    984
   TO     91
 STAY   3974
   AT     79
 HOME   5120
   SO     31
   AS     73
   TO     91
  SEE    300
   OR     18
 HEAR   5078
  THE    950
 SAME   3720
  ONE    160
  MAN    276
  TRY    984
   TO     91
 MEET   2009
  THE    950
 TEAM   9072
   ON     16
  THE    950
 MOON   2116
   AS     73
   HE     50
  HAS    573
   AT     79
  THE    950
OTHER  19508
+ TEN    906

TESTS  90393

Apparently this appeared in the Journal of Recreational Mathematics in 1972; I found the reference in the April 1983 issue of Crux Mathematicorum, which confirmed (by computer) that the solution is unique.

French mathematician Joseph Bertrand offered this observation in his Calcul des probabilités (1889). Inscribe an equilateral triangle in a circle, and then choose a chord of the circle at random. What is the probability that this chord is longer than a side of the triangle? There seem to be three different answers:

1. Choose two random points on the circle and join them, then rotate the triangle until one of its vertices coincides with one of these points. Now the chord is longer than a side of the triangle when its farther end falls on the arc between the other two vertices of the triangle. That arc is one third of the total circumference of the circle, so by this argument the probability is 1/3.

2. Choose a radius of the circle, choose a point on that radius, and draw a chord through that point that’s perpendicular to the radius. Now imagine rotating the triangle so that one of its sides also intersects the radius perpendicularly. Our chord will be longer than a side of the triangle if the point we chose is closer to the circle’s center than the point where the triangle’s side intersects the radius. The triangle’s side bisects the radius, so by this argument the probability is 1/2.

3. Choose a point anywhere in the circle and draw the chord for which this is the midpoint. This chord will be longer than a side of the triangle if the point we chose falls within a concentric circle whose radius is half the radius of the larger circle. That smaller circle has one-fourth the area of the larger circle, so by this argument the probability is 1/4.

Further methods yield still further solutions. After more than a century, the implications of Bertrand’s conundrum are still being discussed.