In 2011, Monash University mathematician Burkard Polster set out to answer a practical question: How precariously can you place a laptop computer on a crowded bedside table so that it will take up minimal space without falling off?

Assuming that both the table and the laptop are rectangular, and that the laptop’s center of gravity is its midpoint, it turns out that the optimal placement occurs when the laptop’s midpoint coincides with one of the table’s corners and the footprint is an isosceles right triangle, as above.

This also assumes that the table is reasonably sized. But then, if it’s tiny, then balancing a laptop on it probably isn’t your biggest problem.

(Burkard Polster, “Mathematical Laptops and Bedside Tables,” Mathematical Intelligencer 33:2 [July 2011], 33-35.)

A book lover is thinking of buying six books from a group of eight. The price of each book is a whole number of dollars, and none is less than $2. The prices are such that each possible selection of six books would cost the buyer a different sum. In the end he can’t make up his mind and buys all eight books. What is the smallest amount he must pay?

Each choice of six books from the group of eight leaves two behind. The price of each possible omitted pair must be unique, or else the corresponding sextets would cost the same, which we know is not the case. So we can solve the problem by working out the lowest possible price for each book that ensures that every possible pair has a distinct price.

The three lowest possible prices are $2, $3, and $4. That’s fine so far, but the next book can’t cost $5, because then we’d have two pairs with the same value (5 + 2 = 3 + 4). So we jump to 6 and see if that works. By looking always for the smallest possible next higher price for each volume, we’ll arrive at 2, 3, 4, 6, 9, 14, 22, 31, which gives a total price of $91.

But, interestingly, that’s not the answer! After Roland Sprague published this puzzle in his 1963 book Recreation in Mathematics, he found the solution 2, 3, 4, 6, 10, 15, 20, 30, which totals 90. And Fritz Düball later found 2, 3, 4, 6, 10, 16, 21, 26, which totals 88. Is that the lowest sum possible? Sprague doesn’t claim that it is, and I’ve not seen this problem elsewhere than in his book.

06/19/2022 UPDATE: Reader Michael Küll wrote a program to search for all solutions in a reasonable range. There are four solutions with a total amount less than or equal to $91:

88 :   2   3   4   6   10   16   21   26
90 :   2   3   4   6   10   15   20   30
91 :   2   3   4   6    9   14   22   31  
91 :   2   3   4   6   10   17   22   27

So Düball’s solution is indeed the best possible. (Thanks, Michael.)

A self-working card curiosity by Shippensburg University mathematician Douglas E. Ensley:

I give you the four aces from a deck of cards and turn my back. Then I ask you to stack the four cards face up with the heart at the bottom, then the club, the diamond, and the spade. Now turn the uppermost card, the spade, face down.

Now you’re invited to perform any of these operations as many times and in any order that you wish:

• Cut any number of cards from the top of the stack to the bottom.
• Turn the top two cards over as one.
• Turn the entire stack over.

When you’ve finished, I ask you to turn the topmost card over, then turn the top two cards over as one, then turn the top three cards over as one. I predict that the club is the only card facing the opposite way from the others, and as long as you’ve followed the directions above, it always will be.

The answer is explained by group theory — see the article below for the details.

(Douglas E. Ensley, “Invariants Under Group Actions to Amaze Your Friends,” Mathematics Magazine 72:5 [December 1999], 383-387.)

In 2006, Math Horizons challenged its readers to pose a problem in such a way that it contained its own answer. Rheta Rubenstein of the University of Michigan-Dearborn offered a pair of questions that answer one another:

1. What fraction of the letters in three-eighths are vowels?
2. What fraction of the letters in one-third are vowels?

(“Self-Answering Problems,” Math Horizons 13:4 [April 2006], 19.)

Oxford zoologists Tim Guilford and Dora Biro discovered a surprise in 2004: Homing pigeons sometimes just follow roads like the rest of us. Although the birds have inbuilt magnetic compasses, they fall back on the known landscape when they’re in familiar territory, following the lines of motorways and trunk roads.

Guilford and Biro strapped cameras and GPS devices to pigeons’ backs and watched them follow the A34 Oxford Bypass, turning at traffic lights and curving around roundabouts. They write, “One dominant linear feature, the A34 Oxford Bypass, appears to be associated with low entropy for much of its length, even where individual birds fly along or over it for a relatively short distance.”

“In fact, you don’t need a mini-GPS to find the circumstantial evidence” of this phenomenon, writes Joe Moran in On Roads. “You will often see seagulls in landlocked Birmingham because they have flown up the Bristol Channel and followed the M5, mistaking it for a river.”

(Tim Guilford et al., “Positional Entropy During Pigeon Homing Ii: Navigational Interpretation of Bayesian Latent State Models,” Journal of Theoretical Biology 227:1 [2004], 25-38.)

This style of compound magic square was first devised by Kenneth Kelsey of Great Britain. The numbers 70-94 appear in the blue boxes, making a pandiagonal magic square. The numbers 95-110 appear in the yellow circles, making a pandiagonal magic square of their own. And embedding one in the other produces a compound square — the numbers in the circles can be added to the numbers in the squares in either of the adjoining sections. So, for example, the second row, 87 + 79 + 91 + 83 + 70 = 410, can include the row of circles above it (87 + 108 + 79 + 105 + 91 + 99 + 83 + 98 + 70 = 820) or below it (87 + 95 + 79 + 102 + 91 + 104 + 83 + 109 + 70 = 820).

In the finished figure, every number from 70 to 110 appears once, and the blue square and the yellow square have the same magic constant — 410!

https://www.youtube.com/watch?v=aJhiY70KY5o

Stand a bicycle so that one pedal is in its lowest position and one in its highest. Now if we pull backward on the low pedal, will the bicycle move forward or backward?

Intuition suggests it will move forward: We’re turning the pedals in the same direction that a rider would, and normally this motion drives the rear wheel to propel the bike forward.

Surprisingly, though, in the experiment the bike moves backward. That’s because (in most bikes, in most gears) each pedal is constantly moving forward with respect to the ground. So pulling backward on the pedal doesn’t produce the intuitive result — it moves the pedal backward relative to the ground, and so produces the opposite result to the one we expect.

An exception: If the bike is in a sufficiently low gear, pulling backward on the pedal will drive the bike forward. But the sprocket ratio must be so low that really we’re betraying intuition rather than the reverse (see the video).