Poles Apart

A problem by Russian mathematician Viktor Prasolov:

A circle is divided by points into 3k arcs such that there are k arcs each of length 1, 2, and 3. Prove that two of the division points are diametrically opposite one another.

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Suppose that this is not the case, that none of the division points are diametrically opposite one another. This means that every arc of length 1 must stand opposite an arc of length 3. Remove one such pair, an arc of length 1 and its opposing arc of length 3. Now we’ve divided the circle into two large arcs. If one of these large arcs contains m arcs of length 1 and n arcs of length 3, then the other contains n arcs of length 1 and m arcs of length 3. The total number of arcs of length 1 and 3 in the two large arcs is 2(k– 1), so n + m = k – 1. Apart from arcs of length 1 and 3, each large arc contains only arcs of length 2, which don’t change the parity (odd or even) of its total length. This means that the parity of the length of each large arc will be the same as the parity of k – 1. But the length of each large arc is (6k – 1 – 3)/2 = 3k – 2. That’s a contradiction: k – 1 and 3k – 2 have opposite parities. So our assumption, that none of the division points on the original circle are diametrically opposite one another, cannot be the case. (From Prasolov’s Problems in Plane and Solid Geometry, 1988.)

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October 31, 2026October 30, 2026 | Puzzles

Turnabout

An “enigma in praise of women,” from an 1820 miscellany:

Happy that man may pass his life,
Who’s free from matrimonial chains;
Who is directed by a wife,
Is sure to suffer for his pains.

What tongue is able to unfold
The falsehood which in women dwell?
The worth in women you behold
Is almost imperceptible.

Adam could find no solid peace
When Eve was given him for a mate;
Till he beheld a woman’s face,
Adam was in a happy state.

See, in the female race appear
Hypocrisy, deceit, and pride;
Truth, darling of a heart sincere,
In women never can reside.

They’re always studying to employ
Their time in malice and in lies;
Their leisure hours in virtuous joy
To spend, ne’er in their thoughts arise.

Destruction take the men, I say,
Who make the fair their chief delight;
Who no regard to women pay,
Keep reason always in their sight.

“By reading the first and third lines together the Enigma is explained.”

Black and White

baird chess problem

By Mrs. W.J. Baird. White to mate in two moves.

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White wins with the quiet. 1. Qe2! Then: 1. … Kd5 2. Nb4# 1. … b4 2. Nf4# 1. … d5 2. Qc2# 1. … N any 2. Qe4# From the Illustrated Sporting & Dramatic News, 1902.

October 26, 2026October 25, 2026 | Puzzles

An Orderly Vacation

A puzzle from Peter Winkler’s excellent Mathematical Puzzles (2021):

A certain island contains a network of roads. Each road segment (between one intersection and the next) has its own name. If d is the average number of road segments that meet at an intersection, show that it’s possible to take a drive on the island that traverses at least d road segments and hits them in alphabetical order.

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Each road has two ends, so the total number of road ends observed at the island’s intersections is twice the number of roads on the island. The number we’re calling d, the average number of road segments that meet at an intersection, is thus 2m/n, where m is the number of roads on the island and n is the number of intersections. Station a pedestrian at each intersection. At time 1, let the pedestrians at either end of the alphabetically first road traverse it, changing places. At time 2, let the pedestrians who find themselves at either end of the alphabetically second road do the same. Continue this procedure until it’s been applied to all the roads. At the end of this exercise each pedestrian will have followed an alphabetical path (because we addressed the roads in alphabetical order). Each road is traversed twice, and the number of pedestrians equals the number of intersections on the island. So the average number of roads that each pedestrian traverses is 2m/n = d.

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October 13, 2026 | Puzzles

Road Thoughts

A puzzle by Sam Loyd:

Here is a pretty problem which I figured out during a ride from Bixley to Quixley astride a razor-back mule. I asked Don Pedro, a native guide who walked ahead of me pulling the mule forward by its reins, if my steed had another gait. He said it had but that it was much slower, so I pursued my journey at uniform speed. To encourage Don Pedro, who was my chief propelling power, I said we would pass through Pixley, so as to get some liquid refreshments; and from that moment he could think of nothing but Pixley.

After we had been traveling for forty minutes I asked how far we had gone. Don Pedro replied: ‘Just half as far as it is to Pixley.’

After creeping along for seven miles more I asked: ‘How far is it to Quixley?’ He replied as before: ‘Just half as far as it is to Pixley.’

We arrived at Quixley in another hour, which induces me to ask you to determine the distance from Bixley to Quixley.

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Two time intervals are given, but they’re not needed to solve the problem. The first question is asked between Bixley and Pixley; call that point x. The second question is asked between Pixley and Quixley; call that point y. The distance between x and y is 7 miles. The distance from x to Pixley is two-thirds the distance from Bixley to Pixley, and the distance from y to Pixley is two-thirds the distance between Pixley and Quixley. So the distance from x to y, 7 miles, is two-thirds the total distance, which must be 10.5 miles.

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October 10, 2026October 9, 2026 | Puzzles

Succinct

A puzzle by Emily Cox and Henry Rathvon, from the October 1991 issue of Games magazine:

You’re visiting a familiar island on which each resident either invariably lies or invariably tells the truth. Three residents appear.

Resident 1: We’re all liars.
Resident 2: Just one of us is a truth teller.

The third resident doesn’t speak. Which of the three are telling the truth?

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Residents 1 and 3 are liars. Resident 2 is a truth teller. Resident 1’s statement can’t be true, because that would entail a contradiction. So Resident 1 is a liar, and at least one of the three tells the truth. Resident 2 can’t be a liar — if she were, then Resident 3 would have to be a truth teller, and Resident 2’s statement would then be true, another contradiction. So Resident 2 is a truth teller, and by her statement Resident 3 must be a liar.

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October 8, 2026October 7, 2026 | Puzzles

Scrap Mettle

In 1882, Lewis Carroll invented a game he called MischMash:

The essence of this game consists in one player proposing a ‘nucleus’ (i.e., a set of two or more letters, such as GP, EMO, IMSE), and in the other trying to find a ‘lawful word’ (i.e., a word known in ordinary society and not a proper name), containing it. Thus, ‘maGPie,’ ‘lEMOn,’ and ‘hIMSElf’ are lawful words containing the nuclei GP, EMO, and IMSE.

He sent these nuclei to his godson. What English words contain these fragments?

OLS
VY
NGU
EWH
STEN

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BOLSTER, NAVY, TONGUE, SOMEWHERE, LISTEN. Bonus Lewis Carroll puzzle: “A and B began the year with only £1,000 a piece. They borrowed naught; they stole naught. On the New Year’s Day they had £60,000 between them. How did they do it?” “They went that day to the Bank of England. A stood in front of it, while B went round and stood behind it.”

October 5, 2026October 4, 2026 | Puzzles

“Tracking the Fugitives”

From the Strand, December 1903: These tracks were left by a man in flight leading a child. “In order to throw his pursuers off the trail the man changed his means of locomotion at each of the places marked in numbers on the diagram. The problem consists in discovering the manner of locomotion between each of these stations and the next; what vehicles the man took; what happened to him in the course of his flight; and what he did with the child at the different stages of the journey.”

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“The solution of this amusing problem is as follows: The fugitive started from station No. 1 on foot, carrying the child; at station No. 2 he mounted a bicycle and, still carrying the child, rode to No. 3; there he placed the child in a wheelbarrow; as indicated by the marks of the legs of the wheelbarrow, he stopped before reaching No. 4 and put down the child, who walked by his side to the station; thence he continued his journey on a tricycle, which also carried the child; at No. 5 he changed his tricycle for a monocycle (that is, a single-wheeled cycle, such as is used by trick-riders), but the child which he was carrying caused him to lose his blance and he fell; he then took the child in his arms and carried it to No. 6; thence he started holding the child by the hand, but farther on he again took it in his arms and so completed the journey at No. 7.”

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September 27, 2026September 26, 2026 | Puzzles

Turnabout

triangle reversal puzzle

A puzzle by A. Khalamaizer: Move three of these circles to make the triangle point downward.

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From the July-August 1996 issue of Quantum. Similar: Wildlife Management and Horse Sense.

September 25, 2026September 24, 2026 | Puzzles

Black and White

expedient chess puzzle 1

I don’t know who devised this; I found it in The Batsford Chess Puzzle Book (2002). Black has only a king and a bishop, while White still has all 16 of his men. But Black threatens to take White’s knight, giving mate immediately. If each side uses every legal expedient, what will be the outcome? For a start, White can stave off mate with 1. Bh2!

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1. Bh2 Bxh2 2. g3 Bxg3 3. Rf4 Bxf4 4. d6 Bxd6 5. Qh2! Bxh2 6. Re5 Bxe5 7. Nb5 Bf4 8. f6 (the only legal alternative to moving the needed b5 knight) Bxc7 9. Nxc7 and the game is drawn by stalemate!

September 24, 2026September 28, 2026 | Puzzles