Suppose you have three identical Dewar flasks labeled A, B, and C. (A Thermos is a Dewar flask.) You also have an empty container labeled D, which has thermally perfect conducting walls and which fits inside a Dewar flask.
Pour 1 liter of 80°C water into flask A and 1 liter of 20°C water into flask B. Now, using all four containers, is it possible to use the hot water to heat the cold water so that the final temperature of the cold water is higher than the final temperature of the hot water? How? (You can’t actually mix the hot water with the cold.)
Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.
The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.
The first die has 5 red faces and 1 blue face. What are the colors on the second die?
From the Second All Soviet Union Mathematical Competition, Leningrad 1968:
On a teacher’s desk sits a balance scale, on which are a set of weights. On each weight is the name of at least one student. As each student enters the classroom, she moves all the weights that bear her name to the other side of the scale.
Before any students enter, the scale is tipped to the right. Prove that there’s some set of students that you can let into the room that will will tip the scale to the left.
Two circles intersect. A line AC is drawn through one of the intersection points, B. AC can pivot around point B — what position will maximize its length?
A puzzle by Pierre Berloquin:
Timothy, Urban, and Vincent are digging identical holes in a field.
- When Timothy and Urban work together, they dig 1 hole in 4 days.
- When Timothy and Vincent work together, they dig 1 hole in 3 days.
- When Urban and Vincent work together, they dig 1 hole in 2 days.
Working alone, how long does it take Timothy to dig one hole?
On Oct. 25, 1875, Lewis Carroll sent this verse to Mrs. J. Chataway, mother of one of his child-friends, Gertrude. “They embody, as you will see, some of my recollections of pleasant days at Sandown”:
Girt with a boyish garb for boyish task,
Eager she wields her spade — yet loves as well
Rest on a friendly knee, the tale to ask
That he delights to tell.
Rude spirits of the seething outer strife,
Unmeet to read her pure and simple spright,
Deem, if you list, such hours a waste of life,
Empty of all delight!
Chat on, sweet maid, and rescue from annoy
Hearts that by wiser talk are unbeguiled!
Ah, happy he who owns that tenderest joy,
The heart-love of a child!
Away, fond thoughts, and vex my soul no more!
Work claims my wakeful nights, my busy days:
Albeit bright memories of that sunlit shore
Yet haunt my dreaming gaze!
He asked her leave to have it published. The child in the verse is not named — why should he feel obliged to ask permission?
You’re standing in a room with an uneven floor. Before you is a square table with four legs. The table wobbles, but by turning it gradually you manage to find a position in which all four feet are supported, eliminating the wobble (though now the tabletop isn’t level).
You wonder: Is this always possible? Assuming that the four legs are of equal length and that the surface of the floor varies smoothly, is it always possible to position a four-legged table so that all four legs are supported?
Three positions, “left,” “middle,” and “right,” are marked on a table. Three cards, an ace, a king, and a queen, lie face up in some or all three of the positions. If more than one card occupies a given position then only the top card is visible, and a hidden card is completely hidden; that is, if only two cards are visible then you don’t know which of them conceals the missing card.
Your goal is to have the cards stacked in the left position with the ace on top, the king in the middle, and the queen on the bottom. To do this you can move one card at a time from the top of one stack to the top of another stack (which may be empty).
The problem is that you have no short-term memory, so you must design an algorithm that tells you what to do based only on what is currently visible. You can’t recall what you’ve done in the past, and you can’t count moves. An observer will tell you when you’ve succeeded. Can you devise a policy that will meet the goal in a bounded number of steps, regardless of the initial position?
“It’s tricky to design an algorithm that makes progress, avoids cycling, and doesn’t do something stupid when it’s about to win,” wrote Dartmouth mathematician Peter Winkler in sharing this puzzle in his book Mathematical Puzzles: A Connoisseur’s Collection (2003). It’s called “The Conway Immobilizer” because it originated with legendary Princeton mathematician John H. Conway and because it’s said to have immobilized one solver in his chair for six hours.
A man presents himself as the The Amazing Sand Counter. He claims that if you put some quantity of sand into a bucket, he will know at a glance how many grains there are, but he won’t tell you the number. Can you devise a test that can verify this ability without telling you anything that you don’t already know? You can ask the Sand Counter to leave the room or turn away, for example, and you can ask him questions. How can you convince yourself that he knows how many grains of sand are in the bucket when he won’t actually tell you the number?
A mother is 21 years older than her son. Six years from now, she will be five times his age. Where’s the father?
I won’t give the answer to this one — if you do the math, you’ll know precisely where he is.
From Stephen Barr’s Experiments in Topology (1989) via Miodrag Petkovic’s Mathematics and Chess (1997):
This apartment contains eight rooms, each measuring 9 square meters, except for the top one, which measures 18 square meters. You have enough red paint to cover 27 square meters, enough yellow paint to cover 27 square meters, enough green paint to cover 18 square meters, and enough blue paint to cover 9 square meters. Can you paint the eight floors in four colors so that each room neighbors rooms of the other three colors?
Can two dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?
A bisecting arc is one that bisects the area of a given region. “What is the shortest bisecting arc of a circle?” Murray Klamkin asked D.J. Newman. Newman supposed that it was a diameter. “What is the shortest bisecting arc of a square?” Newman answered that it was an altitude through the center. Finally Klamkin asked, “And what is the shortest bisecting arc of an equilateral triangle?”
“By this time, Newman had suspected that I was setting him up (and I was) and almost was going to say the angle bisector,” Klamkin writes. “But he hesitated and said let me consider a chord parallel to the base and since this turns out to be shorter than an angle bisector, he gave this as his answer.”
Was he right?
A and B are playing a simple game. Between them are nine tiles numbered 1 to 9. They take tiles alternately from the pile, and the first to collect three tiles that sum to 15 wins the game. Does the first player have a winning strategy?
A conduit carries 50 identical wires under a river, but their ends have not been labeled — you don’t know which ends on the west bank correspond to which on the east bank. To identify them, you can tie together the wires in pairs on the west bank, then row across the river and test the wires on the east bank to discover which pairs close a circuit and are thus connected.
Testing wires is easy, but rowing is hard. How can you plan the work to minimize your trips across the river?
Lori has an icky problem: Worms keep crawling onto her bed. She knows that worms can’t swim, so she puts each leg of the bed into a pail of water, but now the worms crawl up the walls of the room and drop onto her bed from the ceiling. She suspends a large canopy over the bed, but worms drop from the ceiling onto the canopy, creep over its edge to the underside, crawl over the bed, and drop.
Desperate, Lori installs a water-filled gutter around the perimeter of the canopy, but the worms drop from the ceiling onto the outer edge of the gutter, then crawl beneath. (The worms are very determined.) What can Lori do?
Tim Krabbé calls this “one of the funniest chess problems I ever saw.” Its composer, M. Kirtley, won first prize with it in a Problemist tourney in 1986.
It’s a selfmate in 8, which means that White must force Black to checkmate him 8 moves, despite Black’s best efforts to avoid doing so.
The solution is a single line — all of Black’s moves are forced:
1. Nb1+ Kb3 2. Qd1+ Rc2 3. Bc1 axb6 4. Ra1 b5 5. Rh1 bxc4 6. Ke1 c3 7. Ng1 f3 8. Bf1 f2#
All of White’s pieces have returned to their starting squares!