Puzzles
Star Power
A puzzle by A. Korshkov, from the Russian science magazine Kvant:
It’s easy to show that the five acute angles in the points of a regular star, like the one at left, total 180°.
Can you show that the sum of these angles in an irregular star, like the one at right, is also 180°?
Hoop Dreams
A memorably phrased puzzle from The Graham Dial: “Consider a vertical girl whose waist is circular, not smooth, and temporarily at rest. Around the waist rotates a hula hoop of twice its diameter. Show that after one revolution of the hoop, the point originally in contact with the girl has traveled a distance equal to the perimeter of a square circumscribing the girl’s waist.”
Quickie
University of Strathclyde mathematician Adam McBride recalls that in his student days a particular teacher used to present a weekly puzzle. One of these baffled him:
Find positive integers a, b, and c, all different, such that a3 + b3 = c4.
“The previous puzzles had been relatively easy but this one had me stumped,” he wrote later. He created three columns headed a3, b3, and c4 and spent hours looking for a sum that would work. On the night before the deadline, he found one: 703 + 1053 = 354.
“This shows how sad a person I was! However, I then realised also how stupid I had been. I had totally missed the necessary insight.” What was it?
All Relative
1. A puzzle from J.A.H. Hunter’s Fun With Figures, 1956:
Tom and Tim are brothers; their combined ages make up seventeen years. When Tom was as old as Tim was when Tim was twice as old as Tom was when Tom was fifteen years younger than Tim will be when Tim is twice his present age, Tom was two years younger than Tim was when Tim was three years older than Tom was when Tom was a third as old as Tim was when Tim was a year older than Tom was seven years ago. So how old is Tim?
2. Another, by Sam Loyd:
“How fast those children grow!” remarked Grandpa. “Tommy is now twice as old as Maggie was when Tommy was six years older than Maggie is now, and when Maggie is six years older than Tommy is now their combined ages will equal their mother’s age then, although she is now but forty-six.” How old is Maggie?
3. According to Wirt Howe’s New York at the Turn of the Century, 1899-1916, this question inspired an ongoing national debate when it appeared in the New York Press in 1903:
Brooklyn, October 12
Dear Tip:
Mary is 24 years old. She is twice as old as Anne was when she was as old as Anne is now. How old is Anne now? A says the answer is 16; B says 12. Which is correct?
John Mahon
Three Chess Puzzles
1. By A.F. Rockwell. White to mate in two moves. (Solutions are below.)
2. In 1866 Sam Loyd asked: Suppose you’re playing White against an opponent who’s required to mirror every move you make — if you play 1. Nf3 he must play 1. … Nf6, and so on. Can you design a game in which your eighth move forces your opponent to checkmate you with a nonmirror move?
3. An endgame study by J.A. Miles. “White to play and draw the game.”
Overheard
A puzzle by Princeton mathematician John Horton Conway:
Last night I sat behind two wizards on a bus, and overheard the following:
A: I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.
B: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
A: No.
B: Aha! AT LAST I know how old you are!
“This is an incredible puzzle,” writes MIT research affiliate Tanya Khovanova. “This is also an underappreciated puzzle. It is more interesting than it might seem. When someone announces the answer, it is not clear whether they have solved it completely.”
We can start by auditioning various bus numbers. For example, the number of the bus cannot have been 5, because in each possible case the wizard’s age and the number of his children would then uniquely determine their ages — if the wizard is 3 years old and has 3 children, then their ages must be 1, 1, and 3 and he cannot have said “No.” So the bus number cannot be 5.
As we work our way into higher bus numbers this uniqueness disappears, but it’s replaced by another problem — the second wizard must be able to deduce the first wizard’s age despite the ambiguity. For example, if the bus number is 21 and the first wizard tells us that he’s 96 years old and has three children, then it’s true that we can’t work out the children’s ages: They might be 1, 8, and 12 or 2, 3, and 16. But when the wizard informs us of this, we can’t declare triumphantly that at last we know how old he is, because we don’t — he might be 96, but he might also be 240, with children aged 4, 5, and 12 or 3, 8, and 10. So the dialogue above cannot have taken place.
But notice that if we increase the bus number by 1, to 22, then all the math above will still work if we give the wizard an extra 1-year-old child: He might now be 96 years old with four children ages 1, 1, 8, and 12 or 1, 2, 3, and 16; or he might be 240 with four children ages 1, 4, 5, and 12 or 1, 3, 8, and 10. The number of children increases by 1, the sum of their ages increases by 1, and the product remains the same. So if bus number b produces two possible ages for Wizard A, then so will bus number b + 1 — which means that we don’t have to check any bus numbers larger than 21.
This limits the problem to a manageable size, and it turns out that the bus number is 12 and Wizard A is 48 — that’s the only age for which the bus number and the number of children do not uniquely determine the children’s ages (they might be 2, 2, 2, and 6 or 1, 3, 4, and 4).
(Tanya Khovanova, “Conway’s Wizards,” The Mathematical Intelligencer, December 2013.)
Two similar puzzles: A Curious Conversation and A Curious Exchange.
Mileage
I leave my front door, run on a level road for some distance, then run to the top of a hill and return home by the same route. I run 8 mph on level ground, 6 mph uphill, and 12 mph downhill. If my total trip took 2 hours, how far did I run?
Black and White
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Looking Up
A problem from the 2000 Moscow Mathematical Olympiad:
Some of the cards in a deck are face down and some are face up. From time to time Pete draws out a group of one or more contiguous cards in which the first and last are both face down. He turns over this group as a unit and returns it to the deck in the position from which he drew it. Prove that eventually all the cards in the deck will be face up, no matter how Pete proceeds.