Join two of the star’s adjacent vertices, such as A and E. Triangles BMD and AME have an angle in common (the congruent angles at M), so the sum of angles B and D in triangle BMD equals the sum of angles A and E in triangle AME. This means that the sum of the star’s five angles equals the sum of the angles of triangle ACE — which is 180°.

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.”

Hold the hoop steady and let the girl roll around inside it:

Since the ratio of the diameters is 2:1, so is the ratio of the circumferences. This means that a point on the girl oscillates back and forth between two opposite points on the hoop, passing through the hoop’s center on the way and producing a straight line (the “Tusi couple”).

“The perimeter of circumscribing square equals four girl diameters or two hula hoop diameters which is the total displacement of initial point of contact between hula hoop and the aforementioned vertical girl.”

From L.A. Graham, The Surprise Attack in Mathematical Problems, 1968.

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 a^{3} + b^{3} = c^{4}.

“The previous puzzles had been relatively easy but this one had me stumped,” he wrote later. He created three columns headed a^{3}, b^{3}, and c^{4} and spent hours looking for a sum that would work. On the night before the deadline, he found one: 70^{3} + 105^{3} = 35^{4}.

“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?

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?

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.”

White is already nearly stalemated — if he can get rid of his queen he’ll have his draw. But the only line I can find is the ridiculous and ungainly 1. Qf1+ Nc4 2. Qxc4+ Kb6 3. Qa6+ Kc7 4. Qc8+ Kb6 5. Qb7+, where Black must take the queen. All Black’s moves are forced if he wants to avoid a draw, but this is such an unbeautiful line that I feel sure it can’t be the intended solution. Any ideas?

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.)

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?

The length of the hill is immaterial — if I average 6 mph climbing and 12 mph descending, then my average speed on the hill is 8 mph, just as on level ground. So in 2 hours I ran 16 miles.

(A few people have asked why my average speed on the hill isn’t (6 + 12) / 2 = 9 mph. Remember that I’ll spend more time climbing than descending — for example, if the hill segment is 4 miles long, then I’ll spend 4/6 hours up + 4/12 hours down = 1 hour covering 8 miles.)

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.

Assign a digit to each card in the deck, 1 if face up and 2 if face down, working from top to bottom. If these digits are assembled in order, they produce one long number that encodes the arrangement of the cards. For example, if all the cards were face down then this number would be 2222 … 222.

After each of Pete’s operations the value of this number decreases, because the leftmost, or highest-value, digit changes from 2 to 1.

Because there are finitely many n-digit numbers available, this process cannot continue forever. The encoded number must eventually arrive at 1111 … 111, where all the cards are face up.

In 1831 this riddle appeared in a British publication titled Drawing Room Scrap Sheet No. 17:

In the morn when I rise, / I open my eyes, / Tho’ I ne’er sleep a wink all night;
If I wake e’er so soon, / I still lie till noon, / And pay no regard to the light.

I have loss, I have gain, / I have pleasure, and pain; / And am punished with many a stripe;
To diminish my woe, / I burn friend and foe, / And my evenings I end with a pipe.

I travel abroad. / And ne’er miss my road, / Unless I am met by a stranger;
If you come in my way, / Which you very well may, / You will always be subject to danger.

I am chaste, I am young, / I am lusty, and strong, / And my habits oft change in a day;
To court I ne’er go, / Am no lady nor beau, / Yet as frail and fantastic as they.

I live a short time, / I die in my prime, / Lamented by all who possess me;
If I add any more, / To what’s said before / I’m afraid you will easily guess me.

It was headed “For Which a Solution Is Required,” perhaps meaning that the editor himself did not know the solution. I think he may have found the riddle in The Lady’s Magazine, which had published it anonymously in September 1780 without giving the answer. Unfortunately he seems to have been disappointed — the Drawing Room Scrap Sheet never printed a solution either.

A century and a half later, in 1981, Faith Eckler challenged the readers of Word Ways: The Journal of Recreational Linguistics to think of an answer, offering a year’s subscription to the journal as a reward. When no one had claimed the prize by February 2010, Ross Eckler renewed his wife’s challenge, noting that the National Puzzlers’ League had also failed to find a solution.

That’s understandable — it’s tricky. “The author of the riddle cleverly uses ambiguous phrases to mislead the solver,” Ross Eckler notes. “I still lie till noon (inert, or continue to?); evenings I end with a pipe (a tobacco holder, or a thin reedy sound?); to court I ne’er go (a royal venue, a legal venue, or courtship?).”

To date, so far as I know, the riddle remains unsolved. Answers proposed by Word Ways readers have included fame, gossip, chessmen, a hot air balloon, and the Star and Stripes, though none of these seems beyond question. I offer it here for what it’s worth.

UPDATE: A solution has been found! Apparently The Lady’s Diary published the solution in 1783, which Ronnie Kon intrepidly ran to earth in the University of Illinois Rare Book and Manuscript Library. He published it in Word Ways in November 2012 (PDF). I’ll omit the solution here in case you’d still like to guess; be warned that it’s not particularly compelling. (Thanks, Ronnie.)