## The Jeweler’s Observation

Prove that every convex polyhedron has at least two faces with the same number of sides.

## Ride Sharing

You and I have to travel from Startville to Endville, but we have only one bicycle between us. So we decide to leapfrog: We’ll leave Startville at the same time, you walking and I riding. I’ll ride for 1 mile, and then I’ll leave the bicycle at the side of the road and continue on foot. When you reach the bike you’ll ride it for 1 mile, passing me at some point, then leave the bike and continue walking. And so on — we’ll continue in this way until we’ve both reached the destination.

Will this save any time? You say yes: Each of us is riding for part of the distance, and riding is faster than walking, so using the bike must increase our average speed.

I say no: One or the other of us is always walking; ultimately every inch of the distance between Startville and Endville is traversed by someone on foot. So the total time is unchanged — leapfrogging with the bike is no better than walking the whole distance on foot.

Who’s right?

## A Passing Wave

A puzzle from J.A.H. Hunter’s *Fun With Figures* (1956):

A man paddling a canoe upstream sees a glove in the water as he passes under a bridge. Fifteen minutes later, he turns around and paddles downstream. He passes under the bridge and travels another mile before reaching the rock from which he started, which the glove is just passing. If he paddled at the same speed the whole time and lost no time in turning around, what is the speed of the current?

## Roll Call

A problem from the 2002 Moscow Mathematical Olympiad:

A group of recruits stand in a line facing their corporal. They are, unfortunately, rather poorly trained: At the command “Left turn!”, some of them turn left, some turn right, and some turn to face away from the corporal. Is it always possible for the corporal to insert himself in the line so that an equal number of recruits are facing him on his left and on his right?

## Podcast Episode 56: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles to test your wits! Solve along with us as we explore some strange scenarios using only yes-or-no questions. Many were submitted by listeners, and most are based on real events.

A few associated links — these spoil the puzzles, so don’t click until you’ve listened to the episode:

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If you have any questions or comments you can reach us at podcast@futilitycloset.com. And you can finally follow us on Facebook and Twitter. Thanks for listening!

## 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 *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?

## 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?

## 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.

## What Am I?

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.)

## End State

Starting in Delaware, you must tour the 48 contiguous United States, visiting each state exactly once.

Where will you finish?

## Another Car Puzzle

A puzzle from Oswald Jacoby’s *Mathematics for Pleasure* (1962):

The MacDonalds are planning a long car journey of 27,000 miles. If they use tires that last 12,000 miles each, how many tires will they need, and how can they make the best use of them?

## Portions of Pi

A quickie submitted by John Astolfi to *MIT Technology Review’*s Puzzle Corner, July/August 2013:

Consider the expansion of π (3.14159 …) in base 2. Does it contain more 0s than 1s, more 1s than 0s, or an equal number of both? Or is it impossible to tell?

## The Fenn Treasure

In March 2013, New Mexico art dealer Forrest Fenn announced that he had hidden a bronze treasure chest in the Rocky Mountains north of Santa Fe. In the chest, he says, are gold coins, artifacts, and jewelry worth more than $1 million.

Fenn said he’d conceived the idea when diagnosed with cancer in 1988, planning to bury the treasure as a legacy. The cancer went into remission, but he decided to bury the chest anyway. In a self-published memoir he offered the following poem, which he says contains nine clues:

As I have gone alone in there

And with my treasures bold,

I can keep my secret where,

And hint of riches new and old.

Begin it where warm waters halt

And take it in the canyon down,

Not far, but too far to walk.

Put in below the home of Brown.

From there it’s no place for the meek,

The end is ever drawing nigh;

There’ll be no paddle up your creek,

Just heavy loads and water high.

If you’ve been wise and found the blaze,

Look quickly down, your quest to cease,

But tarry scant with marvel gaze,

Just take the chest and go in peace.

So why is it that I must go

And leave my trove for all to seek?

The answers I already know,

I’ve done it tired and now I’m weak.

So hear me all and listen good,

Your effort will be worth the cold.

If you are brave and in the wood

I give you title to the gold.

Fenn has been releasing further clues periodically as he follows the search (“No need to dig up the old outhouses, the treasure is not associated with any structure”). A number of people claim to have found the chest, but none has provided evidence, and Fenn says that to the best of his knowledge it remains undiscovered. There’s much background and discussion about the treasure at ttotc.com.

If you find it, I figure you owe me 75%.