## Knife Act

I have just baked a rectangular cake when my wife comes home and barbarically cuts out a piece for herself. The piece she cuts is rectangular, but it’s not in any convenient proportion to the rest of the cake, and its sides aren’t even parallel to the cake’s sides. I want to divide the remaining cake into two equal-sized halves with a single straight cut. How can I do it?

## Close Dosage

You’re on a drug regimen that requires you to take one pill a day from each of two bottles, A and B. One day you tap one pill into your palm from the A bottle and, inadvertently, two pills from the B bottle. Unfortunately the A and B pills are indistinguishable, and taking more than one B pill per day is fatal. And the pills are very expensive, so you can’t afford to throw out the handful and start over. How can you arrange to take the correct dose without wasting any pills?

## Hour of Babel

A problem from the second Balkan Mathematical Olympiad, 1985:

Of the 1985 people attending an international meeting, no one speaks more than five languages, and in any subset of three attendees, at least two speak a common language. Prove that some language is spoken by at least 200 of the attendees.

## Cube Route

You’re planning to make the wire skeleton of a cube by arranging 12 equal lengths of wire as shown and soldering them at the corners.

It occurs to you that you might be able to simplify the job by using one or more longer lengths of wire and bending them into right angles at the cube’s corners.

If you adopt that plan, what’s the smallest number of corners where soldering will still be necessary?

## Weighing in Verse

How can you find one bad coin among 12 using only three weighings in a pan balance? We published the classic solution in November 2013, but a clever alternative appeared in *Eureka* in October 1950. Here the aspiring counterfeiter is named Felix Fiddlesticks, and he’s trying to find the coin for this mother:

F set the coins out in a row

And chalked on each a letter, so,

To form the words: “F AM NOT LICKED”

(An idea in his brain had clicked).

And now his mother he’ll enjoin:

This plan will reveal the bad coin no matter which it is. For instance, if the coin marked O is heavy, then in the first two weighings the left pan will drop, and in the third weighing it will rise; no other bad coin will produce this result. The plan will also reveal reliably whether the bad coin is heavy or light.

Such cases number twenty-five,

And by F’s scheme we so contrive

No two agree in their effect,

As is with pen and patience checked:

And so the dud is found. Be as it may

It only goes to show CRIME DOES NOT PAY.

(Cedric Smith, “The Twelve Coin Problem,” *The Mathematical Gazette*, Vol. 89, No. 515 [July 2005], pp. 280-281. A generalized algorithm for solving such a problem for any number of coins is given by Michael Weatherfield in the same issue, pp. 275-279.)

## Riddles in the Dark

In 1904, a 12-year-old J.R.R. Tolkien sent this rebus to a family friend, Father Francis Morgan. What does it say?

## Three Coins

Three coins are lying on a table: a quarter, a half dollar, and a silver dollar. You claim one coin, I’ll claim the other two, and then we’ll toss all three. A coin that lands tails counts zero, and a coin that lands heads wins its value (in cents, 25, 50, or 100) for its owner. Whichever of us has the larger score wins all three coins. If all three coins land tails then we toss again.

Which coin should you claim to make the game fair — that is, so that each of us has an expected win of zero?

## Digit Spans

A puzzle by A. Savin: Using each of the digits 1, 2, 3, and 4 twice, write out an eight-digit number in which there is one digit between the 1s, two digits between the 2s, three digits between the 3s, and four digits between the 4s.

## 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:

You can listen using the player above, download this episode directly, or subscribe on iTunes or via the RSS feed at http://feedpress.me/futilitycloset.

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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