Author Archives: Jim Stevenson

The Pearl Necklace Problem

This problem comes from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008:

“S2. In Tiffany’s, a world famous jewellery store, there is a string necklace of 33 pearls. The middle one is the largest and most valuable. The pearls are arranged so that starting from one end, each pearl is worth $100 more than the preceding one, up to [and including] the middle one; and starting from the other end, each pearl is worth $150 more than the preceding one, up to [and including] the middle one. If the total value of the necklace is $65,000 what is the value of the largest pearl?”

I included the words in brackets to erase any ambiguity.

See the Pearl Necklace Problem

Fashion Puzzle

Again we have a puzzle from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).

“On one occasion, Holmes and I were asked to, solve the robbery of a number of dresses from the workshop of a recently deceased ladies’ tailor to the upper echelons of society. Holmes took a short look at the particulars of the case, and sent them all back to the gown-maker’s son with a scribbled note to the effect that it could only be one particular seamstress, with the help of her husband.

However, glancing through my observations some period later, I observed certain facts about the robbery which led me to an interesting little exercise. The stock at the workshop had been very recently valued at the princely sum of £1,800, and when examined after the theft, comprised of precisely 100 completed dresses in a range of styles, but of equal valuation. However, there was no remaining record of how many dresses had been there beforehand. The son did recall his father stating, of the valuation, that if he’d had thirty. dresses more, then a valuation of £1,800 would have meant £3 less per dress.

Are you able to calculate how many dresses were stolen?”

See the Fashion Puzzle

Puzzles and Problems: Tim Dedopulos, algebra

Puzzle of the Purloined Papers

Ian Stewart has a nice logic problem in his Casebook of Mathematical Mysteries, which includes a pastiche of Sherlock Holmes in the form of Herlock Soames and Dr. Watsup, along with brother Spycraft and nemesis Dr. Mogiarty.

“An important document was accidentally mislaid, and then stolen,” Spycraft said. “It is essential to the security of the British Empire that it be recovered without delay. If it gets into the hands of our enemies, careers will be ruined and parts of the Empire may fall. Fortunately, a local constable caught a glimpse of the thief, enough to narrow it down to precisely one of four men.”

“Petty thieves?”

“No, all four are gentlemen of high repute. Admiral Arbuthnot, Bishop Burlington, Captain Charlesworth, and Doctor Dashingham.”

Soames sat bolt upright. “Mogiarty has a hand in this, then.”

Not following his reasoning, I asked him to explain.

“All four are spies, Watsup. Working for Mogiarty.”

“Then … Spycraft must be engaged in counter-espionage!” I cried.

“Yes.” He glanced at his brother. “But you did not hear that from me.”

“Have these traitors been questioned?” I asked.

Spycraft handed me a dossier, and I read it aloud for Soames’s benefit. “Under interrogation Arbuthnot said ‘Burlington did it.’ Burlington said ‘Arbuthnot is lying.’ Charlesworth said ‘It was not I.’ Dashingham said ‘Arbuthnot did it.’ That is all.”

“Not quite all. We know from another source that exactly one of them was telling the truth.”

“You have an informer in Mogiarty’s inner circle, Spycraft?”

“We had an informer, Hemlock. He was garrotted with his own necktie before he could tell us the actual name. Very sad—it was an Old Etonian tie, totally ruined. However, all is not lost. If we can deduce who was the thief, we can obtain a search warrant and recover the document. All four men are being watched; they will have no opportunity to pass the document to Mogiarty. But our hands are tied; we must stick to the letter of the law. Moreover, if we raid the wrong premises, Mogiarty’s lawyers will publicise the mistake and cause irreparable damage.”

Which man was the thief?

See the Purloined Papers Puzzle.

Bottema’s Theorem

This seemingly magical result from Futility Closet defies proof at first.  Go to the Wolfram demo by Jay Warendorff and then …

“Grab point B above and drag it to a new location. Surprisingly, M, the midpoint of RS, doesn’t move.

This works for any triangle — draw squares on two of its sides, note their common vertex, and draw a line that connects the vertices of the respective squares that lie opposite that point. Now changing the location of the common vertex does not change the location of the midpoint of the line.

It was discovered by Dutch mathematician Oene Bottema.”

As we shall see, Bottema’s Theorem has shown up in other guises as well.

See Bottema’s Theorem

Painting Lampposts

This is another simple problem from H. E. Dudeney.

“103. PAINTING THE LAMP-POSTS.

Tim Murphy and Pat Donovan were engaged by the local authorities to paint the lamp-posts in a certain street. Tim, who was an early riser, arrived first on the job, and had painted three on the south side when Pat turned up and pointed out that Tim’s contract was for the north side. So Tim started afresh on the north side and Pat continued on the south. When Pat had finished his side he went across the street and painted six posts for Tim, and then the job was finished. As there was an equal number of lamp-posts on each side of the street, the simple question is: Which man painted the more lamp-posts, and just how many more?”

See Painting Lampposts

Snooker Puzzle

This is a nice puzzle from Alex Bellos’s Monday Puzzle column in the Guardian.

“My cultural highlight of recent weeks has been the brilliant BBC documentary Gods of Snooker, about the time in the 1980s when the sport was a national obsession. Today’s puzzle describes a shot to malfunction the Romford Robot … and put the Whirlwind … in a spin.

Baize theorem

A square snooker table has three corner pockets, as [shown]. A ball is placed at the remaining corner (bottom left). Show that there is no way you can hit the ball so that it returns to its starting position.

The arrows represent one possible shot and how it would rebound around the table.

The table is a mathematical one, which means friction, damping, spin and napping do not exist. In other words, when the ball is hit, it moves in a straight line. The ball changes direction when it bounces off a cushion, with the outgoing angle equal to the incoming angle. The ball and the pockets are infinitely small (i.e. are points), and the ball does not lose momentum, so that its path can include any number of cushion bounces.

Thanks to Dr Pierre Chardaire, associate professor of computing science at the University of East Anglia, who devised today’s puzzle.”

See the Snooker Puzzle

Area vs. Perimeter Puzzle

This surprising, but simple, puzzle is from the 12 April MathsMonday offering by MEI, an independent curriculum development body for mathematics education in the UK.

“In the diagram various regular polygons, P, have been drawn whose sides are tangents to a circle, C.  Show that for any regular polygon drawn in this way:”

(Given that the polygons approximate the circle in the limit, it would not be surprising that this relationship would hold—in the limit.  It is surprising that it should be true for every regular polygon that circumscribes the circle.)

See the Area vs. Perimeter Puzzle

The Hose Knows

This is a fairly straight-forward Brainteaser from the Quantum magazine.

“A man is filling two tanks with water using two hoses. The first hose delivers water at the rate of 2.9 liters per minute, the second at a rate of 8.7 liters per minute. When the smaller tank is half full, he switches hoses. He keeps filling the tanks, and they both fill up completely at the same moment. What is the volume of the larger tank if the volume of the smaller tank is 12.6 liters?”

See The Hose Knows

Bixley to Quixley Puzzle

I braved another attempt at a Sam Loyd puzzle.

“Here is a pretty problem which I figured out during a ride from Bixley to Quixley astride of a razor-back mule. I asked Don Pedro if my steed had another gait, and he said it had but that it was much slower, so I pursued my journey at the uniform speed as shown in the sketch.

To encourage Don Pedro, who was my chief propelling power, I said we would pass through Pixley, so as to get some liquid refreshments; and from that moment he could think of nothing but Pixley. After we had been traveling for forty minutes I asked how far we had gone, and he replied: “Just half as far as it is to Pixley.”  After creeping along for seven miles more I asked: “How far is it to Quixley?” and he replied as before: “Just half as far as it is to Pixley.”

We arrived at Quixley in another hour, which induces me to ask you to figure out the distance from Bixley to Quixley.”

I was disconcerted by what I thought was extraneous information and wondered if I had misunderstood his narrative again.

See the Bixley to Quixley Puzzle