Category Archives: Puzzles and Problems

Refabulating Widgets

This is a work problem from Geoffrey Mott-Smith from 1954.

“ ‘If a man can do a job in one day, how long will it take two men to do the job?’

No book of puzzles, I take it, is complete without such a question. I will not blame the reader in the least if he hastily turns the page, for I, too, was annoyed by “If a man” conundrums in my schooldays. Besides, the answer in the back of the book was always wrong. Everybody knows it will take the two men two days to do the job, because they will talk about women and the weather, they will argue about how the job is to be done, they will negotiate as to which is to do it. In schoolbooks the masons and bricklayers are not men, they are robots.

Strictly on the understanding that I am really talking about robots, I will put it to you:

If a tinker and his helper can refabulate a widget in 2 days, and if the tinker working with the apprentice instead would take 3 days, while the helper and the apprentice would take 6 days to do the job, how long would it take each working alone to refabulate the widget?”

See Refabulating Widgets

Skating Rendezvous Problem

This is a fun problem from the 1989 American Invitational Mathematics Exam (AIME).

“Two skaters, Allie and Billie, are at points A and B, respectively, on a flat, frozen lake. The distance between A and B is 100 meters. Allie leaves A and skates at a speed of 8 meters per second on a straight line that makes a 60° angle with AB. At the same time Allie leaves A, Billie leaves B at a speed of 7 meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?”

See the Skating Rendezvous Problem

Mountain Climbing Puzzle

This puzzle from the Scottish Mathematical Council (SMC) Middle Mathematics Challenge has an interesting twist to it.

“Two young mountaineers were descending a mountain quickly at 6 miles per hour. They had left the hostel late in the day, had climbed to the top of the mountain and were returning by the same route.  One said to the other “It was three o’clock when we left the hostel. I am not sure if we will be back before nine o’clock.” His companion replied “Our pace on the level was 4 miles per hour and we climbed at 3 miles per hour. We will just make it.” What is the total distance they would cover from leaving the hostel to getting back there?”

See the Mountain Climbing Puzzle.

Two Squares Problem

Via Alex Bellos I found another Russian math magazine with fun problems.  It is called Kvantik and Tanya Khovanova has a description (2015):

Kvant [Quantum] was a very popular science magazine in Soviet Russia. It was targeted to high-school children and I was a subscriber. Recently I discovered that a new magazine appeared in Russia. It is called Kvantik, which means Little Kvant. It is a science magazine for middle-school children. The previous years’ archives are available online in Russian. I looked at 2012, the first publication year, and loved it.”

Unfortunately, the magazine is in Russian and the later issues are only partially given online.  To get the full magazine you need to subscribe.  I used Google Translate and the mathematical context to render the English.   Here is an interesting geometric problem that I would have thought to be quite challenging for middle schoolers.

“The vertices of the two squares are joined by two segments, as in the figure. It is given that these segments are equal. Find the angle between them.

Egor Bakaev”

See the Two Squares Problem

(Update 8/22/2022, 9/1/2022)  Simpler Solution, Simplest Solution!
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Two and a Half Circles

Here is a problem from the 2022 Math Calendar.

“Two small circles of radius 4 are inscribed in a large semicircle as shown.  Find the radius of the large semicircle.”

As before, recall that all the answers are integer days of the month.

As seemed to be implied by the original Math Calendar diagram, I made explicit that the upper circle was tangent to the midpoint of the chord.  Otherwise, the problem is insufficiently constrained.

See Two and a Half Circles

The Maths of Lviv

Unfortunately Ukraine has receded from our attention under the threat from our own anti-democratic forces, but this Monday Puzzle from Alex Bellos in March is a timely reminder of the mathematical significance of that country.

“Like many of you I’ve hardly been able to think about anything else these past ten days apart from the war in Ukraine. So today’s puzzles are a celebration of Lviv, Ukraine’s western city, which played an important role in the history of 20th century mathematics. During the 1930s, a remarkable group of scholars came up with new ideas, methods and theorems that helped shape the subject for decades.

The Lwów school of mathematics – at that time, the city was in Poland – was a closely-knit circle of Polish mathematicians, including Stefan Banach, Stanisław Ulam and Hugo Steinhaus, who made important contributions to areas including set-theory, topology and analysis. …

Of the many ideas introduced by the Lwów school, one of the best known is the “ham sandwich theorem,” posed by Steinhaus and solved by Banach using a result of Ulam’s. It states that it is possible to slice a ham sandwich in two with a single slice that cuts each slice of bread and the ham into two equal sizes, whatever the size and positions of the bread and the ham.

Today’s puzzles are also about dividing food. The first is from Hugo Steinhaus’ One Hundred Problems in Elementary Mathematics, published in 1938. The second uses a method involved in the proof of the ham sandwich theorem.

  1. Three friends each contribute £4 to buy a £12 ham. The first friend divides it into three parts, asserting the weights are equal. The second friend, distrustful of the first, reweighs the pieces and judges them to be worth £3, £4 and £5. The third, distrustful of them both, weighs the ham on their own scales, getting another result. If each friend insists that their weighings are correct, how can they share the pieces (without cutting them anew) in such a way that each of them would have to admit they got at least £4 of ham?_
  2. Ten plain and 14 seeded rolls are randomly arranged in a circle, equidistantly spaced, as below. Show that using a straight line it is possible to divide the circle into two halves such that there are an equal number of plain and seeded rolls on either side of the line.

Show there is always a diameter that cuts the circle into two batches of 12 rolls with an equal number of plain and seeded.

Question 2 is adapted from Mathematical Puzzles by Peter Winkler, who gives as a reference Alon and West, The Borsuk-Ulam Theorem and bisection of necklaces, Proceedings of the American Mathematical Society 98 (1986).

See The Maths of Lviv

Missing Interval Puzzle

Henk Reuling posted a deceptively simple-looking geometric problem on Twitter.

“I found this old one cleaning up my ‘downloads’ [source unknown] I haven’t been able to solve it, so help!

According to the given information in the figure, what is the length of the missing interval on the diagonal of the square?”

See the Missing Interval Puzzle.

Pinocchio’s Hats

This problem in logic from Presh Talwalkar recalled an article I wrote a while ago but did not publish.  So I thought I would post it as part of the solution.

“Assume that both of the following sentences are true:

  1. Pinocchio always lies;
  2. Pinocchio says, “All my hats are green.”

We can conclude from these two sentences that:

  • (A) Pinocchio has at least one hat.
  • (B) Pinocchio has only one green hat.
  • (C) Pinocchio has no hats.
  • (D) Pinocchio has at least one green hat.
  • (E) Pinocchio has no green hats.”

Actually, the question is which, none or more, of statements (A) – (E) follow from the two sentences?

See Pinocchio’s Hats

15 Degree Triangle Puzzle

This math problem from Colin Hughes’s Maths Challenge website (mathschallenge.net) is a bit more challenging.

“In the diagram, AB represents the diameter, C lies on the circumference of the circle, and you are given that

(Area of Circle) / (Area of Triangle) = 2π.

Prove that the two smaller angles in the triangle are exactly 15° and 75° respectively.”

See the 15 Degree Triangle Puzzle