Category Archives: Puzzles and Problems

Hard Time Conundrum

This problem comes from the “Problems Drive” section of the Eureka magazine published in 1955 by the Archimedeans at Cambridge University, England.  (“The problems drive is a competition conducted annually by the Archimedeans. Competitors work in pairs and are allowed five minutes per question ….”)

“There are ten times as many seconds remaining in the hour as there are minutes remaining in the day. There are half as many minutes remaining in the day as there will be hours remaining in the week at the end of the day.  What time is it on what day?”

One of the hardest parts of the problem is just being able to translate the statements into mathematical terms.  Solvable in 5 minutes?!!!

Answer.

See the Hard Time Conundrum for a solution.

Tethered Triangle Puzzle

This is another problem from MEI’s MathsMonday.

“Two equilateral triangles share a common vertex. Show that the lengths marked a and b are equal for any such arrangement.”

This seems quite amazing at first.  One can picture the small triangle swinging back and forth with red bungee chords tethering its bottom vertices to the bottom vertices of the large triangle.  It would seem remarkable that the lengths of the chords would remain equal to each other throughout.

See the Tethered Triangle Puzzle

Wandering Epicycle

Here is an intriguing problem from the 2021 Math Calendar.

“If the smaller circle of diameter 7 rotates without slipping within the larger circle, what is the length of the path of P?”

The problem did not state clearly how far the smaller circle should rotate.  Its answer implied it should complete just one full (360°) rotation within the larger circle.

Recall that all the answers are integer days of the month.

Answer.

See the Wandering Epicycle for a solution.

(Update 1/3/2022)  First, this problem is dealt with in more detail and more expansively on the Mathologer Youtube website by Burkard Polster in his 7 December 2018 post on the “Secrets of the Nothing Grinder” (Figure 1).  A further, deeper discussion of epicycles is given in the Mathologer’s 6 July 2018 post on “Epicycles, complex Fourier and Homer Simpson’s orbit” (Figure 2).  And finally, a panoply of related puzzles is given in the 30 December 2021 Mathologer post “The 3-4-7 miracle. Why is this one not super famous” (Figure 3).

This last post reveals the ambiguity of the idea of “one full (360°) rotation” I disingenuously added to the problem to try to get the answer given in Math Calendar version.

For a complete explanation see the Wandering Epicycle Addendum.

Polygon Rings

This is a nice geometric problem from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008.

“Mahti has cut some regular pentagons out of card and is joining them together in a ring. How many pentagons will there be when the ring is complete?

She then decides to join the pentagons with squares which have the same edge length and wants to make a ring as before. Is it possible? If so, determine how many pentagons and squares make up the ring and if not, explain why.”

Answer.

See the Polygon Rings for solutions.

MoMath Mazes

I found these mazes on Twitter and thought they might make a relaxing puzzle interlude.  They come from photographs of the street in front of the Museum of Mathematics (MoMath) in New York.   The idea is to traverse the mazes from the Start to the Goal making only right turns.  It was difficult working out the pattern of the green maze, especially the upper right corner.

See the MoMath Mazes

Puzzles and Problems: MoMath

A Question of Time

This turns out to be an unambiguous, doable problem from the 19th century puzzle master Sam Loyd.  It is based on an observation about jewelers’ signs of the times.  I thought I would include Loyd’s narrative in its entirety.

“A CURIOUS paragraph has been going the rounds of the press which attempts to explain why the signs of the big watches in front of jewelry stores are always alike. They are painted upon the dial, apparently in a haphazard sort of a way, and yet they invariably indicate a certain number of minutes past eight. It cannot be attributable to chance, for it would tax one’s credulity to believe that such a coincidence could occur all over the civilized world.

There is no accepted rule or agreement established with the jewelers or sign painters, for careful inquiry proves that few of them are aware of the fact or ever noticed that any two are alike, it would be a marvelous case of unconscious imitation if it is looked upon as a mere custom, accidentally following a pattern set by the originator of the device of the sign of a big watch. In London, where they take pride in such things, I saw several big watches, looking as if they had hung in front of the stores for countless centuries, all indicating the same mysterious time, accompanied by the announcement that the firms were established a couple of hundred years ago. I do not doubt for a moment that some such similar sign can be found at Nuremberg, where the watch originated during the Fifteenth Century.

The discussion seems to have brought out a recognition of the fact that from an artistic point of view, symmetry requires that the hands should be evenly balanced, as it were, on both sides of the face of the watch.

If they are raised too much there is a certain “exasperating, declamatory effect,” which is not altogether pleasing.

The time would be incorrect if the hands pointed at 9 and 3, and at other points would be too low, so, as a matter of fact, and from an artistic point of view, the position is well selected and is one of the points which, with the aid of a watch, can be shown to be possible. It is a fact however, that the mere puzzle of telling what time the watch indicates, has been held up to public gaze for all these centuries without being thought of or solved?

Take your watch and set it to the time indicated, with the hands at equal distances from the six hour, which shows it to he a possible position, and then tell what time of the day it is! …”

Answer.

See A Question of Time for solutions.

Twisting Beam Problem

Here is a slightly different kind of problem from the Polish Mathematical Olympiads.

“106. A beam of length a is suspended horizontally by its ends by means of two parallel ropes of lengths b.  We twist the beam through an angle φ about the vertical axis passing through the centre of the beam.  How far will the beam rise?”

Answer.

See the Twisting Beam Problem for solutions.

  1. Pages:
  3. 1
  4. ...
  5. 9
  6. 10
  7. 11
  8. 12
  9. 13
  10. 14
  11. 15
  12. ...
  13. 28