If you are up at the Edinburgh Festival, I shall be talking about the strange science of sleep and dreaming on August 14th.  Do come along, much fun will be had. Details here.

Here is this weeks puzzle.  Two cyclists are racing around a circular track.  Jon can race around the track in 6 minutes and Eric can race round in 4 minutes.  They start the race on opposite sides of the track.  How many minutes will it take before Eric overtakes Jon?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


  1. It all depends what mood Eric is in. Whilst he “can” race round in 4 minutes, if he is feeling kind hearted he may decide to go slower so as not to humiliate Jon, in which case it could be a long time before he catches him.

  2. We should be wary of upsetting @Steve and @Simon by discussing issues that they forbid. As they are weekday problem solvers they deem it improper to discuss puzzles on a weekend.

    Yet I can say that the puzzle as set is formally unsolvable. There are four possible answers depending on each case of Jon and Eric racing either clockwise or anticlockwise. Without knowing the directions of each we have to consider each case separately.

    DAve I am sure will say 22 or 7 depending on the direction of travel of Eric.Yet Jon’s direction also matters.

    The average of the four answers is perhaps what RW wants.

    1. Since when was Friday morning part of the weekend?
      And I’m not sure if two people are racing that going in opposite directions is a sensible approach, although Percy’s observation about ‘can’ (as opposed to ‘does’ or ‘will’) is a fair one.
      But without any of that, considering symmetry there are only two options not four – both going the same way or both going in opposite directions.

    2. @ChrisR

      You are forgetting that the cyclists have quite distinct skills. This is shown by their disparate speeds in circling the track. Thus one going to the left while the other goes to the right is quite different to the other way around.

      If you cannot see that consider this case:

      Two men start a race on the edge of a lake. One is a champion marathon runner. The other is a champion swimmer. If they choose to go in opposite directions (ie not both into the lake or both away from the lake) then the race the develops will be quite different depending on which man chooses the watery direction first.

      The difficulty you may be having is in seeing all the options available. Widen your mind and the puzzle and its solutions will be apparent.

    3. @BG – my mind does tend to be fairly wide – my job involves trying to anticipate the unexpected (I didn’t succeed with the details of your response though).
      However I fail to see the relevance in translating a puzzle about a race in a constrained environment (a circular track) to an open situation on the edge of a lake (with a marathon runner attempting to run on water and a swimmer crawling away in the opposite direction – perhaps never to meet again!).
      And whereas the two cyclists may have ‘quite distinct skills’, in the context of the question they share the skill of being able to ride a bike.
      A mind can be too open – it needs to close in on truth and facts when they are clear.

  3. I worked out the answer the complicated way – assuming a 100m track, working out their speeds, then using formulae to work out how far they’d travelled in each minute (including an IF statement to cope with laps). Amending the length of the track produced the same result, thus confirming my calculations were (probably) correct.

    It wouldn’t surprise me if on Monday’s post, someone produces a neat little algebraic formula that can easily work out the answer for any given pair of speeds.

    1. not really a spoiler, more a hint

      Let T be the answer (in minutes).
      The number of laps that Jon has travelled in time T = T/6
      The number of laps that Eric has travelled in time T = T/4
      Eric is cycling faster than Jon and looking to overtake him, but he has 1/2 a lap extra to cycle.
      So, when Eric catches up with Jon (at time T), (T/6)+1/2 = (T/4)
      Go from there.

      (replace 6 and 4 with the relevant figures for other relative speeds – or more precisely, lap times)

    2. Or you could reason another way. If the two cyclists started out together, they would meet again in the least common multiple of 4 and 6 minutes. Halfway through this number of minutes they would be on opposite sides of the circle. So, from opposite sides of the circle to back together would take one-half the least-common multiple number of minutes.

  4. It took me a couple of minutes with a pen and a post-it…but I’ve probably got it wrong (though it feels right, I must say).

    1. I solved it using the Praeter Method and the Drans Effect. Rather a case of using a sledgehammer to crack a walnut, I must say. The answer was staring one in the physog.

    2. @Eddie. I can play the cello a bit but no, I have never worked with the singer from Led Zeppelin.

    3. I solved it using two simultaneous equations. Not because that is the best method but because it will annoy Barry.

    4. I did it by tuning into the cosmos, whilst pondering the time each cyclist would take over 1/2 a lap (which then disappeared when I stood up).
      Admittedly the last time I tuned into the cosmos I was at a football game and couldn’t work out why the ball was getting larger – then it hit me.

    5. @ChrisR

      Given that (undeniably) you exist in the cosmos and (equally undeniably) the puzzle answer exists in the cosmos then simply opening yourself up to the possibility of each is a major and undeniable success factor.

      Congratulations on using my methods so succinctly.

      For those puzzled by the answers given: they all depend on both cyclists going clockwise. Other answers are needed for other other starting assumptions.

      The fast cyclist goes around once in the time that the slow cyclist goes around one and a half times. The fast cyclist is one half lap “behind”. That means the fast cyclist catches the slow cyclist in the same time it takes the slow one to perform one circuit.

      (I have removed the names as gender must not be allowed to sway this problem).

      I also solved this by simply reaching out for the solution. I suppose the hard way with similtaneous equations would eventually produce a result. Yet it would reveal no insight into the understanding of the dynamic.

    6. I think it is pretty clear by the use of the term “overtakes” that the two cyclists are racing in the same direction as each other, but that the actual direction is irrelevant for determining the solution.

      It is an unnecessary assumption that Jon and Eric are male, or even human. There is no point in complicating things by following this false line of thinking or removing labels.

    1. Nice thinking outside the box but actually, it doesn’t. The question asks how long it will take for Eric to overtake John not catch his back wheel.

  5. I figure 6 minutes. After Jons first complete lap around the track Eric will have ran 1 1/2 laps thus catching up to Jon because half of 4 minutes is 2 minutes which is how much faster Eric runs the track than Jon.

  6. An elegant way of answering is by using relativity. Assume Jon is stationary and the track itself is rotating in the opposite direction to Eric. Then Eric is going forward at the stated speed minus Jon’s speed. And he simply needs to do 1/2 a lap.
    This gives rise to a simple formula, and a clear solution.

  7. I totally dig this idea, but . . . who writes the first letters you guys or us? (above says, you guys, even so the email I got just said we write them.) What is the subject matter? Can we write about anything? I can imagine this is often a lot cheaper than therapy or medication for some men and women. Can we include footage of our cats? How about recipes for vegan cupcakes? There is a chance I could have interesting things to write about. What is the statistical probability that the reader will treatment? Wonderful idea! You are all great. Please advise.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: