Puzzle #155: Bugs on a Square

This is a very famous puzzle that many of you would have seen in the past as well. I saw it recently in one of the Martin Gardner books again, but am reproducing it here from a New York Times article from September 8, 2014.

screenshot-2016-09-24-18-10-50

Please look at the figure above. Four bugs are at the corners of a square which has each side measuring 10 inches. Each bug moves directly towards its neighbour as indicated in the arrows. How long does each bug travel before they meet? Please note that you do not need to know calculus to be able to compute the answer.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy bugging!

 

 

 

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7 Responses to Puzzle #155: Bugs on a Square

  1. Amit Mittal says:

    At time 1, each bug has travelled one unit. Thus each bug changes direction of travel at each such time unit . traversing a square of side easily received from pythagoras in case ever required, at t=2 they would have traveled one ninth of the new square and t=3, one eighth of the third square. Thus at time t=10 they will meet at the center of the spiral made. (1. Confessions: Google did help! 2. This puzzle is just awesome, i shudder to think how much time it will take for the calculus to sink in 3. Many Thanks for introducing Martin Gardner to me! 4. I know right!)

  2. Karan says:

    The movement will be an infinite spiral. Haven’t found a way to compute the length yet.

  3. Prakhar says:

    Alok – I asked my son and he thought that the bugs will keep spiraling into an inner loop which will become smaller and smaller but will never meet. My response was 10 inches which I sent to you in a separate email.

  4. Note that by symmetry, bugs always move perpendicular to each other, so when they meet at the center they have traveled a distance equivalent to their initial separation i.e. 10 inches. So the time taken is 10 inches / (speed of the bug).

    Also, from the symmetry argument, they would meet at the centre of the square.

    Cheers!

  5. Abhinav Jain says:

    Distance travelled by each bug will be 10 inches

  6. Vishal Poddar says:

    Considering a pair of bug. The separation between them is 10 inches which has to be covered by travelling along the line. By looking at the symmetry, all bugs will travel 10 inches and will meet at the centre of the square.

  7. Batman says:

    Suppose you are one of the bugs. According to you one bug is directly moving towards you at all times with a constant speed. Since you are moving perpendicular to the bug following you at all times, there is no component of your velocity that is moving towards or away from the bug following you. Hence this means that the time taken for the bugs to meet will be the same as the time taken for one bug to just travel 10 cm (the side of the square). Hence the answer will be 10/v where v is the speed of the bug. A more rigorous solution can be found using calculus, but this also suffices for this problem. I have tried the problem with unsymmetrical shapes(trapezium, rectangle, etc.)..In this I believe calculus is required.

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