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Posers and Puzzles

Posers and Puzzles

  1. Standard member CalJust
    It is what it is
    23 Oct '06 05:19
    Four bugs sit on the corners of a square of side one meter.
    I posted this originally as a continuation of the "fly and the cysclists". Maybe someone else besides gastel can have a go at it - the former is slightly preoccupied with his last few marriages. Whilst I hae a lot of sympathy for his position, seems to me his mind is not on the job - in a mkanner of speakeing, that is.

    Here goes:

    Four bugs (flies, dogs, you name it) are situated on the four corners of a square of side length one meter.

    The bugs are alternating male and female, i.e. if the corners of the square are clockwisw A,B,C & D, then A and C are male, B and D are female.

    Each bug now starts to stalk the bug directly in front of it. As they do so, deviating neither to the right nor the left, they each describe a spiral that meets in the centre of the square.

    I leave it to your imagination what happens at this spot. The question is, however, what is the length of the spiral that the bugs travelled?
  2. Standard member PBE6
    Bananarama
    23 Oct '06 15:36
    Originally posted by CalJust
    Four bugs sit on the corners of a square of side one meter.
    I posted this originally as a continuation of the "fly and the cysclists". Maybe someone else besides gastel can have a go at it - the former is slightly preoccupied with his last few marriages. Whilst I hae a lot of sympathy for his position, seems to me his mind is not on the job - in a mkanner ...[text shortened]... ot. The question is, however, what is the length of the spiral that the bugs travelled?
    Here's my solution:

    At the beginning of the problem, the bugs sit at the four corners of a square. Because the bugs are all doing the same thing, the path of each bug is the same just rotated 90 degrees from the last one. Due to this symmetry, the bugs stay in a square configuration. The square itself just rotates and shrinks.

    To simplify matters, we'll deal with the first bug only. It's velocity V will always be along the side length connected to the bug in front of it. We can break V into two components: one directed towards the centre of the square (Vr); and the other perpendicular to this (Vtheta). (By breaking the velocity into two parts, it's easy to see that Vr shrinks the square, while Vtheta rotates it.) Because of the square geometry, the values of Vr and Vtheta are as follows:

    Vr = Vtheta = V/SQRT(2)

    Now we can set up some equations:

    (1) dr/dt = -Vr (this is negative because the square is shrinking)

    (2) dtheta/dt = Vtheta/r (this is not simply Vtheta, because the rotational effect increases with decreasing r)

    Solving (1), we get r = r0 - Vr*t. Subbing this result into (2), we get:

    (3) dtheta/dt = Vtheta/(r0 - Vr*t)

    Solving (3), we get:

    (theta-theta0)/Vtheta = -(1/Vr)*ln[(r0 - Vr*t)/r0]

    Since r0 - Vr*t = r, we can simplify and solve for r:

    r = r0*e^(-Vr/Vtheta)*(theta-theta0)

    Recall that Vr = Vtheta = V/SQRT(2). Subbing this into the above equation we get:

    r = r0*e^(theta0-theta)

    This gives us the path of the first bug in polar coordinates. The length of this path is the integral of r*dtheta from theta0 to theta, as follows:

    L = int[r0*e^(theta0-theta)]dtheta

    = -r0*e^(theta0-theta)] from: theta0...theta

    = -r0*[e^(theta0-theta) - e^(theta0-theta0)]

    = r0*[1 - e^(theta0-theta)]

    = r0*[1 - 1/e^(theta-theta0)]

    The path length is the limit of this expression as theta tends towards infinity, which becomes:

    lim (theta--}inf) r = r0*[1 - 1/inf] = r0*[1 - 0] = r0

    Therefore, the path length is equal to r0. Interestingly, the bugs will spiral an infinite number of times, but will do so in a finite amount of time and with a bounded path length.
  3. 23 Oct '06 17:41
    My solution:

    By symmetry, bug B's direction of movement is always perpendicular to A's. Therefore B's velocity has no component in the direction of A's velocity. Therefore A approaches B at the same speed as if B wasn't moving at all. So A will walk 1 meter.
  4. Standard member CalJust
    It is what it is
    24 Oct '06 05:30
    Originally posted by David113
    My solution:

    By symmetry, bug B's direction of movement is always perpendicular to A's. Therefore B's velocity has no component in the direction of A's velocity. Therefore A approaches B at the same speed as if B wasn't moving at all. So A will walk 1 meter.
    Congrats, David! Good thinking!

    And congrats, PBE6, good maths!
  5. Standard member TheMaster37
    Kupikupopo!
    24 Oct '06 08:02
    Originally posted by David113
    My solution:

    By symmetry, bug B's direction of movement is always perpendicular to A's. Therefore B's velocity has no component in the direction of A's velocity. Therefore A approaches B at the same speed as if B wasn't moving at all. So A will walk 1 meter.
    I love this solution
  6. Standard member PBE6
    Bananarama
    24 Oct '06 19:47
    Originally posted by TheMaster37
    I love this solution
    I agree, it's very intuitive and clever. And unfortunately for me, it also makes my answer wrong... The path length I came up with is too short, as r0 = SQRT(side length). I just tried putting my model into Excel, along with a different method using dx/dt and dy/dt, and the points came out the same, so I think the model is right, but something is definitely amiss...

    However, just to make myself feel better the derived solution (once it's fixed) will work for any regular polygon arrangement of bugs.