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Math Help - Maximum distance between two stations

  1. #1
    MHF Contributor Unknown008's Avatar
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    [Solved] Maximum distance between two stations

    Ok, I didn't know where exactly to put my question... but since it's not about school, but a math problem I came across in a competition, I'll put it here. Someone told me the answer was 174 km but that person did it using matlab. The answer was later confirmed after the results of that competition, but I still don't know how is one supposed to do it using only pen and paper and nothing else

    Original question:

    The country of Big Wally has a railway which runs in a loop of 1080 km long. Three companies, A, B and C run trains on the track and plan to build stations. Company A will build three stations, equally spaced at 360 km intervals. Company B will build four stations at 270 km intervals and Company C will build five stations at 216 km intervals.

    The government tells them to space their stations so that the longest distance between consecutive stations is as small as possible. What is this distance in kilometres?

    [An approximate sketch I made of the problem]



    Thank you in advance
    Last edited by Unknown008; July 22nd 2010 at 08:19 AM.
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  2. #2
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    Change your problem to :
    A railway runs in a loop of 60 km long.
    3 companies, A, B and C run trains on the track and plan to build stations.
    Company A will build 3 stations at 20 km intervals.
    Company B will build 4 stations at 15 km intervals.
    Company C will build 5 stations at 12 km intervals.

    Will be much easier to "see and handle"; like clock face with 60 minutes...kapish?
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Okay... if i put them on a clock face, and start all of them at 'noon', I get the longest distance as 12 km, when scaled up gives 216 km, which has to be decreased.

    But the problem is when I try to rotate a certain set of stations, I also tend to increase the distance somewhere else. So, I don't know which stations to move, where to minimise the separation ...

    If I consider only the ones with total stations 3 and 4, I get the maximum distance on the 'clock' as 15 km, which is 270 km apart because there is always an empty quadrant. But including the third set just stumps me...
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  4. #4
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    Another way to consider this, is since the A stations are equally spaced,
    they will make angles of 120 degrees in a circle.
    The B stations are equally spaced, so they make angles of 90 degrees in a circle.
    The C stations are equally spaced, so they make angles of 72 degrees in a circle.

    Maximum distance of track between stations is to be minimized,
    therefore the maximum angle between stations in a circle needs to be minimized.
    Angle is directly proportional to arc length..

    \frac{\theta}{360^o}=\frac{arc}{1080}

    arc=\frac{1080\theta}{360^o}

    Then, it does not matter what shape the track is.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Okay... I have tried to think about this way too, but the same question remains, how to find the largest angle while trying to minimise all the angles?

    From the angle's perspective, the angle that has to be found is 58 degrees (174/3).

    I see no relation with the three angles, 120, 90 or 72 with 58. =/
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  6. #6
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    If you draw the circle with stations B at north, south, east and west.
    One A at north and the other 2 at 120 and 240 degrees.
    One C at north and the others at 72 degrees, 144 degrees, 216 degrees, 288 degrees.....

    then the circle has lines seperated in clockwise fashion from north by 72, 18, 30, 24, 36, 36, 24, 30, 18 and 72 degrees.

    Now it's possible to turn C stations "upside-down" to get the maximum angle at 60 degrees in the lower semi-circle
    instead of 72 degrees in the upper one.

    Is there another step that can reduce the angle to 58 degrees?
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  7. #7
    MHF Contributor Unknown008's Avatar
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    I don't know... since the arrangement for the time being is symmetrical about a straight vertical line passing along north-south, rotating one of the stations might reduce one angle, but increase the size of another angle...

    Or do we rotate the B stations this time round?
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  8. #8
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    Quote Originally Posted by Unknown008 View Post
    I don't know... since the arrangement for the time being is symmetrical about a straight vertical line passing along north-south, rotating one of the stations might reduce one angle, but increase the size of another angle...

    Or do we rotate the B stations this time round?
    If we draw the original circle with an A, B and C all at north,
    we will have a maximum angle of 72 degrees between stations.
    Then, if we turn C "upside down, there is a maximum angle of 60 degrees between stations.

    Starting at north, if we label the points where the lines from the circle centre meet the circumference,
    we now have

    A,B at north.

    C is 36 degrees clockwise.
    Another 54 degrees to B facing east.
    18 degrees to the next C.
    12 degrees to the next A.
    60 degrees to B,C facing south.
    60 degrees to the next A.
    12 degrees to the next C.
    18 degrees to B facing west.
    54 degrees to the final C.

    2 stations cannot be together, so we need to rotate B or C.

    What happens if we rotate the B stations in unison (as though attached to the circumference of the wheel)
    by 2 degrees anticlockwise,
    and also rotate the C stations by 2 degrees clockwise ?

    and is that the best we can achieve?
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  9. #9
    MHF Contributor Unknown008's Avatar
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    If we do that, I find that the largest angle becomes 62 degrees, the one between the A station in the South, and the B station on the left...

    EDIT: wait, let me rethink about it...

    no, it's the B station on the right.

    Ah! If I had some sort of thingy like the map in Pirates of the Carribean... I'll keep on rotating these rings until I find the smallest value of the largest angle...
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  10. #10
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    Quote Originally Posted by Unknown008 View Post
    If we do that, I find that the largest angle becomes 62 degrees, the one between the A station in the South, and the B station on the left...

    EDIT: wait, let me rethink about it...

    no, it's the B station on the right.

    Ah! If I had some sort of thingy like the map in Pirates of the Carribean... I'll keep on rotating these rings until I find the smallest value of the largest angle...
    While you are rotating the B's, remember one of the C's is still facing south.
    This is because when we flipped the C's initially, the C pointing north is now pointing south.

    Hence you will have 60 degrees to the left and 58 degrees to the right of the sole C now facing south.
    Now rotate all the C's clockwise, realising that the 58 degrees to the right of direct south is between stationary A and B.
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  11. #11
    MHF Contributor Unknown008's Avatar
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    Oh shoo... I was mistaking B for As and vice versa. Ok, now I get 58 on both sides.

    I'm seeing something else too... in the north west, an angle is getting bigger. And luckily, it's 54 degrees + 4 degrees = 58 degrees.

    Phew, I think that's it. Three maximum angles, with the minimum value possible.

    Thank you!
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