Closest football team

Discussion in 'Recreational Math' started by Peter Finan, Jan 21, 2009.

  1. Peter Finan

    Peter Finan Guest

    I live on the moors near Haworth in West Yorkshire, UK. There's a cup
    semi-final going on right now, and Burnley came back from 4-1 down
    against Spurs in the first leg of a two-legged affair to win the second
    leg 3-0 (in regulation time as I write, extra time in the League Cup, no
    away goals rule. Pfft). Anyway, I said to my wife that Burnley must be
    the closest team to where we live. Rubbish, she said - Halifax and
    Bradford must be closer. I disagreed. Anyway - it got me
    thinking...which team is the closest to where I live? Is it Burnley,
    Halifax, or Bradford? Given where I live, it must be a very close thing.
    So, my question is - given three locations on a map (in this case -
    Burnley, Halifax and Bradford), how do I look at a map and figure out
    where I must live for those football clubs to be equidistant? I thought
    about it, and wondered if there was a simple way to do it using a
    compass and pencil. You could experiment with different circle radii,
    but given my enthusiatic but limited maths knowledge, I can't figure out
    how to do it. Can anyone help?

    Many Thanks,

    Peter Finan
     
    Peter Finan, Jan 21, 2009
    #1
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  2. Ignoring the curvature of the Earth and other issues, of course....

    Take two points, P and Q. Draw the line segment between P and Q,
    PQ. Now draw a circle with center at P and radius |PQ|, and draw a
    circle with center at Q and radius |PQ|. These two circles will
    intersect in two points. The line through those two points bisects PQ,
    meets PQ perpendicularly, and consists exactly of all points that are
    equidistant from P and from Q.

    Now, if you have three points, P, Q, and R, and that they are not
    colinear. Find the line L1 of all points equidistant from P and from
    Q; the line L2 of all points equidistant from P and R. A point which
    is equidistant from P, Q, and R must lie in both L1 and L2, i.e., in
    L1/\L2. This intersection consists of exactly one point, since L1 is
    perpendicular to PQ and L2 is perpendicular to PR, and PQ is not
    parallel to PR. This point is, by construction, equally distant from P
    and from Q, and equally distant from P and from R, hence equidistant
    from all three points.

    If the points are colinear then no such point exists.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes" by Bill Watterson)
    ======================================================================

    Arturo Magidin
    magidin-at-member-ams-org
     
    Arturo Magidin, Jan 21, 2009
    #2
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  3. Peter Finan

    se16 Guest

    Arturo's answer is correct, but it might be easier to search for
    Circumcentre (or Circumcenter), which is where the three perpendicular
    bisectors meet, easily drawn with ruler and compass.

    These perpendicular bisectors divide the space into six zones, which
    in pairs show the areas closest to each of the three points. (All
    this ignores hills and wobbly Yorkshire/Lancashire roads). At a
    glance, somewhere north of Hebden Bridge and west of Haworth looks as
    if it is about 10 miles away from all three.
     
    se16, Jan 22, 2009
    #3
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