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Challenge 10a: Temperatures of the Earth

  1. Dec 11, 2013 #1

    Office_Shredder

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    For the sake of this challenge, you can assume that the world behaves nicely (everything is continuous, differentiable etc, the world is a sphere blah blah blah) but figurative bonus points if you assume fewer things. If you're looking for a bigger challenge check out Challenge 10b.

    The Challenge: Show that at any point in time there exist two points on exact opposite sides of the Earth which have the exact same temperature.
     
  2. jcsd
  3. Dec 11, 2013 #2
    If I've seen a very similar problem, should I still post an answer?
     
  4. Dec 11, 2013 #3

    jbunniii

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    Apologies for any typesetting issues as I'm attempting an answer from my phone.

    Take any great circle and choose a point on the circle to be the origin. Parameterize the circle uniformly from ##x=0## to ##x=1## where the origin is at both endpoints.

    Let ##f:[0,1]\rightarrow \mathbb{R}## be the temperature function. Define ##d(x) = f(x+1/2) - f(x)##. If ##f(1/2)=f(0)## then we're done. Otherwise ##d(0)## and ##d(1/2)## have opposite signs , so (assuming continuity of ##f##) the intermediate value theorem gives us a point ##x## with ##d(x) =0##, hence ##f(x+1/2) = f(x)##.
     
  5. Dec 11, 2013 #4

    Office_Shredder

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    I would lean towards saying that you can give answers as long as you aren't just copying it from someone else.

    Nice solution jbunniiii, anyone have an alternate way of approaching the problem?
     
  6. Dec 13, 2013 #5

    MarneMath

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    Not a proof but...this results follows from Borsuk-Ulam theorem.
     
  7. Dec 18, 2013 #6

    ssd

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    An alternative heuristic approach: See attached image. Let ADBC be a great circle on earth surface.
    AB diameter. At A, temperature is t1, at B temperature is t2. Let us imagine two men moving at same speed from A to B along paths ACB and ADB respectively. The distance from A to B =d. Now the men are recording temperatures continuously. We draw the graphs of the temperatures vs distance on the same axes.
    Now if we draw a straight line corresponding to any temperature t=t0, (between t1 and t2) parallel to distance axis. Both the temperature curves shall cut this line at least once (intermediate value theorem).
    This two intersection points are our answer.
    [/PLAIN] [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Dec 18, 2013 #7

    jbunniii

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    ssd - Maybe I am misunderstanding your argument, but I don't see why the two intersection points ##(d_1, t_0)## and ##(d_2, t_0)## correspond to points on "exact opposite sides of the Earth" as the problem requires. In other words, what guarantees that ##d_1 + d_2## equals the arc length from ##A## to ##B##?
     
    Last edited: Dec 18, 2013
  9. Dec 18, 2013 #8

    ssd

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    Oh, that is right. My answer is wrong. I missed the "exact" opposite part.

    Well, let us consider that the first man is moving from A to B via C and the second man is moving from B to A via D. As before the are recording temperature continuously. They start at same time and move at same constant speed. This implies that they are always exactly on opposite points of a diameter.


    Consider the graph of the difference in the temperature readings (of same instant) of two men vs the distance. Let, t1<t2. The graph starts with -ve value and ends with +ve value. According to the given conditions it must intersect the distance axis (that is, will attain value 0) somewhere between distance 0 to d. That point is the answer. If t1=t2 then the start and end point is one answer.
     
    Last edited: Dec 18, 2013
  10. Dec 20, 2013 #9

    Office_Shredder

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    As far as I can tell this is the solution that jbunniii gave, just with a lot less notation.
     
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