Comments on: Antipodal points http://puzzletweeter.com/2015/09/06/antipodal-points/ Tweet your puzzles! Mon, 27 Jun 2016 18:15:01 +0000 hourly 1 https://wordpress.org/?v=4.4.2 By: Jeetesh M http://puzzletweeter.com/2015/09/06/antipodal-points/#comment-7684 Sun, 07 Feb 2016 06:04:16 +0000 http://puzzletweeter.com/?p=1926#comment-7684 Consider a continuous function, P(x) = f(x + T/2) - f(x)
Then,
P(x+T/2) = f(x+T) - f(x+T/2)
So, P(x) + P(x+T/2) = f(x+T) - f(x) = 0
which means P(x+T/2) = -P(x)
If P(x) = 0 for all x, then we are done.
Otherwise, by Intermediate Value Theorem, there exists y such that y > x, y < x+T/2 & P(y) = 0 f(y+T/2) = f(y)

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By: Shaleen http://puzzletweeter.com/2015/09/06/antipodal-points/#comment-7068 Tue, 08 Sep 2015 00:52:12 +0000 http://puzzletweeter.com/?p=1926#comment-7068 Consider a function L from R->R where L(x) = f(x+T) - f(x).
Since f is continuous over R, so is L.

If L(0)=0, we are done and have our antipodal point.
If not, there exists some x for which L(0) < 0 0 > L(T/2)). Therefore for some value of x between 0 and T/2, L(x)=0 which becomes our antipodal point.

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