Antipodal points and temperature and other angles

[srfriggen]

TL;DR Summary

Can this popular problem be generalized to all angles?

Hello,

I am a high school math teacher and recently presented my students with an intriguing problem: "At any given moment, there are two antipodal points on Earth (180 degrees apart) that have the same temperature." This can be demonstrated using one great circle with two opposite points. If we assume they begin at differing temperatures and rotate them around the circle at the same rate, they will have encountered all intermediate temperatures by the time they've completed 180 degrees. Since they remain 180 degrees apart and the temperature changes continuously, they will have identical temperatures simultaneously at least once.

One of my students proposed extending this idea to points separated by different angles, such as 90 degrees, using a graphical approach, and it appears to be valid. I'm curious if this is a known extension or if anyone has explored similar generalizations.

Thank you.

 

[Erland]

Actually, a stronger result than your original can be proved. The Borsuk-Ulam Theorem says that if $f: S^n\to \Bbb R^n$ (where $S^n$ is the $n$-sphere) is a continuous map, then there exists a point $x\in S^n$ such that $f(-x)=f(x)$. In the case $n=2$, an example is that at any given moment, there are two antipodal points on Earth that have not just the same temperature, but at the same time the same air pressure, provided that temperature and air pressure vary continuously on the Earth's surface.

For your original problem, the theorem (for $n=1$) implies that the conclusion (about temperature only) holds when restricted to any great circle on the Earth. As you noted, this is simple to prove. It is much harder to prove the theorem for $n=2$ and in general.

I don't know if the result (for $n=1$) for other angles than 180 degrees has been studied. But it is easy to prove that this is true. The simplest is to extend the function to a periodic function (going several laps around the circle):

Claim: Let $f:\Bbb R\to \Bbb R$ be a continuous periodic function, with period $2\pi$ (so that $f(x+2\pi)=f(x)$, for all $x\in \Bbb R$).
Then, for every $a\in \Bbb R$, there is an $x\in [0,2\pi]$ such that $f(x+a)=f(x)$.

Proof: $f(x)$ attains its maximum at some $c\in[0,2\pi]$. Because of the periodicity, $f(c)$ is the maximum of $f(x)$ on all of $\Bbb R$.
Put $g(x)=f(x+a)-f(a)$. Then $g(c-a)=f(c)-f(c-a)\ge 0$ and $g(c)=f(c+a)-f(c)\le 0$. Since $g$ is continuous on $[c-a,c]$, the Intermediate Value Theorem gives an $x\in [c-a,c]$ such that $g(x)=0$. Hence, $f(x+a)=f(x)$, so such an $x$ exists, Q. E. D.

 

[Hill]

Typo, I think:

g(x)=f(x+a)−f(a).

meant to say $g(x)=f(x+a)-f(x)$.

 

[Erland]

Typo, I think:

meant to say $g(x)=f(x+a)-f(x)$.

Yes of course. Thank you and sorry for the mistake!