There are always 2 opposite points on the Earth's equator that have exactly the same temperature.
It has been known for millennia that the Earth's shape can be approximated by a sphere. Greek philosophers as early as Pythagoras made claims of the Earth's spherical shape. Perhaps the most notable arguments from antiquities for a spherical Earth were given by Aristotle who observed, among other things, that the Earth cast a round shadow on the Moon during a lunar eclipse.
By the Hellenistic era (~300BC), the spherical Earth hypothesis was an established given. Moreover, the circumference of the Earth was estimated (with astounding accuracy) by Eratosthenes during this time period. Eratosthenes knew that the Sun was directly overhead during the Summer solstice in the ancient Egyptian city Syene. Aware of the distance from Alexandria to Syene, he was able to approximate the Earth's circumference using trigonometry and the length of a shadow cast in Alexandria during the Summer solstice. His mathematics was sound; had Eratosthenes been working with accurate measurements, his estimate of the Earth's meridional circumference would have been nearly exact.
It is now known that the approximate shape of the Earth is an oblate spheroid. Because the Earth rotates, its spheroidal shape compresses at the poles and bulges at the equator. For convenience, we will stick with the assumption that the Earth is a sphere, but the same arguments can be applied if we acknowledge the oblate structure.
The Earth's equator is an example of a great circle. Two points that lie opposite each other on a great circle are said to be diametrically opposed (or antipodal points). Let $C$ be any one of the Earth's great circles. Let $\tau$ be a continuous function mapping each point on the Earth to the temperature at that point. We will generalize the above claim by arguing that some pair of diametrically opposed points $p,q$ on $C$ satisfy $\tau(p) = \tau(q)$.
To prove this claim, we will introduce a new function $\delta(p,q) = \tau(p) - \tau(q)$, defined for all pairs of diametrically opposed points $p,q$ on $C$. Starting with any pair of diametrically opposed points $p,q$, consider the value of $\delta(p,q)$. If $\delta(p,q) = 0$, we have found the desired pair of points since $\delta(p,q) = 0$ only if $\tau(p) = \tau(q)$. Otherwise, observe that $\delta(p,q)$ must take on the opposite sign of $\delta(q,p)$. However, we can move from $p$ along the great circle in a continuous manner, evaluating $\delta$ at all intermediate pairs of diametrically opposed points until we reach $q$. Since $\delta$ changes sign during this process, there must be some intermediate pair of points $p^*, q^*$ for which $\delta(p^*, q^*) = 0$. This argument follows by the continuity of $\delta$ and can be formalized with the intermediate value theorem. Furthermore, observe that the claim holds for any continuous function over the Earth's surface (like barometric pressure or altitude).
A shrewd reader may realize that the above argument can easily be extended to show that there are infinitely many pairs of diametrically opposed points on the Earth's surface with the same temperature. In fact, the Borsuk-Ulam theorem states that, for any continuous mapping $f$ from the sphere onto $R^2$, there is some pair of diametrically opposed points on the sphere that $f$ maps to the same point in $R^2$. This result can be summarized as follows.
At any point in time, there are two antipodal points on the Earth's surface with the same temperature and barometric pressure.
This claim works for any continuous function mapping points on the Earth's surface to points in $R^2$. In fact, the theorem generalizes to higher dimensional spheres, but the two-dimensional version will suffice to prove the next claim.
Ham Sandwich Theorem. A ham and cheese sandwich can always be cut in half such that each half contains the same amount of bread, ham, and cheese.
We will formalize this problem as follows. Let $B$, $H$, and $C$ be bounded subsets of $R^3$ corresponding respectively to the bread, ham, and cheese. Our goal is to find a plane that bisects $B$, $H$, and $C$ simultaneously.
Let $p$ be a point on the unit sphere embedded in $R^3$ centered at the origin. Consider the set of all planes parallel to the tangent plane at $p$. We will imagine these planes as being oriented in the direction of the normal vector from the origin to $p$. Thus, we can refer to points as being right of the plane or left of the plane. By the intermediate value theorem, one of these planes must bisect $B$. This follows since if we choose a far off plane in one direction, all of the bread will be left of the plane, and if we choose a far off plane in the opposite direction, all of the bread will be right of the plane.
We can therefore map every point $p$ on the sphere to a plane $\pi(p)$ that bisects $B$. Moreover, if we define $h(p)$ and $c(p)$ to be the amount of ham and cheese, respectively, right of the plane $\pi(p)$, then $f(p) = (h(p), c(p))$ is a continuous function from each point on the sphere to $R^2$. Indeed, we can always choose the plane $\pi(p)$ in an unambiguous manner (say, preferring the middle-most plane bisecting $H$) to guarantee continuity.
By the Borsuk-Ulam theorem, there exists two antipodal points $p$ and $q$ on the unit sphere for which $f(p) = f(q)$. However, since $p$ and $q$ are antipodal, it follows that $\pi(p) = \pi(q)$ (they have the same set of planes, and thus our choice of the middle-most plane is identical). But, since $\pi(p)$ and $\pi(q)$ are oriented in opposite directions, the fact that $f(p) = f(q)$ implies that $\pi(p)$ bisects both $H$ and $C$. This concludes the proof as $\pi(p)$ also bisects $B$ by definition.
As mentioned above, these results generalize to higher dimensions. We will leave it as an exercise to come up with some interesting four-dimensional example of the Ham Sandwich theorem. Be festive! It's Thanksgiving today in the United States, so try your best to imagine a four-dimensional turkey draped in four-dimensional cranberry sauce.
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