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Is there a meridian that has exactly as much emerged land area 180° on its East than 180° on its West (on the assumption that it could exist...)? If there is no meridian, is there any other great circle with that property?

Edit: Since David Hammen proved that such a median exists, which one is it?

And what is the latitude that has as much emerged land on its North than on its South (that one necessarily exists)?

I found this Wikipedia article but it doesn't really answers the question.

xenoid
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    The Ham Sandwich theorem guarantees that such a meridian exists. Which meridian it is can only be approximated, the accuracy depensing on the quality of the coastline data you have. – Spencer Apr 16 '20 at 21:34
  • Nice, but it doesn't tell me where they are, which is the real question (otherwise I would have just asked in MSE...) – xenoid Apr 17 '20 at 09:17
  • @xenoid: we're not doing homework here ;-) But data is publicly available and so are GIS programs. I could imagine that, given the data, the outcome is just a few mouseclicks. I'd write a program if I a had to because it's fun. https://stackoverflow.com/questions/4681737/how-to-calculate-the-area-of-a-polygon-on-the-earths-surface-using-python#4682656 –  Apr 17 '20 at 09:19
  • This isn't my homework. I've googled the thing and did not find an answer. I can't believe I'm the first one to wonder about this. If you know where the answer is, please share... – xenoid Apr 17 '20 at 09:53
  • @Spencer - The ham sandwich theorem does not apply, for a couple of reasons. (1) While the theorem guarantees that a hyperplane exists if the conditions of the theorem are met, it says nothing about the orientation of that hyperplane. This question places conditions on the hyperplane. (2) The surface of a sphere is a non-Euclidean space. – David Hammen Apr 17 '20 at 10:37
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    The thing is, this now becomes a GIS/math problem rather than an Earth Science one. Get the data, project to an equal-area projection, do the vector math. – Spencer Apr 17 '20 at 12:37
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    @DavidHammen With a suitable equal-area transformation (e.g. Gall Cylindrical Equal-area projection) it becomes an equivalent problem in $E^2$. – Spencer Apr 17 '20 at 12:40
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    Actually, the question has already been asked (but not answered) on GIS SE: https://gis.stackexchange.com/questions/211888/what-latitude-has-equal-land-areas-north-and-south Links in question and comments might be useful. – Jean-Marie Prival Apr 17 '20 at 15:40

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TL;DR: It depends on how one defines land area. With a sensible definition, such a meridian (along with its completion on the other side of the globe) must necessarily exist by virtue of the intermediate value theorem. The same goes for the line of latitude.

Defining the functions

  • $\operatorname{A}_{\text{E}}(\lambda)$ is the area of the land projected onto the reference ellipsoid of the hemisphere to the east of the line of longitude $\lambda$.

  • $\operatorname{A}_{\text{W}}(\lambda)$ is the area of the land projected onto the reference ellipsoid of the hemisphere to the west of the line of longitude $\lambda$.

  • $\operatorname{\Delta A}_{\text{lon}}(\lambda)$ is the difference between $\operatorname{A}_{\text{E}}(\lambda)$ and $\operatorname{A}_{\text{W}}(\lambda)$.

  • $\operatorname{A}_{\text{N}}(\phi)$ is the area of the land projected onto the reference ellipsoid of the hemisphere to the north of the line of latitude $\phi$.

  • $\operatorname{A}_{\text{S}}(\phi)$ is the area of the land projected onto the reference ellipsoid of the hemisphere to the west of the line of longitude $\phi$.

  • $\operatorname{\Delta A}_{\text{lat}}(\phi)$ is the difference between $\operatorname{A}_{\text{N}}(\phi)$ and $\operatorname{A}_{\text{S}}(\phi)$.

My definitions make each of these functions continuous. Note well: Other definitions of land area might result in non-continuous functions. Imagine a perfectly vertical cliff that runs north to south. If the area of that cliff face counted as land area then $\operatorname{\Delta A}_{\text{lon}}(\lambda)$ will not be continuous. The same goes for $\operatorname{\Delta A}_{\text{lat}}(\phi)$ for a perfectly vertical cliff that run east to west.

Proving that the latitude line must exist is easy, so I'll do that first. All of the Earth's land area is north of 90° south latitude, making $\operatorname{\Delta A}_{\text{lat}}(-90)$ a large positive number. All of the Earth's land area is south of 90° north latitude, making $\operatorname{\Delta A}_{\text{lat}}(90)$ a large negative number. Because zero is between this large negative number and large positive number, and because $\operatorname{\Delta A}_{\text{lat}}(\phi)$ is continuous, there must necessarily exist at least one line of latitude $\phi$ for which $\operatorname{\Delta A}_{\text{lat}}(\phi)$ is zero.

Regarding longitude, Pick an arbitrary longitude $\lambda$. If $\operatorname{\Delta A}_{\text{lon}}(\lambda)$ is zero we have a winner. If it's not zero, then since $\operatorname{\Delta A}_{\text{lon}}(\lambda +180°) = -\operatorname{\Delta A}_{\text{lon}}(\lambda)$, there exists at least one longitude $\lambda_0$ between $\lambda$ and $\lambda+180°$ where $\operatorname{\Delta A}_{\text{lon}}(\lambda_0)$ is zero.

David Hammen
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