I think it’s possible to prove, under the right conditions. One assumption is, that Humans are not point like, so that depending on the exact position the person may be half on two sides. For simplicity, I assume a continuos finite population density everywhere, which can be a small peak where a person is and zero everywhere else.
In this case, it is obvious that for any angle, we can draw a line splitting the population in half. Imagine just shifting the line until the population on both sides is the same. This means we can, for any angle, draw a cross with each line splitting the population in half. This can be written as the following condition, considering the colors above and R, G, Y, B as populations in the Red, Green, Yellow, and Blue quarters, respectively: R+G=B+Y and R+B=Y+G. So G=B and R=Y. What we still need to prove is that it is always possible to have G=R. Now we can do this continuously for each angle of the cross, so starting with an arbitrary cross, we rotate it slowly 1/4 turn counterclockwise. Now R is where G was before. Due to the conditions it can be exactly the same cross but with colors switched. So if R>G before, we now have R<B=G, so during this path, and everything being continuous, there must be an angle for which R=G holds and so all four quarters are at equal size.
The real question is, does this hold on a globe with great circles is splitting lines?
I think it’s possible to prove, under the right conditions. One assumption is, that Humans are not point like, so that depending on the exact position the person may be half on two sides. For simplicity, I assume a continuos finite population density everywhere, which can be a small peak where a person is and zero everywhere else.
In this case, it is obvious that for any angle, we can draw a line splitting the population in half. Imagine just shifting the line until the population on both sides is the same. This means we can, for any angle, draw a cross with each line splitting the population in half. This can be written as the following condition, considering the colors above and R, G, Y, B as populations in the Red, Green, Yellow, and Blue quarters, respectively: R+G=B+Y and R+B=Y+G. So G=B and R=Y. What we still need to prove is that it is always possible to have G=R. Now we can do this continuously for each angle of the cross, so starting with an arbitrary cross, we rotate it slowly 1/4 turn counterclockwise. Now R is where G was before. Due to the conditions it can be exactly the same cross but with colors switched. So if R>G before, we now have R<B=G, so during this path, and everything being continuous, there must be an angle for which R=G holds and so all four quarters are at equal size.
The real question is, does this hold on a globe with great circles is splitting lines?