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The questions in my textbook asks:

"Given a sphere $S$, a great circle of $S$ is the intersection of $S$ with a plane through its center. Every great circle divides $S$ into two parts. A hemisphere is the union of the great circle and one of these two parts. Show that if five points are placed arbitrarily on $S$, then there is a hemisphere that contains four of them."

I don't mean for this to be a low-quality question, but I have genuinely been trying this question for a very long time and have little idea as to how to approach it. This question is in the section of the book concerning the division principle and the pigeonhole principle, so I am assuming they are relevant.

Again, I am sorry this question is so low-quality, but I truly need help. Thank you.

TreeGuy
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  • This is a pretty standard example of the pigeonhole principle. I expect that there is a a good answer to this question on Math SE already... – Xander Henderson Apr 10 '22 at 21:25
  • @XanderHenderson Ok, I will go check – TreeGuy Apr 10 '22 at 21:25
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    If you made a great circle that contained 2 points, 3 would remain. If there were 2-3 points to one side of the circle, that would be your hemisphere. Otherwise, there are 0-1 points to that side, and at least 2 points to the other side. Then that side is your hemisphere. – John P Apr 10 '22 at 21:29

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