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I‘m not entirely sure, how to show the following statement:

An arc B with a circumference of $24$ is drawn between K1 and K2 on a circle in the plane with a circumference of $42. 21$ people mark circular arcs of length $4$ on B clockwise. Prove that there must at least be one point on B that belongs at least $5$ different arcs of length $4$ marked by the people.

Now, according to my understanding the statement I’m supposed to show, is wrong.

$\frac{24}{4} = 6, 3 \lt \frac{21}{6} \lt 4$

Therefore, it should be possible for the $21$ persons to mark their arcs in such a way, that every point on the arc is marked by at most $4$ people. But then the statement I’m supposed to prove would be false. Maybe the situation would be different, if no two arcs marked by the people could be completely identical, but I don’t think that it says so anywhere in the problem. We are supposed to prove this using the pigeonhole-principle.

I hope someone can help me.

user1176409
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User1
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    What is the "circumference" of an arc? Do you mean "an arc B with a length of 24" or "an arc B on a circle with a circumference of 24"? And what are K1 and K2? Are these point? An image or at least a sketch would also help a lot to understand your problem. – Hubert Schölnast Jan 15 '23 at 18:53
  • I was wondering about that also. This is all the information given. In the question it talks about the circumference of an arc. I‘m very sure they are talking about an arc with a length of 24. The same goes for K1 and K2, unfortunately that is all the information given. I think they may not even be “related” to the problem, that is, I believe this may just be additional but unneeded information, so that you have to find out what you need to solve the problem first. – User1 Jan 15 '23 at 19:17

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