I need help with the following puzzle:
Consider a round table which hosts $2n$ knights. At each seat, there is a namecard of one knight. The knights don't necessarily sit in front of their own namecard. Prove that we can rotate the table such that at least 2 knights sit in front of their own nameplate.
Using the pigeon hole principle, it's quite easy to see that this is equivalent with:
- Proving that there exists a rotation where no knights sit in front of their own namecard.
- Proving that it is impossible that in every rotation, exactly one knight sits in front of his own namecard.
After going over some examples, I noted that if for knight number $k$, the angle needed to get this knight in front of his nameplate is $\frac{a_k \pi}{n}$, then the sum $a_1 + ... + a_{2n}$ is a multiple of n. If one could prove this, then this would solve the problem: if in every rotation, there is exactly one person sitting in front of his namecard, then $$a_1 + .. . + a_{2n} = 0+1+...+ (2n-1) = n(2n - 1),$$ which is not a multiple of $2n$.
Another method is also welcome of course. Thank you in advance!