Question 124
There are $11$ men sitting around a circular table with equal distances and $11$ cards with numbers $1,2,\ldots,11$ on them are dealt among them. It is possible that one has no cards and the other has more than one. In each step one can give one of his cards to his adjacent individual if the card number $i$ has the following property: before and after this stage, the places of the cards with numbers $i-1$, $i$ and $i+1$ are **not** the vertices of an acute-angled triangle. (card $0$ is the same as card $11$, and card $12$ is the same as card $1$) At the beginning the cards $1$ to $11$ are dealt among them counter-clockwise. (everyone has exactly one card) Prove that there will never be a man who has all the cards.