$11$ people are sitting around a circle table, orderly (means that the distance between two adjacent persons is equal to others) and $11$ cards with numbers $1$ to $11$ are given to them. Some may have no card and some may have more than $1$ card. In each round, one [and only one] can give one of his cards with number $ i $ to his adjacent person if after and before the round, the locations of the cards with numbers $ i-1,i,i+1 $ don’t make an acute-angled triangle. (Card with number $0$ means the card with number $11$ and card with number $12$ means the card with number $1$!) Suppose that the cards are given to the persons regularly clockwise. (Mean that the number of the cards in the clockwise direction is increasing.) Prove that the cards can’t be gathered at one person.
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Tags: topology, geometry, circumcircle, calculus, integration, invariant, combinatorics proposed
21.09.2010 00:19
It has a more difficult part : Prove that after any operations , the cards will not be in opposite order (counterclockwise) . This nice problem proposed by Dr.Naghshine , He has got the main idea from topology !! .
02.10.2010 15:46
Unexpected!because It seems completely combinatorial (we have seen so many problems of this type),And Couldn't be solved without mahamath's hint! Anyway here is the unexpected solution: persons and cards are points.Suppose circumcenter is the origin .By segment $i$ I mean the segment between cards $i$ and $i+1$ and by triangle $i$ ,the triangle made by cards $i,i+1,i+2$.let $C$ be the closed directed path consisting of the 11 consecutive segments mentioned.let $s$ be the winding number of path around the origin and $s_{i}$ the value of the integral over segment $i$.we claim that $s$ is invariant .At first $s=-1$ and at last is $0$ in part (a) and $1$ in part (b). So both parts will be solved. Suppose in a step we have moved card $i$.By assumption the triangle $i-1$ won't contain origin before and after operation.let $X,Y,Z,T$ be respectively:card $i-1$,card $i$ before move,card $i$ after move,card $i+1$.By an illustration it will be easy to see that origin is outside closed path $XYTZ$ and so $s_{i-1}+s_{i}$ will be invariant in this step and so will be $s$.so we are done two notes: 1.If you imagine each step as a continuous move of a vertex along a line (and so segments and path will be transformed continuously) .if it can be proved that $C$ will never touch origin (*)then the rest can be done by topology/analysis .But if in a step that you move card $i$ , the cards $i-1$ and $i+1$ be on each other then (*) may not be true .So if we want to solve in this way we must think what to do when cards $i-1$ and $i+1$ are on each other 2.It can be shown that (b) implies (a) by combinatorial reasoning.