I try to solve the following problem (Moscow Mathematical Olympiad, 1978)
There is a town with six streets: four sides of a square and two its middle lines. Policeman tries to catch a gangster. If policeman and gangster are on the same street, then gangster will be surrender. Speed of the policeman equals 2.1 speed of the gangster. Prove that policeman can catch the gangster.
In fact, all algorithms that I can come up don't work even if speed of policeman equals 100000 speed of the gangster.
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Not really a solution, but an observation. For simplicity assume each side of the square is 2 units long, call vertices of the square corner vertices and center of the square plus middle of each side "middle" vertices. All are crossroads. Side roads are sides of the square and other are mid-roads.
Let's say gangster at some point is at the side street, and at some point he will be located at the middle vertex. If not, he will stuck moving between two sides, and will be caught by policeman. Now assume he is at the middle vertex. WLOG it can be middle of the top side. He has to move somewhere (otherwise distance between him and police only gets smaller), and WLOG assume moving towards left side street (hence needs to cover 1 unit of distance). Hence it means that the policeman needs to be both at least (2+e) units away from both left and top sides, which is impossible.
P.S: First math problem attempted 8 years after finishing my math olympiad career :P btw, I am "Erken" on AoPS.
I belive that the following text is a solution of this problem. If someone find a mistake, it will be good (and bad, simultaneously).
So, let the square be $ABCD$ (clockwise), the midpoints of $AB$, $BC$, $CD$, $DA$ are $K$, $L$, $M$, $N$, respectevly, $O=KM\cap LN$. Let's ask the policeman use the following strategy: $$ \ldots ONAKOKBLOLCMOMDNONA \ldots, $$ i.e he goes clockwise around the square, runs in to check the center each time he reaches the midpoint of a side, then returns to the same midpoint and continues going clockwise.
First of all, we can observe that the gangster will never be at the point $O$. In fact, to get from perimeter to $O$ and back to perimeter takes 2 minutes but the policeman visits $O$ every $\frac{4}{2.1}<2$ minutes.
So, let us consider the set of points where can be the gangster every time the policeman visits $O$. If the policeman went $ONAKO$ then the gangster can be on the interval from point $X$ to point $Y$, where $X$ lies on $KB$ and $XB=\frac{1}{2.1}$ and $Y$ lies on $DN$ and $YD=\frac{2}{2.1}$. In fact, gangster can appear on $AB$ only when the policeman left $K$, and can appear on $AD$ only when the policeman left $A$.
By the next visit of $O$, point $X$ shifted by 2 clockwise and point $Y$ can shifted less then by $\frac{4}{2.1}<2$. So, the legth of the interval decreases every time by $2-\frac{4}{2.1}>0$ and sooner or later the length will become negative, which is impossible.