Two circles of different radii are cut out of cardboard. Each circle is subdivided into $200$ equal sectors. On each circle $100$ sectors are painted white and the other $100$ are painted black. The smaller circle is then placed on top of the larger circle, so that their centers coincide. Show that one can rotate the small circle so that the sectors on the two circles line up and at least $100$ sectors on the small circle lie over sectors of the same color on the big circle.
I know at least three solution (pigeon hole, probabilistic and some double counting, they are basicly the same, see:
of this problem and now I'm intersted if someone has idea how to finish this one:
So we can think this $200$ sectors as vectors $u$ and $v$ in $\mathbb{Z}^{200}$ with entries $100$ times $-1$ (for black) and $100$ times $1$ (for white). Now there is a shift operator $R$ which moves all entries of vector for one place to the right, so we can think every power $R^k$, $k\in \{0,1,2,...,199\}$ as some rotation. Now we have to prove there is such $k$ that standard $$\langle R^kv,u\rangle \; \geq\; 0$$
Any idea how to achieve that?
This operator in $\mathbb{Z}^{4}$ is like: $$R =\pmatrix{0&0&0&1\\ 1&0&0&0\\0&1&0&0\\ 0&0&1&0\\ } $$
Eh, I think I found the way out my self.
Say we have for each $k$: $$\langle R^kv,u\rangle \; <\; 0$$ Then $$ \langle\sum _{k=0}^{199} R^kv,u\rangle = \sum _{k=0}^{199} \langle R^kv,u\rangle \;<\; 0$$
But $\sum _{k=0}^{199} R^k =E$ where $E$ is a matrix with all entries $1$, so $$\sum _{k=0}^{199} R^kv = Ev = {\bf 0}= (0,0,0,....,0)$$ So we have $$ 0=\langle Ev,u\rangle =\langle\sum _{k=0}^{199} R^kv,u\rangle \;<\; 0$$ A contradiction.