Pigeonhole principle on two coloured circle Suppose a circle is divided into 200 congruent sectors, with 100 of them coloured red and the other 100 blue. A smaller concentric circle is placed on the larger circle and also so divided and coloured. 
Prove that no matter how the 100 red sectors are chosen for each circle, the smaller circle can be rotated so that at least 100 sectors of the two circles match in colour.
 A: Consider all $200$ rotations of the smaller circle. Through all of these, each sector matches exactly $100,$ so that the total number of matches is $20,000.$ But then one of the rotations must match at least $100$ sectors.
A: Count in two ways the number of pairs $(S,T)$ where $S$ is a sector of the larger circle, $T$ is sector of the smaller circle and $\operatorname{colour}(S) = \operatorname{colour}(T)$. Directly, there are 
$$ \underbrace{100 \times 100}_{\operatorname{colour}(S) = \operatorname{colour}(T) =\text{ red}} + \underbrace{100 \times 100}_{\operatorname{colour}(S) = \operatorname{colour}(T) = \text{ blue }} = 20000$$
such pairs. Indirectly, let $r_1,\ldots,r_{200}$ denote the number of adjacent equal colours between the large and small circle when the inner circle has been rotated by $360 \times \frac{i}{200}$ degrees.  Since each pair $(S,T)$ becomes adjacent for precisely one rotate of the inner circle, we have
$$ 20000 = \sum_{i=1}^{200} r_i$$
or
$$ \operatorname{average}\{r_1,\ldots,r_{200})  = \frac{1}{200} \sum_{i=1}^{200} r_i= 100.$$
Since the average of the $r_i$ is $100$, at least one particular $r_i$ must be $100$ or greater. 
