Problem 1452

Suppose there do not exist $27$ circles such that every pair of circles intersects.

Consider a coordinate plane such that the line joining any pair of centres of the circles is not parallel to the coordinate axes. We label the circles as $\Gamma_1, \Gamma_2, \dots, \Gamma_{2016}$ such that the $x$-coordinate of the centre of $\Gamma_i$ is less than that of $\Gamma_j$ if $i < j$.

Consider one of the circles $\Gamma_k$. Let $\Gamma$ be the circle with radius $2$ and having the same centre as $\Gamma_k$. Partition the left semicircle of $\Gamma$ into $3$ sectors of $60^\circ$. Then the distance between any two points in the same sector is at most $2$. If there are $26$ centres of the other circles belonging to the same sector, then these circles together with $\Gamma_k$ satisfy the condition, since the distance between any two of these centres is at most $2$, which shows the two circles intersect. Thus, we may assume there are at most $25$ centres in each sector. This implies at most $75$ centres among $\Gamma_1, \Gamma_2, \dots, \Gamma_{k-1}$ lie in $\Gamma$, or equivalently at most $75$ of these circles intersect $\Gamma_k$.

Now, we colour the circles $\Gamma_1, \Gamma_2, \dots, \Gamma_{2016}$ one by one in $76$ colours. Suppose we have coloured $\Gamma_1, \Gamma_2, \dots, \Gamma_k$. Since $\Gamma_{k+1}$ intersects at most $75$ of the previous circles, we can colour it in a way such that its colour is different from the colours of all circles among $\Gamma_1, \Gamma_2, \dots, \Gamma_k$ which intersect $\Gamma_{k+1}$. By the pigeonhole principle, we can find
$$\left\lfloor \frac{2016}{76} \right\rfloor = 27$$
circles having the same colour. This means these $27$ circles are pairwise disjoint. So we are done.